Optimal. Leaf size=138 \[ -\frac{2 a^2 \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{2 a^2 \sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e}+\frac{a^2 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{d \sqrt{\sin (c+d x)}} \]
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Rubi [A] time = 0.307368, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3872, 2873, 2640, 2639, 2564, 329, 298, 203, 206, 2571} \[ -\frac{2 a^2 \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{2 a^2 \sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e}+\frac{a^2 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{d \sqrt{\sin (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 3872
Rule 2873
Rule 2640
Rule 2639
Rule 2564
Rule 329
Rule 298
Rule 203
Rule 206
Rule 2571
Rubi steps
\begin{align*} \int (a+a \sec (c+d x))^2 \sqrt{e \sin (c+d x)} \, dx &=\int (-a-a \cos (c+d x))^2 \sec ^2(c+d x) \sqrt{e \sin (c+d x)} \, dx\\ &=\int \left (a^2 \sqrt{e \sin (c+d x)}+2 a^2 \sec (c+d x) \sqrt{e \sin (c+d x)}+a^2 \sec ^2(c+d x) \sqrt{e \sin (c+d x)}\right ) \, dx\\ &=a^2 \int \sqrt{e \sin (c+d x)} \, dx+a^2 \int \sec ^2(c+d x) \sqrt{e \sin (c+d x)} \, dx+\left (2 a^2\right ) \int \sec (c+d x) \sqrt{e \sin (c+d x)} \, dx\\ &=\frac{a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e}-\frac{1}{2} a^2 \int \sqrt{e \sin (c+d x)} \, dx+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{1-\frac{x^2}{e^2}} \, dx,x,e \sin (c+d x)\right )}{d e}+\frac{\left (a^2 \sqrt{e \sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{\sqrt{\sin (c+d x)}}\\ &=\frac{2 a^2 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{d \sqrt{\sin (c+d x)}}+\frac{a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e}+\frac{\left (4 a^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{x^4}{e^2}} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d e}-\frac{\left (a^2 \sqrt{e \sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{2 \sqrt{\sin (c+d x)}}\\ &=\frac{a^2 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{d \sqrt{\sin (c+d x)}}+\frac{a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e}+\frac{\left (2 a^2 e\right ) \operatorname{Subst}\left (\int \frac{1}{e-x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d}-\frac{\left (2 a^2 e\right ) \operatorname{Subst}\left (\int \frac{1}{e+x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d}\\ &=-\frac{2 a^2 \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{2 a^2 \sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{a^2 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{d \sqrt{\sin (c+d x)}}+\frac{a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e}\\ \end{align*}
Mathematica [C] time = 1.93746, size = 168, normalized size = 1.22 \[ -\frac{2 a^2 \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \sqrt{e \sin (c+d x)} \sec ^4\left (\frac{1}{2} \sin ^{-1}(\sin (c+d x))\right ) \left (\sin ^{\frac{3}{2}}(c+d x) \sqrt{\cos ^2(c+d x)} \text{Hypergeometric2F1}\left (\frac{3}{4},\frac{3}{2},\frac{7}{4},\sin ^2(c+d x)\right )-3 \sin ^{\frac{3}{2}}(c+d x)+3 \sqrt{\cos ^2(c+d x)} \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )-3 \sqrt{\cos ^2(c+d x)} \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )\right )}{3 d \sqrt{\sin (c+d x)}} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 2.126, size = 219, normalized size = 1.6 \begin{align*}{\frac{{a}^{2}}{2\,d\cos \left ( dx+c \right ) } \left ( \sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},{\frac{\sqrt{2}}{2}} \right ) e-2\,\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticE} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},1/2\,\sqrt{2} \right ) e+4\,\cos \left ( dx+c \right ) \sqrt{e\sin \left ( dx+c \right ) }\sqrt{e}{\it Artanh} \left ({\frac{\sqrt{e\sin \left ( dx+c \right ) }}{\sqrt{e}}} \right ) -4\,\cos \left ( dx+c \right ) \sqrt{e\sin \left ( dx+c \right ) }\sqrt{e}\arctan \left ({\frac{\sqrt{e\sin \left ( dx+c \right ) }}{\sqrt{e}}} \right ) -2\,e \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2\,e \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}} \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} \sqrt{e \sin \left (d x + c\right )}\,{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 )} \sqrt{e \sin \left (d x + c\right )}, 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} \sqrt{e \sin \left (d x + c\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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