Optimal. Leaf size=183 \[ -\frac{a \left (3 b^2-a^2 (1-m)\right ) \tan (e+f x) \sec ^2(e+f x)^{m/2} (d \cos (e+f x))^m \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{3}{2};-\tan ^2(e+f x)\right )}{f (1-m)}+\frac{b (d \cos (e+f x))^m \left (2 (1-m) \left (b^2-a^2 (3-m)\right )+a b (4-m) m \tan (e+f x)\right )}{f m \left (m^2-3 m+2\right )}+\frac{b (a+b \tan (e+f x))^2 (d \cos (e+f x))^m}{f (2-m)} \]
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Rubi [A] time = 0.285021, antiderivative size = 175, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3515, 3512, 743, 780, 245} \[ \frac{a \left (a^2-\frac{3 b^2}{1-m}\right ) \tan (e+f x) \sec ^2(e+f x)^{m/2} (d \cos (e+f x))^m \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{3}{2};-\tan ^2(e+f x)\right )}{f}+\frac{b (d \cos (e+f x))^m \left (2 (1-m) \left (b^2-a^2 (3-m)\right )+a b (4-m) m \tan (e+f x)\right )}{f m \left (m^2-3 m+2\right )}+\frac{b (a+b \tan (e+f x))^2 (d \cos (e+f x))^m}{f (2-m)} \]
Antiderivative was successfully verified.
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Rule 3515
Rule 3512
Rule 743
Rule 780
Rule 245
Rubi steps
\begin{align*} \int (d \cos (e+f x))^m (a+b \tan (e+f x))^3 \, dx &=\left ((d \cos (e+f x))^m (d \sec (e+f x))^m\right ) \int (d \sec (e+f x))^{-m} (a+b \tan (e+f x))^3 \, dx\\ &=\frac{\left ((d \cos (e+f x))^m \sec ^2(e+f x)^{m/2}\right ) \operatorname{Subst}\left (\int (a+x)^3 \left (1+\frac{x^2}{b^2}\right )^{-1-\frac{m}{2}} \, dx,x,b \tan (e+f x)\right )}{b f}\\ &=\frac{b (d \cos (e+f x))^m (a+b \tan (e+f x))^2}{f (2-m)}+\frac{\left (b (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2}\right ) \operatorname{Subst}\left (\int (a+x) \left (-2+\frac{a^2 (2-m)}{b^2}+\frac{a (4-m) x}{b^2}\right ) \left (1+\frac{x^2}{b^2}\right )^{-1-\frac{m}{2}} \, dx,x,b \tan (e+f x)\right )}{f (2-m)}\\ &=\frac{b (d \cos (e+f x))^m (a+b \tan (e+f x))^2}{f (2-m)}+\frac{b (d \cos (e+f x))^m \left (2 \left (b^2-a^2 (3-m)\right ) (1-m)+a b (4-m) m \tan (e+f x)\right )}{f m \left (2-3 m+m^2\right )}-\frac{\left (a \left (3 b^2-a^2 (1-m)\right ) (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2}\right ) \operatorname{Subst}\left (\int \left (1+\frac{x^2}{b^2}\right )^{-1-\frac{m}{2}} \, dx,x,b \tan (e+f x)\right )}{b f (1-m)}\\ &=-\frac{a \left (3 b^2-a^2 (1-m)\right ) (d \cos (e+f x))^m \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{3}{2};-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{m/2} \tan (e+f x)}{f (1-m)}+\frac{b (d \cos (e+f x))^m (a+b \tan (e+f x))^2}{f (2-m)}+\frac{b (d \cos (e+f x))^m \left (2 \left (b^2-a^2 (3-m)\right ) (1-m)+a b (4-m) m \tan (e+f x)\right )}{f m \left (2-3 m+m^2\right )}\\ \end{align*}
Mathematica [A] time = 59.0595, size = 215, normalized size = 1.17 \[ \frac{\cos ^3(e+f x) (a+b \tan (e+f x))^3 (d \cos (e+f x))^m \left (b \left (\left (3 a^2 (m-2)+2 b^2\right ) \left (\sec ^2(e+f x)^{m/2}-1\right )+a b (m-2) m \tan ^3(e+f x) \sec ^2(e+f x)^{m/2} \, _2F_1\left (\frac{3}{2},\frac{m}{2}+1;\frac{5}{2};-\tan ^2(e+f x)\right )-b^2 m \tan ^2(e+f x)\right )+a^3 (m-2) m \tan (e+f x) \sec ^2(e+f x)^{m/2} \, _2F_1\left (\frac{1}{2},\frac{m}{2}+1;\frac{3}{2};-\tan ^2(e+f x)\right )\right )}{f (m-2) m (a \cos (e+f x)+b \sin (e+f x))^3} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.644, size = 0, normalized size = 0. \begin{align*} \int \left ( d\cos \left ( fx+e \right ) \right ) ^{m} \left ( a+b\tan \left ( fx+e \right ) \right ) ^{3}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right ) + a\right )}^{3} \left (d \cos \left (f x + e\right )\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{3} \tan \left (f x + e\right )^{3} + 3 \, a b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} b \tan \left (f x + e\right ) + a^{3}\right )} \left (d \cos \left (f x + e\right )\right )^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right ) + a\right )}^{3} \left (d \cos \left (f x + e\right )\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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