Optimal. Leaf size=75 \[ \frac {1}{2} a b^2 \sin (x)+\frac {1}{4} (a+2 b) (a-b)^2 \log (\sin (x)+1)-\frac {1}{4} (a-2 b) (a+b)^2 \log (1-\sin (x))+\frac {1}{2} \sec ^2(x) (a \sin (x)+b) (a+b \sin (x))^2 \]
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Rubi [A] time = 0.14, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {4391, 2668, 739, 774, 633, 31} \[ \frac {1}{2} a b^2 \sin (x)+\frac {1}{4} (a+2 b) (a-b)^2 \log (\sin (x)+1)-\frac {1}{4} (a-2 b) (a+b)^2 \log (1-\sin (x))+\frac {1}{2} \sec ^2(x) (a \sin (x)+b) (a+b \sin (x))^2 \]
Antiderivative was successfully verified.
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Rule 31
Rule 633
Rule 739
Rule 774
Rule 2668
Rule 4391
Rubi steps
\begin {align*} \int (a \sec (x)+b \tan (x))^3 \, dx &=\int \sec ^3(x) (a+b \sin (x))^3 \, dx\\ &=b^3 \operatorname {Subst}\left (\int \frac {(a+x)^3}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (x)\right )\\ &=\frac {1}{2} \sec ^2(x) (b+a \sin (x)) (a+b \sin (x))^2-\frac {1}{2} b \operatorname {Subst}\left (\int \frac {(a+x) \left (-a^2+2 b^2+a x\right )}{b^2-x^2} \, dx,x,b \sin (x)\right )\\ &=\frac {1}{2} a b^2 \sin (x)+\frac {1}{2} \sec ^2(x) (b+a \sin (x)) (a+b \sin (x))^2+\frac {1}{2} b \operatorname {Subst}\left (\int \frac {-a b^2-a \left (-a^2+2 b^2\right )-2 b^2 x}{b^2-x^2} \, dx,x,b \sin (x)\right )\\ &=\frac {1}{2} a b^2 \sin (x)+\frac {1}{2} \sec ^2(x) (b+a \sin (x)) (a+b \sin (x))^2+\frac {1}{4} \left ((a-2 b) (a+b)^2\right ) \operatorname {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (x)\right )-\frac {1}{4} \left ((a-b)^2 (a+2 b)\right ) \operatorname {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (x)\right )\\ &=-\frac {1}{4} (a-2 b) (a+b)^2 \log (1-\sin (x))+\frac {1}{4} (a-b)^2 (a+2 b) \log (1+\sin (x))+\frac {1}{2} a b^2 \sin (x)+\frac {1}{2} \sec ^2(x) (b+a \sin (x)) (a+b \sin (x))^2\\ \end {align*}
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Mathematica [A] time = 0.57, size = 123, normalized size = 1.64 \[ \frac {2 a^4 b \sec ^2(x)+\left (a^2-b^2\right ) \left ((a-2 b) (a+b)^2 \log (1-\sin (x))-(a-b)^2 (a+2 b) \log (\sin (x)+1)\right )+\left (-8 a^4 b+4 a^2 b^3+2 b^5\right ) \tan ^2(x)-2 a \left (a^4+2 a^2 b^2-3 b^4\right ) \tan (x) \sec (x)}{4 \left (b^2-a^2\right )} \]
Antiderivative was successfully verified.
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fricas [A] time = 2.07, size = 85, normalized size = 1.13 \[ \frac {{\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \cos \relax (x)^{2} \log \left (\sin \relax (x) + 1\right ) - {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \cos \relax (x)^{2} \log \left (-\sin \relax (x) + 1\right ) + 6 \, a^{2} b + 2 \, b^{3} + 2 \, {\left (a^{3} + 3 \, a b^{2}\right )} \sin \relax (x)}{4 \, \cos \relax (x)^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.14, size = 86, normalized size = 1.15 \[ \frac {1}{4} \, {\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \log \left (\sin \relax (x) + 1\right ) - \frac {1}{4} \, {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \log \left (-\sin \relax (x) + 1\right ) - \frac {b^{3} \sin \relax (x)^{2} + a^{3} \sin \relax (x) + 3 \, a b^{2} \sin \relax (x) + 3 \, a^{2} b}{2 \, {\left (\sin \relax (x)^{2} - 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 82, normalized size = 1.09 \[ \frac {a^{3} \sec \relax (x ) \tan \relax (x )}{2}+\frac {a^{3} \ln \left (\sec \relax (x )+\tan \relax (x )\right )}{2}+\frac {3 a^{2} b}{2 \cos \relax (x )^{2}}+\frac {3 a \,b^{2} \left (\sin ^{3}\relax (x )\right )}{2 \cos \relax (x )^{2}}+\frac {3 a \,b^{2} \sin \relax (x )}{2}-\frac {3 a \,b^{2} \ln \left (\sec \relax (x )+\tan \relax (x )\right )}{2}+\frac {b^{3} \left (\tan ^{2}\relax (x )\right )}{2}+b^{3} \ln \left (\cos \relax (x )\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.35, size = 95, normalized size = 1.27 \[ \frac {3}{2} \, a^{2} b \tan \relax (x)^{2} - \frac {3}{4} \, a b^{2} {\left (\frac {2 \, \sin \relax (x)}{\sin \relax (x)^{2} - 1} + \log \left (\sin \relax (x) + 1\right ) - \log \left (\sin \relax (x) - 1\right )\right )} - \frac {1}{4} \, a^{3} {\left (\frac {2 \, \sin \relax (x)}{\sin \relax (x)^{2} - 1} - \log \left (\sin \relax (x) + 1\right ) + \log \left (\sin \relax (x) - 1\right )\right )} - \frac {1}{2} \, b^{3} {\left (\frac {1}{\sin \relax (x)^{2} - 1} - \log \left (\sin \relax (x)^{2} - 1\right )\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.50, size = 126, normalized size = 1.68 \[ \frac {\left (a^3+3\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\left (6\,a^2\,b+2\,b^3\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+\left (a^3+3\,a\,b^2\right )\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}-b^3\,\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )\,{\left (a+b\right )}^2\,\left (a-2\,b\right )}{2}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )\,{\left (a-b\right )}^2\,\left (a+2\,b\right )}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 4.94, size = 122, normalized size = 1.63 \[ - \frac {a^{3} \log {\left (\sin {\relax (x )} - 1 \right )}}{4} + \frac {a^{3} \log {\left (\sin {\relax (x )} + 1 \right )}}{4} - \frac {a^{3} \sin {\relax (x )}}{2 \sin ^{2}{\relax (x )} - 2} + \frac {3 a^{2} b \sec ^{2}{\relax (x )}}{2} + \frac {3 a b^{2} \log {\left (\sin {\relax (x )} - 1 \right )}}{4} - \frac {3 a b^{2} \log {\left (\sin {\relax (x )} + 1 \right )}}{4} - \frac {3 a b^{2} \sin {\relax (x )}}{2 \sin ^{2}{\relax (x )} - 2} - \frac {b^{3} \log {\left (\sec ^{2}{\relax (x )} \right )}}{2} + \frac {b^{3} \sec ^{2}{\relax (x )}}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
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