Optimal. Leaf size=257 \[ \frac{e^2 \sqrt{\sin (2 c+2 d x)} \sec (c+d x) \text{EllipticF}\left (c+d x-\frac{\pi }{4},2\right )}{a d \sqrt{e \tan (c+d x)}}+\frac{e^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a d}-\frac{e^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} a d}+\frac{e^{3/2} \log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a d}-\frac{e^{3/2} \log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a d} \]
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Rubi [A] time = 0.291912, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.52, Rules used = {3888, 3884, 3476, 329, 211, 1165, 628, 1162, 617, 204, 2614, 2573, 2641} \[ \frac{e^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a d}-\frac{e^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} a d}+\frac{e^{3/2} \log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a d}-\frac{e^{3/2} \log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a d}+\frac{e^2 \sqrt{\sin (2 c+2 d x)} \sec (c+d x) F\left (\left .c+d x-\frac{\pi }{4}\right |2\right )}{a d \sqrt{e \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 3888
Rule 3884
Rule 3476
Rule 329
Rule 211
Rule 1165
Rule 628
Rule 1162
Rule 617
Rule 204
Rule 2614
Rule 2573
Rule 2641
Rubi steps
\begin{align*} \int \frac{(e \tan (c+d x))^{3/2}}{a+a \sec (c+d x)} \, dx &=\frac{e^2 \int \frac{-a+a \sec (c+d x)}{\sqrt{e \tan (c+d x)}} \, dx}{a^2}\\ &=-\frac{e^2 \int \frac{1}{\sqrt{e \tan (c+d x)}} \, dx}{a}+\frac{e^2 \int \frac{\sec (c+d x)}{\sqrt{e \tan (c+d x)}} \, dx}{a}\\ &=-\frac{e^3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (e^2+x^2\right )} \, dx,x,e \tan (c+d x)\right )}{a d}+\frac{\left (e^2 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{\sin (c+d x)}} \, dx}{a \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}}\\ &=-\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \frac{1}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{a d}+\frac{\left (e^2 \sec (c+d x) \sqrt{\sin (2 c+2 d x)}\right ) \int \frac{1}{\sqrt{\sin (2 c+2 d x)}} \, dx}{a \sqrt{e \tan (c+d x)}}\\ &=\frac{e^2 F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{a d \sqrt{e \tan (c+d x)}}-\frac{e^2 \operatorname{Subst}\left (\int \frac{e-x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{a d}-\frac{e^2 \operatorname{Subst}\left (\int \frac{e+x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{a d}\\ &=\frac{e^2 F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{a d \sqrt{e \tan (c+d x)}}+\frac{e^{3/2} \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{e}+2 x}{-e-\sqrt{2} \sqrt{e} x-x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a d}+\frac{e^{3/2} \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{e}-2 x}{-e+\sqrt{2} \sqrt{e} x-x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a d}-\frac{e^2 \operatorname{Subst}\left (\int \frac{1}{e-\sqrt{2} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 a d}-\frac{e^2 \operatorname{Subst}\left (\int \frac{1}{e+\sqrt{2} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 a d}\\ &=\frac{e^{3/2} \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a d}-\frac{e^{3/2} \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a d}+\frac{e^2 F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{a d \sqrt{e \tan (c+d x)}}-\frac{e^{3/2} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a d}+\frac{e^{3/2} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a d}\\ &=\frac{e^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a d}-\frac{e^{3/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a d}+\frac{e^{3/2} \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a d}-\frac{e^{3/2} \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} a d}+\frac{e^2 F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{a d \sqrt{e \tan (c+d x)}}\\ \end{align*}
Mathematica [C] time = 12.5936, size = 1211, normalized size = 4.71 \[ -\frac{4 \sqrt [4]{-1} \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt{\tan (c+d x)}\right ),-1\right ) (e \tan (c+d x))^{3/2} \sec ^4(c+d x)}{d (\sec (c+d x) a+a) \tan ^{\frac{3}{2}}(c+d x) \left (\tan ^2(c+d x)+1\right )^{3/2}}-\frac{2 e^{-i (c+d x)} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec (2 c) (e \tan (c+d x))^{3/2} \sec (c+d x)}{d (\sec (c+d x) a+a) \tan ^{\frac{3}{2}}(c+d x)}-\frac{e^{-2 i c} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (e^{4 i c} \sqrt{-1+e^{4 i (c+d x)}} \tan ^{-1}\left (\sqrt{-1+e^{4 i (c+d x)}}\right )+2 \sqrt{-1+e^{2 i (c+d x)}} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right ) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec (2 c) (e \tan (c+d x))^{3/2} \sec (c+d x)}{2 d \left (-1+e^{2 i (c+d x)}\right ) (\sec (c+d x) a+a) \tan ^{\frac{3}{2}}(c+d x)}-\frac{e^{-2 i c} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (\sqrt{-1+e^{4 i (c+d x)}} \tan ^{-1}\left (\sqrt{-1+e^{4 i (c+d x)}}\right )+2 e^{4 i c} \sqrt{-1+e^{2 i (c+d x)}} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right ) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec (2 c) (e \tan (c+d x))^{3/2} \sec (c+d x)}{2 d \left (-1+e^{2 i (c+d x)}\right ) (\sec (c+d x) a+a) \tan ^{\frac{3}{2}}(c+d x)}+\frac{e^{-i (2 c+d x)} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (3 \left (-1+e^{4 i (c+d x)}\right )+e^{4 i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt{1-e^{4 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},e^{4 i (c+d x)}\right )\right ) \sec (2 c) (e \tan (c+d x))^{3/2} \sec (c+d x)}{3 d \left (-1+e^{2 i (c+d x)}\right ) (\sec (c+d x) a+a) \tan ^{\frac{3}{2}}(c+d x)}-\frac{e^{-i d x} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (e^{2 i (c+2 d x)} \left (-1+e^{2 i c}\right ) \sqrt{1-e^{4 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},e^{4 i (c+d x)}\right )-3 e^{4 i (c+d x)}+3\right ) \sec (2 c) (e \tan (c+d x))^{3/2} \sec (c+d x)}{3 d \left (-1+e^{2 i (c+d x)}\right ) (\sec (c+d x) a+a) \tan ^{\frac{3}{2}}(c+d x)}+\frac{\cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \csc (c+d x) \left (\frac{8 \cos (c) \cos (d x) \sec (2 c) \sin ^2\left (\frac{c}{2}\right )}{d}-\frac{16 \cos \left (\frac{c}{2}\right ) \sec (2 c) \sin ^3\left (\frac{c}{2}\right ) \sin (d x)}{d}\right ) (e \tan (c+d x))^{3/2}}{\sec (c+d x) a+a} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.226, size = 319, normalized size = 1.2 \begin{align*}{\frac{\sqrt{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) }{2\,da \left ( \sin \left ( dx+c \right ) \right ) ^{4}} \left ( i{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -i{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) +{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) +{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -4\,{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},1/2\,\sqrt{2} \right ) \right ) \sqrt{{\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{-1+\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}} \left ({\frac{e\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}}} \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 \frac{\left (e \tan \left (d x + c\right )\right )^{\frac{3}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [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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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\left (e \tan{\left (c + d x \right )}\right )^{\frac{3}{2}}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \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 \frac{\left (e \tan \left (d x + c\right )\right )^{\frac{3}{2}}}{a \sec \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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