Optimal. Leaf size=160 \[ \frac{2 b \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{\left (a^2-6 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{3 b \tan (c+d x)}{a^3 d}+\frac{3 \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac{\tan (c+d x) \sec (c+d x)}{a d (a+b \cos (c+d x))} \]
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Rubi [A] time = 0.672492, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3056, 3055, 3001, 3770, 2659, 205} \[ \frac{2 b \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{\left (a^2-6 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{3 b \tan (c+d x)}{a^3 d}+\frac{3 \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac{\tan (c+d x) \sec (c+d x)}{a d (a+b \cos (c+d x))} \]
Antiderivative was successfully verified.
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Rule 3056
Rule 3055
Rule 3001
Rule 3770
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{\left (1-\cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=-\frac{\sec (c+d x) \tan (c+d x)}{a d (a+b \cos (c+d x))}+\frac{\int \frac{\left (3 \left (a^2-b^2\right )-2 \left (a^2-b^2\right ) \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{3 \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac{\sec (c+d x) \tan (c+d x)}{a d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-6 b \left (a^2-b^2\right )-a \left (a^2-b^2\right ) \cos (c+d x)+3 b \left (a^2-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{3 b \tan (c+d x)}{a^3 d}+\frac{3 \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac{\sec (c+d x) \tan (c+d x)}{a d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-a^4+7 a^2 b^2-6 b^4+3 a b \left (a^2-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{3 b \tan (c+d x)}{a^3 d}+\frac{3 \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac{\sec (c+d x) \tan (c+d x)}{a d (a+b \cos (c+d x))}-\frac{\left (a^2-6 b^2\right ) \int \sec (c+d x) \, dx}{2 a^4}+\frac{\left (b \left (2 a^2-3 b^2\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^4}\\ &=-\frac{\left (a^2-6 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{3 b \tan (c+d x)}{a^3 d}+\frac{3 \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac{\sec (c+d x) \tan (c+d x)}{a d (a+b \cos (c+d x))}+\frac{\left (2 b \left (2 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 d}\\ &=\frac{2 b \left (2 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 \sqrt{a-b} \sqrt{a+b} d}-\frac{\left (a^2-6 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{3 b \tan (c+d x)}{a^3 d}+\frac{3 \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac{\sec (c+d x) \tan (c+d x)}{a d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 3.42272, size = 271, normalized size = 1.69 \[ \frac{\frac{8 b \left (3 b^2-2 a^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+\frac{a^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a^2}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+2 a^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 a^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-\frac{4 a b^2 \sin (c+d x)}{a+b \cos (c+d x)}-8 a b \tan (c+d x)-12 b^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{4 a^4 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.065, size = 364, normalized size = 2.3 \begin{align*} -2\,{\frac{{b}^{2}\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}+4\,{\frac{b}{d{a}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-6\,{\frac{{b}^{3}}{d{a}^{4}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+{\frac{1}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{1}{2\,d{a}^{2}} \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\,d{a}^{2}}\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\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{2\,d{a}^{2}} \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\,d{a}^{2}}\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}}} \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 = 3.34888, size = 1688, normalized size = 10.55 \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(-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 = 1.81437, size = 363, normalized size = 2.27 \begin{align*} -\frac{\frac{4 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\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 )} a^{3}} + \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{4 \,{\left (2 \, a^{2} b - 3 \, b^{3}\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 )}}{\sqrt{a^{2} - b^{2}} 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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