Optimal. Leaf size=196 \[ \frac {2 e^{\frac {1}{2} (2 d+2 e x+i \pi )} F^{c (a+b x)} (e-b c \log (F)) \, _2F_1\left (2,\frac {b c \log (F)}{e}+1;\frac {b c \log (F)}{e}+2;-e^{\frac {1}{2} (2 d+2 e x+i \pi )}\right )}{3 e^2 f^2}+\frac {b c \log (F) \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}+\frac {i \pi }{4}\right ) F^{c (a+b x)}}{6 e^2 f^2}+\frac {\tanh \left (\frac {d}{2}+\frac {e x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}+\frac {i \pi }{4}\right ) F^{c (a+b x)}}{6 e f^2} \]
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Rubi [A] time = 0.11, 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 = {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.
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Rule 5490
Rule 5492
Rule 5496
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 = 3.34, size = 255, normalized size = 1.30 \[ \frac {F^{c (a+b x)} \left (\cosh \left (\frac {1}{2} (d+e x)\right )+i \sinh \left (\frac {1}{2} (d+e x)\right )\right ) \left ((1-i) \left (e^2-b^2 c^2 \log ^2(F)\right ) \left (\cosh \left (\frac {1}{2} (d+e x)\right )+i \sinh \left (\frac {1}{2} (d+e x)\right )\right )^3 \left (-1+(1+i) \, _2F_1\left (1,\frac {b c \log (F)}{e};\frac {b c \log (F)}{e}+1;-i (\cosh (d+e x)+\sinh (d+e x))\right )\right )+2 \sinh \left (\frac {1}{2} (d+e x)\right ) \left (e^2-b^2 c^2 \log ^2(F)\right ) \left (\cosh \left (\frac {1}{2} (d+e x)\right )+i \sinh \left (\frac {1}{2} (d+e x)\right )\right )^2+e (b c \log (F)+i e) \left (\cosh \left (\frac {1}{2} (d+e x)\right )+i \sinh \left (\frac {1}{2} (d+e x)\right )\right )+2 e^2 \sinh \left (\frac {1}{2} (d+e x)\right )\right )}{3 e^3 (f+i f \sinh (d+e x))^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.49, size = 0, normalized size = 0.00 \[ \frac {{\left (6 \, e^{2} e^{\left (e x + d\right )} + {\left (-2 i \, b^{2} c^{2} e^{\left (2 \, e x + 2 \, d\right )} - 4 \, b^{2} c^{2} e^{\left (e x + d\right )} + 2 i \, b^{2} c^{2}\right )} \log \relax (F)^{2} - 2 i \, e^{2} - 2 \, {\left (i \, b c e e^{\left (2 \, e x + 2 \, d\right )} + b c e e^{\left (e x + d\right )}\right )} \log \relax (F)\right )} F^{b c x + a c} + {\left (3 \, e^{3} f^{2} e^{\left (3 \, e x + 3 \, d\right )} - 9 i \, e^{3} f^{2} e^{\left (2 \, e x + 2 \, d\right )} - 9 \, e^{3} f^{2} e^{\left (e x + d\right )} + 3 i \, e^{3} f^{2}\right )} {\rm integral}\left (\frac {{\left (2 i \, b^{3} c^{3} \log \relax (F)^{3} - 2 i \, b c e^{2} \log \relax (F)\right )} F^{b c x + a c}}{3 \, e^{3} f^{2} e^{\left (e x + d\right )} - 3 i \, e^{3} f^{2}}, x\right )}{3 \, e^{3} f^{2} e^{\left (3 \, e x + 3 \, d\right )} - 9 i \, e^{3} f^{2} e^{\left (2 \, e x + 2 \, d\right )} - 9 \, e^{3} f^{2} e^{\left (e x + d\right )} + 3 i \, e^{3} f^{2}} \]
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 \[ \int \frac {F^{{\left (b x + a\right )} c}}{{\left (i \, f \sinh \left (e x + d\right ) + f\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.59, 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 \[ \text {result too large to display} \]
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 \[ \int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (f+f\,\mathrm {sinh}\left (d+e\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]
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 \[ - \frac {\int \frac {F^{a c} F^{b c x}}{\sinh ^{2}{\left (d + e x \right )} - 2 i \sinh {\left (d + e x \right )} - 1}\, dx}{f^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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