Optimal. Leaf size=161 \[ \frac{a^2 (e \tan (c+d x))^{m+1} \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )}{d e (m+1)}+\frac{2 a^2 \sec (c+d x) \cos ^2(c+d x)^{\frac{m+2}{2}} (e \tan (c+d x))^{m+1} \text{Hypergeometric2F1}\left (\frac{m+1}{2},\frac{m+2}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)}+\frac{a^2 (e \tan (c+d x))^{m+1}}{d e (m+1)} \]
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Rubi [A] time = 0.169426, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3886, 3476, 364, 2617, 2607, 32} \[ \frac{a^2 (e \tan (c+d x))^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\tan ^2(c+d x)\right )}{d e (m+1)}+\frac{2 a^2 \sec (c+d x) \cos ^2(c+d x)^{\frac{m+2}{2}} (e \tan (c+d x))^{m+1} \, _2F_1\left (\frac{m+1}{2},\frac{m+2}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)}+\frac{a^2 (e \tan (c+d x))^{m+1}}{d e (m+1)} \]
Antiderivative was successfully verified.
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Rule 3886
Rule 3476
Rule 364
Rule 2617
Rule 2607
Rule 32
Rubi steps
\begin{align*} \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^m \, dx &=\int \left (a^2 (e \tan (c+d x))^m+2 a^2 \sec (c+d x) (e \tan (c+d x))^m+a^2 \sec ^2(c+d x) (e \tan (c+d x))^m\right ) \, dx\\ &=a^2 \int (e \tan (c+d x))^m \, dx+a^2 \int \sec ^2(c+d x) (e \tan (c+d x))^m \, dx+\left (2 a^2\right ) \int \sec (c+d x) (e \tan (c+d x))^m \, dx\\ &=\frac{2 a^2 \cos ^2(c+d x)^{\frac{2+m}{2}} \, _2F_1\left (\frac{1+m}{2},\frac{2+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{1+m}}{d e (1+m)}+\frac{a^2 \operatorname{Subst}\left (\int (e x)^m \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (a^2 e\right ) \operatorname{Subst}\left (\int \frac{x^m}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{d}\\ &=\frac{a^2 (e \tan (c+d x))^{1+m}}{d e (1+m)}+\frac{a^2 \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{1+m}}{d e (1+m)}+\frac{2 a^2 \cos ^2(c+d x)^{\frac{2+m}{2}} \, _2F_1\left (\frac{1+m}{2},\frac{2+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{1+m}}{d e (1+m)}\\ \end{align*}
Mathematica [F] time = 0.999945, size = 0, normalized size = 0. \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^m \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.532, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sec \left ( dx+c \right ) \right ) ^{2} \left ( e\tan \left ( dx+c \right ) \right ) ^{m}\, 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 (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \tan \left (d x + c\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 (a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}\right )} \left (e \tan \left (d x + c\right )\right )^{m}, 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*} a^{2} \left (\int \left (e \tan{\left (c + d x \right )}\right )^{m}\, dx + \int 2 \left (e \tan{\left (c + d x \right )}\right )^{m} \sec{\left (c + d x \right )}\, dx + \int \left (e \tan{\left (c + d x \right )}\right )^{m} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \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 (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \tan \left (d x + c\right )\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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