Optimal. Leaf size=266 \[ \frac{\left (-7 a^2 b^2+12 a^4-2 b^4\right ) \sin (c+d x)}{3 b^4 d \left (a^2-b^2\right )}+\frac{2 a^4 \left (4 a^2-5 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\left (4 a^2-b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d \left (a^2-b^2\right )}-\frac{a \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b^3 d \left (a^2-b^2\right )}-\frac{a x \left (4 a^2+b^2\right )}{b^5} \]
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Rubi [A] time = 0.722972, antiderivative size = 266, 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 = {2792, 3049, 3023, 2735, 2659, 205} \[ \frac{\left (-7 a^2 b^2+12 a^4-2 b^4\right ) \sin (c+d x)}{3 b^4 d \left (a^2-b^2\right )}+\frac{2 a^4 \left (4 a^2-5 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\left (4 a^2-b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d \left (a^2-b^2\right )}-\frac{a \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{b^3 d \left (a^2-b^2\right )}-\frac{a x \left (4 a^2+b^2\right )}{b^5} \]
Antiderivative was successfully verified.
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Rule 2792
Rule 3049
Rule 3023
Rule 2735
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{\cos ^5(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{\cos ^2(c+d x) \left (3 a^2-a b \cos (c+d x)-\left (4 a^2-b^2\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{\left (4 a^2-b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{\cos (c+d x) \left (-2 a \left (4 a^2-b^2\right )+b \left (a^2+2 b^2\right ) \cos (c+d x)+6 a \left (2 a^2-b^2\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=-\frac{a \left (2 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (4 a^2-b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{6 a^2 \left (2 a^2-b^2\right )-2 a b \left (2 a^2+b^2\right ) \cos (c+d x)-2 \left (12 a^4-7 a^2 b^2-2 b^4\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )}\\ &=\frac{\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}-\frac{a \left (2 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (4 a^2-b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{6 a^2 b \left (2 a^2-b^2\right )+6 a \left (a^2-b^2\right ) \left (4 a^2+b^2\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^4 \left (a^2-b^2\right )}\\ &=-\frac{a \left (4 a^2+b^2\right ) x}{b^5}+\frac{\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}-\frac{a \left (2 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (4 a^2-b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (a^4 \left (4 a^2-5 b^2\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^5 \left (a^2-b^2\right )}\\ &=-\frac{a \left (4 a^2+b^2\right ) x}{b^5}+\frac{\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}-\frac{a \left (2 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (4 a^2-b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (2 a^4 \left (4 a^2-5 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 )}{b^5 \left (a^2-b^2\right ) d}\\ &=-\frac{a \left (4 a^2+b^2\right ) x}{b^5}+\frac{2 a^4 \left (4 a^2-5 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)^{3/2} b^5 (a+b)^{3/2} d}+\frac{\left (12 a^4-7 a^2 b^2-2 b^4\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}-\frac{a \left (2 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac{\left (4 a^2-b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [C] time = 0.879634, size = 176, normalized size = 0.66 \[ \frac{9 b \left (4 a^2+b^2\right ) \sin (c+d x)+\frac{24 a^4 \left (4 a^2-5 b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac{12 a^5 b \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}-6 a b^2 \sin (2 (c+d x))-12 a (2 a-i b) (2 a+i b) (c+d x)+b^3 \sin (3 (c+d x))}{12 b^5 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.098, size = 504, normalized size = 1.9 \begin{align*} \text{result too large to display} \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 = 2.6388, size = 1644, normalized size = 6.18 \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 = 1.31688, size = 450, normalized size = 1.69 \begin{align*} \frac{\frac{6 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{2} b^{4} - b^{6}\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{6 \,{\left (4 \, a^{6} - 5 \, a^{4} b^{2}\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^{2} b^{5} - b^{7}\right )} \sqrt{a^{2} - b^{2}}} - \frac{3 \,{\left (4 \, a^{3} + a b^{2}\right )}{\left (d x + c\right )}}{b^{5}} + \frac{2 \,{\left (9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, 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} b^{4}}}{3 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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