Optimal. Leaf size=391 \[ \frac {3 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^5 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 a^5 \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a^4 b \sec ^5(c+d x)}{d}-\frac {5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {5 a^3 b^2 \tan (c+d x) \sec ^5(c+d x)}{3 d}-\frac {5 a^3 b^2 \tan (c+d x) \sec ^3(c+d x)}{12 d}-\frac {5 a^3 b^2 \tan (c+d x) \sec (c+d x)}{8 d}+\frac {10 a^2 b^3 \sec ^7(c+d x)}{7 d}-\frac {2 a^2 b^3 \sec ^5(c+d x)}{d}+\frac {15 a b^4 \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac {5 a b^4 \tan ^3(c+d x) \sec ^5(c+d x)}{8 d}-\frac {5 a b^4 \tan (c+d x) \sec ^5(c+d x)}{16 d}+\frac {5 a b^4 \tan (c+d x) \sec ^3(c+d x)}{64 d}+\frac {15 a b^4 \tan (c+d x) \sec (c+d x)}{128 d}+\frac {b^5 \sec ^9(c+d x)}{9 d}-\frac {2 b^5 \sec ^7(c+d x)}{7 d}+\frac {b^5 \sec ^5(c+d x)}{5 d} \]
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Rubi [A] time = 0.39, antiderivative size = 391, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 8, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3090, 3768, 3770, 2606, 30, 2611, 14, 270} \[ \frac {10 a^2 b^3 \sec ^7(c+d x)}{7 d}-\frac {2 a^2 b^3 \sec ^5(c+d x)}{d}-\frac {5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {5 a^3 b^2 \tan (c+d x) \sec ^5(c+d x)}{3 d}-\frac {5 a^3 b^2 \tan (c+d x) \sec ^3(c+d x)}{12 d}-\frac {5 a^3 b^2 \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a^4 b \sec ^5(c+d x)}{d}+\frac {3 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^5 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 a^5 \tan (c+d x) \sec (c+d x)}{8 d}+\frac {15 a b^4 \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac {5 a b^4 \tan ^3(c+d x) \sec ^5(c+d x)}{8 d}-\frac {5 a b^4 \tan (c+d x) \sec ^5(c+d x)}{16 d}+\frac {5 a b^4 \tan (c+d x) \sec ^3(c+d x)}{64 d}+\frac {15 a b^4 \tan (c+d x) \sec (c+d x)}{128 d}+\frac {b^5 \sec ^9(c+d x)}{9 d}-\frac {2 b^5 \sec ^7(c+d x)}{7 d}+\frac {b^5 \sec ^5(c+d x)}{5 d} \]
Antiderivative was successfully verified.
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Rule 14
Rule 30
Rule 270
Rule 2606
Rule 2611
Rule 3090
Rule 3768
Rule 3770
Rubi steps
\begin {align*} \int \sec ^{10}(c+d x) (a \cos (c+d x)+b \sin (c+d x))^5 \, dx &=\int \left (a^5 \sec ^5(c+d x)+5 a^4 b \sec ^5(c+d x) \tan (c+d x)+10 a^3 b^2 \sec ^5(c+d x) \tan ^2(c+d x)+10 a^2 b^3 \sec ^5(c+d x) \tan ^3(c+d x)+5 a b^4 \sec ^5(c+d x) \tan ^4(c+d x)+b^5 \sec ^5(c+d x) \tan ^5(c+d x)\right ) \, dx\\ &=a^5 \int \sec ^5(c+d x) \, dx+\left (5 a^4 b\right ) \int \sec ^5(c+d x) \tan (c+d x) \, dx+\left (10 a^3 b^2\right ) \int \sec ^5(c+d x) \tan ^2(c+d x) \, dx+\left (10 a^2 b^3\right ) \int \sec ^5(c+d x) \tan ^3(c+d x) \, dx+\left (5 a b^4\right ) \int \sec ^5(c+d x) \tan ^4(c+d x) \, dx+b^5 \int \sec ^5(c+d x) \tan ^5(c+d x) \, dx\\ &=\frac {a^5 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {5 a^3 b^2 \sec ^5(c+d x) \tan (c+d x)}{3 d}+\frac {5 a b^4 \sec ^5(c+d x) \tan ^3(c+d x)}{8 d}+\frac {1}{4} \left (3 a^5\right ) \int \sec ^3(c+d x) \, dx-\frac {1}{3} \left (5 a^3 b^2\right ) \int \sec ^5(c+d x) \, dx-\frac {1}{8} \left (15 a b^4\right ) \int \sec ^5(c+d x) \tan ^2(c+d x) \, dx+\frac {\left (5 a^4 b\right ) \operatorname {Subst}\left (\int x^4 \, dx,x,\sec (c+d x)\right )}{d}+\frac {\left (10 a^2 b^3\right ) \operatorname {Subst}\left (\int x^4 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{d}+\frac {b^5 \operatorname {Subst}\left (\int x^4 \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {a^4 b \sec ^5(c+d x)}{d}+\frac {3 a^5 