Optimal. Leaf size=343 \[ -\frac {2 a^2 b^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {4 a^2 b \left (a^2+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))^2}-\frac {(a-b) \sin (c+d x)}{4 (a+b)^3 d (1-\cos (c+d x))}-\frac {\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))}+\frac {\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))^2}+\frac {\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))}+\frac {(a+b) \sin (c+d x)}{4 (a-b)^3 d (1+\cos (c+d x))}+\frac {a^3 b^2 \sin (c+d x)}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))} \]
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Rubi [A]
time = 0.39, antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps
used = 15, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3957, 2976,
2729, 2727, 2743, 12, 2738, 214} \begin {gather*} -\frac {2 a^2 b^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac {4 a^2 b \left (a^2+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {a^3 b^2 \sin (c+d x)}{d \left (a^2-b^2\right )^3 (a \cos (c+d x)+b)}-\frac {\sin (c+d x)}{12 d (a+b)^2 (1-\cos (c+d x))}-\frac {(a-b) \sin (c+d x)}{4 d (a+b)^3 (1-\cos (c+d x))}+\frac {(a+b) \sin (c+d x)}{4 d (a-b)^3 (\cos (c+d x)+1)}+\frac {\sin (c+d x)}{12 d (a-b)^2 (\cos (c+d x)+1)}-\frac {\sin (c+d x)}{12 d (a+b)^2 (1-\cos (c+d x))^2}+\frac {\sin (c+d x)}{12 d (a-b)^2 (\cos (c+d x)+1)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 214
Rule 2727
Rule 2729
Rule 2738
Rule 2743
Rule 2976
Rule 3957
Rubi steps
\begin {align*} \int \frac {\csc ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{(-b-a \cos (c+d x))^2} \, dx\\ &=\int \left (\frac {1}{4 (a-b)^2 (-1-\cos (c+d x))^2}+\frac {-a-b}{4 (a-b)^3 (-1-\cos (c+d x))}+\frac {1}{4 (a+b)^2 (1-\cos (c+d x))^2}+\frac {a-b}{4 (a+b)^3 (1-\cos (c+d x))}+\frac {a^2 b^2}{\left (a^2-b^2\right )^2 (-b-a \cos (c+d x))^2}+\frac {2 a^2 b \left (a^2+b^2\right )}{\left (a^2-b^2\right )^3 (-b-a \cos (c+d x))}\right ) \, dx\\ &=\frac {\int \frac {1}{(-1-\cos (c+d x))^2} \, dx}{4 (a-b)^2}+\frac {(a-b) \int \frac {1}{1-\cos (c+d x)} \, dx}{4 (a+b)^3}+\frac {\int \frac {1}{(1-\cos (c+d x))^2} \, dx}{4 (a+b)^2}-\frac {(a+b) \int \frac {1}{-1-\cos (c+d x)} \, dx}{4 (a-b)^3}+\frac {\left (a^2 b^2\right ) \int \frac {1}{(-b-a \cos (c+d x))^2} \, dx}{\left (a^2-b^2\right )^2}+\frac {\left (2 a^2 b \left (a^2+b^2\right )\right ) \int \frac {1}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^3}\\ &=-\frac {\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))^2}-\frac {(a-b) \sin (c+d x)}{4 (a+b)^3 d (1-\cos (c+d x))}+\frac {\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))^2}+\frac {(a+b) \sin (c+d x)}{4 (a-b)^3 d (1+\cos (c+d x))}+\frac {a^3 b^2 \sin (c+d x)}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}-\frac {\int \frac {1}{-1-\cos (c+d x)} \, dx}{12 (a-b)^2}+\frac {\int \frac {1}{1-\cos (c+d x)} \, dx}{12 (a+b)^2}+\frac {\left (a^2 b^2\right ) \int \frac {b}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^3}+\frac {\left (4 a^2 b \left (a^2+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac {4 a^2 b \left (a^2+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))^2}-\frac {(a-b) \sin (c+d x)}{4 (a+b)^3 d (1-\cos (c+d x))}-\frac {\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))}+\frac {\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))^2}+\frac {\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))}+\frac {(a+b) \sin (c+d x)}{4 (a-b)^3 d (1+\cos (c+d x))}+\frac {a^3 b^2 \sin (c+d x)}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac {\left (a^2 b^3\right ) \int \frac {1}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^3}\\ &=-\frac {4 a^2 b \left (a^2+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))^2}-\frac {(a-b) \sin (c+d x)}{4 (a+b)^3 d (1-\cos (c+d x))}-\frac {\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))}+\frac {\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))^2}+\frac {\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))}+\frac {(a+b) \sin (c+d x)}{4 (a-b)^3 d (1+\cos (c+d x))}+\frac {a^3 b^2 \sin (c+d x)}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}+\frac {\left (2 a^2 b^3\right ) \text {Subst}\left (\int \frac {1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac {2 a^2 b^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {4 a^2 b \left (a^2+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))^2}-\frac {(a-b) \sin (c+d x)}{4 (a+b)^3 d (1-\cos (c+d x))}-\frac {\sin (c+d x)}{12 (a+b)^2 d (1-\cos (c+d x))}+\frac {\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))^2}+\frac {\sin (c+d x)}{12 (a-b)^2 d (1+\cos (c+d x))}+\frac {(a+b) \sin (c+d x)}{4 (a-b)^3 d (1+\cos (c+d x))}+\frac {a^3 b^2 \sin (c+d x)}{\left (a^2-b^2\right )^3 d (b+a \cos (c+d x))}\\ \end {align*}
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Mathematica [A]
time = 0.75, size = 281, normalized size = 0.82 \begin {gather*} \frac {(b+a \cos (c+d x)) \sec ^2(c+d x) \left (\frac {48 a^2 b \left (2 a^2+3 b^2\right ) \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (c+d x))}{\left (a^2-b^2\right )^{7/2}}-\frac {4 (2 a-b) (b+a \cos (c+d x)) \cot \left (\frac {1}{2} (c+d x)\right )}{(a+b)^3}-\frac {(b+a \cos (c+d x)) \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{(a+b)^2}+\frac {24 a^3 b^2 \sin (c+d x)}{(a-b)^3 (a+b)^3}+\frac {4 (2 a+b) (b+a \cos (c+d x)) \tan \left (\frac {1}{2} (c+d x)\right )}{(a-b)^3}+\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2}\right )}{24 d (a+b \sec (c+d x))^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.25, size = 242, normalized size = 0.71
method | result | size |
derivativedivides | \(\frac {\frac {\frac {a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 \left (a^{2}-2 b a +b^{2}\right ) \left (a -b \right )}-\frac {1}{24 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {3 a -b}{8 \left (a +b \right )^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 b \,a^{2} \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b}-\frac {\left (2 a^{2}+3 b^{2}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\) | \(242\) |
default | \(\frac {\frac {\frac {a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 \left (a^{2}-2 b a +b^{2}\right ) \left (a -b \right )}-\frac {1}{24 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {3 a -b}{8 \left (a +b \right )^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 b \,a^{2} \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b}-\frac {\left (2 a^{2}+3 b^{2}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\) | \(242\) |
risch | \(\frac {2 i \left (-6 a^{4} b \,{\mathrm e}^{7 i \left (d x +c \right )}-9 a^{2} b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-6 a^{3} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-9 a \,b^{4} {\mathrm e}^{6 i \left (d x +c \right )}+14 a^{4} b \,{\mathrm e}^{5 i \left (d x +c \right )}+25 a^{2} b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+6 b^{5} {\mathrm e}^{5 i \left (d x +c \right )}-6 a^{5} {\mathrm e}^{4 i \left (d x +c \right )}+40 a^{3} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+11 a \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+2 a^{4} b \,{\mathrm e}^{3 i \left (d x +c \right )}-47 a^{2} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-4 a^{5} {\mathrm e}^{2 i \left (d x +c \right )}-30 a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-11 a \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-2 a^{4} b \,{\mathrm e}^{i \left (d x +c \right )}+15 a^{2} b^{3} {\mathrm e}^{i \left (d x +c \right )}+2 b^{5} {\mathrm e}^{i \left (d x +c \right )}+2 a^{5}+12 a^{3} b^{2}+a \,b^{4}\right )}{3 \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right ) \left (-a^{2}+b^{2}\right ) d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {2 a^{4} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {3 a^{2} b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {2 a^{4} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {3 a^{2} b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\) | \(715\) |
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 [A]
