Optimal. Leaf size=372 \[ -\frac {\left (3 a^2 A+2 a b (6 A+B)+b^2 (15 A+8 B)\right ) \log (1-\sin (c+d x))}{16 d (a+b)^4}+\frac {\left (3 a^2 A-2 a b (6 A-B)+b^2 (15 A-8 B)\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^4}-\frac {\sec ^4(c+d x) (-(a A-b B) \sin (c+d x)-a B+A b)}{4 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {b^4 \left (-5 a^2 B+6 a A b-b^2 B\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^4}+\frac {\sec ^2(c+d x) \left (b \left (a^2 A-6 a b B+5 A b^2\right )+\left (3 a^3 A+2 a^2 b B-9 a A b^2+4 b^3 B\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac {b \left (3 a^4 A+2 a^3 b B-12 a^2 A b^2+22 a b^3 B-15 A b^4\right )}{8 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))} \]
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Rubi [A] time = 0.56, antiderivative size = 372, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2837, 823, 801} \[ -\frac {b \left (-12 a^2 A b^2+3 a^4 A+2 a^3 b B+22 a b^3 B-15 A b^4\right )}{8 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {b^4 \left (-5 a^2 B+6 a A b-b^2 B\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^4}-\frac {\left (3 a^2 A+2 a b (6 A+B)+b^2 (15 A+8 B)\right ) \log (1-\sin (c+d x))}{16 d (a+b)^4}+\frac {\left (3 a^2 A-2 a b (6 A-B)+b^2 (15 A-8 B)\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^4}-\frac {\sec ^4(c+d x) (-(a A-b B) \sin (c+d x)-a B+A b)}{4 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac {\sec ^2(c+d x) \left (\left (3 a^3 A+2 a^2 b B-9 a A b^2+4 b^3 B\right ) \sin (c+d x)+b \left (a^2 A-6 a b B+5 A b^2\right )\right )}{8 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))} \]
Antiderivative was successfully verified.
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Rule 801
Rule 823
Rule 2837
Rubi steps
\begin {align*} \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {A+\frac {B x}{b}}{(a+x)^2 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {\sec ^4(c+d x) (A b-a B-(a A-b B) \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {b^3 \operatorname {Subst}\left (\int \frac {-3 a^2 A+5 A b^2-2 a b B-4 (a A-b B) x}{(a+x)^2 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 \left (a^2-b^2\right ) d}\\ &=-\frac {\sec ^4(c+d x) (A b-a B-(a A-b B) \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\sec ^2(c+d x) \left (b \left (a^2 A+5 A b^2-6 a b B\right )+\left (3 a^3 A-9 a A b^2+2 a^2 b B+4 b^3 B\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {b \operatorname {Subst}\left (\int \frac {3 a^4 A-6 a^2 A b^2+15 A b^4+2 a^3 b B-14 a b^3 B+2 \left (3 a^3 A-9 a A b^2+2 a^2 b B+4 b^3 B\right ) x}{(a+x)^2 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac {\sec ^4(c+d x) (A b-a B-(a A-b B) \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\sec ^2(c+d x) \left (b \left (a^2 A+5 A b^2-6 a b B\right )+\left (3 a^3 A-9 a A b^2+2 a^2 b B+4 b^3 B\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {b \operatorname {Subst}\left (\int \left (\frac {(a-b)^2 \left (3 a^2 A+2 a b (6 A+B)+b^2 (15 A+8 B)\right )}{2 b (a+b)^2 (b-x)}+\frac {3 a^4 A-12 a^2 A b^2-15 A b^4+2 a^3 b B+22 a b^3 B}{\left (a^2-b^2\right ) (a+x)^2}+\frac {8 b^3 \left (-6 a A b+5 a^2 B+b^2 B\right )}{\left (-a^2+b^2\right )^2 (a+x)}+\frac {(a+b)^2 \left (3 a^2 A+b^2 (15 A-8 B)-2 a b (6 A-B)\right )}{2 (a-b)^2 b (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac {\left (3 a^2 A+2 a b (6 A+B)+b^2 (15 A+8 B)\right ) \log (1-\sin (c+d x))}{16 (a+b)^4 d}+\frac {\left (3 a^2 A+b^2 (15 A-8 B)-2 a b (6 A-B)\right ) \log (1+\sin (c+d x))}{16 (a-b)^4 d}-\frac {b^4 \left (6 a A b-5 a^2 B-b^2 B\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4 d}-\frac {b \left (3 a^4 A-12 a^2 A b^2-15 A b^4+2 a^3 b B+22 a b^3 B\right )}{8 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac {\sec ^4(c+d x) (A b-a B-(a A-b B) \sin (c+d x))}{4 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\sec ^2(c+d x) \left (b \left (a^2 A+5 A b^2-6 a b B\right )+\left (3 a^3 A-9 a A b^2+2 a^2 b B+4 b^3 B\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}\\ \end {align*}
