Optimal. Leaf size=406 \[ \frac{\sqrt{2} (c-d) \cos (e+f x) (a \sin (e+f x)+a)^m (2 c (2 C m+C)-d (-A (2 m+7)+2 B m+7 B+2 C m-5 C)) \sqrt{c+d \sin (e+f x)} F_1\left (m+\frac{1}{2};\frac{1}{2},-\frac{3}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (2 m+7) \sqrt{1-\sin (e+f x)} \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}-\frac{\sqrt{2} (c-d) \cos (e+f x) (2 c C (m+1)-d (B (2 m+7)+2 C m)) (a \sin (e+f x)+a)^{m+1} \sqrt{c+d \sin (e+f x)} F_1\left (m+\frac{3}{2};\frac{1}{2},-\frac{3}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3) (2 m+7) \sqrt{1-\sin (e+f x)} \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}-\frac{2 C \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{5/2}}{d f (2 m+7)} \]
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Rubi [A] time = 1.03439, antiderivative size = 403, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 6, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {3045, 2987, 2788, 140, 139, 138} \[ \frac{\sqrt{2} (c-d) \cos (e+f x) (a \sin (e+f x)+a)^m (2 c (2 C m+C)-d (-A (2 m+7)+2 B m+7 B+2 C m-5 C)) \sqrt{c+d \sin (e+f x)} F_1\left (m+\frac{1}{2};\frac{1}{2},-\frac{3}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (2 m+7) \sqrt{1-\sin (e+f x)} \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}+\frac{\sqrt{2} (c-d) \cos (e+f x) (B d (2 m+7)-2 c C (m+1)+2 C d m) (a \sin (e+f x)+a)^{m+1} \sqrt{c+d \sin (e+f x)} F_1\left (m+\frac{3}{2};\frac{1}{2},-\frac{3}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3) (2 m+7) \sqrt{1-\sin (e+f x)} \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}-\frac{2 C \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{5/2}}{d f (2 m+7)} \]
Antiderivative was successfully verified.
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Rule 3045
Rule 2987
Rule 2788
Rule 140
Rule 139
Rule 138
Rubi steps
\begin{align*} \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{5/2}}{d f (7+2 m)}+\frac{2 \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \left (\frac{1}{2} a \left (2 A d \left (\frac{7}{2}+m\right )+2 C \left (\frac{5 d}{2}+c m\right )\right )+\frac{1}{2} a (2 C d m-2 c C (1+m)+B d (7+2 m)) \sin (e+f x)\right ) \, dx}{a d (7+2 m)}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{5/2}}{d f (7+2 m)}+\frac{(2 C d m-2 c C (1+m)+B d (7+2 m)) \int (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^{3/2} \, dx}{a d (7+2 m)}+\frac{(2 c (C+2 C m)-d (7 B-5 C+2 B m+2 C m-A (7+2 m))) \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \, dx}{d (7+2 m)}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{5/2}}{d f (7+2 m)}+\frac{(a (2 C d m-2 c C (1+m)+B d (7+2 m)) \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m} (c+d x)^{3/2}}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{d f (7+2 m) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}+\frac{\left (a^2 (2 c (C+2 C m)-d (7 B-5 C+2 B m+2 C m-A (7+2 m))) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m} (c+d x)^{3/2}}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{d f (7+2 m) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{5/2}}{d f (7+2 m)}+\frac{\left (a (2 C d m-2 c C (1+m)+B d (7+2 m)) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m} (c+d x)^{3/2}}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (7+2 m) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}+\frac{\left (a^2 (2 c (C+2 C m)-d (7 B-5 C+2 B m+2 C m-A (7+2 m))) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m} (c+d x)^{3/2}}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (7+2 m) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{5/2}}{d f (7+2 m)}+\frac{\left ((a c-a d) (2 C d m-2 c C (1+m)+B d (7+2 m)) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} \sqrt{c+d \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m} \left (\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}\right )^{3/2}}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (7+2 m) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)} \sqrt{\frac{a (c+d \sin (e+f x))}{a c-a d}}}+\frac{\left (a (a c-a d) (2 c (C+2 C m)-d (7 B-5 C+2 B m+2 C m-A (7+2 m))) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} \sqrt{c+d \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m} \left (\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}\right )^{3/2}}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (7+2 m) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)} \sqrt{\frac{a (c+d \sin (e+f x))}{a c-a d}}}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{5/2}}{d f (7+2 m)}+\frac{\sqrt{2} (c-d) (2 c (C+2 C m)-d (7 B-5 C+2 B m+2 C m-A (7+2 m))) F_1\left (\frac{1}{2}+m;\frac{1}{2},-\frac{3}{2};\frac{3}{2}+m;\frac{1}{2} (1+\sin (e+f x)),-\frac{d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{c+d \sin (e+f x)}}{d f (1+2 m) (7+2 m) \sqrt{1-\sin (e+f x)} \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}+\frac{\sqrt{2} (c-d) (2 C d m-2 c C (1+m)+B d (7+2 m)) F_1\left (\frac{3}{2}+m;\frac{1}{2},-\frac{3}{2};\frac{5}{2}+m;\frac{1}{2} (1+\sin (e+f x)),-\frac{d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) \sqrt{1-\sin (e+f x)} (a+a \sin (e+f x))^{1+m} \sqrt{c+d \sin (e+f x)}}{d f (3+2 m) (7+2 m) (a-a \sin (e+f x)) \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}\\ \end{align*}
Mathematica [B] time = 9.74848, size = 6591, normalized size = 16.23 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.727, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \left ( A+B\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} + B \sin \left (f x + e\right ) + A\right )}{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}{\left (a \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 ({\left (C c + B d\right )} \cos \left (f x + e\right )^{2} -{\left (A + C\right )} c - B d +{\left (C d \cos \left (f x + e\right )^{2} - B c -{\left (A + C\right )} d\right )} \sin \left (f x + e\right )\right )} \sqrt{d \sin \left (f x + e\right ) + c}{\left (a \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(-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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