Optimal. Leaf size=93 \[ -\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt{2} b c^{3/2}}-\frac{\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}} \]
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Rubi [A] time = 0.236104, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {4397, 3797, 3795, 207} \[ -\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{4 \sqrt{2} b c^{3/2}}-\frac{\tan (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 4397
Rule 3797
Rule 3795
Rule 207
Rubi steps
\begin{align*} \int \frac{\sec ^2(2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx &=\int \frac{\sec ^2(2 a+2 b x)}{(-c+c \sec (2 a+2 b x))^{3/2}} \, dx\\ &=-\frac{\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}+\frac{3 \int \frac{\sec (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}} \, dx}{4 c}\\ &=-\frac{\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{-2 c+x^2} \, dx,x,-\frac{c \tan (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}}\right )}{4 b c}\\ &=-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{-c+c \sec (2 a+2 b x)}}\right )}{4 \sqrt{2} b c^{3/2}}-\frac{\tan (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}\\ \end{align*}
Mathematica [A] time = 0.609946, size = 84, normalized size = 0.9 \[ \frac{\tan (2 (a+b x)) \left (3 \sin ^2(a+b x) \tan ^{-1}\left (\sqrt{\tan ^2(a+b x)-1}\right ) \sqrt{\tan ^2(a+b x)-1} \sec (2 (a+b x))-1\right )}{4 b (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.423, size = 433, normalized size = 4.7 \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{\sec \left (2 \, b x + 2 \, a\right )^{2}}{\left (c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.79174, size = 706, normalized size = 7.59 \begin{align*} \left [\frac{3 \, \sqrt{2} \sqrt{c} \log \left (\frac{c \tan \left (b x + a\right )^{3} - 2 \, \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}{\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt{c} - 2 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) \tan \left (b x + a\right )^{3} + 2 \, \sqrt{2} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}{\left (\tan \left (b x + a\right )^{2} - 1\right )}}{16 \, b c^{2} \tan \left (b x + a\right )^{3}}, -\frac{3 \, \sqrt{2} \sqrt{-c} \arctan \left (\frac{\sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}{\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt{-c}}{c \tan \left (b x + a\right )}\right ) \tan \left (b x + a\right )^{3} - \sqrt{2} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}{\left (\tan \left (b x + a\right )^{2} - 1\right )}}{8 \, b c^{2} \tan \left (b x + a\right )^{3}}\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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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