Optimal. Leaf size=139 \[ \frac {2 \sqrt {2} F_1\left (\frac {1}{2}+m;-\frac {1}{2},1+m;\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \sec (e+f x) \sqrt {1-\sin (e+f x)} (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m}{(c-d) f (1+2 m)} \]
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Rubi [A]
time = 0.15, antiderivative size = 139, 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 = {3087, 145, 144,
143} \begin {gather*} \frac {2 \sqrt {2} \sqrt {1-\sin (e+f x)} \sec (e+f x) (a \sin (e+f x)+a)^{m+1} (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m F_1\left (m+\frac {1}{2};-\frac {1}{2},m+1;m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) (c-d)} \end {gather*}
Antiderivative was successfully verified.
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Rule 143
Rule 144
Rule 145
Rule 3087
Rubi steps
\begin {align*} \int (a-a \sin (e+f x)) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx &=\frac {\left (\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}\right ) \text {Subst}\left (\int \sqrt {a-a x} (a+a x)^{-\frac {1}{2}+m} (c+d x)^{-1-m} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\left (\sqrt {2} \sec (e+f x) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}\right ) \text {Subst}\left (\int \sqrt {\frac {1}{2}-\frac {x}{2}} (a+a x)^{-\frac {1}{2}+m} (c+d x)^{-1-m} \, dx,x,\sin (e+f x)\right )}{f \sqrt {\frac {a-a \sin (e+f x)}{a}}}\\ &=\frac {\left (\sqrt {2} a \sec (e+f x) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{-m} \left (\frac {a (c+d \sin (e+f x))}{a c-a d}\right )^m\right ) \text {Subst}\left (\int \sqrt {\frac {1}{2}-\frac {x}{2}} (a+a x)^{-\frac {1}{2}+m} \left (\frac {a c}{a c-a d}+\frac {a d x}{a c-a d}\right )^{-1-m} \, dx,x,\sin (e+f x)\right )}{(a c-a d) f \sqrt {\frac {a-a \sin (e+f x)}{a}}}\\ &=\frac {2 \sqrt {2} F_1\left (\frac {1}{2}+m;-\frac {1}{2},1+m;\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \sec (e+f x) \sqrt {1-\sin (e+f x)} (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m}{(c-d) f (1+2 m)}\\ \end {align*}
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Mathematica [F]
time = 5.88, size = 0, normalized size = 0.00 \begin {gather*} \int (a-a \sin (e+f x)) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.50, size = 0, normalized size = 0.00 \[\int \left (a -a \sin \left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{-1-m}\, dx\]
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+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (a-a\,\sin \left (e+f\,x\right )\right )}{{\left (c+d\,\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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