Optimal. Leaf size=217 \[ -\frac{2 b^3 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{b \left (2 a^2-3 b^2\right ) \tan (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac{\left (a^2+6 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac{\left (a^2-3 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a^2 d \left (a^2-b^2\right )}+\frac{b^2 \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
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Rubi [A] time = 0.681478, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2802, 3055, 3001, 3770, 2659, 205} \[ -\frac{2 b^3 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{b \left (2 a^2-3 b^2\right ) \tan (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac{\left (a^2+6 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac{\left (a^2-3 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a^2 d \left (a^2-b^2\right )}+\frac{b^2 \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
Antiderivative was successfully verified.
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Rule 2802
Rule 3055
Rule 3001
Rule 3770
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{\sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=\frac{b^2 \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (a^2-3 b^2-a b \cos (c+d x)+2 b^2 \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{\left (a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-2 b \left (2 a^2-3 b^2\right )+a \left (a^2+b^2\right ) \cos (c+d x)+b \left (a^2-3 b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=-\frac{b \left (2 a^2-3 b^2\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (a^4+5 a^2 b^2-6 b^4+a b \left (a^2-3 b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=-\frac{b \left (2 a^2-3 b^2\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (b^3 \left (4 a^2-3 b^2\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^4 \left (a^2-b^2\right )}+\frac{\left (a^2+6 b^2\right ) \int \sec (c+d x) \, dx}{2 a^4}\\ &=\frac{\left (a^2+6 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{b \left (2 a^2-3 b^2\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (2 b^3 \left (4 a^2-3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 \left (a^2-b^2\right ) d}\\ &=-\frac{2 b^3 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2} d}+\frac{\left (a^2+6 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{b \left (2 a^2-3 b^2\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 5.53266, size = 285, normalized size = 1.31 \[ \frac{\frac{8 b^3 \left (3 b^2-4 a^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac{a^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a^2}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-2 a^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 a^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{4 a b^4 \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}-8 a b \tan (c+d x)-12 b^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{4 a^4 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.123, size = 401, normalized size = 1.9 \begin{align*}{\frac{1}{2\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{1}{2\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+2\,{\frac{b}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }}-{\frac{1}{2\,{a}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ){b}^{2}}{d{a}^{4}}}-{\frac{1}{2\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{2\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+2\,{\frac{b}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}+{\frac{1}{2\,{a}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ){b}^{2}}{d{a}^{4}}}+2\,{\frac{{b}^{4}\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ({a}^{2}-{b}^{2} \right ) \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 ) }}-8\,{\frac{{b}^{3}}{{a}^{2}d \left ( a-b \right ) \left ( a+b \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) }+6\,{\frac{{b}^{5}}{d{a}^{4} \left ( a-b \right ) \left ( a+b \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \right ) }{\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 = 8.45971, size = 2009, normalized size = 9.26 \begin{align*} \text{result too large to display} \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 ^{3}{\left (c + d x \right )}}{\left (a + b \cos{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.4547, size = 396, normalized size = 1.82 \begin{align*} \frac{\frac{4 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{5} - a^{3} b^{2}\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 )}} + \frac{4 \,{\left (4 \, a^{2} b^{3} - 3 \, b^{5}\right )}{\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 (a^{6} - a^{4} b^{2}\right )} \sqrt{a^{2} - b^{2}}} + \frac{{\left (a^{2} + 6 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac{{\left (a^{2} + 6 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac{2 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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