Optimal. Leaf size=182 \[ -\frac {3 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \sqrt {a^2-b^2} d}+\frac {3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2}-\frac {\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))} \]
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Rubi [A]
time = 0.30, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2969, 3134,
3080, 3855, 2739, 632, 210} \begin {gather*} \frac {3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {3 \left (a^2-2 b^2\right ) \text {ArcTan}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 d \sqrt {a^2-b^2}}-\frac {\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 210
Rule 632
Rule 2739
Rule 2969
Rule 3080
Rule 3134
Rule 3855
Rubi steps
\begin {align*} \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2}+\frac {\int \frac {\csc (c+d x) \left (-6 b^2-2 a b \sin (c+d x)+\left (a^2+3 b^2\right ) \sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2} \, dx}{2 a^2 b}\\ &=\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2}-\frac {\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))}+\frac {\int \frac {\csc (c+d x) \left (-6 b^2 \left (a^2-b^2\right )-3 a b \left (a^2-b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 a^3 b \left (a^2-b^2\right )}\\ &=\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2}-\frac {\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))}-\frac {(3 b) \int \csc (c+d x) \, dx}{a^4}-\frac {\left (3 \left (a^2-2 b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{2 a^4}\\ &=\frac {3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2}-\frac {\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))}-\frac {\left (3 \left (a^2-2 b^2\right )\right ) \text {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^4 d}\\ &=\frac {3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2}-\frac {\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))}+\frac {\left (6 \left (a^2-2 b^2\right )\right ) \text {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^4 d}\\ &=-\frac {3 \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 )}{a^4 \sqrt {a^2-b^2} d}+\frac {3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))^2}-\frac {\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))}\\ \end {align*}
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Mathematica [A]
time = 1.90, size = 184, normalized size = 1.01 \begin {gather*} \frac {-\frac {6 \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 )}{\sqrt {a^2-b^2}}-a \cot \left (\frac {1}{2} (c+d x)\right )+6 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-6 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2 \left (a^2-b^2\right ) \cos (c+d x)}{b (a+b \sin (c+d x))^2}-\frac {a \left (a^2+4 b^2\right ) \cos (c+d x)}{b (a+b \sin (c+d x))}+a \tan \left (\frac {1}{2} (c+d x)\right )}{2 a^4 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.70, size = 214, normalized size = 1.18
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{3}}-\frac {1}{2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}}-\frac {4 \left (\frac {-\frac {a \left (a^{2}-6 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 b \left (a^{2}+2 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {a \left (a^{2}+14 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {5 a^{2} b}{4}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {3 \left (a^{2}-2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{4 \sqrt {a^{2}-b^{2}}}\right )}{a^{4}}}{d}\) | \(214\) |
default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{3}}-\frac {1}{2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}}-\frac {4 \left (\frac {-\frac {a \left (a^{2}-6 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 b \left (a^{2}+2 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {a \left (a^{2}+14 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {5 a^{2} b}{4}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {3 \left (a^{2}-2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{4 \sqrt {a^{2}-b^{2}}}\right )}{a^{4}}}{d}\) | \(214\) |
risch | \(\frac {i \left (-2 i a^{3} b \,{\mathrm e}^{5 i \left (d x +c \right )}-3 i a \,b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+4 i a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}+24 i a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+2 a^{4} {\mathrm e}^{4 i \left (d x +c \right )}+9 a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-2 i a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}-21 i a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}-2 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-18 b^{2} a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+a^{2} b^{2}+6 b^{4}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} b^{2} a^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, d \,a^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, d \,a^{4}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, d \,a^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, d \,a^{4}}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{4} d}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{4} d}\) | \(582\) |
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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 490 vs.
