Optimal. Leaf size=388 \[ -\frac {3 a^5 \cos (c+d x)}{2 b d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {a^4 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{7/2}}+\frac {4 a^4 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{7/2}}-\frac {a^4 \cos (c+d x)}{2 b d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}-\frac {2 a^2 \left (a^4-3 a^2 b^2+6 b^4\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{7/2}}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a+b)^3 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 d (a-b)^3 (\sin (c+d x)+1)} \]
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Rubi [A] time = 0.58, antiderivative size = 388, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2897, 2648, 2664, 2754, 12, 2660, 618, 204} \[ -\frac {a^4 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{7/2}}+\frac {4 a^4 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{7/2}}-\frac {2 a^2 \left (-3 a^2 b^2+a^4+6 b^4\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{7/2}}-\frac {3 a^5 \cos (c+d x)}{2 b d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {a^4 \cos (c+d x)}{2 b d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a+b)^3 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 d (a-b)^3 (\sin (c+d x)+1)} \]
Antiderivative was successfully verified.
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Rule 12
Rule 204
Rule 618
Rule 2648
Rule 2660
Rule 2664
Rule 2754
Rule 2897
Rubi steps
\begin {align*} \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\int \left (-\frac {1}{2 (a+b)^3 (-1+\sin (c+d x))}+\frac {1}{2 (a-b)^3 (1+\sin (c+d x))}+\frac {a^4}{b^2 \left (-a^2+b^2\right ) (a+b \sin (c+d x))^3}-\frac {2 \left (-a^5+2 a^3 b^2\right )}{b^2 \left (-a^2+b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {a^6-3 a^4 b^2+6 a^2 b^4}{b^2 \left (-a^2+b^2\right )^3 (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac {\int \frac {1}{1+\sin (c+d x)} \, dx}{2 (a-b)^3}-\frac {\int \frac {1}{-1+\sin (c+d x)} \, dx}{2 (a+b)^3}+\frac {\left (2 a^3 \left (a^2-2 b^2\right )\right ) \int \frac {1}{(a+b \sin (c+d x))^2} \, dx}{b^2 \left (a^2-b^2\right )^2}-\frac {a^4 \int \frac {1}{(a+b \sin (c+d x))^3} \, dx}{b^2 \left (a^2-b^2\right )}-\frac {\left (a^2 \left (a^4-3 a^2 b^2+6 b^4\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )^3}\\ &=\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {a^4 \cos (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {\left (2 a^3 \left (a^2-2 b^2\right )\right ) \int \frac {a}{a+b \sin (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )^3}+\frac {a^4 \int \frac {-2 a+b \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{2 b^2 \left (a^2-b^2\right )^2}-\frac {\left (2 a^2 \left (a^4-3 a^2 b^2+6 b^4\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 )}{b^2 \left (a^2-b^2\right )^3 d}\\ &=\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {a^4 \cos (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 a^5 \cos (c+d x)}{2 b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {a^4 \int \frac {2 a^2+b^2}{a+b \sin (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )^3}+\frac {\left (2 a^4 \left (a^2-2 b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )^3}+\frac {\left (4 a^2 \left (a^4-3 a^2 b^2+6 b^4\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 )}{b^2 \left (a^2-b^2\right )^3 d}\\ &=-\frac {2 a^2 \left (a^4-3 a^2 b^2+6 b^4\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {a^4 \cos (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 a^5 \cos (c+d x)}{2 b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {\left (a^4 \left (2 a^2+b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )^3}+\frac {\left (4 a^4 \left (a^2-2 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 )}{b^2 \left (a^2-b^2\right )^3 d}\\ &=-\frac {2 a^2 \left (a^4-3 a^2 b^2+6 b^4\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {a^4 \cos (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 a^5 \cos (c+d x)}{2 b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {\left (8 a^4 \left (a^2-2 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 )}{b^2 \left (a^2-b^2\right )^3 d}-\frac {\left (a^4 \left (2 a^2+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 )}{b^2 \left (a^2-b^2\right )^3 d}\\ &=\frac {4 a^4 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} d}-\frac {2 a^2 \left (a^4-3 a^2 b^2+6 b^4\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {a^4 \cos (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 a^5 \cos (c+d x)}{2 b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {\left (2 a^4 \left (2 a^2+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 )}{b^2 \left (a^2-b^2\right )^3 d}\\ &=\frac {4 a^4 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} d}-\frac {a^4 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} d}-\frac {2 a^2 \left (a^4-3 a^2 b^2+6 b^4\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{7/2} d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}-\frac {a^4 \cos (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}-\frac {3 a^5 \cos (c+d x)}{2 b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {2 a^3 \left (a^2-2 b^2\right ) \cos (c+d x)}{b \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}\\ \end {align*}
