Optimal. Leaf size=51 \[ \frac {a^2-b^2}{2 b^3 (a+b \sin (x))^2}-\frac {2 a}{b^3 (a+b \sin (x))}-\frac {\log (a+b \sin (x))}{b^3} \]
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Rubi [A] time = 0.08, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4391, 2668, 697} \[ \frac {a^2-b^2}{2 b^3 (a+b \sin (x))^2}-\frac {2 a}{b^3 (a+b \sin (x))}-\frac {\log (a+b \sin (x))}{b^3} \]
Antiderivative was successfully verified.
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Rule 697
Rule 2668
Rule 4391
Rubi steps
\begin {align*} \int \frac {1}{(a \sec (x)+b \tan (x))^3} \, dx &=\int \frac {\cos ^3(x)}{(a+b \sin (x))^3} \, dx\\ &=\frac {\operatorname {Subst}\left (\int \frac {b^2-x^2}{(a+x)^3} \, dx,x,b \sin (x)\right )}{b^3}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{-a-x}+\frac {-a^2+b^2}{(a+x)^3}+\frac {2 a}{(a+x)^2}\right ) \, dx,x,b \sin (x)\right )}{b^3}\\ &=-\frac {\log (a+b \sin (x))}{b^3}+\frac {a^2-b^2}{2 b^3 (a+b \sin (x))^2}-\frac {2 a}{b^3 (a+b \sin (x))}\\ \end {align*}
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Mathematica [A] time = 0.17, size = 40, normalized size = 0.78 \[ -\frac {\frac {3 a^2+4 a b \sin (x)+b^2}{2 (a+b \sin (x))^2}+\log (a+b \sin (x))}{b^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.95, size = 83, normalized size = 1.63 \[ \frac {4 \, a b \sin \relax (x) + 3 \, a^{2} + b^{2} - 2 \, {\left (b^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}\right )} \log \left (b \sin \relax (x) + a\right )}{2 \, {\left (b^{5} \cos \relax (x)^{2} - 2 \, a b^{4} \sin \relax (x) - a^{2} b^{3} - b^{5}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.14, size = 43, normalized size = 0.84 \[ -\frac {\log \left ({\left | b \sin \relax (x) + a \right |}\right )}{b^{3}} + \frac {3 \, b \sin \relax (x)^{2} + 2 \, a \sin \relax (x) - b}{2 \, {\left (b \sin \relax (x) + a\right )}^{2} b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.17, size = 57, normalized size = 1.12 \[ -\frac {2 a}{b^{3} \left (a +b \sin \relax (x )\right )}-\frac {\ln \left (a +b \sin \relax (x )\right )}{b^{3}}+\frac {a^{2}}{2 b^{3} \left (a +b \sin \relax (x )\right )^{2}}-\frac {1}{2 b \left (a +b \sin \relax (x )\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.51, size = 201, normalized size = 3.94 \[ \frac {2 \, {\left (\frac {{\left (a^{3} + a b^{2}\right )} \sin \relax (x)}{\cos \relax (x) + 1} + \frac {{\left (3 \, a^{2} b + b^{3}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {{\left (a^{3} + a b^{2}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}}\right )}}{a^{4} b^{2} + \frac {4 \, a^{3} b^{3} \sin \relax (x)}{\cos \relax (x) + 1} + \frac {4 \, a^{3} b^{3} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {a^{4} b^{2} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {2 \, {\left (a^{4} b^{2} + 2 \, a^{2} b^{4}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}} - \frac {\log \left (a + \frac {2 \, b \sin \relax (x)}{\cos \relax (x) + 1} + \frac {a \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )}{b^{3}} + \frac {\log \left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right )}{b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.70, size = 106, normalized size = 2.08 \[ \frac {2\,a^3\,b\,\sin \relax (x)+3\,a^2\,b^2\,{\sin \relax (x)}^2+2\,a\,b^3\,\sin \relax (x)+b^4\,{\sin \relax (x)}^2}{2\,a^4\,b^3+4\,a^3\,b^4\,\sin \relax (x)+2\,a^2\,b^5\,{\sin \relax (x)}^2}-\frac {2\,\mathrm {atanh}\left (\frac {b^2+a\,\sin \relax (x)\,b}{2\,a^2+\sin \relax (x)\,a\,b-b^2}\right )}{b^3} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 2.76, size = 503, normalized size = 9.86 \[ \begin {cases} - \frac {2 a^{2} \log {\left (\frac {a \sec {\relax (x )}}{b} + \tan {\relax (x )} \right )} \sec ^{2}{\relax (x )}}{2 a^{2} b^{3} \sec ^{2}{\relax (x )} + 4 a b^{4} \tan {\relax (x )} \sec {\relax (x )} + 2 b^{5} \tan ^{2}{\relax (x )}} + \frac {a^{2} \log {\left (\tan ^{2}{\relax (x )} + 1 \right )} \sec ^{2}{\relax (x )}}{2 a^{2} b^{3} \sec ^{2}{\relax (x )} + 4 a b^{4} \tan {\relax (x )} \sec {\relax (x )} + 2 b^{5} \tan ^{2}{\relax (x )}} - \frac {a^{2} \sec ^{2}{\relax (x )}}{2 a^{2} b^{3} \sec ^{2}{\relax (x )} + 4 a b^{4} \tan {\relax (x )} \sec {\relax (x )} + 2 b^{5} \tan ^{2}{\relax (x )}} - \frac {4 a b \log {\left (\frac {a \sec {\relax (x )}}{b} + \tan {\relax (x )} \right )} \tan {\relax (x )} \sec {\relax (x )}}{2 a^{2} b^{3} \sec ^{2}{\relax (x )} + 4 a b^{4} \tan {\relax (x )} \sec {\relax (x )} + 2 b^{5} \tan ^{2}{\relax (x )}} + \frac {2 a b \log {\left (\tan ^{2}{\relax (x )} + 1 \right )} \tan {\relax (x )} \sec {\relax (x )}}{2 a^{2} b^{3} \sec ^{2}{\relax (x )} + 4 a b^{4} \tan {\relax (x )} \sec {\relax (x )} + 2 b^{5} \tan ^{2}{\relax (x )}} - \frac {2 b^{2} \log {\left (\frac {a \sec {\relax (x )}}{b} + \tan {\relax (x )} \right )} \tan ^{2}{\relax (x )}}{2 a^{2} b^{3} \sec ^{2}{\relax (x )} + 4 a b^{4} \tan {\relax (x )} \sec {\relax (x )} + 2 b^{5} \tan ^{2}{\relax (x )}} + \frac {b^{2} \log {\left (\tan ^{2}{\relax (x )} + 1 \right )} \tan ^{2}{\relax (x )}}{2 a^{2} b^{3} \sec ^{2}{\relax (x )} + 4 a b^{4} \tan {\relax (x )} \sec {\relax (x )} + 2 b^{5} \tan ^{2}{\relax (x )}} + \frac {b^{2} \tan ^{2}{\relax (x )}}{2 a^{2} b^{3} \sec ^{2}{\relax (x )} + 4 a b^{4} \tan {\relax (x )} \sec {\relax (x )} + 2 b^{5} \tan ^{2}{\relax (x )}} - \frac {b^{2}}{2 a^{2} b^{3} \sec ^{2}{\relax (x )} + 4 a b^{4} \tan {\relax (x )} \sec {\relax (x )} + 2 b^{5} \tan ^{2}{\relax (x )}} & \text {for}\: b \neq 0 \\\frac {\frac {2 \tan ^{3}{\relax (x )}}{3 \sec ^{3}{\relax (x )}} + \frac {\tan {\relax (x )}}{\sec ^{3}{\relax (x )}}}{a^{3}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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