Optimal. Leaf size=299 \[ -\frac {a^3 \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac {5 a^3 \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac {5 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a^3 x}{16}-\frac {3 a^2 b \sin ^5(c+d x)}{5 d}-\frac {a^2 b \sin ^3(c+d x)}{d}-\frac {3 a^2 b \sin (c+d x)}{d}+\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {45 a b^2 \tan (c+d x)}{8 d}-\frac {3 a b^2 \sin ^4(c+d x) \tan (c+d x)}{4 d}-\frac {15 a b^2 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac {45}{8} a b^2 x+\frac {5 b^3 \sin ^3(c+d x)}{6 d}+\frac {5 b^3 \sin (c+d x)}{2 d}+\frac {b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}-\frac {5 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d} \]
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Rubi [A] time = 0.34, antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 11, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3872, 2912, 2635, 8, 2592, 302, 206, 2591, 288, 321, 203} \[ -\frac {3 a^2 b \sin ^5(c+d x)}{5 d}-\frac {a^2 b \sin ^3(c+d x)}{d}-\frac {3 a^2 b \sin (c+d x)}{d}+\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a^3 \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac {5 a^3 \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac {5 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {5 a^3 x}{16}+\frac {45 a b^2 \tan (c+d x)}{8 d}-\frac {3 a b^2 \sin ^4(c+d x) \tan (c+d x)}{4 d}-\frac {15 a b^2 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac {45}{8} a b^2 x+\frac {5 b^3 \sin ^3(c+d x)}{6 d}+\frac {5 b^3 \sin (c+d x)}{2 d}+\frac {b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}-\frac {5 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 203
Rule 206
Rule 288
Rule 302
Rule 321
Rule 2591
Rule 2592
Rule 2635
Rule 2912
Rule 3872
Rubi steps
\begin {align*} \int (a+b \sec (c+d x))^3 \sin ^6(c+d x) \, dx &=-\int (-b-a \cos (c+d x))^3 \sin ^3(c+d x) \tan ^3(c+d x) \, dx\\ &=-\int \left (-a^3 \sin ^6(c+d x)-3 a^2 b \sin ^5(c+d x) \tan (c+d x)-3 a b^2 \sin ^4(c+d x) \tan ^2(c+d x)-b^3 \sin ^3(c+d x) \tan ^3(c+d x)\right ) \, dx\\ &=a^3 \int \sin ^6(c+d x) \, dx+\left (3 a^2 b\right ) \int \sin ^5(c+d x) \tan (c+d x) \, dx+\left (3 a b^2\right ) \int \sin ^4(c+d x) \tan ^2(c+d x) \, dx+b^3 \int \sin ^3(c+d x) \tan ^3(c+d x) \, dx\\ &=-\frac {a^3 \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac {1}{6} \left (5 a^3\right ) \int \sin ^4(c+d x) \, dx+\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \frac {x^6}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}+\frac {\left (3 a b^2\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{d}+\frac {b^3 \operatorname {Subst}\left (\int \frac {x^6}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac {5 a^3 \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {a^3 \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac {3 a b^2 \sin ^4(c+d x) \tan (c+d x)}{4 d}+\frac {b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}+\frac {1}{8} \left (5 a^3\right ) \int \sin ^2(c+d x) \, dx+\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \left (-1-x^2-x^4+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}+\frac {\left (15 a b^2\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{4 d}-\frac {\left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\sin (c+d x)\right )}{2 d}\\ &=-\frac {3 a^2 b \sin (c+d x)}{d}-\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a^2 b \sin ^3(c+d x)}{d}-\frac {5 a^3 \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {3 a^2 b \sin ^5(c+d x)}{5 d}-\frac {a^3 \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac {15 a b^2 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac {3 a b^2 \sin ^4(c+d x) \tan (c+d x)}{4 d}+\frac {b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}+\frac {1}{16} \left (5 a^3\right ) \int 1 \, dx+\frac {\left (3 a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}+\frac {\left (45 a b^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 d}-\frac {\left (5 b^3\right ) \operatorname {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{2 d}\\ &=\frac {5 a^3 x}{16}+\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {3 a^2 b \sin (c+d x)}{d}+\frac {5 b^3 \sin (c+d x)}{2 d}-\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a^2 b \sin ^3(c+d x)}{d}+\frac {5 b^3 \sin ^3(c+d x)}{6 d}-\frac {5 a^3 \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {3 a^2 b \sin ^5(c+d x)}{5 d}-\frac {a^3 \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac {45 a b^2 \tan (c+d x)}{8 d}-\frac {15 a b^2 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac {3 a b^2 \sin ^4(c+d x) \tan (c+d x)}{4 d}+\frac {b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}-\frac {\left (45 a b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 d}-\frac {\left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{2 d}\\ &=\frac {5 a^3 x}{16}-\frac {45}{8} a b^2 x+\frac {3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {5 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {3 a^2 b \sin (c+d x)}{d}+\frac {5 b^3 \sin (c+d x)}{2 d}-\frac {5 a^3 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {a^2 b \sin ^3(c+d x)}{d}+\frac {5 b^3 \sin ^3(c+d x)}{6 d}-\frac {5 a^3 \cos (c+d x) \sin ^3(c+d x)}{24 d}-\frac {3 a^2 b \sin ^5(c+d x)}{5 d}-\frac {a^3 \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac {45 a b^2 \tan (c+d x)}{8 d}-\frac {15 a b^2 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac {3 a b^2 \sin ^4(c+d x) \tan (c+d x)}{4 d}+\frac {b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}\\ \end {align*}
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Mathematica [B] time = 6.26, size = 818, normalized size = 2.74 \[ \frac {\left (5 b^3-6 a^2 b\right ) \cos ^3(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^3}{2 d (b+a \cos (c+d x))^3}+\frac {\left (6 a^2 b-5 b^3\right ) \cos ^3(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^3}{2 d (b+a \cos (c+d x))^3}+\frac {3 a b^2 \cos ^3(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \sec (c+d x))^3}{d (b+a \cos (c+d x))^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {3 b \left (6 b^2-11 a^2\right ) \cos ^3(c+d x) \sin (c+d x) (a+b \sec (c+d x))^3}{8 d (b+a \cos (c+d x))^3}-\frac {3 a \left (5 a^2-32 b^2\right ) \cos ^3(c+d x) \sin (2 (c+d x)) (a+b \sec (c+d x))^3}{64 d (b+a \cos (c+d x))^3}-\frac {b \left (4 b^2-21 a^2\right ) \cos ^3(c+d x) \sin (3 (c+d x)) (a+b \sec (c+d x))^3}{48 d (b+a \cos (c+d x))^3}+\frac {3 a \left (a^2-2 b^2\right ) \cos ^3(c+d x) \sin (4 (c+d x)) (a+b \sec (c+d x))^3}{64 d (b+a \cos (c+d x))^3}-\frac {3 a^2 