Optimal. Leaf size=195 \[ \frac{2 a^2 (e \sin (c+d x))^{m+1} \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)}+\frac{a^2 \cos (c+d x) (e \sin (c+d x))^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1) \sqrt{\cos ^2(c+d x)}}+\frac{a^2 \sqrt{\cos ^2(c+d x)} \sec (c+d x) (e \sin (c+d x))^{m+1} \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)} \]
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Rubi [A] time = 0.285148, antiderivative size = 195, 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 = {3872, 2873, 2643, 2564, 364, 2577} \[ \frac{2 a^2 (e \sin (c+d x))^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)}+\frac{a^2 \cos (c+d x) (e \sin (c+d x))^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1) \sqrt{\cos ^2(c+d x)}}+\frac{a^2 \sqrt{\cos ^2(c+d x)} \sec (c+d x) (e \sin (c+d x))^{m+1} \, _2F_1\left (\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)} \]
Antiderivative was successfully verified.
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Rule 3872
Rule 2873
Rule 2643
Rule 2564
Rule 364
Rule 2577
Rubi steps
\begin{align*} \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^m \, dx &=\int (-a-a \cos (c+d x))^2 \sec ^2(c+d x) (e \sin (c+d x))^m \, dx\\ &=\int \left (a^2 (e \sin (c+d x))^m+2 a^2 \sec (c+d x) (e \sin (c+d x))^m+a^2 \sec ^2(c+d x) (e \sin (c+d x))^m\right ) \, dx\\ &=a^2 \int (e \sin (c+d x))^m \, dx+a^2 \int \sec ^2(c+d x) (e \sin (c+d x))^m \, dx+\left (2 a^2\right ) \int \sec (c+d x) (e \sin (c+d x))^m \, dx\\ &=\frac{a^2 \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m) \sqrt{\cos ^2(c+d x)}}+\frac{a^2 \sqrt{\cos ^2(c+d x)} \, _2F_1\left (\frac{3}{2},\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{x^m}{1-\frac{x^2}{e^2}} \, dx,x,e \sin (c+d x)\right )}{d e}\\ &=\frac{a^2 \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m) \sqrt{\cos ^2(c+d x)}}+\frac{2 a^2 \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\frac{a^2 \sqrt{\cos ^2(c+d x)} \, _2F_1\left (\frac{3}{2},\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) (e \sin (c+d x))^{1+m}}{d e (1+m)}\\ \end{align*}
Mathematica [F] time = 0.93236, size = 0, normalized size = 0. \[ \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^m \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.79, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sec \left ( dx+c \right ) \right ) ^{2} \left ( e\sin \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 \sin \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 \sin \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(-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 (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \sin \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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