Optimal. Leaf size=169 \[ -\frac{e^3 (e \tan (c+d x))^{m-3} \text{Hypergeometric2F1}\left (1,\frac{m-3}{2},\frac{m-1}{2},-\tan ^2(c+d x)\right )}{a^2 d (3-m)}+\frac{2 e^3 \sec (c+d x) \cos ^2(c+d x)^{\frac{m-2}{2}} (e \tan (c+d x))^{m-3} \text{Hypergeometric2F1}\left (\frac{m-3}{2},\frac{m-2}{2},\frac{m-1}{2},\sin ^2(c+d x)\right )}{a^2 d (3-m)}-\frac{e^3 (e \tan (c+d x))^{m-3}}{a^2 d (3-m)} \]
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Rubi [A] time = 0.272374, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{e^3 (e \tan (c+d x))^{m-3} \, _2F_1\left (1,\frac{m-3}{2};\frac{m-1}{2};-\tan ^2(c+d x)\right )}{a^2 d (3-m)}+\frac{2 e^3 \sec (c+d x) \cos ^2(c+d x)^{\frac{m-2}{2}} (e \tan (c+d x))^{m-3} \, _2F_1\left (\frac{m-3}{2},\frac{m-2}{2};\frac{m-1}{2};\sin ^2(c+d x)\right )}{a^2 d (3-m)}-\frac{e^3 (e \tan (c+d x))^{m-3}}{a^2 d (3-m)} \]
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))^2} \, dx &=\frac{e^4 \int (-a+a \sec (c+d x))^2 (e \tan (c+d x))^{-4+m} \, dx}{a^4}\\ &=\frac{e^4 \int \left (a^2 (e \tan (c+d x))^{-4+m}-2 a^2 \sec (c+d x) (e \tan (c+d x))^{-4+m}+a^2 \sec ^2(c+d x) (e \tan (c+d x))^{-4+m}\right ) \, dx}{a^4}\\ &=\frac{e^4 \int (e \tan (c+d x))^{-4+m} \, dx}{a^2}+\frac{e^4 \int \sec ^2(c+d x) (e \tan (c+d x))^{-4+m} \, dx}{a^2}-\frac{\left (2 e^4\right ) \int \sec (c+d x) (e \tan (c+d x))^{-4+m} \, dx}{a^2}\\ &=\frac{2 e^3 \cos ^2(c+d x)^{\frac{1}{2} (-2+m)} \, _2F_1\left (\frac{1}{2} (-3+m),\frac{1}{2} (-2+m);\frac{1}{2} (-1+m);\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-3+m}}{a^2 d (3-m)}+\frac{e^4 \operatorname{Subst}\left (\int (e x)^{-4+m} \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac{e^5 \operatorname{Subst}\left (\int \frac{x^{-4+m}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a^2 d}\\ &=-\frac{e^3 (e \tan (c+d x))^{-3+m}}{a^2 d (3-m)}-\frac{e^3 \, _2F_1\left (1,\frac{1}{2} (-3+m);\frac{1}{2} (-1+m);-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-3+m}}{a^2 d (3-m)}+\frac{2 e^3 \cos ^2(c+d x)^{\frac{1}{2} (-2+m)} \, _2F_1\left (\frac{1}{2} (-3+m),\frac{1}{2} (-2+m);\frac{1}{2} (-1+m);\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-3+m}}{a^2 d (3-m)}\\ \end{align*}
Mathematica [F] time = 0.858509, size = 0, normalized size = 0. \[ \int \frac{(e \tan (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.359, 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 ) ^{2}}}\, 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 )}^{2}}\,{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^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, 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 ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx}{a^{2}} \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 )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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