Optimal. Leaf size=154 \[ \frac{2 a e^2 \sqrt{\sin (c+d x)} \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right )}{3 d \sqrt{e \sin (c+d x)}}+\frac{a e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{a e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}-\frac{2 a e \sqrt{e \sin (c+d x)}}{d}-\frac{2 a e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d} \]
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Rubi [A] time = 0.199885, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 11, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.478, Rules used = {3872, 2838, 2564, 321, 329, 212, 206, 203, 2635, 2642, 2641} \[ \frac{a e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{a e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{2 a e^2 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 d \sqrt{e \sin (c+d x)}}-\frac{2 a e \sqrt{e \sin (c+d x)}}{d}-\frac{2 a e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d} \]
Antiderivative was successfully verified.
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Rule 3872
Rule 2838
Rule 2564
Rule 321
Rule 329
Rule 212
Rule 206
Rule 203
Rule 2635
Rule 2642
Rule 2641
Rubi steps
\begin{align*} \int (a+a \sec (c+d x)) (e \sin (c+d x))^{3/2} \, dx &=-\int (-a-a \cos (c+d x)) \sec (c+d x) (e \sin (c+d x))^{3/2} \, dx\\ &=a \int (e \sin (c+d x))^{3/2} \, dx+a \int \sec (c+d x) (e \sin (c+d x))^{3/2} \, dx\\ &=-\frac{2 a e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{a \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 e^2\right ) \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx\\ &=-\frac{2 a e \sqrt{e \sin (c+d x)}}{d}-\frac{2 a e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{(a e) \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 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{2 a 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{2 a e \sqrt{e \sin (c+d x)}}{d}-\frac{2 a e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{(2 a e) \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^4}{e^2}} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d}\\ &=\frac{2 a 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{2 a e \sqrt{e \sin (c+d x)}}{d}-\frac{2 a e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}+\frac{\left (a e^2\right ) \operatorname{Subst}\left (\int \frac{1}{e-x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d}+\frac{\left (a e^2\right ) \operatorname{Subst}\left (\int \frac{1}{e+x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d}\\ &=\frac{a e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{a e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d}+\frac{2 a 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{2 a e \sqrt{e \sin (c+d x)}}{d}-\frac{2 a e \cos (c+d x) \sqrt{e \sin (c+d x)}}{3 d}\\ \end{align*}
Mathematica [A] time = 0.574349, size = 170, normalized size = 1.1 \[ \frac{a (e \sin (c+d x))^{3/2} \left (-8 \text{EllipticF}\left (\frac{1}{4} (-2 c-2 d x+\pi ),2\right )-24 \sqrt{\sin (c+d x)}-3 \log \left (1-\sqrt{\sin (c+d x)}\right )+3 \log \left (\sqrt{\sin (c+d x)}+1\right )+12 \tan ^{-1}\left (\sqrt{\sin (c+d x)}\right )+6 \tanh ^{-1}\left (\sqrt{\sin (c+d x)}\right )+16 \sin ^{\frac{5}{2}}(c+d x) \cos (c+d x) \sec (2 (c+d x))-8 \sqrt{\sin (c+d x)} \cos (c+d x) \sec (2 (c+d x))\right )}{12 d \sin ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Maple [A] time = 1.127, size = 210, normalized size = 1.4 \begin{align*}{\frac{a}{d}{e}^{{\frac{3}{2}}}{\it Artanh} \left ({\sqrt{e\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{e}}}} \right ) }+{\frac{a}{d}{e}^{{\frac{3}{2}}}\arctan \left ({\sqrt{e\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{e}}}} \right ) }-2\,{\frac{ae\sqrt{e\sin \left ( dx+c \right ) }}{d}}-{\frac{a{e}^{2}}{3\,d\cos \left ( dx+c \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 ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}}+{\frac{2\,a{e}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}}-{\frac{2\,a{e}^{2}\sin \left ( dx+c \right ) }{3\,d\cos \left ( dx+c \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 )} \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 e \sec \left (d x + c\right ) + a 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 )} \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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