Optimal. Leaf size=196 \[ \frac {2 e^{\frac {1}{2} (2 d+i \pi +2 e x)} F^{c (a+b x)} \, _2F_1\left (2,1+\frac {b c \log (F)}{e};2+\frac {b c \log (F)}{e};-e^{\frac {1}{2} (2 d+i \pi +2 e x)}\right ) (e-b c \log (F))}{3 e^2 f^2}+\frac {b c F^{c (a+b x)} \log (F) \text {sech}^2\left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right )}{6 e^2 f^2}+\frac {F^{c (a+b x)} \text {sech}^2\left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right ) \tanh \left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right )}{6 e f^2} \]
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Rubi [A]
time = 0.08, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5604, 5598,
5600} \begin {gather*} \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} \end {gather*}
Antiderivative was successfully verified.
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Rule 5598
Rule 5600
Rule 5604
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*}
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Mathematica [A]
time = 1.66, size = 255, normalized size = 1.30 \begin {gather*} \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 (e (i e+b c \log (F)) \left (\cosh \left (\frac {1}{2} (d+e x)\right )+i \sinh \left (\frac {1}{2} (d+e x)\right )\right )+(1-i) \left (-1+(1+i) \, _2F_1\left (1,\frac {b c \log (F)}{e};1+\frac {b c \log (F)}{e};-i (\cosh (d+e x)+\sinh (d+e x))\right )\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 )^3+2 e^2 \sinh \left (\frac {1}{2} (d+e x)\right )+2 \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 \sinh \left (\frac {1}{2} (d+e x)\right )\right )}{3 e^3 (f+i f \sinh (d+e x))^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 2.89, size = 0, normalized size = 0.00 \[\int \frac {F^{c \left (b x +a \right )}}{\left (f +i f \sinh \left (e x +d \right )\right )^{2}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {F^{a c} F^{b c x}}{\sinh ^{2}{\left (d + e x \right )} - 2 i \sinh {\left (d + e x \right )} - 1}\, dx}{f^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (f+f\,\mathrm {sinh}\left (d+e\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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