Optimal. Leaf size=215 \[ -\frac{2 \sqrt{2} (A-C) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};-\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt{\sin (e+f x)+1}}-\frac{4 \sqrt{2} C \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};-\frac{3}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt{\sin (e+f x)+1}} \]
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Rubi [A] time = 0.234322, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3017, 2755, 139, 138, 2784} \[ -\frac{2 \sqrt{2} (A-C) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};-\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt{\sin (e+f x)+1}}-\frac{4 \sqrt{2} C \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};-\frac{3}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt{\sin (e+f x)+1}} \]
Antiderivative was successfully verified.
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Rule 3017
Rule 2755
Rule 139
Rule 138
Rule 2784
Rubi steps
\begin{align*} \int (a+b \sin (e+f x))^m \left (A+(A+C) \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx &=(A-C) \int (1+\sin (e+f x)) (a+b \sin (e+f x))^m \, dx+C \int (1+\sin (e+f x))^2 (a+b \sin (e+f x))^m \, dx\\ &=\frac{((A-C) \cos (e+f x)) \operatorname{Subst}\left (\int \frac{\sqrt{1+x} (a+b x)^m}{\sqrt{1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}+\frac{(C \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(1+x)^{3/2} (a+b x)^m}{\sqrt{1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}\\ &=\frac{\left ((A-C) \cos (e+f x) (a+b \sin (e+f x))^m \left (-\frac{a+b \sin (e+f x)}{-a-b}\right )^{-m}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+x} \left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^m}{\sqrt{1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}+\frac{\left (C \cos (e+f x) (a+b \sin (e+f x))^m \left (-\frac{a+b \sin (e+f x)}{-a-b}\right )^{-m}\right ) \operatorname{Subst}\left (\int \frac{(1+x)^{3/2} \left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^m}{\sqrt{1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}\\ &=-\frac{4 \sqrt{2} C F_1\left (\frac{1}{2};-\frac{3}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m}}{f \sqrt{1+\sin (e+f x)}}-\frac{2 \sqrt{2} (A-C) F_1\left (\frac{1}{2};-\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m}}{f \sqrt{1+\sin (e+f x)}}\\ \end{align*}
Mathematica [F] time = 12.6691, size = 0, normalized size = 0. \[ \int (a+b \sin (e+f x))^m \left (A+(A+C) \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.44, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+ \left ( A+C \right ) \sin \left ( fx+e \right ) +C \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sin \left (f x + e\right )^{2} +{\left (A + C\right )} \sin \left (f x + e\right ) + A\right )}{\left (b \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \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 (-{\left (C \cos \left (f x + e\right )^{2} -{\left (A + C\right )} \sin \left (f x + e\right ) - A - C\right )}{\left (b \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sin \left (f x + e\right )^{2} +{\left (A + C\right )} \sin \left (f x + e\right ) + A\right )}{\left (b \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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