Optimal. Leaf size=133 \[ \frac{\left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{3 a b \tan (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac{b \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2} \]
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Rubi [A] time = 0.175449, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3833, 4003, 12, 3831, 2659, 208} \[ \frac{\left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{3 a b \tan (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac{b \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 3833
Rule 4003
Rule 12
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx &=-\frac{b \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\int \frac{\sec (c+d x) (-2 a+b \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac{b \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{3 a b \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\int \frac{\left (2 a^2+b^2\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac{b \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{3 a b \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (2 a^2+b^2\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac{b \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{3 a b \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (2 a^2+b^2\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^2}\\ &=-\frac{b \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{3 a b \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (2 a^2+b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b \left (a^2-b^2\right )^2 d}\\ &=\frac{\left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac{b \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{3 a b \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end{align*}
Mathematica [A] time = 0.45444, size = 115, normalized size = 0.86 \[ \frac{\frac{b \sin (c+d x) \left (\left (b^2-4 a^2\right ) \cos (c+d x)-3 a b\right )}{(a \cos (c+d x)+b)^2}-\frac{2 \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{2 d (a-b)^2 (a+b)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.061, size = 186, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{ \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) ^{2}} \left ( -1/2\,{\frac{ \left ( 4\,a+b \right ) b \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{ \left ( a-b \right ) \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+1/2\,{\frac{ \left ( 4\,a-b \right ) b\tan \left ( 1/2\,dx+c/2 \right ) }{ \left ( a+b \right ) \left ({a}^{2}-2\,ab+{b}^{2} \right ) }} \right ) }+{\frac{2\,{a}^{2}+{b}^{2}}{{a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4}}{\it Artanh} \left ({(a-b)\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ) } \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 = 1.88728, size = 1296, normalized size = 9.74 \begin{align*} \left [\frac{{\left (2 \, a^{2} b^{2} + b^{4} +{\left (2 \, a^{4} + a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 2 \,{\left (3 \, a^{3} b^{2} - 3 \, a b^{4} +{\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \,{\left ({\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d\right )}}, \frac{{\left (2 \, a^{2} b^{2} + b^{4} +{\left (2 \, a^{4} + a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) -{\left (3 \, a^{3} b^{2} - 3 \, a b^{4} +{\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d\right )}}\right ] \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{\sec{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.37075, size = 343, normalized size = 2.58 \begin{align*} -\frac{\frac{{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}{\left (2 \, a^{2} + b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{4 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}^{2}}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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