Optimal. Leaf size=121 \[ \frac {2^{\frac {9}{4}-m} c (g \cos (e+f x))^{5/2} \, _2F_1\left (\frac {1}{4} (-1+4 m),\frac {1}{4} (5+4 m);\frac {1}{4} (9+4 m);\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{-\frac {1}{4}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f g (5+4 m)} \]
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Rubi [A]
time = 0.21, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2932, 2768, 72,
71} \begin {gather*} \frac {c 2^{\frac {9}{4}-m} (g \cos (e+f x))^{5/2} (1-\sin (e+f x))^{m-\frac {1}{4}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{-m-1} \, _2F_1\left (\frac {1}{4} (4 m-1),\frac {1}{4} (4 m+5);\frac {1}{4} (4 m+9);\frac {1}{2} (\sin (e+f x)+1)\right )}{f g (4 m+5)} \end {gather*}
Antiderivative was successfully verified.
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Rule 71
Rule 72
Rule 2768
Rule 2932
Rubi steps
\begin {align*} \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m} \, dx &=\left ((g \cos (e+f x))^{-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int (g \cos (e+f x))^{\frac {3}{2}+2 m} (c-c \sin (e+f x))^{-2 m} \, dx\\ &=\frac {\left (c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{\frac {1}{2} \left (-\frac {5}{2}-2 m\right )+m} (c+c \sin (e+f x))^{\frac {1}{2} \left (-\frac {5}{2}-2 m\right )}\right ) \text {Subst}\left (\int (c-c x)^{-2 m+\frac {1}{2} \left (\frac {1}{2}+2 m\right )} (c+c x)^{\frac {1}{2} \left (\frac {1}{2}+2 m\right )} \, dx,x,\sin (e+f x)\right )}{f g}\\ &=\frac {\left (2^{\frac {1}{4}-m} c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{\frac {1}{4}+\frac {1}{2} \left (-\frac {5}{2}-2 m\right )} \left (\frac {c-c \sin (e+f x)}{c}\right )^{-\frac {1}{4}+m} (c+c \sin (e+f x))^{\frac {1}{2} \left (-\frac {5}{2}-2 m\right )}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{-2 m+\frac {1}{2} \left (\frac {1}{2}+2 m\right )} (c+c x)^{\frac {1}{2} \left (\frac {1}{2}+2 m\right )} \, dx,x,\sin (e+f x)\right )}{f g}\\ &=\frac {2^{\frac {9}{4}-m} c (g \cos (e+f x))^{5/2} \, _2F_1\left (\frac {1}{4} (-1+4 m),\frac {1}{4} (5+4 m);\frac {1}{4} (9+4 m);\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{-\frac {1}{4}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f g (5+4 m)}\\ \end {align*}
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Mathematica [F]
time = 23.34, size = 0, normalized size = 0.00 \begin {gather*} \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-m} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-m}\, dx\]
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(-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 = -0.01 \begin {gather*} \int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^m} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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