Optimal. Leaf size=218 \[ \frac{3 b \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^5 d \sqrt{a^2-b^2}}-\frac{\left (a^2-12 b^2\right ) \cot (c+d x)}{2 a^4 b d}+\frac{3 \left (a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^5 d}+\frac{\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac{3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2} \]
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Rubi [A] time = 0.771971, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2890, 3055, 3001, 3770, 2660, 618, 204} \[ \frac{3 b \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^5 d \sqrt{a^2-b^2}}-\frac{\left (a^2-12 b^2\right ) \cot (c+d x)}{2 a^4 b d}+\frac{3 \left (a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^5 d}+\frac{\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac{3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 2890
Rule 3055
Rule 3001
Rule 3770
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{\cos (c+d x) \cot ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\frac{\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac{\int \frac{\csc ^2(c+d x) \left (2 \left (a^2-6 b^2\right )-2 a b \sin (c+d x)+8 b^2 \sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2} \, dx}{4 a^2 b}\\ &=\frac{\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2}-\frac{3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}+\frac{\int \frac{\csc ^2(c+d x) \left (2 \left (a^4-13 a^2 b^2+12 b^4\right )-4 a b \left (a^2-b^2\right ) \sin (c+d x)+12 b^2 \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{4 a^3 b \left (a^2-b^2\right )}\\ &=-\frac{\left (a^2-12 b^2\right ) \cot (c+d x)}{2 a^4 b d}+\frac{\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2}-\frac{3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}+\frac{\int \frac{\csc (c+d x) \left (-6 b \left (a^4-5 a^2 b^2+4 b^4\right )+12 a b^2 \left (a^2-b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{4 a^4 b \left (a^2-b^2\right )}\\ &=-\frac{\left (a^2-12 b^2\right ) \cot (c+d x)}{2 a^4 b d}+\frac{\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2}-\frac{3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}-\frac{\left (3 \left (a^2-4 b^2\right )\right ) \int \csc (c+d x) \, dx}{2 a^5}+\frac{\left (12 a^2 b^2 \left (a^2-b^2\right )+6 b^2 \left (a^4-5 a^2 b^2+4 b^4\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{4 a^5 b \left (a^2-b^2\right )}\\ &=\frac{3 \left (a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^5 d}-\frac{\left (a^2-12 b^2\right ) \cot (c+d x)}{2 a^4 b d}+\frac{\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2}-\frac{3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}+\frac{\left (3 b \left (3 a^2-4 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^5 d}\\ &=\frac{3 \left (a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^5 d}-\frac{\left (a^2-12 b^2\right ) \cot (c+d x)}{2 a^4 b d}+\frac{\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2}-\frac{3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}-\frac{\left (6 b \left (3 a^2-4 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^5 d}\\ &=\frac{3 b \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^5 \sqrt{a^2-b^2} d}+\frac{3 \left (a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^5 d}-\frac{\left (a^2-12 b^2\right ) \cot (c+d x)}{2 a^4 b d}+\frac{\left (a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d (a+b \sin (c+d x))^2}-\frac{3 b \cot (c+d x)}{a^3 d (a+b \sin (c+d x))}\\ \end{align*}
Mathematica [A] time = 6.18532, size = 319, normalized size = 1.46 \[ -\frac{3 \left (a^2-4 b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^5 d}+\frac{3 \left (a^2-4 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^5 d}+\frac{6 b^2 \cos (c+d x)-a^2 \cos (c+d x)}{2 a^4 d (a+b \sin (c+d x))}+\frac{b^2 \cos (c+d x)-a^2 \cos (c+d x)}{2 a^3 d (a+b \sin (c+d x))^2}+\frac{3 b \left (3 a^2-4 b^2\right ) \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (a \sin \left (\frac{1}{2} (c+d x)\right )+b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^5 d \sqrt{a^2-b^2}}-\frac{3 b \tan \left (\frac{1}{2} (c+d x)\right )}{2 a^4 d}+\frac{3 b \cot \left (\frac{1}{2} (c+d x)\right )}{2 a^4 d}-\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right )}{8 a^3 d}+\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right )}{8 a^3 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.213, size = 642, normalized size = 2.9 \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 [B] time = 3.7479, size = 3416, normalized size = 15.67 \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 = 1.42662, size = 533, normalized size = 2.44 \begin{align*} -\frac{\frac{12 \,{\left (a^{2} - 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{5}} - \frac{24 \,{\left (3 \, a^{2} b - 4 \, b^{3}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}} a^{5}} - \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}} - \frac{6 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 24 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 12 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 32 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 5 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 48 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 16 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 4 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 112 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 76 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 8 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{4}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{2} a^{5}}}{8 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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