Optimal. Leaf size=192 \[ -\frac{a^2 e^2 \sqrt{\sin (c+d x)} \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right )}{3 d \sqrt{e \sin (c+d x)}}+\frac{2 a^2 e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{2 a^2 e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}-\frac{4 a^2 e \sqrt{e \sin (c+d x)}}{d}-\frac{2 a^2 e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{a^2 e \sec (c+d x) \sqrt{e \sin (c+d x)}}{d} \]
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Rubi [A] time = 0.380084, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {3872, 2873, 2635, 2642, 2641, 2564, 321, 329, 212, 206, 203, 2566} \[ \frac{2 a^2 e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{2 a^2 e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}-\frac{a^2 e^2 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 d \sqrt{e \sin (c+d x)}}-\frac{4 a^2 e \sqrt{e \sin (c+d x)}}{d}-\frac{2 a^2 e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{a^2 e \sec (c+d x) \sqrt{e \sin (c+d x)}}{d} \]
Antiderivative was successfully verified.
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Rule 3872
Rule 2873
Rule 2635
Rule 2642
Rule 2641
Rule 2564
Rule 321
Rule 329
Rule 212
Rule 206
Rule 203
Rule 2566
Rubi steps
\begin{align*} \int (a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2} \, dx &=\int (-a-a \cos (c+d x))^2 \sec ^2(c+d x) (e \sin (c+d x))^{3/2} \, dx\\ &=\int \left (a^2 (e \sin (c+d x))^{3/2}+2 a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}+a^2 \sec ^2(c+d x) (e \sin (c+d x))^{3/2}\right ) \, dx\\ &=a^2 \int (e \sin (c+d x))^{3/2} \, dx+a^2 \int \sec ^2(c+d x) (e \sin (c+d x))^{3/2} \, dx+\left (2 a^2\right ) \int \sec (c+d x) (e \sin (c+d x))^{3/2} \, dx\\ &=-\frac{2 a^2 e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{a^2 e \sec (c+d x) \sqrt{e \sin (c+d x)}}{d}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{x^{3/2}}{1-\frac{x^2}{e^2}} \, dx,x,e \sin (c+d x)\right )}{d e}+\frac{1}{3} \left (a^2 e^2\right ) \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx-\frac{1}{2} \left (a^2 e^2\right ) \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx\\ &=-\frac{4 a^2 e \sqrt{e \sin (c+d x)}}{d}-\frac{2 a^2 e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{a^2 e \sec (c+d x) \sqrt{e \sin (c+d x)}}{d}+\frac{\left (2 a^2 e\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1-\frac{x^2}{e^2}\right )} \, dx,x,e \sin (c+d x)\right )}{d}+\frac{\left (a^2 e^2 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{3 \sqrt{e \sin (c+d x)}}-\frac{\left (a^2 e^2 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{2 \sqrt{e \sin (c+d x)}}\\ &=-\frac{a^2 e^2 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{3 d \sqrt{e \sin (c+d x)}}-\frac{4 a^2 e \sqrt{e \sin (c+d x)}}{d}-\frac{2 a^2 e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{a^2 e \sec (c+d x) \sqrt{e \sin (c+d x)}}{d}+\frac{\left (4 a^2 e\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^4}{e^2}} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d}\\ &=-\frac{a^2 e^2 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{3 d \sqrt{e \sin (c+d x)}}-\frac{4 a^2 e \sqrt{e \sin (c+d x)}}{d}-\frac{2 a^2 e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{a^2 e \sec (c+d x) \sqrt{e \sin (c+d x)}}{d}+\frac{\left (2 a^2 e^2\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^2\right ) \operatorname{Subst}\left (\int \frac{1}{e+x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d}\\ &=\frac{2 a^2 e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{2 a^2 e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}-\frac{a^2 e^2 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{3 d \sqrt{e \sin (c+d x)}}-\frac{4 a^2 e \sqrt{e \sin (c+d x)}}{d}-\frac{2 a^2 e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{a^2 e \sec (c+d x) \sqrt{e \sin (c+d x)}}{d}\\ \end{align*}
Mathematica [C] time = 14.6744, size = 204, normalized size = 1.06 \[ \frac{16 a^2 e \sin ^4\left (\frac{1}{2} \sin ^{-1}(\sin (c+d x))\right ) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \sqrt{e \sin (c+d x)} \left (-\sqrt{\sin (c+d x)} \sqrt{\cos ^2(c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},\sin ^2(c+d x)\right )+2 \sin ^{\frac{5}{2}}(c+d x)+\sqrt{\sin (c+d x)}-12 \sqrt{\sin (c+d x)} \sqrt{\cos ^2(c+d x)}+6 \sqrt{\cos ^2(c+d x)} \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )+6 \sqrt{\cos ^2(c+d x)} \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )\right )}{3 d \sin ^{\frac{9}{2}}(c+d x)} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 2.076, size = 201, normalized size = 1.1 \begin{align*}{\frac{{a}^{2}}{6\,d\cos \left ( dx+c \right ) } \left ( 12\,\cos \left ( dx+c \right ){e}^{3/2}\sqrt{e\sin \left ( dx+c \right ) }\arctan \left ({\frac{\sqrt{e\sin \left ( dx+c \right ) }}{\sqrt{e}}} \right ) +12\,\cos \left ( dx+c \right ){e}^{3/2}\sqrt{e\sin \left ( dx+c \right ) }{\it Artanh} \left ({\frac{\sqrt{e\sin \left ( dx+c \right ) }}{\sqrt{e}}} \right ) +\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}-4\,{e}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}-24\,{e}^{2}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +6\,{e}^{2}\sin \left ( dx+c \right ) \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} \left (e \sin \left (d x + c\right )\right )^{\frac{3}{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 ({\left (a^{2} e \sec \left (d x + c\right )^{2} + 2 \, a^{2} e \sec \left (d x + c\right ) + a^{2} e\right )} \sqrt{e \sin \left (d x + c\right )} \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} \left (e \sin \left (d x + c\right )\right )^{\frac{3}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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