\(2 (173) = 346\).
time = 0.47, size = 1064, normalized size = 5.85 \begin {gather*} \left [-\frac {2 \, {\left (a^{5} + 5 \, a^{3} b^{2} - 6 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} - 18 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, {\left (2 \, a^{3} b - 4 \, a b^{3} - 2 \, {\left (a^{3} b - 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{4} - a^{2} b^{2} - 2 \, b^{4} - {\left (a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \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 ) - 6 \, {\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right ) + 6 \, {\left (2 \, a^{3} b^{2} - 2 \, a b^{4} - 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{4} b - b^{5} - {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 6 \, {\left (2 \, a^{3} b^{2} - 2 \, a b^{4} - 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{4} b - b^{5} - {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (2 \, {\left (a^{7} b - a^{5} b^{3}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{7} b - a^{5} b^{3}\right )} d + {\left ({\left (a^{6} b^{2} - a^{4} b^{4}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{8} - a^{4} b^{4}\right )} d\right )} \sin \left (d x + c\right )\right )}}, -\frac {{\left (a^{5} + 5 \, a^{3} b^{2} - 6 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} - 9 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, {\left (2 \, a^{3} b - 4 \, a b^{3} - 2 \, {\left (a^{3} b - 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{4} - a^{2} b^{2} - 2 \, b^{4} - {\left (a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \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 ) - 3 \, {\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right ) + 3 \, {\left (2 \, a^{3} b^{2} - 2 \, a b^{4} - 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{4} b - b^{5} - {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (2 \, a^{3} b^{2} - 2 \, a b^{4} - 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{4} b - b^{5} - {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (2 \, {\left (a^{7} b - a^{5} b^{3}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{7} b - a^{5} b^{3}\right )} d + {\left ({\left (a^{6} b^{2} - a^{4} b^{4}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{8} - a^{4} b^{4}\right )} d\right )} \sin \left (d x + c\right )\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{4}{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.50, size = 273, normalized size = 1.50 \begin {gather*} -\frac {\frac {6 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} + \frac {6 \, {\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 )} {\left (a^{2} - 2 \, b^{2}\right )}}{\sqrt {a^{2} - b^{2}} a^{4}} - \frac {6 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 14 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{2} b\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} a^{4}}}{2 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 9.88, size = 956, normalized size = 5.25 \begin {gather*} \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (4\,a^2+32\,b^2\right )+a^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-12\,b^2\right )+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,a^2\,b+10\,b^3\right )}{a}+14\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (2\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (4\,a^5+8\,a^3\,b^2\right )+2\,a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+8\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+8\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {3\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {\mathrm {atan}\left (\frac {\frac {\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (a^2-2\,b^2\right )\,\left (\frac {3\,a^6-12\,a^4\,b^2}{a^6}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^4\,b-24\,a^2\,b^3\right )}{a^5}-\frac {3\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2\,b-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^8-8\,a^6\,b^2\right )}{a^5}\right )\,\left (a^2-2\,b^2\right )}{2\,\left (a^6-a^4\,b^2\right )}\right )\,3{}\mathrm {i}}{2\,\left (a^6-a^4\,b^2\right )}+\frac {\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (a^2-2\,b^2\right )\,\left (\frac {3\,a^6-12\,a^4\,b^2}{a^6}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^4\,b-24\,a^2\,b^3\right )}{a^5}+\frac {3\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2\,b-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^8-8\,a^6\,b^2\right )}{a^5}\right )\,\left (a^2-2\,b^2\right )}{2\,\left (a^6-a^4\,b^2\right )}\right )\,3{}\mathrm {i}}{2\,\left (a^6-a^4\,b^2\right )}}{\frac {2\,\left (9\,a^2\,b-18\,b^3\right )}{a^6}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,a^2-18\,b^2\right )}{a^5}+\frac {3\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (a^2-2\,b^2\right )\,\left (\frac {3\,a^6-12\,a^4\,b^2}{a^6}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^4\,b-24\,a^2\,b^3\right )}{a^5}-\frac {3\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2\,b-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^8-8\,a^6\,b^2\right )}{a^5}\right )\,\left (a^2-2\,b^2\right )}{2\,\left (a^6-a^4\,b^2\right )}\right )}{2\,\left (a^6-a^4\,b^2\right )}-\frac {3\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (a^2-2\,b^2\right )\,\left (\frac {3\,a^6-12\,a^4\,b^2}{a^6}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^4\,b-24\,a^2\,b^3\right )}{a^5}+\frac {3\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (2\,a^2\,b-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^8-8\,a^6\,b^2\right )}{a^5}\right )\,\left (a^2-2\,b^2\right )}{2\,\left (a^6-a^4\,b^2\right )}\right )}{2\,\left (a^6-a^4\,b^2\right )}}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (a^2-2\,b^2\right )\,3{}\mathrm {i}}{d\,\left (a^6-a^4\,b^2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
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