Optimal. Leaf size=182 \[ -\frac{\left (-9 a^2 b^2+6 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{a \left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{3 a x}{b^4}+\frac{\sin (c+d x) \cos ^2(c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac{3 \sin (c+d x)}{2 b^3 d} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.458377, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3048, 3032, 3023, 2735, 2659, 205} \[ -\frac{\left (-9 a^2 b^2+6 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{a \left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{3 a x}{b^4}+\frac{\sin (c+d x) \cos ^2(c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac{3 \sin (c+d x)}{2 b^3 d} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3048
Rule 3032
Rule 3023
Rule 2735
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{\cos ^2(c+d x) \left (1-\cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx &=\frac{\cos ^2(c+d x) \sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac{\int \frac{\cos (c+d x) \left (-2 \left (a^2-b^2\right )+3 \left (a^2-b^2\right ) \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac{\cos ^2(c+d x) \sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac{a \left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{b \left (3 a^4-5 a^2 b^2+2 b^4\right )+3 a \left (a^2-b^2\right )^2 \cos (c+d x)-3 b \left (a^2-b^2\right )^2 \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{3 \sin (c+d x)}{2 b^3 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac{a \left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{b^2 \left (3 a^4-5 a^2 b^2+2 b^4\right )+6 a b \left (a^2-b^2\right )^2 \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=\frac{3 a x}{b^4}-\frac{3 \sin (c+d x)}{2 b^3 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac{a \left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (6 a^4-9 a^2 b^2+2 b^4\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )}\\ &=\frac{3 a x}{b^4}-\frac{3 \sin (c+d x)}{2 b^3 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac{a \left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (6 a^4-9 a^2 b^2+2 b^4\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^4 \left (a^2-b^2\right ) d}\\ &=\frac{3 a x}{b^4}-\frac{\left (6 a^4-9 a^2 b^2+2 b^4\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^4 (a+b)^{3/2} d}-\frac{3 \sin (c+d x)}{2 b^3 d}+\frac{\cos ^2(c+d x) \sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac{a \left (3 a^2-2 b^2\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 0.978476, size = 159, normalized size = 0.87 \[ \frac{\frac{a b \left (4 b^2-5 a^2\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}-\frac{2 \left (-9 a^2 b^2+6 a^4+2 b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac{a^2 b \sin (c+d x)}{(a+b \cos (c+d x))^2}+6 a (c+d x)-2 b \sin (c+d x)}{2 b^4 d} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] time = 0.037, size = 576, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Fricas [B] time = 1.8494, size = 1879, normalized size = 10.32 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Giac [A] time = 1.73813, size = 450, normalized size = 2.47 \begin{align*} \frac{\frac{{\left (6 \, a^{4} - 9 \, a^{2} b^{2} + 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 )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt{a^{2} - b^{2}}} - \frac{4 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{2} b^{3} - b^{5}\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}} + \frac{3 \,{\left (d x + c\right )} a}{b^{4}} - \frac{2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} b^{3}}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]