Optimal. Leaf size=740 \[ \frac{e^2 \sqrt{\sin (2 c+2 d x)} \sec (c+d x) \text{EllipticF}\left (c+d x-\frac{\pi }{4},2\right )}{b d \sqrt{e \tan (c+d x)}}-\frac{e^{3/2} \left (a^2-b^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a b^2 d}+\frac{e^{3/2} \left (a^2-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} a b^2 d}-\frac{e^{3/2} \left (a^2-b^2\right ) \log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a b^2 d}+\frac{e^{3/2} \left (a^2-b^2\right ) \log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a b^2 d}-\frac{2 \sqrt{2} e^2 \sqrt{a^2-b^2} \sqrt{\sin (c+d x)} \Pi \left (\frac{b}{a-\sqrt{a^2-b^2}};\left .\sin ^{-1}\left (\frac{\sqrt{-\cos (c+d x)}}{\sqrt{\sin (c+d x)+1}}\right )\right |-1\right )}{a b d \sqrt{-\cos (c+d x)} \sqrt{e \tan (c+d x)}}+\frac{2 \sqrt{2} e^2 \sqrt{a^2-b^2} \sqrt{\sin (c+d x)} \Pi \left (\frac{b}{a+\sqrt{a^2-b^2}};\left .\sin ^{-1}\left (\frac{\sqrt{-\cos (c+d x)}}{\sqrt{\sin (c+d x)+1}}\right )\right |-1\right )}{a b d \sqrt{-\cos (c+d x)} \sqrt{e \tan (c+d x)}}+\frac{a e^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} b^2 d}-\frac{a e^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} b^2 d}+\frac{a e^{3/2} \log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} b^2 d}-\frac{a e^{3/2} \log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} b^2 d} \]
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Rubi [A] time = 1.01498, antiderivative size = 740, normalized size of antiderivative = 1., number of steps used = 35, number of rules used = 19, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.76, Rules used = {3891, 3884, 3476, 329, 211, 1165, 628, 1162, 617, 204, 2614, 2573, 2641, 3892, 2733, 2729, 2907, 1213, 537} \[ -\frac{e^{3/2} \left (a^2-b^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a b^2 d}+\frac{e^{3/2} \left (a^2-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} a b^2 d}-\frac{e^{3/2} \left (a^2-b^2\right ) \log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a b^2 d}+\frac{e^{3/2} \left (a^2-b^2\right ) \log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a b^2 d}-\frac{2 \sqrt{2} e^2 \sqrt{a^2-b^2} \sqrt{\sin (c+d x)} \Pi \left (\frac{b}{a-\sqrt{a^2-b^2}};\left .\sin ^{-1}\left (\frac{\sqrt{-\cos (c+d x)}}{\sqrt{\sin (c+d x)+1}}\right )\right |-1\right )}{a b d \sqrt{-\cos (c+d x)} \sqrt{e \tan (c+d x)}}+\frac{2 \sqrt{2} e^2 \sqrt{a^2-b^2} \sqrt{\sin (c+d x)} \Pi \left (\frac{b}{a+\sqrt{a^2-b^2}};\left .\sin ^{-1}\left (\frac{\sqrt{-\cos (c+d x)}}{\sqrt{\sin (c+d x)+1}}\right )\right |-1\right )}{a b d \sqrt{-\cos (c+d x)} \sqrt{e \tan (c+d x)}}+\frac{a e^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} b^2 d}-\frac{a e^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} b^2 d}+\frac{a e^{3/2} \log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} b^2 d}-\frac{a e^{3/2} \log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} b^2 d}+\frac{e^2 \sqrt{\sin (2 c+2 d x)} \sec (c+d x) F\left (\left .c+d x-\frac{\pi }{4}\right |2\right )}{b d \sqrt{e \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 3891
Rule 3884
Rule 3476
Rule 329
Rule 211
Rule 1165
Rule 628
Rule 1162
Rule 617
Rule 204
Rule 2614
Rule 2573
Rule 2641
Rule 3892
Rule 2733
Rule 2729
Rule 2907
Rule 1213
Rule 537
Rubi steps
\begin{align*} \int \frac{(e \tan (c+d x))^{3/2}}{a+b \sec (c+d x)} \, dx &=-\frac{e^2 \int \frac{a-b \sec (c+d x)}{\sqrt{e \tan (c+d x)}} \, dx}{b^2}+\frac{\left (\left (a^2-b^2\right ) e^2\right ) \int \frac{1}{(a+b \sec (c+d x)) \sqrt{e \tan (c+d x)}} \, dx}{b^2}\\ &=-\frac{\left (a e^2\right ) \int \frac{1}{\sqrt{e \tan (c+d x)}} \, dx}{b^2}+\frac{e^2 \int \frac{\sec (c+d x)}{\sqrt{e \tan (c+d x)}} \, dx}{b}+\frac{\left (\left (a^2-b^2\right ) e^2\right ) \int \frac{1}{\sqrt{e \tan (c+d x)}} \, dx}{a b^2}-\frac{\left (\left (a^2-b^2\right ) e^2\right ) \int \frac{1}{(b+a \cos (c+d x)) \sqrt{e \tan (c+d x)}} \, dx}{a b}\\ &=-\frac{\left (a