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^5 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {5 a^3 b^2 \sec ^3(c+d x) \tan (c+d x)}{12 d}+\frac {5 a^3 b^2 \sec ^5(c+d x) \tan (c+d x)}{3 d}-\frac {5 a b^4 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac {5 a b^4 \sec ^5(c+d x) \tan ^3(c+d x)}{8 d}+\frac {1}{8} \left (3 a^5\right ) \int \sec (c+d x) \, dx-\frac {1}{4} \left (5 a^3 b^2\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{16} \left (5 a b^4\right ) \int \sec ^5(c+d x) \, dx+\frac {\left (10 a^2 b^3\right ) \operatorname {Subst}\left (\int \left (-x^4+x^6\right ) \, dx,x,\sec (c+d x)\right )}{d}+\frac {b^5 \operatorname {Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {3 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^4 b \sec ^5(c+d x)}{d}-\frac {2 a^2 b^3 \sec ^5(c+d x)}{d}+\frac {b^5 \sec ^5(c+d x)}{5 d}+\frac {10 a^2 b^3 \sec ^7(c+d x)}{7 d}-\frac {2 b^5 \sec ^7(c+d x)}{7 d}+\frac {b^5 \sec ^9(c+d x)}{9 d}+\frac {3 a^5 \sec (c+d x) \tan (c+d x)}{8 d}-\frac {5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^5 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {5 a^3 b^2 \sec ^3(c+d x) \tan (c+d x)}{12 d}+\frac {5 a b^4 \sec ^3(c+d x) \tan (c+d x)}{64 d}+\frac {5 a^3 b^2 \sec ^5(c+d x) \tan (c+d x)}{3 d}-\frac {5 a b^4 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac {5 a b^4 \sec ^5(c+d x) \tan ^3(c+d x)}{8 d}-\frac {1}{8} \left (5 a^3 b^2\right ) \int \sec (c+d x) \, dx+\frac {1}{64} \left (15 a b^4\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {3 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^4 b \sec ^5(c+d x)}{d}-\frac {2 a^2 b^3 \sec ^5(c+d x)}{d}+\frac {b^5 \sec ^5(c+d x)}{5 d}+\frac {10 a^2 b^3 \sec ^7(c+d x)}{7 d}-\frac {2 b^5 \sec ^7(c+d x)}{7 d}+\frac {b^5 \sec ^9(c+d x)}{9 d}+\frac {3 a^5 \sec (c+d x) \tan (c+d x)}{8 d}-\frac {5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {15 a b^4 \sec (c+d x) \tan (c+d x)}{128 d}+\frac {a^5 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {5 a^3 b^2 \sec ^3(c+d x) \tan (c+d x)}{12 d}+\frac {5 a b^4 \sec ^3(c+d x) \tan (c+d x)}{64 d}+\frac {5 a^3 b^2 \sec ^5(c+d x) \tan (c+d x)}{3 d}-\frac {5 a b^4 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac {5 a b^4 \sec ^5(c+d x) \tan ^3(c+d x)}{8 d}+\frac {1}{128} \left (15 a b^4\right ) \int \sec (c+d x) \, dx\\ &=\frac {3 a^5 \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {5 a^3 b^2 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {15 a b^4 \tanh ^{-1}(\sin (c+d x))}{128 d}+\frac {a^4 b \sec ^5(c+d x)}{d}-\frac {2 a^2 b^3 \sec ^5(c+d x)}{d}+\frac {b^5 \sec ^5(c+d x)}{5 d}+\frac {10 a^2 b^3 \sec ^7(c+d x)}{7 d}-\frac {2 b^5 \sec ^7(c+d x)}{7 d}+\frac {b^5 \sec ^9(c+d x)}{9 d}+\frac {3 a^5 \sec (c+d x) \tan (c+d x)}{8 d}-\frac {5 a^3 b^2 \sec (c+d x) \tan (c+d x)}{8 d}+\frac {15 a b^4 \sec (c+d x) \tan (c+d x)}{128 d}+\frac {a^5 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {5 a^3 b^2 \sec ^3(c+d x) \tan (c+d x)}{12 d}+\frac {5 a b^4 \sec ^3(c+d x) \tan (c+d x)}{64 d}+\frac {5 a^3 b^2 \sec ^5(c+d x) \tan (c+d x)}{3 d}-\frac {5 a b^4 \sec ^5(c+d x) \tan (c+d x)}{16 d}+\frac {5 a b^4 \sec ^5(c+d x) \tan ^3(c+d x)}{8 d}\\ \end {align*}
