Optimal. Leaf size=375 \[ \frac {\sqrt {2} \cos (e+f x) (d (A (2 m+5)+C (3-2 m))+2 c (2 C m+C)) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)} F_1\left (m+\frac {1}{2};\frac {1}{2},-\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (2 m+5) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}+\frac {2 \sqrt {2} C (d m-c (m+1)) \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \sqrt {c+d \sin (e+f x)} F_1\left (m+\frac {3}{2};\frac {1}{2},-\frac {1}{2};m+\frac {5}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3) (2 m+5) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{3/2}}{d f (2 m+5)} \]
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Rubi [A] time = 0.86, antiderivative size = 374, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3046, 2987, 2788, 140, 139, 138} \[ \frac {\sqrt {2} \cos (e+f x) (A d (2 m+5)+2 c (2 C m+C)+C d (3-2 m)) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)} F_1\left (m+\frac {1}{2};\frac {1}{2},-\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (2 m+5) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}+\frac {2 \sqrt {2} C (d m-c (m+1)) \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \sqrt {c+d \sin (e+f x)} F_1\left (m+\frac {3}{2};\frac {1}{2},-\frac {1}{2};m+\frac {5}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3) (2 m+5) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{3/2}}{d f (2 m+5)} \]
Antiderivative was successfully verified.
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Rule 138
Rule 139
Rule 140
Rule 2788
Rule 2987
Rule 3046
Rubi steps
\begin {align*} \int (a+a \sin (e+f x))^m \sqrt {c+d \sin (e+f x)} \left (A+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))^{3/2}}{d f (5+2 m)}+\frac {2 \int (a+a \sin (e+f x))^m \sqrt {c+d \sin (e+f x)} \left (\frac {1}{2} a \left (2 A d \left (\frac {5}{2}+m\right )+2 C \left (\frac {3 d}{2}+c m\right )\right )+a C (d m-c (1+m)) \sin (e+f x)\right ) \, dx}{a d (5+2 m)}\\ &=-\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2}}{d f (5+2 m)}+\frac {(2 C (d m-c (1+m))) \int (a+a \sin (e+f x))^{1+m} \sqrt {c+d \sin (e+f x)} \, dx}{a d (5+2 m)}+\frac {(C d (3-2 m)+A d (5+2 m)+2 c (C+2 C m)) \int (a+a \sin (e+f x))^m \sqrt {c+d \sin (e+f x)} \, dx}{d (5+2 m)}\\ &=-\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2}}{d f (5+2 m)}+\frac {(2 a C (d m-c (1+m)) \cos (e+f x)) \operatorname {Subst}\left (\int \frac {(a+a x)^{\frac {1}{2}+m} \sqrt {c+d x}}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{d f (5+2 m) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}+\frac {\left (a^2 (C d (3-2 m)+A d (5+2 m)+2 c (C+2 C m)) \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m} \sqrt {c+d x}}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{d f (5+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))^{3/2}}{d f (5+2 m)}+\frac {\left (\sqrt {2} a C (d m-c (1+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 {c+d x}}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{d f (5+2 m) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}+\frac {\left (a^2 (C d (3-2 m)+A d (5+2 m)+2 c (C+2 C 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 {c+d x}}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} d f (5+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))^{3/2}}{d f (5+2 m)}+\frac {\left (\sqrt {2} a C (d m-c (1+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} \sqrt {\frac {a c}{a c-a d}+\frac {a d x}{a c-a d}}}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{d f (5+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^2 (C d (3-2 m)+A d (5+2 m)+2 c (C+2 C 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} \sqrt {\frac {a c}{a c-a d}+\frac {a d x}{a c-a d}}}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} d f (5+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))^{3/2}}{d f (5+2 m)}+\frac {\sqrt {2} (C d (3-2 m)+A d (5+2 m)+2 c (C+2 C 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 {c+d \sin (e+f x)}}{d f (1+2 m) (5+2 m) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}+\frac {2 \sqrt {2} C (d m-c (1+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 {c+d \sin (e+f x)}}{d f (3+2 m) (5+2 m) (a-a \sin (e+f x)) \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}\\ \end {align*}
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Mathematica [B] time = 7.77, size = 1874, normalized size = 5.00 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (C \cos \left (f x + e\right )^{2} - A - C\right )} \sqrt {d \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sin \left (f x + e\right )^{2} + A\right )} \sqrt {d \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 3.57, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {c +d \sin \left (f x +e \right )}\, \left (A +C \left (\sin ^{2}\left (f x +e \right )\right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sin \left (f x + e\right )^{2} + A\right )} \sqrt {d \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (C\,{\sin \left (e+f\,x\right )}^2+A\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\sqrt {c+d\,\sin \left (e+f\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + C \sin ^{2}{\left (e + f x \right )}\right ) \sqrt {c + d \sin {\left (e + f x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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