b \cos ^3(c+d x) \sin (5 (c+d x)) (a+b \sec (c+d x))^3}{80 d (b+a \cos (c+d x))^3}-\frac {a^3 \cos ^3(c+d x) \sin (6 (c+d x)) (a+b \sec (c+d x))^3}{192 d (b+a \cos (c+d x))^3}+\frac {3 a b^2 \cos ^3(c+d x) \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \sec (c+d x))^3}{d (b+a \cos (c+d x))^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {b^3 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d (b+a \cos (c+d x))^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b^3 \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d (b+a \cos (c+d x))^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {5 a \left (a^2-18 b^2\right ) (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{16 d (b+a \cos (c+d x))^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.53, size = 241, normalized size = 0.81 \[ \frac {75 \, {\left (a^{3} - 18 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + 60 \, {\left (6 \, a^{2} b - 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 60 \, {\left (6 \, a^{2} b - 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (40 \, a^{3} \cos \left (d x + c\right )^{7} + 144 \, a^{2} b \cos \left (d x + c\right )^{6} - 10 \, {\left (13 \, a^{3} - 18 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} - 16 \, {\left (33 \, a^{2} b - 5 \, b^{3}\right )} \cos \left (d x + c\right )^{4} - 720 \, a b^{2} \cos \left (d x + c\right ) + 15 \, {\left (11 \, a^{3} - 54 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 120 \, b^{3} + 16 \, {\left (69 \, a^{2} b - 35 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.46, size = 563, normalized size = 1.88 \[ \frac {75 \, {\left (a^{3} - 18 \, a b^{2}\right )} {\left (d x + c\right )} + 120 \, {\left (6 \, a^{2} b - 5 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 120 \, {\left (6 \, a^{2} b - 5 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {240 \, {\left (6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{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 ) - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} + \frac {2 \, {\left (75 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 720 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 630 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 480 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 425 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 4560 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 2610 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 2720 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 990 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12384 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1980 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 5760 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 990 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12384 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1980 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5760 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 425 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4560 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2610 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2720 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 75 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 720 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 630 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 480 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.70, size = 354, normalized size = 1.18 \[ -\frac {a^{3} \cos \left (d x +c \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{6 d}-\frac {5 a^{3} \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{24 d}-\frac {5 a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{16 d}+\frac {5 a^{3} x}{16}+\frac {5 a^{3} c}{16 d}-\frac {3 a^{2} b \left (\sin ^{5}\left (d x +c \right )\right )}{5 d}-\frac {a^{2} b \left (\sin ^{3}\left (d x +c \right )\right )}{d}-\frac {3 a^{2} b \sin \left (d x +c \right )}{d}+\frac {3 a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 b^{2} a \left (\sin ^{7}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {3 b^{2} a \left (\sin ^{5}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{d}+\frac {15 b^{2} a \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{4 d}+\frac {45 a \,b^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}-\frac {45 a \,b^{2} x}{8}-\frac {45 a \,b^{2} c}{8 d}+\frac {b^{3} \left (\sin ^{7}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {b^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{2 d}+\frac {5 b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{6 d}+\frac {5 b^{3} \sin \left (d x +c \right )}{2 d}-\frac {5 b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.91, size = 242, normalized size = 0.81 \[ \frac {5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 96 \, {\left (6 \, \sin \left (d x + c\right )^{5} + 10 \, \sin \left (d x + c\right )^{3} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, \sin \left (d x + c\right )\right )} a^{2} b - 360 \, {\left (15 \, d x + 15 \, c - \frac {9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} a b^{2} + 80 \, {\left (4 \, \sin \left (d x + c\right )^{3} - \frac {6 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 24 \, \sin \left (d x + c\right )\right )} b^{3}}{960 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.69, size = 373, normalized size = 1.25 \[ \frac {7\,b^3\,\sin \left (c+d\,x\right )}{3\,d}+\frac {5\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{8\,d}-\frac {5\,b^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {13\,a^3\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{24\,d}-\frac {a^3\,{\cos \left (c+d\,x\right )}^5\,\sin \left (c+d\,x\right )}{6\,d}+\frac {b^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}-\frac {b^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}-\frac {11\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{16\,d}-\frac {23\,a^2\,b\,\sin \left (c+d\,x\right )}{5\,d}-\frac {45\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}+\frac {6\,a^2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {27\,a\,b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,d}+\frac {3\,a\,b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {11\,a^2\,b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{5\,d}-\frac {3\,a\,b^2\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,d}-\frac {3\,a^2\,b\,{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )}{5\,d} \]
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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