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Mathematica [A] time = 2.14, size = 331, normalized size = 0.85 \[ \frac {1260 a \left (656 a^4+2320 a^2 b^2+845 b^4\right ) \tan (c+d x) \sec ^7(c+d x)-40320 a \left (48 a^4-80 a^2 b^2+15 b^4\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec ^9(c+d x) \left (372960 a^5 \sin (4 (c+d x))+131040 a^5 \sin (6 (c+d x))+15120 a^5 \sin (8 (c+d x))+1935360 a^4 b+453600 a^3 b^2 \sin (4 (c+d x))-218400 a^3 b^2 \sin (6 (c+d x))-25200 a^3 b^2 \sin (8 (c+d x))-184320 a^2 b^3+73728 \left (35 a^4 b-20 a^2 b^3-3 b^5\right ) \cos (2 (c+d x))+129024 \left (5 a^4 b-10 a^2 b^3+b^5\right ) \cos (4 (c+d x))-488250 a b^4 \sin (4 (c+d x))+40950 a b^4 \sin (6 (c+d x))+4725 a b^4 \sin (8 (c+d x))+223232 b^5\right )}{5160960 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 257, normalized size = 0.66 \[ \frac {315 \, {\left (48 \, a^{5} - 80 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{9} \log \left (\sin \left (d x + c\right ) + 1\right ) - 315 \, {\left (48 \, a^{5} - 80 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{9} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8960 \, b^{5} + 16128 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} + 23040 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 210 \, {\left (3 \, {\left (48 \, a^{5} - 80 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{7} + 240 \, a b^{4} \cos \left (d x + c\right ) + 2 \, {\left (48 \, a^{5} - 80 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{5} + 40 \, {\left (16 \, a^{3} b^{2} - 9 \, a b^{4}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{80640 \, d \cos \left (d x + c\right )^{9}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.74, size = 888, normalized size = 2.27 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.28, size = 688, normalized size = 1.76 \[ \frac {8 b^{5} \cos \left (d x +c \right )}{315 d}-\frac {15 a \,b^{4} \sin \left (d x +c \right )}{128 d}+\frac {5 a^{3} b^{2} \sin \left (d x +c \right )}{8 d}+\frac {3 a^{5} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}-\frac {4 a^{2} b^{3} \cos \left (d x +c \right )}{7 d}-\frac {2 \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2} b^{3}}{7 d}+\frac {a^{4} b}{d \cos \left (d x +c \right )^{5}}+\frac {a^{5} \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{9 d \cos \left (d x +c \right )^{9}}-\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{315 d \cos \left (d x +c \right )^{3}}-\frac {5 a^{3} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {5 a^{3} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{6}}+\frac {10 a^{2} b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{7 d \cos \left (d x +c \right )^{7}}+\frac {5 a \,b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{8}}+\frac {5 a^{3} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {2 a^{2} b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{7 d \cos \left (d x +c \right )^{3}}+\frac {5 a \,b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{64 d \cos \left (d x +c \right )^{4}}+\frac {3 a^{5} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {15 a \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{128 d}-\frac {5 a \,b^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{128 d}+\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{105 d \cos \left (d x +c \right )}+\frac {b^{5} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{105 d}+\frac {4 \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) b^{5}}{315 d}+\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{105 d \cos \left (d x +c \right )^{5}}+\frac {b^{5} \left (\sin ^{6}\left (d x +c \right )\right )}{21 d \cos \left (d x +c \right )^{7}}-\frac {5 a \,b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{128 d \cos \left (d x +c \right )^{2}}+\frac {5 a^{3} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {6 a^{2} b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{7 d \cos \left (d x +c \right )^{5}}+\frac {5 a \,b^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{6}}-\frac {2 a^{2} b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{7 d \cos \left (d x +c \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.01, size = 360, normalized size = 0.92 \[ -\frac {1575 \, a b^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{7} - 11 \, \sin \left (d x + c\right )^{5} - 11 \, \sin \left (d x + c\right )^{3} + 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{8} - 4 \, \sin \left (d x + c\right )^{6} + 6 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 8400 \, a^{3} b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 8 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 5040 \, a^{5} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac {80640 \, a^{4} b}{\cos \left (d x + c\right )^{5}} + \frac {23040 \, {\left (7 \, \cos \left (d x + c\right )^{2} - 5\right )} a^{2} b^{3}}{\cos \left (d x + c\right )^{7}} - \frac {256 \, {\left (63 \, \cos \left (d x + c\right )^{4} - 90 \, \cos \left (d x + c\right )^{2} + 35\right )} b^{5}}{\cos \left (d x + c\right )^{9}}}{80640 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.71, size = 675, normalized size = 1.73 \[ \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a^5}{4}-\frac {5\,a^3\,b^2}{4}+\frac {15\,a\,b^4}{64}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,a^5}{4}+\frac {5\,a^3\,b^2}{4}-\frac {15\,a\,b^4}{64}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,\left (40\,a^4\,b-40\,a^2\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}\,\left (\frac {5\,a^5}{4}+\frac {5\,a^3\,b^2}{4}-\frac {15\,a\,b^4}{64}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (-\frac {11\,a^5}{2}+\frac {95\,a^3\,b^2}{6}+\frac {65\,a\,b^4}{32}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}\,\left (-\frac {11\,a^5}{2}+\frac {95\,a^3\,b^2}{6}+\frac {65\,a\,b^4}{32}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {19\,a^5}{2}-\frac {45\,a^3\,b^2}{2}+\frac {775\,a\,b^4}{32}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\left (\frac {19\,a^5}{2}-\frac {45\,a^3\,b^2}{2}+\frac {775\,a\,b^4}{32}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {15\,a^5}{2}+\frac {15\,a^3\,b^2}{2}-\frac {845\,a\,b^4}{32}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (\frac {15\,a^5}{2}+\frac {15\,a^3\,b^2}{2}-\frac {845\,a\,b^4}{32}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (8\,a^4\,b-\frac {72\,a^2\,b^3}{7}+\frac {16\,b^5}{35}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (32\,a^4\,b-\frac {8\,a^2\,b^3}{7}+\frac {64\,b^5}{35}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (80\,a^4\,b-40\,a^2\,b^3+\frac {32\,b^5}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (-120\,a^4\,b+40\,a^2\,b^3+16\,b^5\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (-88\,a^4\,b+56\,a^2\,b^3+\frac {32\,b^5}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (132\,a^4\,b-104\,a^2\,b^3+\frac {112\,b^5}{5}\right )+2\,a^4\,b+\frac {16\,b^5}{315}-\frac {8\,a^2\,b^3}{7}+10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}-9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}+36\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-84\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+126\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-126\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+84\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-36\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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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