Optimal. Leaf size=331 \[ \frac {3 \left (40 a^4-24 a^2 b^2+b^4\right ) x}{8 b^7}-\frac {3 a \left (10 a^4-11 a^2 b^2+2 b^4\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^7 \sqrt {a^2-b^2} d}+\frac {a \left (30 a^2-13 b^2\right ) \cos (c+d x)}{2 b^6 d}-\frac {3 \left (20 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^5 d}+\frac {\left (10 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{2 a b^4 d}-\frac {\left (15 a^2-4 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{4 a^2 b^3 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (7 a^2-2 b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))} \]
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Rubi [A]
time = 0.64, antiderivative size = 331, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2970, 3128,
3102, 2814, 2739, 632, 210} \begin {gather*} \frac {\left (7 a^2-2 b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac {\left (a^2-b^2\right ) \sin ^4(c+d x) \cos (c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {a \left (30 a^2-13 b^2\right ) \cos (c+d x)}{2 b^6 d}-\frac {3 \left (20 a^2-7 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^5 d}+\frac {\left (10 a^2-3 b^2\right ) \sin ^2(c+d x) \cos (c+d x)}{2 a b^4 d}-\frac {\left (15 a^2-4 b^2\right ) \sin ^3(c+d x) \cos (c+d x)}{4 a^2 b^3 d}-\frac {3 a \left (10 a^4-11 a^2 b^2+2 b^4\right ) \text {ArcTan}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^7 d \sqrt {a^2-b^2}}+\frac {3 x \left (40 a^4-24 a^2 b^2+b^4\right )}{8 b^7} \end {gather*}
Antiderivative was successfully verified.
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Rule 210
Rule 632
Rule 2739
Rule 2814
Rule 2970
Rule 3102
Rule 3128
Rubi steps
\begin {align*} \int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (7 a^2-2 b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac {\int \frac {\sin ^3(c+d x) \left (6 \left (4 a^2-b^2\right )-a b \sin (c+d x)-2 \left (15 a^2-4 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 a^2 b^2}\\ &=-\frac {\left (15 a^2-4 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{4 a^2 b^3 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (7 a^2-2 b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac {\int \frac {\sin ^2(c+d x) \left (-6 a \left (15 a^2-4 b^2\right )+6 a^2 b \sin (c+d x)+12 a \left (10 a^2-3 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{8 a^2 b^3}\\ &=\frac {\left (10 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{2 a b^4 d}-\frac {\left (15 a^2-4 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{4 a^2 b^3 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (7 a^2-2 b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac {\int \frac {\sin (c+d x) \left (24 a^2 \left (10 a^2-3 b^2\right )-30 a^3 b \sin (c+d x)-18 a^2 \left (20 a^2-7 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{24 a^2 b^4}\\ &=-\frac {3 \left (20 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^5 d}+\frac {\left (10 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{2 a b^4 d}-\frac {\left (15 a^2-4 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{4 a^2 b^3 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (7 a^2-2 b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac {\int \frac {-18 a^3 \left (20 a^2-7 b^2\right )+6 a^2 b \left (20 a^2-3 b^2\right ) \sin (c+d x)+24 a^3 \left (30 a^2-13 b^2\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{48 a^2 b^5}\\ &=\frac {a \left (30 a^2-13 b^2\right ) \cos (c+d x)}{2 b^6 d}-\frac {3 \left (20 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^5 d}+\frac {\left (10 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{2 a b^4 d}-\frac {\left (15 a^2-4 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{4 a^2 b^3 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (7 a^2-2 b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac {\int \frac {-18 a^3 b \left (20 a^2-7 b^2\right )-18 a^2 \left (40 a^4-24 a^2 b^2+b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{48 a^2 b^6}\\ &=\frac {3 \left (40 a^4-24 a^2 b^2+b^4\right ) x}{8 b^7}+\frac {a \left (30 a^2-13 b^2\right ) \cos (c+d x)}{2 b^6 d}-\frac {3 \left (20 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^5 d}+\frac {\left (10 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{2 a b^4 d}-\frac {\left (15 a^2-4 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{4 a^2 b^3 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (7 a^2-2 b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac {\left (3 a \left (10 a^4-11 a^2 b^2+2 b^4\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{2 b^7}\\ &=\frac {3 \left (40 a^4-24 a^2 b^2+b^4\right ) x}{8 b^7}+\frac {a \left (30 a^2-13 b^2\right ) \cos (c+d x)}{2 b^6 d}-\frac {3 \left (20 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^5 d}+\frac {\left (10 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{2 a b^4 d}-\frac {\left (15 a^2-4 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{4 a^2 b^3 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (7 a^2-2 b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}-\frac {\left (3 a \left (10 a^4-11 a^2 b^2+2 b^4\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 )}{b^7 d}\\ &=\frac {3 \left (40 a^4-24 a^2 b^2+b^4\right ) x}{8 b^7}+\frac {a \left (30 a^2-13 b^2\right ) \cos (c+d x)}{2 b^6 d}-\frac {3 \left (20 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^5 d}+\frac {\left (10 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{2 a b^4 d}-\frac {\left (15 a^2-4 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{4 a^2 b^3 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (7 a^2-2 b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}+\frac {\left (6 a \left (10 a^4-11 a^2 b^2+2 b^4\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 )}{b^7 d}\\ &=\frac {3 \left (40 a^4-24 a^2 b^2+b^4\right ) x}{8 b^7}-\frac {3 a \left (10 a^4-11 a^2 b^2+2 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^7 \sqrt {a^2-b^2} d}+\frac {a \left (30 a^2-13 b^2\right ) \cos (c+d x)}{2 b^6 d}-\frac {3 \left (20 a^2-7 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^5 d}+\frac {\left (10 a^2-3 b^2\right ) \cos (c+d x) \sin ^2(c+d x)}{2 a b^4 d}-\frac {\left (15 a^2-4 b^2\right ) \cos (c+d x) \sin ^3(c+d x)}{4 a^2 b^3 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a b^2 d (a+b \sin (c+d x))^2}+\frac {\left (7 a^2-2 b^2\right ) \cos (c+d x) \sin ^4(c+d x)}{2 a^2 b^2 d (a+b \sin (c+d x))}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1250\) vs. \(2(331)=662\).
