Optimal. Leaf size=113 \[ \frac{7 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f}-\frac{7 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f}+\frac{7 b (b \sec (e+f x))^{3/2}}{6 f}-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{2 b f} \]
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Rubi [A] time = 0.0836008, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2622, 288, 321, 329, 298, 203, 206} \[ \frac{7 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f}-\frac{7 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f}+\frac{7 b (b \sec (e+f x))^{3/2}}{6 f}-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{2 b f} \]
Antiderivative was successfully verified.
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Rule 2622
Rule 288
Rule 321
Rule 329
Rule 298
Rule 203
Rule 206
Rubi steps
\begin{align*} \int \csc ^3(e+f x) (b \sec (e+f x))^{5/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^{9/2}}{\left (-1+\frac{x^2}{b^2}\right )^2} \, dx,x,b \sec (e+f x)\right )}{b^3 f}\\ &=-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{2 b f}+\frac{7 \operatorname{Subst}\left (\int \frac{x^{5/2}}{-1+\frac{x^2}{b^2}} \, dx,x,b \sec (e+f x)\right )}{4 b f}\\ &=\frac{7 b (b \sec (e+f x))^{3/2}}{6 f}-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{2 b f}+\frac{(7 b) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{-1+\frac{x^2}{b^2}} \, dx,x,b \sec (e+f x)\right )}{4 f}\\ &=\frac{7 b (b \sec (e+f x))^{3/2}}{6 f}-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{2 b f}+\frac{(7 b) \operatorname{Subst}\left (\int \frac{x^2}{-1+\frac{x^4}{b^2}} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{2 f}\\ &=\frac{7 b (b \sec (e+f x))^{3/2}}{6 f}-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{2 b f}-\frac{\left (7 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{4 f}+\frac{\left (7 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{4 f}\\ &=\frac{7 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f}-\frac{7 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f}+\frac{7 b (b \sec (e+f x))^{3/2}}{6 f}-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{2 b f}\\ \end{align*}
Mathematica [A] time = 1.95521, size = 109, normalized size = 0.96 \[ \frac{b^3 \left (-12 \csc ^2(e+f x)+16 \sec ^2(e+f x)+21 \sqrt{\sec (e+f x)} \left (\log \left (1-\sqrt{\sec (e+f x)}\right )-\log \left (\sqrt{\sec (e+f x)}+1\right )\right )+42 \sqrt{\sec (e+f x)} \tan ^{-1}\left (\sqrt{\sec (e+f x)}\right )\right )}{24 f \sqrt{b \sec (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.16, size = 699, normalized size = 6.2 \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 [B] time = 2.70897, size = 1133, normalized size = 10.03 \begin{align*} \left [\frac{42 \,{\left (b^{2} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{\frac{b}{\cos \left (f x + e\right )}}{\left (\cos \left (f x + e\right ) + 1\right )}}{2 \, b}\right ) + 21 \,{\left (b^{2} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt{-b} \log \left (\frac{b \cos \left (f x + e\right )^{2} - 4 \,{\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt{-b} \sqrt{\frac{b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \,{\left (7 \, b^{2} \cos \left (f x + e\right )^{2} - 4 \, b^{2}\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{48 \,{\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )}}, -\frac{42 \,{\left (b^{2} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt{b} \arctan \left (\frac{\sqrt{\frac{b}{\cos \left (f x + e\right )}}{\left (\cos \left (f x + e\right ) - 1\right )}}{2 \, \sqrt{b}}\right ) - 21 \,{\left (b^{2} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt{b} \log \left (\frac{b \cos \left (f x + e\right )^{2} - 4 \,{\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt{b} \sqrt{\frac{b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right ) - 8 \,{\left (7 \, b^{2} \cos \left (f x + e\right )^{2} - 4 \, b^{2}\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{48 \,{\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )}}\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 [A] time = 1.15476, size = 173, normalized size = 1.53 \begin{align*} \frac{b^{8}{\left (\frac{6 \, \sqrt{b \cos \left (f x + e\right )}}{{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} b^{4}} + \frac{21 \, \arctan \left (\frac{\sqrt{b \cos \left (f x + e\right )}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{5}} - \frac{21 \, \arctan \left (\frac{\sqrt{b \cos \left (f x + e\right )}}{\sqrt{b}}\right )}{b^{\frac{11}{2}}} + \frac{8}{\sqrt{b \cos \left (f x + e\right )} b^{5} \cos \left (f x + e\right )}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{12 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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