Optimal. Leaf size=196 \[ \frac{2 e^{\frac{1}{2} (2 d+2 e x+i \pi )} F^{c (a+b x)} (e-b c \log (F)) \, _2F_1\left (2,\frac{b c \log (F)}{e}+1;\frac{b c \log (F)}{e}+2;-e^{\frac{1}{2} (2 d+2 e x+i \pi )}\right )}{3 e^2 f^2}+\frac{b c \log (F) \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{i \pi }{4}\right ) F^{c (a+b x)}}{6 e^2 f^2}+\frac{\tanh \left (\frac{d}{2}+\frac{e x}{2}+\frac{i \pi }{4}\right ) \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{i \pi }{4}\right ) F^{c (a+b x)}}{6 e f^2} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.114658, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {5496, 5490, 5492} \[ \frac{2 e^{\frac{1}{2} (2 d+2 e x+i \pi )} F^{c (a+b x)} (e-b c \log (F)) \, _2F_1\left (2,\frac{b c \log (F)}{e}+1;\frac{b c \log (F)}{e}+2;-e^{\frac{1}{2} (2 d+2 e x+i \pi )}\right )}{3 e^2 f^2}+\frac{b c \log (F) \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{i \pi }{4}\right ) F^{c (a+b x)}}{6 e^2 f^2}+\frac{\tanh \left (\frac{d}{2}+\frac{e x}{2}+\frac{i \pi }{4}\right ) \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{i \pi }{4}\right ) F^{c (a+b x)}}{6 e f^2} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 5496
Rule 5490
Rule 5492
Rubi steps
\begin{align*} \int \frac{F^{c (a+b x)}}{(f+i f \sinh (d+e x))^2} \, dx &=\frac{\int F^{c (a+b x)} \text{sech}^4\left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right ) \, dx}{4 f^2}\\ &=\frac{b c F^{c (a+b x)} \log (F) \text{sech}^2\left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right )}{6 e^2 f^2}+\frac{F^{c (a+b x)} \text{sech}^2\left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right ) \tanh \left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right )}{6 e f^2}+\frac{\left (1-\frac{b^2 c^2 \log ^2(F)}{e^2}\right ) \int F^{c (a+b x)} \text{sech}^2\left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right ) \, dx}{6 f^2}\\ &=\frac{2 e^{\frac{1}{2} (2 d+i \pi +2 e x)} F^{c (a+b x)} \, _2F_1\left (2,1+\frac{b c \log (F)}{e};2+\frac{b c \log (F)}{e};-e^{\frac{1}{2} (2 d+i \pi +2 e x)}\right ) (e-b c \log (F))}{3 e^2 f^2}+\frac{b c F^{c (a+b x)} \log (F) \text{sech}^2\left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right )}{6 e^2 f^2}+\frac{F^{c (a+b x)} \text{sech}^2\left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right ) \tanh \left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right )}{6 e f^2}\\ \end{align*}
Mathematica [A] time = 3.01077, size = 255, normalized size = 1.3 \[ \frac{F^{c (a+b x)} \left (\cosh \left (\frac{1}{2} (d+e x)\right )+i \sinh \left (\frac{1}{2} (d+e x)\right )\right ) \left ((1-i) \left (e^2-b^2 c^2 \log ^2(F)\right ) \left (\cosh \left (\frac{1}{2} (d+e x)\right )+i \sinh \left (\frac{1}{2} (d+e x)\right )\right )^3 \left (-1+(1+i) \, _2F_1\left (1,\frac{b c \log (F)}{e};\frac{b c \log (F)}{e}+1;-i (\cosh (d+e x)+\sinh (d+e x))\right )\right )+2 \sinh \left (\frac{1}{2} (d+e x)\right ) \left (e^2-b^2 c^2 \log ^2(F)\right ) \left (\cosh \left (\frac{1}{2} (d+e x)\right )+i \sinh \left (\frac{1}{2} (d+e x)\right )\right )^2+e (b c \log (F)+i e) \left (\cosh \left (\frac{1}{2} (d+e x)\right )+i \sinh \left (\frac{1}{2} (d+e x)\right )\right )+2 e^2 \sinh \left (\frac{1}{2} (d+e x)\right )\right )}{3 e^3 (f+i f \sinh (d+e x))^2} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [F] time = 0.06, size = 0, normalized size = 0. \begin{align*} \int{\frac{{F}^{c \left ( bx+a \right ) }}{ \left ( f+if\sinh \left ( ex+d \right ) \right ) ^{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (6 \, e^{2} e^{\left (e x + d\right )} +{\left (-2 i \, b^{2} c^{2} e^{\left (2 \, e x + 2 \, d\right )} - 4 \, b^{2} c^{2} e^{\left (e x + d\right )} + 2 i \, b^{2} c^{2}\right )} \log \left (F\right )^{2} - 2 i \, e^{2} +{\left (-2 i \, b c e e^{\left (2 \, e x + 2 \, d\right )} - 2 \, b c e e^{\left (e x + d\right )}\right )} \log \left (F\right )\right )} F^{b c x + a c} +{\left (3 \, e^{3} f^{2} e^{\left (3 \, e x + 3 \, d\right )} - 9 i \, e^{3} f^{2} e^{\left (2 \, e x + 2 \, d\right )} - 9 \, e^{3} f^{2} e^{\left (e x + d\right )} + 3 i \, e^{3} f^{2}\right )}{\rm integral}\left (\frac{{\left (2 i \, b^{3} c^{3} \log \left (F\right )^{3} - 2 i \, b c e^{2} \log \left (F\right )\right )} F^{b c x + a c}}{3 \,{\left (e^{3} f^{2} e^{\left (e x + d\right )} - i \, e^{3} f^{2}\right )}}, x\right )}{3 \, e^{3} f^{2} e^{\left (3 \, e x + 3 \, d\right )} - 9 i \, e^{3} f^{2} e^{\left (2 \, e x + 2 \, d\right )} - 9 \, e^{3} f^{2} e^{\left (e x + d\right )} + 3 i \, e^{3} f^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (b x + a\right )} c}}{{\left (i \, f \sinh \left (e x + d\right ) + f\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]