Optimal. Leaf size=252 \[ \frac{e^5 (e \tan (c+d x))^{m-5} \text{Hypergeometric2F1}\left (1,\frac{m-5}{2},\frac{m-3}{2},-\tan ^2(c+d x)\right )}{a^3 d (5-m)}-\frac{e^5 \sec ^3(c+d x) \cos ^2(c+d x)^{\frac{m-2}{2}} (e \tan (c+d x))^{m-5} \text{Hypergeometric2F1}\left (\frac{m-5}{2},\frac{m-2}{2},\frac{m-3}{2},\sin ^2(c+d x)\right )}{a^3 d (5-m)}-\frac{3 e^5 \sec (c+d x) \cos ^2(c+d x)^{\frac{m-4}{2}} (e \tan (c+d x))^{m-5} \text{Hypergeometric2F1}\left (\frac{m-5}{2},\frac{m-4}{2},\frac{m-3}{2},\sin ^2(c+d x)\right )}{a^3 d (5-m)}+\frac{3 e^5 (e \tan (c+d x))^{m-5}}{a^3 d (5-m)} \]
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Rubi [A] time = 0.341305, antiderivative size = 252, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3888, 3886, 3476, 364, 2617, 2607, 32} \[ \frac{e^5 (e \tan (c+d x))^{m-5} \, _2F_1\left (1,\frac{m-5}{2};\frac{m-3}{2};-\tan ^2(c+d x)\right )}{a^3 d (5-m)}-\frac{e^5 \sec ^3(c+d x) \cos ^2(c+d x)^{\frac{m-2}{2}} (e \tan (c+d x))^{m-5} \, _2F_1\left (\frac{m-5}{2},\frac{m-2}{2};\frac{m-3}{2};\sin ^2(c+d x)\right )}{a^3 d (5-m)}-\frac{3 e^5 \sec (c+d x) \cos ^2(c+d x)^{\frac{m-4}{2}} (e \tan (c+d x))^{m-5} \, _2F_1\left (\frac{m-5}{2},\frac{m-4}{2};\frac{m-3}{2};\sin ^2(c+d x)\right )}{a^3 d (5-m)}+\frac{3 e^5 (e \tan (c+d x))^{m-5}}{a^3 d (5-m)} \]
Antiderivative was successfully verified.
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Rule 3888
Rule 3886
Rule 3476
Rule 364
Rule 2617
Rule 2607
Rule 32
Rubi steps
\begin{align*} \int \frac{(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx &=\frac{e^6 \int (-a+a \sec (c+d x))^3 (e \tan (c+d x))^{-6+m} \, dx}{a^6}\\ &=\frac{e^6 \int \left (-a^3 (e \tan (c+d x))^{-6+m}+3 a^3 \sec (c+d x) (e \tan (c+d x))^{-6+m}-3 a^3 \sec ^2(c+d x) (e \tan (c+d x))^{-6+m}+a^3 \sec ^3(c+d x) (e \tan (c+d x))^{-6+m}\right ) \, dx}{a^6}\\ &=-\frac{e^6 \int (e \tan (c+d x))^{-6+m} \, dx}{a^3}+\frac{e^6 \int \sec ^3(c+d x) (e \tan (c+d x))^{-6+m} \, dx}{a^3}+\frac{\left (3 e^6\right ) \int \sec (c+d x) (e \tan (c+d x))^{-6+m} \, dx}{a^3}-\frac{\left (3 e^6\right ) \int \sec ^2(c+d x) (e \tan (c+d x))^{-6+m} \, dx}{a^3}\\ &=-\frac{3 e^5 \cos ^2(c+d x)^{\frac{1}{2} (-4+m)} \, _2F_1\left (\frac{1}{2} (-5+m),\frac{1}{2} (-4+m);\frac{1}{2} (-3+m);\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac{e^5 \cos ^2(c+d x)^{\frac{1}{2} (-2+m)} \, _2F_1\left (\frac{1}{2} (-5+m),\frac{1}{2} (-2+m);\frac{1}{2} (-3+m);\sin ^2(c+d x)\right ) \sec ^3(c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac{\left (3 e^6\right ) \operatorname{Subst}\left (\int (e x)^{-6+m} \, dx,x,\tan (c+d x)\right )}{a^3 d}-\frac{e^7 \operatorname{Subst}\left (\int \frac{x^{-6+m}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a^3 d}\\ &=\frac{3 e^5 (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}+\frac{e^5 \, _2F_1\left (1,\frac{1}{2} (-5+m);\frac{1}{2} (-3+m);-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac{3 e^5 \cos ^2(c+d x)^{\frac{1}{2} (-4+m)} \, _2F_1\left (\frac{1}{2} (-5+m),\frac{1}{2} (-4+m);\frac{1}{2} (-3+m);\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac{e^5 \cos ^2(c+d x)^{\frac{1}{2} (-2+m)} \, _2F_1\left (\frac{1}{2} (-5+m),\frac{1}{2} (-2+m);\frac{1}{2} (-3+m);\sin ^2(c+d x)\right ) \sec ^3(c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}\\ \end{align*}
Mathematica [F] time = 1.47004, size = 0, normalized size = 0. \[ \int \frac{(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.451, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( e\tan \left ( dx+c \right ) \right ) ^{m}}{ \left ( a+a\sec \left ( dx+c \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 \frac{\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}\,{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 (\frac{\left (e \tan \left (d x + c\right )\right )^{m}}{a^{3} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \sec \left (d x + c\right )^{2} + 3 \, a^{3} \sec \left (d x + c\right ) + a^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\left (e \tan{\left (c + d x \right )}\right )^{m}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx}{a^{3}} \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 \frac{\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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