Optimal. Leaf size=155 \[ -\frac{3 a b \sec (c+d x)}{2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{b \sec (c+d x)}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{\left (2 a^2-b^2\right ) \sec (c+d x) \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{2 d \left (a^2+b^2\right )^{5/2} \sqrt{\sec ^2(c+d x)}} \]
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Rubi [A] time = 0.114242, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3512, 745, 807, 725, 206} \[ -\frac{3 a b \sec (c+d x)}{2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{b \sec (c+d x)}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{\left (2 a^2-b^2\right ) \sec (c+d x) \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{2 d \left (a^2+b^2\right )^{5/2} \sqrt{\sec ^2(c+d x)}} \]
Antiderivative was successfully verified.
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Rule 3512
Rule 745
Rule 807
Rule 725
Rule 206
Rubi steps
\begin{align*} \int \frac{\sec (c+d x)}{(a+b \tan (c+d x))^3} \, dx &=\frac{\sec (c+d x) \operatorname{Subst}\left (\int \frac{1}{(a+x)^3 \sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{b d \sqrt{\sec ^2(c+d x)}}\\ &=-\frac{b \sec (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{\sec (c+d x) \operatorname{Subst}\left (\int \frac{-2 a+x}{(a+x)^2 \sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{2 b \left (a^2+b^2\right ) d \sqrt{\sec ^2(c+d x)}}\\ &=-\frac{b \sec (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{3 a b \sec (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\left (\left (2 a^2-b^2\right ) \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{(a+x) \sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{2 b \left (a^2+b^2\right )^2 d \sqrt{\sec ^2(c+d x)}}\\ &=-\frac{b \sec (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{3 a b \sec (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\left (\left (2 a^2-b^2\right ) \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a^2}{b^2}-x^2} \, dx,x,\frac{1-\frac{a \tan (c+d x)}{b}}{\sqrt{\sec ^2(c+d x)}}\right )}{2 b \left (a^2+b^2\right )^2 d \sqrt{\sec ^2(c+d x)}}\\ &=-\frac{\left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{b \left (1-\frac{a \tan (c+d x)}{b}\right )}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right ) \sec (c+d x)}{2 \left (a^2+b^2\right )^{5/2} d \sqrt{\sec ^2(c+d x)}}-\frac{b \sec (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{3 a b \sec (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end{align*}
Mathematica [A] time = 0.894272, size = 110, normalized size = 0.71 \[ \frac{\frac{2 \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac{b \sec (c+d x) \left (4 a^2+3 a b \tan (c+d x)+b^2\right )}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}}{2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.069, size = 280, normalized size = 1.8 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{ \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) ^{2}} \left ( -1/2\,{\frac{{b}^{2} \left ( 5\,{a}^{2}+2\,{b}^{2} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{a \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}-1/2\,{\frac{b \left ( 4\,{a}^{4}-7\,{a}^{2}{b}^{2}-2\,{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ){a}^{2}}}+1/2\,{\frac{{b}^{2} \left ( 11\,{a}^{2}+2\,{b}^{2} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{a \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}+1/2\,{\frac{b \left ( 4\,{a}^{2}+{b}^{2} \right ) }{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}} \right ) }+{\frac{2\,{a}^{2}-{b}^{2}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}{\it Artanh} \left ({\frac{1}{2} \left ( 2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \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 = 2.03376, size = 795, normalized size = 5.13 \begin{align*} -\frac{{\left (2 \, a^{2} b^{2} - b^{4} +{\left (2 \, a^{4} - 3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 2 \,{\left (4 \, a^{4} b + 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) + 6 \,{\left (a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{4 \,{\left ({\left (a^{8} + 2 \, a^{6} b^{2} - 2 \, a^{2} b^{6} - b^{8}\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 ) \sin \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 )}} \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 \tan{\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 = 2.2486, size = 396, normalized size = 2.55 \begin{align*} -\frac{\frac{{\left (2 \, a^{2} - b^{2}\right )} \log \left (\frac{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} + b^{2}}} - \frac{2 \,{\left (5 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 7 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 11 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a^{4} b - a^{2} b^{3}\right )}}{{\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}^{2}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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