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Mathematica [A] time = 4.28, size = 370, normalized size = 0.99 \[ \frac {\frac {2 \left (b^2-a^2\right ) \sec ^4(c+d x) ((b B-a A) \sin (c+d x)-a B+A b)}{a+b \sin (c+d x)}-\frac {\left (3 a^3 A+2 a^2 b B-9 a A b^2+4 b^3 B\right ) ((a-b) \log (1-\sin (c+d x))-(a+b) \log (\sin (c+d x)+1)+2 b \log (a+b \sin (c+d x)))}{(a-b) (a+b)}+\frac {\sec ^2(c+d x) \left (b \left (a^2 A-6 a b B+5 A b^2\right )+\left (3 a^3 A+2 a^2 b B-9 a A b^2+4 b^3 B\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)}+b \left (-3 a^4 A-2 a^3 b B+12 a^2 A b^2-22 a b^3 B+15 A b^4\right ) \left (\frac {1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {\log (1-\sin (c+d x))}{2 b (a+b)^2}+\frac {\log (\sin (c+d x)+1)}{2 b (a-b)^2}-\frac {2 a \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}\right )}{8 d \left (a^2-b^2\right )^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 3.30, size = 881, normalized size = 2.37 \[ \frac {4 \, B a^{7} - 4 \, A a^{6} b - 12 \, B a^{5} b^{2} + 12 \, A a^{4} b^{3} + 12 \, B a^{3} b^{4} - 12 \, A a^{2} b^{5} - 4 \, B a b^{6} + 4 \, A b^{7} - 2 \, {\left (3 \, A a^{6} b + 2 \, B a^{5} b^{2} - 15 \, A a^{4} b^{3} + 20 \, B a^{3} b^{4} - 3 \, A a^{2} b^{5} - 22 \, B a b^{6} + 15 \, A b^{7}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (A a^{6} b - 6 \, B a^{5} b^{2} + 3 \, A a^{4} b^{3} + 12 \, B a^{3} b^{4} - 9 \, A a^{2} b^{5} - 6 \, B a b^{6} + 5 \, A b^{7}\right )} \cos \left (d x + c\right )^{2} + 16 \, {\left ({\left (5 \, B a^{2} b^{5} - 6 \, A a b^{6} + B b^{7}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (5 \, B a^{3} b^{4} - 6 \, A a^{2} b^{5} + B a b^{6}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left ({\left (3 \, A a^{6} b + 2 \, B a^{5} b^{2} - 15 \, A a^{4} b^{3} - 20 \, B a^{3} b^{4} + 5 \, {\left (9 \, A - 8 \, B\right )} a^{2} b^{5} + 6 \, {\left (8 \, A - 5 \, B\right )} a b^{6} + {\left (15 \, A - 8 \, B\right )} b^{7}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (3 \, A a^{7} + 2 \, B a^{6} b - 15 \, A a^{5} b^{2} - 20 \, B a^{4} b^{3} + 5 \, {\left (9 \, A - 8 \, B\right )} a^{3} b^{4} + 6 \, {\left (8 \, A - 5 \, B\right )} a^{2} b^{5} + {\left (15 \, A - 8 \, B\right )} a b^{6}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (3 \, A a^{6} b + 2 \, B a^{5} b^{2} - 15 \, A a^{4} b^{3} - 20 \, B a^{3} b^{4} + 5 \, {\left (9 \, A + 8 \, B\right )} a^{2} b^{5} - 6 \, {\left (8 \, A + 5 \, B\right )} a b^{6} + {\left (15 \, A + 8 \, B\right )} b^{7}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (3 \, A a^{7} + 2 \, B a^{6} b - 15 \, A a^{5} b^{2} - 20 \, B a^{4} b^{3} + 5 \, {\left (9 \, A + 8 \, B\right )} a^{3} b^{4} - 6 \, {\left (8 \, A + 5 \, B\right )} a^{2} b^{5} + {\left (15 \, A + 8 \, B\right )} a b^{6}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{7} - 2 \, B a^{6} b - 6 \, A a^{5} b^{2} + 6 \, B a^{4} b^{3} + 6 \, A a^{3} b^{4} - 6 \, B a^{2} b^{5} - 2 \, A a b^{6} + 2 \, B b^{7} + {\left (3 \, A a^{7} + 2 \, B a^{6} b - 15 \, A a^{5} b^{2} + 21 \, A a^{3} b^{4} - 6 \, B a^{2} b^{5} - 9 \, A a b^{6} + 4 \, B b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left ({\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 )^{4} \sin \left (d x + c\right ) + {\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 )^{4}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.34, size = 761, normalized size = 2.05 \[ \frac {\frac {16 \, {\left (5 \, B a^{2} b^{5} - 6 \, A a b^{6} + B b^{7}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}} - \frac {{\left (3 \, A a^{2} + 12 \, A a b + 2 \, B a b + 15 \, A b^{2} + 8 \, B b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac {{\left (3 \, A a^{2} - 12 \, A a b + 2 \, B a b + 15 \, A b^{2} - 8 \, B b^{2}\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac {16 \, {\left (5 \, B a^{2} b^{5} \sin \left (d x + c\right ) - 6 \, A a b^{6} \sin \left (d x + c\right ) + B b^{7} \sin \left (d x + c\right ) + 6 \, B a^{3} b^{4} - 7 \, A a^{2} b^{5} + A b^{7}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}} + \frac {2 \, {\left (30 \, B a^{2} b^{4} \sin \left (d x + c\right )^{4} - 36 \, A a b^{5} \sin \left (d x + c\right )^{4} + 6 \, B b^{6} \sin \left (d x + c\right )^{4} - 3 \, A a^{6} \sin \left (d x + c\right )^{3} - 2 \, B a^{5} b \sin \left (d x + c\right )^{3} + 15 \, A a^{4} b^{2} \sin \left (d x + c\right )^{3} - 12 \, B a^{3} b^{3} \sin \left (d x + c\right )^{3} - 5 \, A a^{2} b^{4} \sin \left (d x + c\right )^{3} + 14 \, B a b^{5} \sin \left (d x + c\right )^{3} - 7 \, A b^{6} \sin \left (d x + c\right )^{3} + 12 \, B a^{4} b^{2} \sin \left (d x + c\right )^{2} - 16 \, A a^{3} b^{3} \sin \left (d x + c\right )^{2} - 68 \, B a^{2} b^{4} \sin \left (d x + c\right )^{2} + 88 \, A a b^{5} \sin \left (d x + c\right )^{2} - 16 \, B b^{6} \sin \left (d x + c\right )^{2} + 5 \, A a^{6} \sin \left (d x + c\right ) - 2 \, B a^{5} b \sin \left (d x + c\right ) - 17 \, A a^{4} b^{2} \sin \left (d x + c\right ) + 20 \, B a^{3} b^{3} \sin \left (d x + c\right ) + 3 \, A a^{2} b^{4} \sin \left (d x + c\right ) - 18 \, B a b^{5} \sin \left (d x + c\right ) + 9 \, A b^{6} \sin \left (d x + c\right ) + 2 \, B a^{6} - 4 \, A a^{5} b - 14 \, B a^{4} b^{2} + 24 \, A a^{3} b^{3} + 36 \, B a^{2} b^{4} - 56 \, A a b^{5} + 12 \, B b^{6}\right )}}{{\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.82, size = 675, normalized size = 1.81 \[ -\frac {\ln \left (\sin \left (d x +c \right )-1\right ) B a b}{8 d \left (a +b \right )^{4}}+\frac {b^{6} \ln \left (a +b \sin \left (d x +c \right )\right ) B}{d \left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {b^{5} A}{d \left (a +b \right )^{3} \left (a -b \right )^{3} \left (a +b \sin \left (d x +c \right )\right )}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) A a b}{4 d \left (a -b \right )^{4}}+\frac {a B}{16 d \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )}-\frac {5 B b}{16 d \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )}+\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) a^{2} A}{16 d \left (a -b \right )^{4}}+\frac {A}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {B}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {A}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {B}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {6 b^{5} \ln \left (a +b \sin \left (d x +c \right )\right ) A a}{d \left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {5 b^{4} \ln \left (a +b \sin \left (d x +c \right )\right ) B \,a^{2}}{d \left (a +b \right )^{4} \left (a -b \right )^{4}}-\frac {b^{4} a B}{d \left (a +b \right )^{3} \left (a -b \right )^{3} \left (a +b \sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right ) B a b}{8 d \left (a -b \right )^{4}}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) A a b}{4 d \left (a +b \right )^{4}}+\frac {7 A b}{16 d \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )}+\frac {15 \ln \left (1+\sin \left (d x +c \right )\right ) A \,b^{2}}{16 d \left (a -b \right )^{4}}-\frac {\ln \left (1+\sin \left (d x +c \right )\right ) B \,b^{2}}{2 d \left (a -b \right )^{4}}-\frac {3 a A}{16 d \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )}-\frac {7 A b}{16 d \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )}-\frac {a B}{16 d \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )}-\frac {5 B b}{16 d \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) a^{2} A}{16 d \left (a +b \right )^{4}}-\frac {15 \ln \left (\sin \left (d x +c \right )-1\right ) A \,b^{2}}{16 d \left (a +b \right )^{4}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) B \,b^{2}}{2 d \left (a +b \right )^{4}}-\frac {3 a A}{16 d \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 659, normalized size = 1.77 \[ \frac {\frac {16 \, {\left (5 \, B a^{2} b^{4} - 6 \, A a b^{5} + B b^{6}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} + \frac {{\left (3 \, A a^{2} - 2 \, {\left (6 \, A - B\right )} a b + {\left (15 \, A - 8 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac {{\left (3 \, A a^{2} + 2 \, {\left (6 \, A + B\right )} a b + {\left (15 \, A + 8 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} + \frac {2 \, {\left (2 \, B a^{5} - 4 \, A a^{4} b - 12 \, B a^{3} b^{2} + 20 \, A a^{2} b^{3} - 14 \, B a b^{4} + 8 \, A b^{5} - {\left (3 \, A a^{4} b + 2 \, B a^{3} b^{2} - 12 \, A a^{2} b^{3} + 22 \, B a b^{4} - 15 \, A b^{5}\right )} \sin \left (d x + c\right )^{4} - {\left (3 \, A a^{5} + 2 \, B a^{4} b - 12 \, A a^{3} b^{2} + 2 \, B a^{2} b^{3} + 9 \, A a b^{4} - 4 \, B b^{5}\right )} \sin \left (d x + c\right )^{3} + {\left (5 \, A a^{4} b + 10 \, B a^{3} b^{2} - 28 \, A a^{2} b^{3} + 38 \, B a b^{4} - 25 \, A b^{5}\right )} \sin \left (d x + c\right )^{2} + {\left (5 \, A a^{5} - 16 \, A a^{3} b^{2} + 6 \, B a^{2} b^{3} + 11 \, A a b^{4} - 6 \, B b^{5}\right )} \sin \left (d x + c\right )\right )}}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6} + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )^{5} + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \sin \left (d x + c\right )^{4} - 2 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )^{3} - 2 \, {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 13.46, size = 615, normalized size = 1.65 \[ \frac {\frac {B\,a^5-2\,A\,a^4\,b-6\,B\,a^3\,b^2+10\,A\,a^2\,b^3-7\,B\,a\,b^4+4\,A\,b^5}{4\,\left (a^2-b^2\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\sin \left (c+d\,x\right )}^4\,\left (3\,A\,a^4\,b+2\,B\,a^3\,b^2-12\,A\,a^2\,b^3+22\,B\,a\,b^4-15\,A\,b^5\right )}{8\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {\sin \left (c+d\,x\right )\,\left (5\,A\,a^3-11\,A\,a\,b^2+6\,B\,b^3\right )}{8\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (3\,A\,a^3+2\,B\,a^2\,b-9\,A\,a\,b^2+4\,B\,b^3\right )}{8\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\sin \left (c+d\,x\right )}^2\,\left (5\,A\,a^4\,b+10\,B\,a^3\,b^2-28\,A\,a^2\,b^3+38\,B\,a\,b^4-25\,A\,b^5\right )}{8\,\left (a^2-b^2\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left (b\,{\sin \left (c+d\,x\right )}^5+a\,{\sin \left (c+d\,x\right )}^4-2\,b\,{\sin \left (c+d\,x\right )}^3-2\,a\,{\sin \left (c+d\,x\right )}^2+b\,\sin \left (c+d\,x\right )+a\right )}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (5\,B\,a^2\,b^4-6\,A\,a\,b^5+B\,b^6\right )}{d\,\left (a^8-4\,a^6\,b^2+6\,a^4\,b^4-4\,a^2\,b^6+b^8\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (3\,A\,a^2+\left (12\,A+2\,B\right )\,a\,b+\left (15\,A+8\,B\right )\,b^2\right )}{d\,\left (16\,a^4+64\,a^3\,b+96\,a^2\,b^2+64\,a\,b^3+16\,b^4\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (3\,A\,a^2+\left (2\,B-12\,A\right )\,a\,b+\left (15\,A-8\,B\right )\,b^2\right )}{d\,\left (16\,a^4-64\,a^3\,b+96\,a^2\,b^2-64\,a\,b^3+16\,b^4\right )} \]
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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