time = 4.16, size = 1040, normalized size = 3.03 \begin {gather*} \left [-\frac {22 \, a^{5} b^{2} - 14 \, a^{3} b^{4} - 8 \, a b^{6} + 2 \, {\left (2 \, a^{7} + 10 \, a^{5} b^{2} - 11 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{6} b - 7 \, a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - {\left (2 \, a^{5} b + 3 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (2 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, a^{5} b + 3 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 6 \, {\left (a^{7} + 6 \, a^{5} b^{2} - 5 \, a^{3} b^{4} - 2 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} + 10 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \cos \left (d x + c\right )}{6 \, {\left ({\left (a^{9} - 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} - 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right )^{3} + {\left (a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{9} - 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} - 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right ) - {\left (a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} d\right )} \sin \left (d x + c\right )}, -\frac {11 \, a^{5} b^{2} - 7 \, a^{3} b^{4} - 4 \, a b^{6} + {\left (2 \, a^{7} + 10 \, a^{5} b^{2} - 11 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )^{4} - {\left (4 \, a^{6} b - 7 \, a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - {\left (2 \, a^{5} b + 3 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (2 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, a^{5} b + 3 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 3 \, {\left (a^{7} + 6 \, a^{5} b^{2} - 5 \, a^{3} b^{4} - 2 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \cos \left (d x + c\right )}{3 \, {\left ({\left (a^{9} - 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} - 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right )^{3} + {\left (a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{9} - 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} - 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right ) - {\left (a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} d\right )} \sin \left (d x + c\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 {\csc ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.52, size = 457, normalized size = 1.33 \begin {gather*} -\frac {\frac {48 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\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} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}} - \frac {48 \, {\left (2 \, a^{4} b + 3 \, a^{2} b^{3}\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^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, 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} + b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6} - 6 \, a^{5} b + 15 \, a^{4} b^{2} - 20 \, a^{3} b^{3} + 15 \, a^{2} b^{4} - 6 \, a b^{5} + b^{6}} + \frac {9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.62, size = 403, normalized size = 1.17 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d\,{\left (a-b\right )}^2}+\frac {\frac {a^3-3\,a^2\,b+3\,a\,b^2-b^3}{3\,\left (a+b\right )}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (4\,a^4-13\,a^3\,b+15\,a^2\,b^2-7\,a\,b^3+b^4\right )}{3\,{\left (a+b\right )}^2}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a^5-13\,a^4\,b+38\,a^3\,b^2-18\,a^2\,b^3+7\,a\,b^4-b^5\right )}{{\left (a+b\right )}^3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (8\,a^4-32\,a^3\,b+48\,a^2\,b^2-32\,a\,b^3+8\,b^4\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (8\,a^4-16\,a^3\,b+16\,a\,b^3-8\,b^4\right )\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {16\,a^2-16\,b^2}{64\,{\left (a-b\right )}^4}+\frac {1}{8\,{\left (a-b\right )}^2}\right )}{d}+\frac {a^2\,b\,\mathrm {atan}\left (\frac {1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^6-3{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b^2+3{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^4-1{}\mathrm {i}\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^6}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{5/2}}\right )\,\left (2\,a^2+3\,b^2\right )\,2{}\mathrm {i}}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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