Optimal. Leaf size=293 \[ \frac {\sqrt {2} \left (A d (c (1-m)-d m)-B \left (d^2-c^2 m-c d m\right )\right ) F_1\left (\frac {1}{2}+m;\frac {1}{2},1;\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{(c-d)^2 d (c+d) f (1+2 m) \sqrt {1-\sin (e+f x)}}+\frac {2^{\frac {1}{2}+m} (B c-A d) m \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right ) f}-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))} \]
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Rubi [A]
time = 0.42, antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3063, 3065,
2731, 2730, 2867, 142, 141} \begin {gather*} \frac {\sqrt {2} \cos (e+f x) \left (A d (c (1-m)-d m)-B \left (c^2 (-m)-c d m+d^2\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 )}{d f (2 m+1) (c-d)^2 (c+d) \sqrt {1-\sin (e+f x)}}+\frac {2^{m+\frac {1}{2}} m (B c-A d) \cos (e+f x) (\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 )}-\frac {(B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^m}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))} \end {gather*}
Antiderivative was successfully verified.
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Rule 141
Rule 142
Rule 2730
Rule 2731
Rule 2867
Rule 3063
Rule 3065
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))^2} \, dx &=-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}-\frac {\int \frac {(a+a \sin (e+f x))^m (-a (A c-B d+B c m-A d m)+a (B c-A d) m \sin (e+f x))}{c+d \sin (e+f x)} \, dx}{a \left (c^2-d^2\right )}\\ &=-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}-\frac {((B c-A d) m) \int (a+a \sin (e+f x))^m \, dx}{d \left (c^2-d^2\right )}+\frac {\left (A d (c (1-m)-d m)-B \left (d^2-c^2 m-c d m\right )\right ) \int \frac {(a+a \sin (e+f x))^m}{c+d \sin (e+f x)} \, dx}{d \left (c^2-d^2\right )}\\ &=-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac {\left (a^2 \left (A d (c (1-m)-d m)-B \left (d^2-c^2 m-c d m\right )\right ) \cos (e+f x)\right ) \text {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 )}{d \left (c^2-d^2\right ) f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}-\frac {\left ((B c-A d) m (1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (1+\sin (e+f x))^m \, dx}{d \left (c^2-d^2\right )}\\ &=\frac {2^{\frac {1}{2}+m} (B c-A d) m \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 ) f}-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac {\left (a^2 \left (A d (c (1-m)-d m)-B \left (d^2-c^2 m-c d m\right )\right ) \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}}\right ) \text {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 )}{\sqrt {2} d \left (c^2-d^2\right ) f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}\\ &=\frac {\sqrt {2} \left (A d (c (1-m)-d m)-B \left (d^2-c^2 m-c d m\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}{(c-d)^2 d (c+d) f (1+2 m) \sqrt {1-\sin (e+f x)}}+\frac {2^{\frac {1}{2}+m} (B c-A d) m \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 ) f}-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(651\) vs. \(2(293)=586\).
time = 1.63, size = 651, normalized size = 2.22 \begin {gather*} \frac {6 (c+d) \cos ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )^{-\frac {1}{2}+m} \cot \left (\frac {1}{4} (2 e+\pi +2 f x)\right ) (a (1+\sin (e+f x)))^m \left (\frac {(-B c+A d) F_1\left (\frac {1}{2};\frac {1}{2}-m,2;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\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+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )+\left (-8 d F_1\left (\frac {3}{2};\frac {1}{2}-m,3;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )+(c+d) (-1+2 m) F_1\left (\frac {3}{2};\frac {3}{2}-m,2;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )\right ) \cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )}+\frac {B F_1\left (\frac {1}{2};\frac {1}{2}-m,1;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right ) (c+d \sin (e+f x))}{-3 (c+d) F_1\left (\frac {1}{2};\frac {1}{2}-m,1;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )+\left (-4 d F_1\left (\frac {3}{2};\frac {1}{2}-m,2;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )+(c+d) (-1+2 m) F_1\left (\frac {3}{2};\frac {3}{2}-m,1;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )\right ) \cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )}\right ) \sin ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )^{\frac {1}{2}-m}}{d f (c+d \sin (e+f x))^2} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 2.25, size = 0, normalized size = 0.00 \[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )}{\left (c +d \sin \left (f x +e \right )\right )^{2}}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \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 )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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