Optimal. Leaf size=151 \[ \frac {2 e^{d+e x} F^{c (a+b x)} (e-b c \log (F)) \, _2F_1\left (2,\frac {b c \log (F)}{e}+1;\frac {b c \log (F)}{e}+2;-e^{d+e x}\right )}{3 e^2 f^2}+\frac {b c \log (F) \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}\right ) F^{c (a+b x)}}{6 e^2 f^2}+\frac {\tanh \left (\frac {d}{2}+\frac {e x}{2}\right ) \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}\right ) F^{c (a+b x)}}{6 e f^2} \]
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Rubi [A] time = 0.10, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {5497, 5490, 5492} \[ \frac {2 e^{d+e x} F^{c (a+b x)} (e-b c \log (F)) \, _2F_1\left (2,\frac {b c \log (F)}{e}+1;\frac {b c \log (F)}{e}+2;-e^{d+e x}\right )}{3 e^2 f^2}+\frac {b c \log (F) \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}\right ) F^{c (a+b x)}}{6 e^2 f^2}+\frac {\tanh \left (\frac {d}{2}+\frac {e x}{2}\right ) \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}\right ) F^{c (a+b x)}}{6 e f^2} \]
Antiderivative was successfully verified.
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Rule 5490
Rule 5492
Rule 5497
Rubi steps
\begin {align*} \int \frac {F^{c (a+b x)}}{(f+f \cosh (d+e x))^2} \, dx &=\frac {\int F^{c (a+b x)} \text {sech}^4\left (\frac {d}{2}+\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 {e x}{2}\right )}{6 e^2 f^2}+\frac {F^{c (a+b x)} \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}\right ) \tanh \left (\frac {d}{2}+\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 {e x}{2}\right ) \, dx}{6 f^2}\\ &=\frac {2 e^{d+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^{d+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 {e x}{2}\right )}{6 e^2 f^2}+\frac {F^{c (a+b x)} \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}\right ) \tanh \left (\frac {d}{2}+\frac {e x}{2}\right )}{6 e f^2}\\ \end {align*}
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Mathematica [A] time = 0.35, size = 127, normalized size = 0.84 \[ \frac {2 \cosh \left (\frac {1}{2} (d+e x)\right ) F^{c (a+b x)} \left (4 e^{d+e x} \cosh ^3\left (\frac {1}{2} (d+e x)\right ) (e-b c \log (F)) \, _2F_1\left (2,\frac {b c \log (F)}{e}+1;\frac {b c \log (F)}{e}+2;-e^{d+e x}\right )+b c \log (F) \cosh \left (\frac {1}{2} (d+e x)\right )+e \sinh \left (\frac {1}{2} (d+e x)\right )\right )}{3 e^2 f^2 (\cosh (d+e x)+1)^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {F^{b c x + a c}}{f^{2} \cosh \left (e x + d\right )^{2} + 2 \, f^{2} \cosh \left (e x + d\right ) + f^{2}}, x\right ) \]
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 \[ \int \frac {F^{{\left (b x + a\right )} c}}{{\left (f \cosh \left (e x + d\right ) + f\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.39, size = 0, normalized size = 0.00 \[ \int \frac {F^{c \left (b x +a \right )}}{\left (f +f \cosh \left (e x +d \right )\right )^{2}}\, dx \]
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 \[ \int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (f+f\,\mathrm {cosh}\left (d+e\,x\right )\right )}^2} \,d x \]
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 \[ \frac {\int \frac {F^{a c} F^{b c x}}{\cosh ^{2}{\left (d + e x \right )} + 2 \cosh {\left (d + e x \right )} + 1}\, dx}{f^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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