time = 7.04, size = 1250, normalized size = 3.78 \begin {gather*} -\frac {-\frac {6 \left (-8 (c+d x)+\frac {2 a \left (8 a^4-20 a^2 b^2+15 b^4\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {a b \left (4 a^2-3 b^2\right ) \cos (c+d x)}{(a-b) (a+b) (a+b \sin (c+d x))^2}-\frac {3 b \left (4 a^4-7 a^2 b^2+2 b^4\right ) \cos (c+d x)}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}\right )}{b^3}+\frac {6 \left (\frac {6 a b \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}}+\frac {\cos (c+d x) \left (a \left (2 a^2+b^2\right )+b \left (a^2+2 b^2\right ) \sin (c+d x)\right )}{(a+b \sin (c+d x))^2}\right )}{(a-b)^2 (a+b)^2}+\frac {2 \left (-24 \left (-8 a^2+b^2\right ) (c+d x)-\frac {6 a \left (64 a^6-168 a^4 b^2+140 a^2 b^4-35 b^6\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+96 a b \cos (c+d x)+\frac {a b \left (-16 a^4+20 a^2 b^2-5 b^4\right ) \cos (c+d x)}{(a-b) (a+b) (a+b \sin (c+d x))^2}+\frac {b \left (112 a^6-220 a^4 b^2+115 a^2 b^4-10 b^6\right ) \cos (c+d x)}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}-8 b^2 \sin (2 (c+d x))\right )}{b^5}+\frac {\frac {12 a \left (640 a^8-1920 a^6 b^2+2016 a^4 b^4-840 a^2 b^6+105 b^8\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {-3840 a^{10} (c+d x)+7680 a^8 b^2 (c+d x)-2976 a^6 b^4 (c+d x)-1776 a^4 b^6 (c+d x)+960 a^2 b^8 (c+d x)-48 b^{10} (c+d x)-3840 a^9 b \cos (c+d x)+8640 a^7 b^3 \cos (c+d x)-5696 a^5 b^5 \cos (c+d x)+788 a^3 b^7 \cos (c+d x)+114 a b^9 \cos (c+d x)+1920 a^8 b^2 (c+d x) \cos (2 (c+d x))-4800 a^6 b^4 (c+d x) \cos (2 (c+d x))+3888 a^4 b^6 (c+d x) \cos (2 (c+d x))-1056 a^2 b^8 (c+d x) \cos (2 (c+d x))+48 b^{10} (c+d x) \cos (2 (c+d x))+320 a^7 b^3 \cos (3 (c+d x))-760 a^5 b^5 \cos (3 (c+d x))+560 a^3 b^7 \cos (3 (c+d x))-120 a b^9 \cos (3 (c+d x))-8 a^5 b^5 \cos (5 (c+d x))+16 a^3 b^7 \cos (5 (c+d x))-8 a b^9 \cos (5 (c+d x))-7680 a^9 b (c+d x) \sin (c+d x)+19200 a^7 b^3 (c+d x) \sin (c+d x)-15552 a^5 b^5 (c+d x) \sin (c+d x)+4224 a^3 b^7 (c+d x) \sin (c+d x)-192 a b^9 (c+d x) \sin (c+d x)-2880 a^8 b^2 \sin (2 (c+d x))+6880 a^6 b^4 \sin (2 (c+d x))-5182 a^4 b^6 \sin (2 (c+d x))+1221 a^2 b^8 \sin (2 (c+d x))-36 b^{10} \sin (2 (c+d x))-40 a^6 b^4 \sin (4 (c+d x))+88 a^4 b^6 \sin (4 (c+d x))-56 a^2 b^8 \sin (4 (c+d x))+8 b^{10} \sin (4 (c+d x))+2 a^4 b^6 \sin (6 (c+d x))-4 a^2 b^8 \sin (6 (c+d x))+2 b^{10} \sin (6 (c+d x))}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}}{b^7}}{256 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.48, size = 450, normalized size = 1.36 Too large to display
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 = 0.48, size = 1110, normalized size = 3.35 \begin {gather*} \left [-\frac {4 \, {\left (a^{3} b^{5} - a b^{7}\right )} \cos \left (d x + c\right )^{5} - 3 \, {\left (40 \, a^{6} b^{2} - 64 \, a^{4} b^{4} + 25 \, a^{2} b^{6} - b^{8}\right )} d x \cos \left (d x + c\right )^{2} - 2 \, {\left (20 \, a^{5} b^{3} - 27 \, a^{3} b^{5} + 7 \, a b^{7}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (40 \, a^{8} - 24 \, a^{6} b^{2} - 39 \, a^{4} b^{4} + 24 \, a^{2} b^{6} - b^{8}\right )} d x - 6 \, {\left (10 \, a^{7} - a^{5} b^{2} - 9 \, a^{3} b^{4} + 2 \, a b^{6} - {\left (10 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (10 \, a^{6} b - 11 \, a^{4} b^{3} + 2 \, a^{2} b^{5}\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 (20 \, a^{7} b - 22 \, a^{5} b^{3} - a^{3} b^{5} + 3 \, a b^{7}\right )} \cos \left (d x + c\right ) - {\left (2 \, {\left (a^{2} b^{6} - b^{8}\right )} \cos \left (d x + c\right )^{5} - {\left (10 \, a^{4} b^{4} - 11 \, a^{2} b^{6} + b^{8}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (40 \, a^{7} b - 64 \, a^{5} b^{3} + 25 \, a^{3} b^{5} - a b^{7}\right )} d x - 3 \, {\left (60 \, a^{6} b^{2} - 91 \, a^{4} b^{4} + 32 \, a^{2} b^{6} - b^{8}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left ({\left (a^{2} b^{9} - b^{11}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{3} b^{8} - a b^{10}\right )} d \sin \left (d x + c\right ) - {\left (a^{4} b^{7} - b^{11}\right )} d\right )}}, -\frac {4 \, {\left (a^{3} b^{5} - a b^{7}\right )} \cos \left (d x + c\right )^{5} - 3 \, {\left (40 \, a^{6} b^{2} - 64 \, a^{4} b^{4} + 25 \, a^{2} b^{6} - b^{8}\right )} d x \cos \left (d x + c\right )^{2} - 2 \, {\left (20 \, a^{5} b^{3} - 27 \, a^{3} b^{5} + 7 \, a b^{7}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (40 \, a^{8} - 24 \, a^{6} b^{2} - 39 \, a^{4} b^{4} + 24 \, a^{2} b^{6} - b^{8}\right )} d x + 12 \, {\left (10 \, a^{7} - a^{5} b^{2} - 9 \, a^{3} b^{4} + 2 \, a b^{6} - {\left (10 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + 2 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (10 \, a^{6} b - 11 \, a^{4} b^{3} + 2 \, a^{2} b^{5}\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 ) + 6 \, {\left (20 \, a^{7} b - 22 \, a^{5} b^{3} - a^{3} b^{5} + 3 \, a b^{7}\right )} \cos \left (d x + c\right ) - {\left (2 \, {\left (a^{2} b^{6} - b^{8}\right )} \cos \left (d x + c\right )^{5} - {\left (10 \, a^{4} b^{4} - 11 \, a^{2} b^{6} + b^{8}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (40 \, a^{7} b - 64 \, a^{5} b^{3} + 25 \, a^{3} b^{5} - a b^{7}\right )} d x - 3 \, {\left (60 \, a^{6} b^{2} - 91 \, a^{4} b^{4} + 32 \, a^{2} b^{6} - b^{8}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left ({\left (a^{2} b^{9} - b^{11}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{3} b^{8} - a b^{10}\right )} d \sin \left (d x + c\right ) - {\left (a^{4} b^{7} - b^{11}\right )} d\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.56, size = 540, normalized size = 1.63 \begin {gather*} \frac {\frac {3 \, {\left (40 \, a^{4} - 24 \, a^{2} b^{2} + b^{4}\right )} {\left (d x + c\right )}}{b^{7}} - \frac {24 \, {\left (10 \, a^{5} - 11 \, a^{3} b^{2} + 2 \, a b^{4}\right )} {\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 )}}{\sqrt {a^{2} - b^{2}} b^{7}} + \frac {8 \, {\left (9 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 31 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 16 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 10 \, a^{5} - 5 \, a^{3} b^{2}\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} b^{6}} + \frac {2 \, {\left (24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 80 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 48 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 240 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 96 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 240 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 80 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 80 \, a^{3} - 32 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} b^{6}}}{8 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 16.60, size = 2500, normalized size = 7.55 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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