Optimal. Leaf size=163 \[ \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}-\frac {2^{\frac {1}{2}-m} B \cos (e+f x) \, _2F_1\left (\frac {1}{2} (1+2 m),\frac {1}{2} (1+2 m);\frac {1}{2} (3+2 m);\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)} \]
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Rubi [A]
time = 0.21, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3051, 2824,
2768, 72, 71} \begin {gather*} \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1}}{f (2 m+1)}-\frac {B 2^{\frac {1}{2}-m} \cos (e+f x) (1-\sin (e+f x))^{m+\frac {1}{2}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac {1}{2} (2 m+1),\frac {1}{2} (2 m+1);\frac {1}{2} (2 m+3);\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)} \end {gather*}
Antiderivative was successfully verified.
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Rule 71
Rule 72
Rule 2768
Rule 2824
Rule 3051
Rubi steps
\begin {align*} \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-1-m} \, dx &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}-\frac {B \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m} \, dx}{c}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}-\frac {\left (B \cos ^{-2 m}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \cos ^{2 m}(e+f x) (c-c \sin (e+f x))^{-2 m} \, dx}{c}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}-\frac {\left (B c \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{\frac {1}{2} (-1-2 m)+m} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 m)}\right ) \text {Subst}\left (\int (c-c x)^{-2 m+\frac {1}{2} (-1+2 m)} (c+c x)^{\frac {1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}-\frac {\left (2^{-\frac {1}{2}-m} B c \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-\frac {1}{2}+\frac {1}{2} (-1-2 m)} \left (\frac {c-c \sin (e+f x)}{c}\right )^{\frac {1}{2}+m} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 m)}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{-2 m+\frac {1}{2} (-1+2 m)} (c+c x)^{\frac {1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}-\frac {2^{\frac {1}{2}-m} B \cos (e+f x) \, _2F_1\left (\frac {1}{2} (1+2 m),\frac {1}{2} (1+2 m);\frac {1}{2} (3+2 m);\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 (c-c \sin (e+f x))^{-1-m}}{f (1+2 m)}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in
optimal.
time = 11.56, size = 675, normalized size = 4.14 \begin {gather*} -\frac {2^{-m} (-3+2 m) \cos ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right ) \cot \left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right ) \sin ^{-2 m}\left (\frac {1}{2} \left (-e+\frac {\pi }{2}-f x\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^{-2 (-1-m)} (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-1-m} \left (8 B (1+2 m) F_1\left (\frac {1}{2}-m;-2 m,1;\frac {3}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right )\right ) \tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right )-(A+B) \left ((-1+2 m) \, _2F_1\left (-\frac {1}{2}-m,-2 m;\frac {1}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right )\right )+(1+2 m) \, _2F_1\left (\frac {1}{2}-m,-2 m;\frac {3}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right )\right ) \tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right )\right )\right )}{f \left (-1+4 m^2\right ) \left (-64 B m F_1\left (\frac {3}{2}-m;1-2 m,1;\frac {5}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right )\right ) \sin ^4\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right )-32 B F_1\left (\frac {3}{2}-m;-2 m,2;\frac {5}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right )\right ) \sin ^4\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right )+(-3+2 m) \left (-4 B F_1\left (\frac {1}{2}-m;-2 m,1;\frac {3}{2}-m;\tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right ),-\tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right )\right ) \sin ^2\left (\frac {1}{2} \left (-e+\frac {\pi }{2}-f x\right )\right )+(A+B) \left (1+\cos \left (\frac {1}{2} \left (-e+\frac {\pi }{2}-f x\right )\right )\right ) \left (1-\tan ^2\left (\frac {1}{4} \left (-e+\frac {\pi }{2}-f x\right )\right )\right )^{2 m}\right )\right )} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 0.45, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{-1-m}\, dx\]
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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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 = -0.01 \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-c\,\sin \left (e+f\,x\right )\right )}^{m+1}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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