Optimal. Leaf size=240 \[ -\frac {\left (35 a^2+57 a b+24 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac {\left (35 a^2-57 a b+24 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right )}+\frac {\sec ^2(c+d x) \left (4 b \left (4 a^2-3 b^2\right )-a \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}-\frac {a^8 \log (a+b \sin (c+d x))}{b^3 d \left (a^2-b^2\right )^3}+\frac {a \sin (c+d x)}{b^2 d}-\frac {\sin ^2(c+d x)}{2 b d} \]
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Rubi [A] time = 0.62, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2837, 12, 1647, 1629} \[ -\frac {a^8 \log (a+b \sin (c+d x))}{b^3 d \left (a^2-b^2\right )^3}-\frac {\left (35 a^2+57 a b+24 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac {\left (35 a^2-57 a b+24 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 d \left (a^2-b^2\right )}+\frac {\sec ^2(c+d x) \left (4 b \left (4 a^2-3 b^2\right )-a \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac {a \sin (c+d x)}{b^2 d}-\frac {\sin ^2(c+d x)}{2 b d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 1629
Rule 1647
Rule 2837
Rubi steps
\begin {align*} \int \frac {\sin ^3(c+d x) \tan ^5(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {x^8}{b^8 (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^8}{(a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=-\frac {\sec ^4(c+d x) \left (\frac {b}{a^2-b^2}-\frac {a \sin (c+d x)}{a^2-b^2}\right )}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {a^2 b^8}{a^2-b^2}+\frac {3 a b^8 x}{a^2-b^2}-4 b^6 x^2-4 b^4 x^4-4 b^2 x^6}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 b^5 d}\\ &=\frac {\sec ^2(c+d x) \left (4 b \left (4 a^2-3 b^2\right )-a \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}-\frac {\sec ^4(c+d x) \left (\frac {b}{a^2-b^2}-\frac {a \sin (c+d x)}{a^2-b^2}\right )}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {\frac {a^2 b^8 \left (11 a^2-7 b^2\right )}{\left (a^2-b^2\right )^2}-\frac {a b^8 \left (13 a^2-9 b^2\right ) x}{\left (a^2-b^2\right )^2}+16 b^6 x^2+8 b^4 x^4}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 b^7 d}\\ &=\frac {\sec ^2(c+d x) \left (4 b \left (4 a^2-3 b^2\right )-a \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}-\frac {\sec ^4(c+d x) \left (\frac {b}{a^2-b^2}-\frac {a \sin (c+d x)}{a^2-b^2}\right )}{4 d}+\frac {\operatorname {Subst}\left (\int \left (8 a b^4+\frac {b^7 \left (35 a^2+57 a b+24 b^2\right )}{2 (a+b)^3 (b-x)}-8 b^4 x-\frac {8 a^8 b^4}{(a-b)^3 (a+b)^3 (a+x)}+\frac {b^7 \left (35 a^2-57 a b+24 b^2\right )}{2 (a-b)^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^7 d}\\ &=-\frac {\left (35 a^2+57 a b+24 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {\left (35 a^2-57 a b+24 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}-\frac {a^8 \log (a+b \sin (c+d x))}{b^3 \left (a^2-b^2\right )^3 d}+\frac {a \sin (c+d x)}{b^2 d}-\frac {\sin ^2(c+d x)}{2 b d}+\frac {\sec ^2(c+d x) \left (4 b \left (4 a^2-3 b^2\right )-a \left (13 a^2-9 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}-\frac {\sec ^4(c+d x) \left (\frac {b}{a^2-b^2}-\frac {a \sin (c+d x)}{a^2-b^2}\right )}{4 d}\\ \end {align*}
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Mathematica [A] time = 3.01, size = 212, normalized size = 0.88 \[ \frac {-\frac {16 a^8 \log (a+b \sin (c+d x))}{b^3 (a-b)^3 (a+b)^3}-\frac {\left (35 a^2+57 a b+24 b^2\right ) \log (1-\sin (c+d x))}{(a+b)^3}+\frac {\left (35 a^2-57 a b+24 b^2\right ) \log (\sin (c+d x)+1)}{(a-b)^3}+\frac {16 a \sin (c+d x)}{b^2}+\frac {13 a+11 b}{(a+b)^2 (\sin (c+d x)-1)}+\frac {13 a-11 b}{(a-b)^2 (\sin (c+d x)+1)}+\frac {1}{(a+b) (\sin (c+d x)-1)^2}-\frac {1}{(a-b) (\sin (c+d x)+1)^2}-\frac {8 \sin ^2(c+d x)}{b}}{16 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.74, size = 429, normalized size = 1.79 \[ -\frac {16 \, a^{8} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, a^{4} b^{4} - 8 \, a^{2} b^{6} + 4 \, b^{8} - 8 \, {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (d x + c\right )^{6} - {\left (35 \, a^{5} b^{3} + 48 \, a^{4} b^{4} - 42 \, a^{3} b^{5} - 64 \, a^{2} b^{6} + 15 \, a b^{7} + 24 \, b^{8}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (35 \, a^{5} b^{3} - 48 \, a^{4} b^{4} - 42 \, a^{3} b^{5} + 64 \, a^{2} b^{6} + 15 \, a b^{7} - 24 \, b^{8}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \cos \left (d x + c\right )^{4} - 8 \, {\left (4 \, a^{4} b^{4} - 7 \, a^{2} b^{6} + 3 \, b^{8}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{5} b^{3} - 4 \, a^{3} b^{5} + 2 \, a b^{7} + 8 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \cos \left (d x + c\right )^{4} - {\left (13 \, a^{5} b^{3} - 22 \, a^{3} b^{5} + 9 \, a b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} b^{3} - 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} - b^{9}\right )} d \cos \left (d x + c\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.30, size = 403, normalized size = 1.68 \[ -\frac {\frac {16 \, a^{8} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b^{3} - 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} - b^{9}} - \frac {{\left (35 \, a^{2} - 57 \, a b + 24 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (35 \, a^{2} + 57 \, a b + 24 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {8 \, {\left (b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )\right )}}{b^{2}} + \frac {2 \, {\left (36 \, a^{4} b \sin \left (d x + c\right )^{4} - 48 \, a^{2} b^{3} \sin \left (d x + c\right )^{4} + 18 \, b^{5} \sin \left (d x + c\right )^{4} - 13 \, a^{5} \sin \left (d x + c\right )^{3} + 22 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} - 9 \, a b^{4} \sin \left (d x + c\right )^{3} - 56 \, a^{4} b \sin \left (d x + c\right )^{2} + 68 \, a^{2} b^{3} \sin \left (d x + c\right )^{2} - 24 \, b^{5} \sin \left (d x + c\right )^{2} + 11 \, a^{5} \sin \left (d x + c\right ) - 18 \, a^{3} b^{2} \sin \left (d x + c\right ) + 7 \, a b^{4} \sin \left (d x + c\right ) + 22 \, a^{4} b - 24 \, a^{2} b^{3} + 8 \, b^{5}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.49, size = 338, normalized size = 1.41 \[ \frac {1}{2 d \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {13 a}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {11 b}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {35 \ln \left (\sin \left (d x +c \right )-1\right ) a^{2}}{16 d \left (a +b \right )^{3}}-\frac {57 \ln \left (\sin \left (d x +c \right )-1\right ) a b}{16 d \left (a +b \right )^{3}}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) b^{2}}{2 d \left (a +b \right )^{3}}-\frac {\sin ^{2}\left (d x +c \right )}{2 b d}+\frac {a \sin \left (d x +c \right )}{b^{2} d}-\frac {a^{8} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{3} \left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {1}{2 d \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {13 a}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {11 b}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {35 \ln \left (1+\sin \left (d x +c \right )\right ) a^{2}}{16 d \left (a -b \right )^{3}}-\frac {57 \ln \left (1+\sin \left (d x +c \right )\right ) a b}{16 d \left (a -b \right )^{3}}+\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) b^{2}}{2 d \left (a -b \right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.38, size = 316, normalized size = 1.32 \[ -\frac {\frac {16 \, a^{8} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} b^{3} - 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} - b^{9}} - \frac {{\left (35 \, a^{2} - 57 \, a b + 24 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (35 \, a^{2} + 57 \, a b + 24 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left ({\left (13 \, a^{3} - 9 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} + 14 \, a^{2} b - 10 \, b^{3} - 4 \, {\left (4 \, a^{2} b - 3 \, b^{3}\right )} \sin \left (d x + c\right )^{2} - {\left (11 \, a^{3} - 7 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}} + \frac {8 \, {\left (b \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )\right )}}{b^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 14.19, size = 806, normalized size = 3.36 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {b^2}{4\,{\left (a-b\right )}^3}+\frac {13\,b}{8\,{\left (a-b\right )}^2}+\frac {35}{8\,\left (a-b\right )}\right )}{d}-\frac {\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (-3\,a^4+a^2\,b^2+b^4\right )}{b\,{\left (a^2-b^2\right )}^2}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^4-5\,a^2\,b^2+3\,b^4\right )}{b\,{\left (a^2-b^2\right )}^2}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (8\,a^5+7\,a^3\,b^2-11\,a\,b^4\right )}{2\,b^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (8\,a^5+7\,a^3\,b^2-11\,a\,b^4\right )}{2\,b^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (8\,a^5-27\,a^3\,b^2+15\,a\,b^4\right )}{4\,b^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (24\,a^5-45\,a^3\,b^2+25\,a\,b^4\right )}{4\,b^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (24\,a^5-45\,a^3\,b^2+25\,a\,b^4\right )}{4\,b^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^2-3\,b^2\right )}{b\,\left (a^2-b^2\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (2\,a^2-3\,b^2\right )}{b\,\left (a^2-b^2\right )}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (a^4-5\,a^2\,b^2+3\,b^4\right )}{b\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (8\,a^4-27\,a^2\,b^2+15\,b^4\right )}{4\,b^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {35}{8\,\left (a+b\right )}-\frac {13\,b}{8\,{\left (a+b\right )}^2}+\frac {b^2}{4\,{\left (a+b\right )}^3}\right )}{d}+\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (a^2+3\,b^2\right )}{b^3\,d}+\frac {a^8\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (-a^6\,b^3+3\,a^4\,b^5-3\,a^2\,b^7+b^9\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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