e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (e^2+x^2\right )} \, dx,x,e \tan (c+d x)\right )}{b^2 d}+\frac{\left (\left (a^2-b^2\right ) e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (e^2+x^2\right )} \, dx,x,e \tan (c+d x)\right )}{a b^2 d}-\frac{\left (\left (a^2-b^2\right ) e^2\right ) \int \frac{\sqrt{e \cot (c+d x)}}{b+a \cos (c+d x)} \, dx}{a b \sqrt{e \cot (c+d x)} \sqrt{e \tan (c+d x)}}+\frac{\left (e^2 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{\sin (c+d x)}} \, dx}{b \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}}\\ &=-\frac{\left (2 a e^3\right ) \operatorname{Subst}\left (\int \frac{1}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{b^2 d}+\frac{\left (2 \left (a^2-b^2\right ) e^3\right ) \operatorname{Subst}\left (\int \frac{1}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{a b^2 d}-\frac{\left (\left (a^2-b^2\right ) e^2 \sqrt{\sin (c+d x)}\right ) \int \frac{\sqrt{-\cos (c+d x)}}{(b+a \cos (c+d x)) \sqrt{\sin (c+d x)}} \, dx}{a b \sqrt{-\cos (c+d x)} \sqrt{e \tan (c+d x)}}+\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}{b \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)}}{b d \sqrt{e \tan (c+d x)}}-\frac{\left (a e^2\right ) \operatorname{Subst}\left (\int \frac{e-x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{b^2 d}-\frac{\left (a e^2\right ) \operatorname{Subst}\left (\int \frac{e+x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{b^2 d}+\frac{\left (\left (a^2-b^2\right ) e^2\right ) \operatorname{Subst}\left (\int \frac{e-x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{a b^2 d}+\frac{\left (\left (a^2-b^2\right ) e^2\right ) \operatorname{Subst}\left (\int \frac{e+x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{a b^2 d}-\frac{\left (2 \sqrt{2} \left (a^2-b^2\right ) \left (1-\frac{a}{\sqrt{a^2-b^2}}\right ) e^2 \sqrt{\sin (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (-a+\sqrt{a^2-b^2}+b x^2\right ) \sqrt{1-x^4}} \, dx,x,\frac{\sqrt{-\cos (c+d x)}}{\sqrt{1+\sin (c+d x)}}\right )}{a b d \sqrt{-\cos (c+d x)} \sqrt{e \tan (c+d x)}}-\frac{\left (2 \sqrt{2} \left (a^2-b^2\right ) \left (1+\frac{a}{\sqrt{a^2-b^2}}\right ) e^2 \sqrt{\sin (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (-a-\sqrt{a^2-b^2}+b x^2\right ) \sqrt{1-x^4}} \, dx,x,\frac{\sqrt{-\cos (c+d x)}}{\sqrt{1+\sin (c+d x)}}\right )}{a b d \sqrt{-\cos (c+d x)} \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)}}{b d \sqrt{e \tan (c+d x)}}+\frac{\left (a e^{3/2}\right ) \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} b^2 d}+\frac{\left (a e^{3/2}\right ) \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} b^2 d}-\frac{\left (\left (a^2-b^2\right ) e^{3/2}\right ) \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 b^2 d}-\frac{\left (\left (a^2-b^2\right ) e^{3/2}\right ) \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 b^2 d}-\frac{\left (a e^2\right ) \operatorname{Subst}\left (\int \frac{1}{e-\sqrt{2} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 b^2 d}-\frac{\left (a e^2\right ) \operatorname{Subst}\left (\int \frac{1}{e+\sqrt{2} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 b^2 d}+\frac{\left (\left (a^2-b^2\right ) e^2\right ) \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 b^2 d}+\frac{\left (\left (a^2-b^2\right ) e^2\right ) \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 b^2 d}-\frac{\left (2 \sqrt{2} \left (a^2-b^2\right ) \left (1-\frac{a}{\sqrt{a^2-b^2}}\right ) e^2 \sqrt{\sin (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+x^2} \left (-a+\sqrt{a^2-b^2}+b x^2\right )} \, dx,x,\frac{\sqrt{-\cos (c+d x)}}{\sqrt{1+\sin (c+d x)}}\right )}{a b d \sqrt{-\cos (c+d x)} \sqrt{e \tan (c+d x)}}-\frac{\left (2 \sqrt{2} \left (a^2-b^2\right ) \left (1+\frac{a}{\sqrt{a^2-b^2}}\right ) e^2 \sqrt{\sin (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+x^2} \left (-a-\sqrt{a^2-b^2}+b x^2\right )} \, dx,x,\frac{\sqrt{-\cos (c+d x)}}{\sqrt{1+\sin (c+d x)}}\right )}{a b d \sqrt{-\cos (c+d x)} \sqrt{e \tan (c+d x)}}\\ &=\frac{a 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} b^2 d}-\frac{\left (a^2-b^2\right ) 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 b^2 d}-\frac{a 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} b^2 d}+\frac{\left (a^2-b^2\right ) 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 b^2 d}-\frac{2 \sqrt{2} \sqrt{a^2-b^2} e^2 \Pi \left (\frac{b}{a-\sqrt{a^2-b^2}};\left .