Optimal. Leaf size=113 \[ \frac {b^4 \log (a+b \csc (x))}{a \left (a^2-b^2\right )^2}-\frac {1}{4 (a+b) (1-\csc (x))}-\frac {1}{4 (a-b) (\csc (x)+1)}+\frac {(2 a+3 b) \log (1-\csc (x))}{4 (a+b)^2}+\frac {(2 a-3 b) \log (\csc (x)+1)}{4 (a-b)^2}+\frac {\log (\sin (x))}{a} \]
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Rubi [A] time = 0.17, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3885, 894} \[ \frac {b^4 \log (a+b \csc (x))}{a \left (a^2-b^2\right )^2}-\frac {1}{4 (a+b) (1-\csc (x))}-\frac {1}{4 (a-b) (\csc (x)+1)}+\frac {(2 a+3 b) \log (1-\csc (x))}{4 (a+b)^2}+\frac {(2 a-3 b) \log (\csc (x)+1)}{4 (a-b)^2}+\frac {\log (\sin (x))}{a} \]
Antiderivative was successfully verified.
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Rule 894
Rule 3885
Rubi steps
\begin {align*} \int \frac {\tan ^3(x)}{a+b \csc (x)} \, dx &=-\left (b^4 \operatorname {Subst}\left (\int \frac {1}{x (a+x) \left (b^2-x^2\right )^2} \, dx,x,b \csc (x)\right )\right )\\ &=-\left (b^4 \operatorname {Subst}\left (\int \left (\frac {1}{4 b^3 (a+b) (b-x)^2}+\frac {2 a+3 b}{4 b^4 (a+b)^2 (b-x)}+\frac {1}{a b^4 x}-\frac {1}{a (a-b)^2 (a+b)^2 (a+x)}-\frac {1}{4 (a-b) b^3 (b+x)^2}+\frac {-2 a+3 b}{4 (a-b)^2 b^4 (b+x)}\right ) \, dx,x,b \csc (x)\right )\right )\\ &=-\frac {1}{4 (a+b) (1-\csc (x))}-\frac {1}{4 (a-b) (1+\csc (x))}+\frac {(2 a+3 b) \log (1-\csc (x))}{4 (a+b)^2}+\frac {(2 a-3 b) \log (1+\csc (x))}{4 (a-b)^2}+\frac {b^4 \log (a+b \csc (x))}{a \left (a^2-b^2\right )^2}+\frac {\log (\sin (x))}{a}\\ \end {align*}
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Mathematica [A] time = 0.49, size = 115, normalized size = 1.02 \[ \frac {\csc (x) (a \sin (x)+b) \left (\frac {4 b^4 \log (a \sin (x)+b)}{a (a-b)^2 (a+b)^2}-\frac {1}{(a+b) (\sin (x)-1)}+\frac {1}{(a-b) (\sin (x)+1)}+\frac {(2 a+3 b) \log (1-\sin (x))}{(a+b)^2}+\frac {(2 a-3 b) \log (\sin (x)+1)}{(a-b)^2}\right )}{4 (a+b \csc (x))} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.87, size = 144, normalized size = 1.27 \[ \frac {4 \, b^{4} \cos \relax (x)^{2} \log \left (a \sin \relax (x) + b\right ) + 2 \, a^{4} - 2 \, a^{2} b^{2} + {\left (2 \, a^{4} + a^{3} b - 4 \, a^{2} b^{2} - 3 \, a b^{3}\right )} \cos \relax (x)^{2} \log \left (\sin \relax (x) + 1\right ) + {\left (2 \, a^{4} - a^{3} b - 4 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \relax (x)^{2} \log \left (-\sin \relax (x) + 1\right ) - 2 \, {\left (a^{3} b - a b^{3}\right )} \sin \relax (x)}{4 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \relax (x)^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 139, normalized size = 1.23 \[ \frac {b^{4} \log \left ({\left | a \sin \relax (x) + b \right |}\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac {{\left (2 \, a - 3 \, b\right )} \log \left (\sin \relax (x) + 1\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {{\left (2 \, a + 3 \, b\right )} \log \left (-\sin \relax (x) + 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {a^{3} - a b^{2} - {\left (a^{2} b - b^{3}\right )} \sin \relax (x)}{2 \, {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} {\left (\sin \relax (x) + 1\right )} {\left (\sin \relax (x) - 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.51, size = 117, normalized size = 1.04 \[ \frac {b^{4} \ln \left (b +a \sin \relax (x )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a}-\frac {1}{\left (4 a +4 b \right ) \left (-1+\sin \relax (x )\right )}+\frac {a \ln \left (-1+\sin \relax (x )\right )}{2 \left (a +b \right )^{2}}+\frac {3 \ln \left (-1+\sin \relax (x )\right ) b}{4 \left (a +b \right )^{2}}+\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \relax (x )\right )}-\frac {3 \ln \left (1+\sin \relax (x )\right ) b}{4 \left (a -b \right )^{2}}+\frac {a \ln \left (1+\sin \relax (x )\right )}{2 \left (a -b \right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.32, size = 120, normalized size = 1.06 \[ \frac {b^{4} \log \left (a \sin \relax (x) + b\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac {{\left (2 \, a - 3 \, b\right )} \log \left (\sin \relax (x) + 1\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {{\left (2 \, a + 3 \, b\right )} \log \left (\sin \relax (x) - 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {b \sin \relax (x) - a}{2 \, {\left ({\left (a^{2} - b^{2}\right )} \sin \relax (x)^{2} - a^{2} + b^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.14, size = 172, normalized size = 1.52 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )\,\left (2\,a-3\,b\right )}{2\,{\left (a-b\right )}^2}-\frac {\frac {b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{a^2-b^2}-\frac {2\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^2-b^2}+\frac {b\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2-b^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}-\frac {\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )}{a}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )\,\left (2\,a+3\,b\right )}{2\,{\left (a+b\right )}^2}+\frac {b^4\,\ln \left (b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )+b\right )}{a\,{\left (a^2-b^2\right )}^2} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{3}{\relax (x )}}{a + b \csc {\relax (x )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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