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Mathematica [A] time = 3.19, size = 195, normalized size = 0.50 \[ \frac {-\frac {6 a^2 \left (a^2+4 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac {a^3 \cos (c+d x) \left (\left (a^2-8 b^2\right ) \sin (c+d x)-7 a b\right )}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))^2}+\sin \left (\frac {1}{2} (c+d x)\right ) \left (\frac {2}{(a-b)^3 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2}{(a+b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{2 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.52, size = 906, normalized size = 2.34 \[ \left [\frac {4 \, a^{6} b - 12 \, a^{4} b^{3} + 12 \, a^{2} b^{5} - 4 \, b^{7} + 2 \, {\left (11 \, a^{6} b - 5 \, a^{4} b^{3} - 8 \, a^{2} b^{5} + 2 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (a^{4} b^{2} + 4 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{5} b + 4 \, a^{3} b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{6} + 5 \, a^{4} b^{2} + 4 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (2 \, a^{7} - 6 \, a^{5} b^{2} + 6 \, a^{3} b^{4} - 2 \, a b^{6} + {\left (a^{7} - 11 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + 6 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} b^{2} - 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} - 4 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{9} b - 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} - 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{10} - 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} + 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )\right )}}, \frac {2 \, a^{6} b - 6 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - 2 \, b^{7} + {\left (11 \, a^{6} b - 5 \, a^{4} b^{3} - 8 \, a^{2} b^{5} + 2 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (a^{4} b^{2} + 4 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{5} b + 4 \, a^{3} b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{6} + 5 \, a^{4} b^{2} + 4 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - {\left (2 \, a^{7} - 6 \, a^{5} b^{2} + 6 \, a^{3} b^{4} - 2 \, a b^{6} + {\left (a^{7} - 11 \, a^{5} b^{2} + 4 \, a^{3} b^{4} + 6 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{8} b^{2} - 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} - 4 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{9} b - 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} - 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{10} - 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} + 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.35, size = 351, normalized size = 0.90 \[ -\frac {\frac {3 \, {\left (a^{4} + 4 \, a^{2} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2} b - b^{3}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}} + \frac {a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 14 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 22 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, a^{4} b}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.54, size = 590, normalized size = 1.52 \[ -\frac {1}{d \left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{d \left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a^{5} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}-\frac {6 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}-\frac {7 a^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}-\frac {14 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}+\frac {a^{5} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}-\frac {22 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}-\frac {7 a^{4} b}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2}}-\frac {3 a^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \sqrt {a^{2}-b^{2}}}-\frac {12 a^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) b^{2}}{d \left (a -b \right )^{3} \left (a +b \right )^{3} \sqrt {a^{2}-b^{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 18.23, size = 585, normalized size = 1.51 \[ \frac {\frac {13\,a^4\,b+2\,a^2\,b^3}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^5+6\,a^3\,b^2+8\,a\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^4\,b+9\,a^2\,b^3+4\,b^5\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-3\,a^4+40\,a^2\,b^2+8\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (a^2+4\,b^2\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {9\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2+4\,b^2\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-a^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2+4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2+4\,b^2\right )+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {3\,a^2\,\mathrm {atan}\left (\frac {\frac {3\,a^2\,\left (a^2+4\,b^2\right )\,\left (2\,a^6\,b-6\,a^4\,b^3+6\,a^2\,b^5-2\,b^7\right )}{2\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}+\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+4\,b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}}{3\,a^4+12\,a^2\,b^2}\right )\,\left (a^2+4\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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