Optimal. Leaf size=133 \[ \frac{\left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{3 a b \sin (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{b \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2} \]
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Rubi [A] time = 0.109509, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {2664, 2754, 12, 2659, 205} \[ \frac{\left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{3 a b \sin (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{b \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 2664
Rule 2754
Rule 12
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{1}{(a+b \cos (c+d x))^3} \, dx &=-\frac{b \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\int \frac{-2 a+b \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac{b \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{3 a b \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{2 a^2+b^2}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac{b \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{3 a b \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\left (2 a^2+b^2\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac{b \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{3 a b \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\left (2 a^2+b^2\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 )}{\left (a^2-b^2\right )^2 d}\\ &=\frac{\left (2 a^2+b^2\right ) \tan ^{-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 \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{3 a b \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 0.391753, size = 113, normalized size = 0.85 \[ \frac{\frac{b \sin (c+d x) \left (-4 a^2-3 a b \cos (c+d x)+b^2\right )}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))^2}-\frac{2 \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}}{2 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.082, size = 400, normalized size = 3. \begin{align*} -4\,{\frac{a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}b}{d \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 ( a-b \right ) \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}-{\frac{{b}^{2}}{d \left ( a-b \right ) \left ({a}^{2}+2\,ab+{b}^{2} \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a- \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}b+a+b \right ) ^{-2}}-4\,{\frac{a\tan \left ( 1/2\,dx+c/2 \right ) b}{d \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 ( a+b \right ) \left ({a}^{2}-2\,ab+{b}^{2} \right ) }}+{\frac{{b}^{2}}{d \left ( a+b \right ) \left ({a}^{2}-2\,ab+{b}^{2} \right ) }\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a- \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}b+a+b \right ) ^{-2}}+2\,{\frac{{a}^{2}}{d \left ({a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4} \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 ) }+{\frac{{b}^{2}}{d \left ({a}^{4}-2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }\arctan \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 ) }}}} \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 = 2.27104, size = 1276, normalized size = 9.59 \begin{align*} \left [-\frac{{\left (2 \, a^{4} + a^{2} b^{2} +{\left (2 \, 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 )\right )} \sqrt{-a^{2} + b^{2}} \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 ) + 2 \,{\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5} + 3 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \,{\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, 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 ) +{\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d\right )}}, \frac{{\left (2 \, a^{4} + a^{2} b^{2} +{\left (2 \, 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 )\right )} \sqrt{a^{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 ) -{\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5} + 3 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, 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 ) +{\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d\right )}}\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 [B] time = 1.21951, size = 339, normalized size = 2.55 \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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