Optimal. Leaf size=184 \[ -\frac{2 e^{i (d+e x)} F^{c (a+b x)} (e+i b c \log (F)) \text{Hypergeometric2F1}\left (2,1-\frac{i b c \log (F)}{e},2-\frac{i b c \log (F)}{e},i e^{i (d+e x)}\right )}{3 e^2 f^2}-\frac{b c \log (F) \csc ^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right ) F^{c (a+b x)}}{6 e^2 f^2}-\frac{\cot \left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right ) \csc ^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right ) F^{c (a+b x)}}{6 e f^2} \]
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Rubi [A] time = 0.0997605, antiderivative size = 184, normalized size of antiderivative = 1., 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 = {4456, 4448, 4450} \[ -\frac{2 e^{i (d+e x)} F^{c (a+b x)} (e+i b c \log (F)) \, _2F_1\left (2,1-\frac{i b c \log (F)}{e};2-\frac{i b c \log (F)}{e};i e^{i (d+e x)}\right )}{3 e^2 f^2}-\frac{b c \log (F) \csc ^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right ) F^{c (a+b x)}}{6 e^2 f^2}-\frac{\cot \left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right ) \csc ^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{\pi }{4}\right ) F^{c (a+b x)}}{6 e f^2} \]
Antiderivative was successfully verified.
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Rule 4456
Rule 4448
Rule 4450
Rubi steps
\begin{align*} \int \frac{F^{c (a+b x)}}{(f+f \sin (d+e x))^2} \, dx &=\frac{\int F^{c (a+b x)} \sec ^4\left (\frac{d}{2}-\frac{\pi }{4}+\frac{e x}{2}\right ) \, dx}{4 f^2}\\ &=-\frac{F^{c (a+b x)} \cot \left (\frac{d}{2}+\frac{\pi }{4}+\frac{e x}{2}\right ) \csc ^2\left (\frac{d}{2}+\frac{\pi }{4}+\frac{e x}{2}\right )}{6 e f^2}-\frac{b c F^{c (a+b x)} \csc ^2\left (\frac{d}{2}+\frac{\pi }{4}+\frac{e x}{2}\right ) \log (F)}{6 e^2 f^2}+\frac{\left (1+\frac{b^2 c^2 \log ^2(F)}{e^2}\right ) \int F^{c (a+b x)} \sec ^2\left (\frac{d}{2}-\frac{\pi }{4}+\frac{e x}{2}\right ) \, dx}{6 f^2}\\ &=-\frac{F^{c (a+b x)} \cot \left (\frac{d}{2}+\frac{\pi }{4}+\frac{e x}{2}\right ) \csc ^2\left (\frac{d}{2}+\frac{\pi }{4}+\frac{e x}{2}\right )}{6 e f^2}-\frac{b c F^{c (a+b x)} \csc ^2\left (\frac{d}{2}+\frac{\pi }{4}+\frac{e x}{2}\right ) \log (F)}{6 e^2 f^2}-\frac{2 e^{i (d+e x)} F^{c (a+b x)} \, _2F_1\left (2,1-\frac{i b c \log (F)}{e};2-\frac{i b c \log (F)}{e};i e^{i (d+e x)}\right ) (e+i b c \log (F))}{3 e^2 f^2}\\ \end{align*}
Mathematica [A] time = 3.23028, size = 240, normalized size = 1.3 \[ \frac{F^{c (a+b x)} \left (\sin \left (\frac{1}{2} (d+e x)\right )+\cos \left (\frac{1}{2} (d+e x)\right )\right ) \left ((-1+i) \left (b^2 c^2 \log ^2(F)+e^2\right ) \left (\sin \left (\frac{1}{2} (d+e x)\right )+\cos \left (\frac{1}{2} (d+e x)\right )\right )^3 \left (1-(1-i) \text{Hypergeometric2F1}\left (1,-\frac{i b c \log (F)}{e},1-\frac{i b c \log (F)}{e},-\sin (d+e x)+i \cos (d+e x)\right )\right )+2 \sin \left (\frac{1}{2} (d+e x)\right ) \left (b^2 c^2 \log ^2(F)+e^2\right ) \left (\sin \left (\frac{1}{2} (d+e x)\right )+\cos \left (\frac{1}{2} (d+e x)\right )\right )^2-e (b c \log (F)+e) \left (\sin \left (\frac{1}{2} (d+e x)\right )+\cos \left (\frac{1}{2} (d+e x)\right )\right )+2 e^2 \sin \left (\frac{1}{2} (d+e x)\right )\right )}{3 e^3 f^2 (\sin (d+e x)+1)^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.373, size = 0, normalized size = 0. \begin{align*} \int{\frac{{F}^{c \left ( bx+a \right ) }}{ \left ( f+f\sin \left ( ex+d \right ) \right ) ^{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{F^{b c x + a c}}{f^{2} \cos \left (e x + d\right )^{2} - 2 \, f^{2} \sin \left (e x + d\right ) - 2 \, f^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{F^{a c} F^{b c x}}{\sin ^{2}{\left (d + e x \right )} + 2 \sin{\left (d + e x \right )} + 1}\, dx}{f^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (b x + a\right )} c}}{{\left (f \sin \left (e x + d\right ) + f\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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