Optimal. Leaf size=163 \[ \frac {a (4+m) \, _2F_1\left (1,-1+m;m;\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^{-1+m}}{4 f (1-m)}-\frac {a^2 \sin ^2(e+f x) (a+a \sin (e+f x))^{-1+m}}{f m (a-a \sin (e+f x))}+\frac {(a+a \sin (e+f x))^{-1+m} \left (a \left (2-3 m-m^2\right )+2 a m \sin (e+f x)\right )}{2 f (1-m) m (1-\sin (e+f x))} \]
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Rubi [A]
time = 0.10, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2786, 102, 151,
70} \begin {gather*} -\frac {a^2 \sin ^2(e+f x) (a \sin (e+f x)+a)^{m-1}}{f m (a-a \sin (e+f x))}+\frac {a (m+4) (a \sin (e+f x)+a)^{m-1} \, _2F_1\left (1,m-1;m;\frac {1}{2} (\sin (e+f x)+1)\right )}{4 f (1-m)}+\frac {\left (2 a m \sin (e+f x)+a \left (-m^2-3 m+2\right )\right ) (a \sin (e+f x)+a)^{m-1}}{2 f (1-m) m (1-\sin (e+f x))} \end {gather*}
Antiderivative was successfully verified.
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Rule 70
Rule 102
Rule 151
Rule 2786
Rubi steps
\begin {align*} \int (a+a \sin (e+f x))^m \tan ^3(e+f x) \, dx &=\frac {\text {Subst}\left (\int \frac {x^3 (a+x)^{-2+m}}{(a-x)^2} \, dx,x,a \sin (e+f x)\right )}{f}\\ &=-\frac {a^2 \sin ^2(e+f x) (a+a \sin (e+f x))^{-1+m}}{f m (a-a \sin (e+f x))}-\frac {\text {Subst}\left (\int \frac {x (a+x)^{-2+m} \left (-2 a^2-a m x\right )}{(a-x)^2} \, dx,x,a \sin (e+f x)\right )}{f m}\\ &=-\frac {a^2 \sin ^2(e+f x) (a+a \sin (e+f x))^{-1+m}}{f m (a-a \sin (e+f x))}+\frac {(a+a \sin (e+f x))^{-1+m} \left (a \left (2-3 m-m^2\right )+2 a m \sin (e+f x)\right )}{2 f (1-m) m (1-\sin (e+f x))}-\frac {\left (a^2 (4+m)\right ) \text {Subst}\left (\int \frac {(a+x)^{-2+m}}{a-x} \, dx,x,a \sin (e+f x)\right )}{2 f}\\ &=\frac {a (4+m) \, _2F_1\left (1,-1+m;m;\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^{-1+m}}{4 f (1-m)}-\frac {a^2 \sin ^2(e+f x) (a+a \sin (e+f x))^{-1+m}}{f m (a-a \sin (e+f x))}+\frac {(a+a \sin (e+f x))^{-1+m} \left (a \left (2-3 m-m^2\right )+2 a m \sin (e+f x)\right )}{2 f (1-m) m (1-\sin (e+f x))}\\ \end {align*}
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Mathematica [A]
time = 0.17, size = 105, normalized size = 0.64 \begin {gather*} \frac {a (a (1+\sin (e+f x)))^{-1+m} \left (-2 \left (-2+3 m+m^2\right )-m (4+m) \, _2F_1\left (1,-1+m;m;\frac {1}{2} (1+\sin (e+f x))\right ) (-1+\sin (e+f x))+4 m \sin (e+f x)+4 (-1+m) \sin ^2(e+f x)\right )}{4 f (-1+m) m (-1+\sin (e+f x))} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (\tan ^{3}\left (f x +e \right )\right )\, dx\]
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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \tan ^{3}{\left (e + f x \right )}\, dx \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 {\mathrm {tan}\left (e+f\,x\right )}^3\,{\left (a+a\,\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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