Integral $\frac{2}{876} \int_{0}^{T} \left[ x^{-3/8} (1+b\,x)^{3/4} (1+ab\,x)^{-3} e^{(c\,x^2)}+\log{(5/4)}\right]\,\mathrm dx$ I am trying to evaluate definite integral:
$$\frac{2}{876} \int_{0}^{T} \left[ x^{-3/8} (1+b\,x)^{3/4} (1+ab\,x)^{-3} e^{(c\,x^2)}+\log{(5/4)}\right]\,\mathrm dx$$
$a$, $b$, $c$ are all positive ($a=3.8/3777$, $b=3777$, $c=4.8\times 10^{-3}$). I have plotted the whole integrand. In the plot $\tau$ corresponds to $x$ in the integral. The parameter $\eta$ can be ignored. I have tried the Leibniz method (also called Feynman's integration trick) but without any success. Maybe with a clever change of variable it can be expressed in terms of (combinations of) special functions... or at least get some bounds? Actually I have a different way to estimate the integral, since it's a physics problem and I can guess the (upper limit) of it. T should be $T\sim 57$ (months), I hope. Thanks.


"abandon hope all ye who enter here" as wrote Dante Alighieri in "The Divine Comedy" 
Using your numbers, you have
$$f(x)= \frac{1}{438} \left(\frac{ (1+3777 x)^{3/4}}{x^{3/8}
   \left(1+\frac{19 }{5}x\right)^3}\,e^{\frac{3 x^2}{625}}+\log \left(\frac{5}{4}\right)\right)$$ and it seems to me that you want to find the values of $T$ such that
$$F(T)=\int_0^T f(x) \,dx = \eta $$ There is absolutely no hope to get a closed form formula for the result and, for whatever you could need to do, all the work needs numerical methods.
What is interesting is to look at the plot of $G(T)=\log(F(T))$ as a function of $T$; there are two parts : an almost horizontal part as long as $\eta < 1$ and for $\eta >1$ $G(T)$ exhibits an almost parabolic shape.
For solving the equation, Newton method is very effective for finding the zero of the equation
$$G(T)-\log(\eta)=0$$
Using in particular the fundamental theorem of calulus, the iterates will be given by
$$T_{n+1}=T_n-\frac{F(T_n)}{f(T_n)}\log \left(\frac{F(T_n)}{\eta }\right)$$
Let us try with $T_0=123$ and $\eta=1$. The iterates will be
$$\left(
\begin{array}{cc}
 n & T_n \\
 0 & 123.000 \\
 1 & 74.4050 \\
 2 & 56.6963 \\
 3 & 53.1111 \\
 4 & 52.8262 \\
 5 & 52.8235
\end{array}
\right)$$ which is quite fast in spite of a quite poor estimate.
In fact it seems that a rather good estimate could be
$$T(\eta)=52.8235 + a \big[\log(\eta)\big]^b$$ A quick a dirty non linear regression gives (with $R^2=0.999906$)
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & 9.08300 & 0.03959 & \{9.00532,9.16069\} \\
 b & 0.54957 & 0.00055 & \{0.54848,0.55066\} \\
\end{array}$$


*

*For $\eta=10^{100}$, the estimate is $233.304$ while the solution is $227.851$

*For $\eta=10^{1000}$, the estimate is $692.559$ while the solution is $696.057$
