Optimal. Leaf size=389 \[ \frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^m (2 c (2 C m+C)-d (-A (2 m+3)+2 B m+3 B+2 C m-C)) \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{1}{2};\frac{1}{2},\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (2 m+3) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}-\frac{\sqrt{2} \cos (e+f x) (2 c C (m+1)-d (B (2 m+3)+2 C m)) (a \sin (e+f x)+a)^{m+1} \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{3}{2};\frac{1}{2},\frac{1}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3)^2 \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}-\frac{2 C \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt{c+d \sin (e+f x)}}{d f (2 m+3)} \]
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Rubi [A] time = 0.928285, antiderivative size = 386, 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} \[ \frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^m (2 c (2 C m+C)-d (-A (2 m+3)+2 B m+3 B+2 C m-C)) \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{1}{2};\frac{1}{2},\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (2 m+3) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{2} \cos (e+f x) (B d (2 m+3)-2 c C (m+1)+2 C d m) (a \sin (e+f x)+a)^{m+1} \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{3}{2};\frac{1}{2},\frac{1}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3)^2 \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}-\frac{2 C \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt{c+d \sin (e+f x)}}{d f (2 m+3)} \]
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 \frac{(a+a \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right )}{\sqrt{c+d \sin (e+f x)}} \, dx &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{c+d \sin (e+f x)}}{d f (3+2 m)}+\frac{2 \int \frac{(a+a \sin (e+f x))^m \left (\frac{1}{2} a (A d (3+2 m)+C (d+2 c m))+\frac{1}{2} a (2 C d m-2 c C (1+m)+B d (3+2 m)) \sin (e+f x)\right )}{\sqrt{c+d \sin (e+f x)}} \, dx}{a d (3+2 m)}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{c+d \sin (e+f x)}}{d f (3+2 m)}+\frac{(2 C d m-2 c C (1+m)+B d (3+2 m)) \int \frac{(a+a \sin (e+f x))^{1+m}}{\sqrt{c+d \sin (e+f x)}} \, dx}{a d (3+2 m)}+\frac{(2 c (C+2 C m)-d (3 B-C+2 B m+2 C m-A (3+2 m))) \int \frac{(a+a \sin (e+f x))^m}{\sqrt{c+d \sin (e+f x)}} \, dx}{d (3+2 m)}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{c+d \sin (e+f x)}}{d f (3+2 m)}+\frac{(a (2 C d m-2 c C (1+m)+B d (3+2 m)) \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m}}{\sqrt{a-a x} \sqrt{c+d x}} \, dx,x,\sin (e+f x)\right )}{d f (3+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 (3 B-C+2 B m+2 C m-A (3+2 m))) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m}}{\sqrt{a-a x} \sqrt{c+d x}} \, dx,x,\sin (e+f x)\right )}{d f (3+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 \sqrt{c+d \sin (e+f x)}}{d f (3+2 m)}+\frac{\left (a (2 C d m-2 c C (1+m)+B d (3+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}}{\sqrt{\frac{1}{2}-\frac{x}{2}} \sqrt{c+d x}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (3+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 (3 B-C+2 B m+2 C m-A (3+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}}{\sqrt{\frac{1}{2}-\frac{x}{2}} \sqrt{c+d x}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (3+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 \sqrt{c+d \sin (e+f x)}}{d f (3+2 m)}+\frac{\left (a (2 C d m-2 c C (1+m)+B d (3+2 m)) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} \sqrt{\frac{a (c+d \sin (e+f x))}{a c-a d}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m}}{\sqrt{\frac{1}{2}-\frac{x}{2}} \sqrt{\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (3+2 m) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\left (a^2 (2 c (C+2 C m)-d (3 B-C+2 B m+2 C m-A (3+2 m))) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} \sqrt{\frac{a (c+d \sin (e+f x))}{a c-a d}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m}}{\sqrt{\frac{1}{2}-\frac{x}{2}} \sqrt{\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (3+2 m) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{c+d \sin (e+f x)}}{d f (3+2 m)}+\frac{\sqrt{2} (2 c (C+2 C m)-d (3 B-C+2 B m+2 C m-A (3+2 m))) F_1\left (\frac{1}{2}+m;\frac{1}{2},\frac{1}{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{\frac{c+d \sin (e+f x)}{c-d}}}{d f (1+2 m) (3+2 m) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{2} (2 C d m-2 c C (1+m)+B d (3+2 m)) F_1\left (\frac{3}{2}+m;\frac{1}{2},\frac{1}{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{\frac{c+d \sin (e+f x)}{c-d}}}{d f (3+2 m)^2 (a-a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [B] time = 34.5322, size = 9783, normalized size = 25.15 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.748, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) +C \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{c+d\sin \left ( fx+e \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 \frac{{\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt{d \sin \left (f x + e\right ) + c}}\,{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 (-\frac{{\left (C \cos \left (f x + e\right )^{2} - B \sin \left (f x + e\right ) - A - C\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt{d \sin \left (f x + e\right ) + c}}, 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 \frac{{\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt{d \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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