Optimal. Leaf size=155 \[ -\frac{2 b^2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d}-\frac{\left (a^2-3 b^2\right ) \tan (c+d x)}{3 a^3 d}+\frac{b \left (a^2-2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{b \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac{\tan (c+d x) \sec ^2(c+d x)}{3 a d} \]
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Rubi [A] time = 0.569702, antiderivative size = 155, 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} \[ -\frac{2 b^2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d}-\frac{\left (a^2-3 b^2\right ) \tan (c+d x)}{3 a^3 d}+\frac{b \left (a^2-2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{b \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac{\tan (c+d x) \sec ^2(c+d x)}{3 a d} \]
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 ^4(c+d x)}{a+b \cos (c+d x)} \, dx &=\frac{\sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac{\int \frac{\left (-3 b-a \cos (c+d x)+2 b \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx}{3 a}\\ &=-\frac{b \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac{\sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac{\int \frac{\left (-2 \left (a^2-3 b^2\right )+a b \cos (c+d x)-3 b^2 \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^2}\\ &=-\frac{\left (a^2-3 b^2\right ) \tan (c+d x)}{3 a^3 d}-\frac{b \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac{\sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac{\int \frac{\left (3 b \left (a^2-2 b^2\right )-3 a b^2 \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^3}\\ &=-\frac{\left (a^2-3 b^2\right ) \tan (c+d x)}{3 a^3 d}-\frac{b \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac{\sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac{\left (b \left (a^2-2 b^2\right )\right ) \int \sec (c+d x) \, dx}{2 a^4}-\frac{\left (b^2 \left (a^2-b^2\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^4}\\ &=\frac{b \left (a^2-2 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)}{3 a^3 d}-\frac{b \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac{\sec ^2(c+d x) \tan (c+d x)}{3 a d}-\frac{\left (2 b^2 \left (a^2-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 \sqrt{a-b} b^2 \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d}+\frac{b \left (a^2-2 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)}{3 a^3 d}-\frac{b \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac{\sec ^2(c+d x) \tan (c+d x)}{3 a d}\\ \end{align*}
Mathematica [A] time = 2.50952, size = 256, normalized size = 1.65 \[ -\frac{24 b^2 \sqrt{b^2-a^2} \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )+\frac{1}{2} \sec ^3(c+d x) \left (4 a \sin (c+d x) \left (\left (a^2-3 b^2\right ) \cos (2 (c+d x))-a^2+3 a b \cos (c+d x)-3 b^2\right )+9 b \left (a^2-2 b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+3 b \left (a^2-2 b^2\right ) \cos (3 (c+d x)) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{12 a^4 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.057, size = 407, normalized size = 2.6 \begin{align*} -2\,{\frac{{b}^{2}}{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 ) }+2\,{\frac{{b}^{4}}{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}{3\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{b}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{b}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{{b}^{2}}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{b}{2\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{{b}^{3}}{d{a}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{1}{3\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{b}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{b}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{{b}^{2}}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{b}{2\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{{b}^{3}}{d{a}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \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 [A] time = 2.04623, size = 1045, normalized size = 6.74 \begin{align*} \left [\frac{6 \, \sqrt{-a^{2} + b^{2}} b^{2} \cos \left (d x + c\right )^{3} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 3 \,{\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (3 \, a^{2} b \cos \left (d x + c\right ) - 2 \, a^{3} + 2 \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, a^{4} d \cos \left (d x + c\right )^{3}}, -\frac{12 \, \sqrt{a^{2} - b^{2}} b^{2} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} - 3 \,{\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (a^{2} b - 2 \, b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (3 \, a^{2} b \cos \left (d x + c\right ) - 2 \, a^{3} + 2 \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, a^{4} d \cos \left (d x + c\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 [A] time = 1.24251, size = 359, normalized size = 2.32 \begin{align*} \frac{\frac{3 \,{\left (a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac{3 \,{\left (a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac{12 \,{\left (a^{2} b^{2} - b^{4}\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 (3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 8 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, b^{2} \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 )}^{3} a^{3}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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