Optimal. Leaf size=163 \[ -\frac {a^2 \sin ^2(e+f x) (a \sin (e+f x)+a)^{m-1}}{f m (a-a \sin (e+f x))}+\frac {a (m+4) (a \sin (e+f x)+a)^{m-1} \, _2F_1\left (1,m-1;m;\frac {1}{2} (\sin (e+f x)+1)\right )}{4 f (1-m)}+\frac {\left (2 a m \sin (e+f x)+a \left (-m^2-3 m+2\right )\right ) (a \sin (e+f x)+a)^{m-1}}{2 f (1-m) m (1-\sin (e+f x))} \]
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Rubi [A] time = 0.15, 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 = {2707, 100, 146, 68} \[ -\frac {a^2 \sin ^2(e+f x) (a \sin (e+f x)+a)^{m-1}}{f m (a-a \sin (e+f x))}+\frac {a (m+4) (a \sin (e+f x)+a)^{m-1} \, _2F_1\left (1,m-1;m;\frac {1}{2} (\sin (e+f x)+1)\right )}{4 f (1-m)}+\frac {\left (2 a m \sin (e+f x)+a \left (-m^2-3 m+2\right )\right ) (a \sin (e+f x)+a)^{m-1}}{2 f (1-m) m (1-\sin (e+f x))} \]
Antiderivative was successfully verified.
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Rule 68
Rule 100
Rule 146
Rule 2707
Rubi steps
\begin {align*} \int (a+a \sin (e+f x))^m \tan ^3(e+f x) \, dx &=\frac {\operatorname {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 {\operatorname {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 ) \operatorname {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 = 0.26, size = 105, normalized size = 0.64 \[ \frac {a (a (\sin (e+f x)+1))^{m-1} \left (-m (m+4) (\sin (e+f x)-1) \, _2F_1\left (1,m-1;m;\frac {1}{2} (\sin (e+f x)+1)\right )+4 (m-1) \sin ^2(e+f x)+4 m \sin (e+f x)-2 \left (m^2+3 m-2\right )\right )}{4 f (m-1) m (\sin (e+f x)-1)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{m} \tan \left (f x + e\right )^{3}, x\right ) \]
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 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \tan \left (f x + e\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.26, 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 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \tan \left (f x + e\right )^{3}\,{d x} \]
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 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^3\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]
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 \[ \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \tan ^{3}{\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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