Optimal. Leaf size=467 \[ \frac {\cos (e+f x) \left (B \left (c^3 (1-m) m+2 c^2 d (1-m) m-c d^2 \left (m^2-3 m+3\right )+2 d^3 m\right )-A d \left (-\left (c^2 \left (m^2-3 m+2\right )\right )+2 c d (2-m) m-d^2 \left (m^2-m+1\right )\right )\right ) (a \sin (e+f x)+a)^m F_1\left (m+\frac {1}{2};\frac {1}{2},1;m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{\sqrt {2} d f (2 m+1) (c-d)^3 (c+d)^2 \sqrt {1-\sin (e+f x)}}-\frac {2^{m-\frac {1}{2}} m \cos (e+f x) \left (A d (c (3-m)-d m)-B \left (c^2 (1-m)-c d m+2 d^2\right )\right ) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{d f \left (c^2-d^2\right )^2}+\frac {\cos (e+f x) \left (A d (c (3-m)-d m)-B \left (c^2 (1-m)-c d m+2 d^2\right )\right ) (a \sin (e+f x)+a)^m}{2 f \left (c^2-d^2\right )^2 (c+d \sin (e+f x))}-\frac {(B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^m}{2 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2} \]
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Rubi [A] time = 1.35, antiderivative size = 467, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2984, 2986, 2652, 2651, 2788, 137, 136} \[ \frac {\cos (e+f x) \left (B \left (2 c^2 d (1-m) m+c^3 (1-m) m-c d^2 \left (m^2-3 m+3\right )+2 d^3 m\right )-A d \left (c^2 \left (-\left (m^2-3 m+2\right )\right )+2 c d (2-m) m-d^2 \left (m^2-m+1\right )\right )\right ) (a \sin (e+f x)+a)^m F_1\left (m+\frac {1}{2};\frac {1}{2},1;m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{\sqrt {2} d f (2 m+1) (c-d)^3 (c+d)^2 \sqrt {1-\sin (e+f x)}}-\frac {2^{m-\frac {1}{2}} m \cos (e+f x) \left (A d (c (3-m)-d m)-B \left (c^2 (1-m)-c d m+2 d^2\right )\right ) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{d f \left (c^2-d^2\right )^2}+\frac {\cos (e+f x) \left (A d (c (3-m)-d m)-B \left (c^2 (1-m)-c d m+2 d^2\right )\right ) (a \sin (e+f x)+a)^m}{2 f \left (c^2-d^2\right )^2 (c+d \sin (e+f x))}-\frac {(B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^m}{2 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 136
Rule 137
Rule 2651
Rule 2652
Rule 2788
Rule 2984
Rule 2986
Rubi steps
\begin {align*} \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx &=-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac {\int \frac {(a+a \sin (e+f x))^m (-a (2 A c-2 B d+B c m-A d m)-a (B c-A d) (1-m) \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx}{2 a \left (c^2-d^2\right )}\\ &=-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac {\int \frac {(a+a \sin (e+f x))^m \left (-a^2 ((B c-A d) (1-m) (d-c m)-(c-d m) (2 A c-2 B d+B c m-A d m))+a^2 m \left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \sin (e+f x)\right )}{c+d \sin (e+f x)} \, dx}{2 a^2 \left (c^2-d^2\right )^2}\\ &=-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac {\left (m \left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right )\right ) \int (a+a \sin (e+f x))^m \, dx}{2 d \left (c^2-d^2\right )^2}+\frac {\left (B \left (2 d^3 m+c^3 (1-m) m+2 c^2 d (1-m) m-c d^2 \left (3-3 m+m^2\right )\right )-A d \left (2 c d (2-m) m-c^2 \left (2-3 m+m^2\right )-d^2 \left (1-m+m^2\right )\right )\right ) \int \frac {(a+a \sin (e+f x))^m}{c+d \sin (e+f x)} \, dx}{2 d \left (c^2-d^2\right )^2}\\ &=-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac {\left (a^2 \left (B \left (2 d^3 m+c^3 (1-m) m+2 c^2 d (1-m) m-c d^2 \left (3-3 m+m^2\right )\right )-A d \left (2 c d (2-m) m-c^2 \left (2-3 m+m^2\right )-d^2 \left (1-m+m^2\right )\right )\right ) \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m}}{\sqrt {a-a x} (c+d x)} \, dx,x,\sin (e+f x)\right )}{2 d \left (c^2-d^2\right )^2 f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}+\frac {\left (m \left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) (1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (1+\sin (e+f x))^m \, dx}{2 d \left (c^2-d^2\right )^2}\\ &=-\frac {2^{-\frac {1}{2}+m} m \left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right )^2 f}-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac {\left (a^2 \left (B \left (2 d^3 m+c^3 (1-m) m+2 c^2 d (1-m) m-c d^2 \left (3-3 m+m^2\right )\right )-A d \left (2 c d (2-m) m-c^2 \left (2-3 m+m^2\right )-d^2 \left (1-m+m^2\right )\right )\right ) \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}} (c+d x)} \, dx,x,\sin (e+f x)\right )}{2 \sqrt {2} d \left (c^2-d^2\right )^2 f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}\\ &=\frac {\left (B \left (2 d^3 m+c^3 (1-m) m+2 c^2 d (1-m) m-c d^2 \left (3-3 m+m^2\right )\right )-A d \left (2 c d (2-m) m-c^2 \left (2-3 m+m^2\right )-d^2 \left (1-m+m^2\right )\right )\right ) F_1\left (\frac {1}{2}+m;\frac {1}{2},1;\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 {2} (c-d)^3 d (c+d)^2 f (1+2 m) \sqrt {1-\sin (e+f x)}}-\frac {2^{-\frac {1}{2}+m} m \left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right )^2 f}-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}\\ \end {align*}
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Mathematica [A] time = 2.53, size = 651, normalized size = 1.39 \[ \frac {6 (c+d) \cot \left (\frac {1}{4} (2 e+2 f x+\pi )\right ) \sin ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right )^{\frac {1}{2}-m} \cos ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )^{m-\frac {1}{2}} (a (\sin (e+f x)+1))^m \left (\frac {(A d-B c) F_1\left (\frac {1}{2};\frac {1}{2}-m,3;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ) \left ((2 m-1) (c+d) F_1\left (\frac {3}{2};\frac {3}{2}-m,3;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )-12 d F_1\left (\frac {3}{2};\frac {1}{2}-m,4;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )-3 (c+d) F_1\left (\frac {1}{2};\frac {1}{2}-m,3;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}+\frac {B (c+d \sin (e+f x)) F_1\left (\frac {1}{2};\frac {1}{2}-m,2;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ) \left ((2 m-1) (c+d) F_1\left (\frac {3}{2};\frac {3}{2}-m,2;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )-8 d F_1\left (\frac {3}{2};\frac {1}{2}-m,3;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )-3 (c+d) F_1\left (\frac {1}{2};\frac {1}{2}-m,2;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}\right )}{d f (c+d \sin (e+f x))^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{3 \, c d^{2} \cos \left (f x + e\right )^{2} - c^{3} - 3 \, c d^{2} + {\left (d^{3} \cos \left (f x + e\right )^{2} - 3 \, c^{2} d - d^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]
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 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 12.57, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )}{\left (c +d \sin \left (f x +e \right )\right )^{3}}\, dx \]
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 \[ \int \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{3}}\,{d x} \]
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 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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