Optimal. Leaf size=282 \[ -\frac{a \sqrt{\sin (2 c+2 d x)} \sec (c+d x) \text{EllipticF}\left (c+d x-\frac{\pi }{4},2\right )}{3 d e^2 \sqrt{e \tan (c+d x)}}+\frac{a \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{5/2}}-\frac{a \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} d e^{5/2}}+\frac{a \log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{a \log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{2 (a \sec (c+d x)+a)}{3 d e (e \tan (c+d x))^{3/2}} \]
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Rubi [A] time = 0.275203, antiderivative size = 282, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.565, Rules used = {3882, 3884, 3476, 329, 211, 1165, 628, 1162, 617, 204, 2614, 2573, 2641} \[ \frac{a \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{5/2}}-\frac{a \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} d e^{5/2}}+\frac{a \log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{a \log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{a \sqrt{\sin (2 c+2 d x)} \sec (c+d x) F\left (\left .c+d x-\frac{\pi }{4}\right |2\right )}{3 d e^2 \sqrt{e \tan (c+d x)}}-\frac{2 (a \sec (c+d x)+a)}{3 d e (e \tan (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3882
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{a+a \sec (c+d x)}{(e \tan (c+d x))^{5/2}} \, dx &=-\frac{2 (a+a \sec (c+d x))}{3 d e (e \tan (c+d x))^{3/2}}+\frac{2 \int \frac{-\frac{3 a}{2}-\frac{1}{2} a \sec (c+d x)}{\sqrt{e \tan (c+d x)}} \, dx}{3 e^2}\\ &=-\frac{2 (a+a \sec (c+d x))}{3 d e (e \tan (c+d x))^{3/2}}-\frac{a \int \frac{\sec (c+d x)}{\sqrt{e \tan (c+d x)}} \, dx}{3 e^2}-\frac{a \int \frac{1}{\sqrt{e \tan (c+d x)}} \, dx}{e^2}\\ &=-\frac{2 (a+a \sec (c+d x))}{3 d e (e \tan (c+d x))^{3/2}}-\frac{a \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (e^2+x^2\right )} \, dx,x,e \tan (c+d x)\right )}{d e}-\frac{\left (a \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{\sin (c+d x)}} \, dx}{3 e^2 \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}}\\ &=-\frac{2 (a+a \sec (c+d x))}{3 d e (e \tan (c+d x))^{3/2}}-\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{d e}-\frac{\left (a \sec (c+d x) \sqrt{\sin (2 c+2 d x)}\right ) \int \frac{1}{\sqrt{\sin (2 c+2 d x)}} \, dx}{3 e^2 \sqrt{e \tan (c+d x)}}\\ &=-\frac{2 (a+a \sec (c+d x))}{3 d e (e \tan (c+d x))^{3/2}}-\frac{a F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{3 d e^2 \sqrt{e \tan (c+d x)}}-\frac{a \operatorname{Subst}\left (\int \frac{e-x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{d e^2}-\frac{a \operatorname{Subst}\left (\int \frac{e+x^2}{e^2+x^4} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{d e^2}\\ &=-\frac{2 (a+a \sec (c+d x))}{3 d e (e \tan (c+d x))^{3/2}}-\frac{a F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{3 d e^2 \sqrt{e \tan (c+d x)}}+\frac{a \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} d e^{5/2}}+\frac{a \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} d e^{5/2}}-\frac{a \operatorname{Subst}\left (\int \frac{1}{e-\sqrt{2} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 d e^2}-\frac{a \operatorname{Subst}\left (\int \frac{1}{e+\sqrt{2} \sqrt{e} x+x^2} \, dx,x,\sqrt{e \tan (c+d x)}\right )}{2 d e^2}\\ &=\frac{a \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{a \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{2 (a+a \sec (c+d x))}{3 d e (e \tan (c+d x))^{3/2}}-\frac{a F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{3 d e^2 \sqrt{e \tan (c+d x)}}-\frac{a \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} d e^{5/2}}+\frac{a \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} d e^{5/2}}\\ &=\frac{a \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{5/2}}-\frac{a \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} d e^{5/2}}+\frac{a \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{a \log \left (\sqrt{e}+\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}\right )}{2 \sqrt{2} d e^{5/2}}-\frac{2 (a+a \sec (c+d x))}{3 d e (e \tan (c+d x))^{3/2}}-\frac{a F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{3 d e^2 \sqrt{e \tan (c+d x)}}\\ \end{align*}
Mathematica [C] time = 1.62, size = 200, normalized size = 0.71 \[ -\frac{a \csc (c+d x) \sqrt{e \tan (c+d x)} \left (\sqrt{\sec ^2(c+d x)} \left (2 \cot \left (\frac{1}{2} (c+d x)\right )-3 \sqrt{\sin (2 (c+d x))} \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+2 \cos \left (\frac{3}{2} (c+d x)\right ) \csc \left (\frac{1}{2} (c+d x)\right )+3 \sqrt{\sin (2 (c+d x))} \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )\right )-4 \sqrt [4]{-1} \sqrt{\tan (c+d x)} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt{\tan (c+d x)}\right ),-1\right )\right )}{6 d e^3 \sqrt{\sec ^2(c+d x)}} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.216, size = 658, normalized size = 2.3 \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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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(-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 \frac{a \sec \left (d x + c\right ) + a}{\left (e \tan \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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