Optimal. Leaf size=311 \[ -\frac{\sqrt{2} \cos (e+f x) \left (a d (a d-2 b c (m+2))+b^2 \left (c^2 (m+2)+d^2 (m+1)\right )\right ) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt{\sin (e+f x)+1}}+\frac{\sqrt{2} d (a+b) \cos (e+f x) (a d-2 b c (m+2)) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m-1;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt{\sin (e+f x)+1}}-\frac{d^2 \cos (e+f x) (a+b \sin (e+f x))^{m+1}}{b f (m+2)} \]
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Rubi [A] time = 0.439, antiderivative size = 311, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2791, 2756, 2665, 139, 138} \[ -\frac{\sqrt{2} \cos (e+f x) \left (a d (a d-2 b c (m+2))+b^2 \left (c^2 (m+2)+d^2 (m+1)\right )\right ) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt{\sin (e+f x)+1}}+\frac{\sqrt{2} d (a+b) \cos (e+f x) (a d-2 b c (m+2)) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m-1;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt{\sin (e+f x)+1}}-\frac{d^2 \cos (e+f x) (a+b \sin (e+f x))^{m+1}}{b f (m+2)} \]
Antiderivative was successfully verified.
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Rule 2791
Rule 2756
Rule 2665
Rule 139
Rule 138
Rubi steps
\begin{align*} \int (a+b \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx &=-\frac{d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}+\frac{\int (a+b \sin (e+f x))^m \left (b \left (d^2 (1+m)+c^2 (2+m)\right )-d (a d-2 b c (2+m)) \sin (e+f x)\right ) \, dx}{b (2+m)}\\ &=-\frac{d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}-\frac{(d (a d-2 b c (2+m))) \int (a+b \sin (e+f x))^{1+m} \, dx}{b^2 (2+m)}+\frac{\left (a d (a d-2 b c (2+m))+b^2 \left (d^2 (1+m)+c^2 (2+m)\right )\right ) \int (a+b \sin (e+f x))^m \, dx}{b^2 (2+m)}\\ &=-\frac{d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}-\frac{(d (a d-2 b c (2+m)) \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(a+b x)^{1+m}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sin (e+f x)\right )}{b^2 f (2+m) \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}+\frac{\left (\left (a d (a d-2 b c (2+m))+b^2 \left (d^2 (1+m)+c^2 (2+m)\right )\right ) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^m}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sin (e+f x)\right )}{b^2 f (2+m) \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}\\ &=-\frac{d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}+\frac{\left ((-a-b) d (a d-2 b c (2+m)) \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{\left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^{1+m}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sin (e+f x)\right )}{b^2 f (2+m) \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}+\frac{\left (\left (a d (a d-2 b c (2+m))+b^2 \left (d^2 (1+m)+c^2 (2+m)\right )\right ) \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{\left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^m}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sin (e+f x)\right )}{b^2 f (2+m) \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}\\ &=-\frac{d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}+\frac{\sqrt{2} (a+b) d (a d-2 b c (2+m)) F_1\left (\frac{1}{2};\frac{1}{2},-1-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}}{b^2 f (2+m) \sqrt{1+\sin (e+f x)}}-\frac{\sqrt{2} \left (a d (a d-2 b c (2+m))+b^2 \left (d^2 (1+m)+c^2 (2+m)\right )\right ) 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}}{b^2 f (2+m) \sqrt{1+\sin (e+f x)}}\\ \end{align*}
Mathematica [F] time = 17.9353, size = 0, normalized size = 0. \[ \int (a+b \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.369, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sin \left ( fx+e \right ) \right ) ^{m} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{2}\, 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 (d \sin \left (f x + e\right ) + c\right )}^{2}{\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 (d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}\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 (d \sin \left (f x + e\right ) + c\right )}^{2}{\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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