\sin ^{-1}\left (\frac{\sqrt{-\cos (c+d x)}}{\sqrt{1+\sin (c+d x)}}\right )\right |-1\right ) \sqrt{\sin (c+d x)}}{a b d \sqrt{-\cos (c+d x)} \sqrt{e \tan (c+d x)}}+\frac{2 \sqrt{2} \sqrt{a^2-b^2} e^2 \Pi \left (\frac{b}{a+\sqrt{a^2-b^2}};\left .\sin ^{-1}\left (\frac{\sqrt{-\cos (c+d x)}}{\sqrt{1+\sin (c+d x)}}\right )\right |-1\right ) \sqrt{\sin (c+d x)}}{a b d \sqrt{-\cos (c+d x)} \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)}}{b d \sqrt{e \tan (c+d x)}}-\frac{\left (a e^{3/2}\right ) \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} b^2 d}+\frac{\left (a e^{3/2}\right ) \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} b^2 d}+\frac{\left (\left (a^2-b^2\right ) e^{3/2}\right ) \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 b^2 d}-\frac{\left (\left (a^2-b^2\right ) e^{3/2}\right ) \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 b^2 d}\\ &=\frac{a e^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} b^2 d}-\frac{\left (a^2-b^2\right ) e^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a b^2 d}-\frac{a e^{3/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} b^2 d}+\frac{\left (a^2-b^2\right ) e^{3/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a b^2 d}+\frac{a 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} b^2 d}-\frac{\left (a^2-b^2\right ) 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 b^2 d}-\frac{a 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} b^2 d}+\frac{\left (a^2-b^2\right ) 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 b^2 d}-\frac{2 \sqrt{2} \sqrt{a^2-b^2} e^2 \Pi \left (\frac{b}{a-\sqrt{a^2-b^2}};\left .\sin ^{-1}\left (\frac{\sqrt{-\cos (c+d x)}}{\sqrt{1+\sin (c+d x)}}\right )\right |-1\right ) \sqrt{\sin (c+d x)}}{a b d \sqrt{-\cos (c+d x)} \sqrt{e \tan (c+d x)}}+\frac{2 \sqrt{2} \sqrt{a^2-b^2} e^2 \Pi \left (\frac{b}{a+\sqrt{a^2-b^2}};\left .\sin ^{-1}\left (\frac{\sqrt{-\cos (c+d x)}}{\sqrt{1+\sin (c+d x)}}\right )\right |-1\right ) \sqrt{\sin (c+d x)}}{a b d \sqrt{-\cos (c+d x)} \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)}}{b d \sqrt{e \tan (c+d x)}}\\ \end{align*}
Mathematica [C] time = 6.92316, size = 202, normalized size = 0.27 \[ \frac{2 e \sqrt{1-\tan \left (\frac{1}{2} (c+d x)\right )} \cot \left (\frac{1}{2} (c+d x)\right ) \sqrt{e \tan (c+d x)} \sqrt{\frac{-\sin (c+d x)+\cos (c+d x)-1}{\cos (c+d x)+1}} \left (-\Pi \left (-\frac{\sqrt{a-b}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\sqrt{-\tan \left (\frac{1}{2} (c+d x)\right )}\right )\right |-1\right )-\Pi \left (\frac{\sqrt{a-b}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\sqrt{-\tan \left (\frac{1}{2} (c+d x)\right )}\right )\right |-1\right )+\Pi \left (-i;\left .\sin ^{-1}\left (\sqrt{-\tan \left (\frac{1}{2} (c+d x)\right )}\right )\right |-1\right )+\Pi \left (i;\left .\sin ^{-1}\left (\sqrt{-\tan \left (\frac{1}{2} (c+d x)\right )}\right )\right |-1\right )\right )}{a d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.252, size = 1801, normalized size = 2.4 \begin{align*} \text{result too large to display} \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}}}{b \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*} \int \frac{\left (e \tan{\left (c + d x \right )}\right )^{\frac{3}{2}}}{a + b \sec{\left (c + d x \right )}}\, dx \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}}}{b \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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