Optimal. Leaf size=109 \[ -\frac {\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}-\frac {\sec (x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \sin (x)\right )}{3 \left (a^2-b^2\right )^2}+\frac {2 a^3 b \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}} \]
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Rubi [A] time = 0.24, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3872, 2866, 12, 2660, 618, 206} \[ \frac {2 a^3 b \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}-\frac {\sec (x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \sin (x)\right )}{3 \left (a^2-b^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 206
Rule 618
Rule 2660
Rule 2866
Rule 3872
Rubi steps
\begin {align*} \int \frac {\sec ^4(x)}{a+b \csc (x)} \, dx &=\int \frac {\sec ^3(x) \tan (x)}{b+a \sin (x)} \, dx\\ &=-\frac {\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}+\frac {\int \frac {\sec ^2(x) \left (-a b+2 a^2 \sin (x)\right )}{b+a \sin (x)} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac {\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}-\frac {\sec (x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \sin (x)\right )}{3 \left (a^2-b^2\right )^2}+\frac {\int -\frac {3 a^3 b}{b+a \sin (x)} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=-\frac {\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}-\frac {\sec (x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \sin (x)\right )}{3 \left (a^2-b^2\right )^2}-\frac {\left (a^3 b\right ) \int \frac {1}{b+a \sin (x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac {\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}-\frac {\sec (x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \sin (x)\right )}{3 \left (a^2-b^2\right )^2}-\frac {\left (2 a^3 b\right ) \operatorname {Subst}\left (\int \frac {1}{b+2 a x+b x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{\left (a^2-b^2\right )^2}\\ &=-\frac {\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}-\frac {\sec (x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \sin (x)\right )}{3 \left (a^2-b^2\right )^2}+\frac {\left (4 a^3 b\right ) \operatorname {Subst}\left (\int \frac {1}{4 \left (a^2-b^2\right )-x^2} \, dx,x,2 a+2 b \tan \left (\frac {x}{2}\right )\right )}{\left (a^2-b^2\right )^2}\\ &=\frac {2 a^3 b \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}-\frac {\sec (x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \sin (x)\right )}{3 \left (a^2-b^2\right )^2}\\ \end {align*}
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Mathematica [A] time = 0.81, size = 121, normalized size = 1.11 \[ \frac {\sec ^3(x) \left (6 a^3 \sin (x)+2 a^3 \sin (3 x)-6 a^2 b \cos (2 x)-10 a^2 b-3 a b^2 \sin (x)+a b^2 \sin (3 x)+4 b^3\right )-\frac {24 a^3 b \tan ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}}{12 (a-b)^2 (a+b)^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.56, size = 395, normalized size = 3.62 \[ \left [\frac {3 \, \sqrt {a^{2} - b^{2}} a^{3} b \cos \relax (x)^{3} \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \relax (x)^{2} + 2 \, a b \sin \relax (x) + a^{2} + b^{2} + 2 \, {\left (b \cos \relax (x) \sin \relax (x) + a \cos \relax (x)\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) - 2 \, a^{4} b + 4 \, a^{2} b^{3} - 2 \, b^{5} - 6 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \relax (x)^{2} + 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)}{6 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \relax (x)^{3}}, \frac {3 \, \sqrt {-a^{2} + b^{2}} a^{3} b \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \relax (x) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \relax (x)}\right ) \cos \relax (x)^{3} - a^{4} b + 2 \, a^{2} b^{3} - b^{5} - 3 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \relax (x)^{2} + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)}{3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \relax (x)^{3}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.73, size = 187, normalized size = 1.72 \[ -\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a^{3} b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {2 \, {\left (3 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{5} - 6 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{4} + 3 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{4} - 2 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} - 4 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 6 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, a^{3} \tan \left (\frac {1}{2} \, x\right ) - 4 \, a^{2} b + b^{3}\right )}}{3 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} - 1\right )}^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.54, size = 200, normalized size = 1.83 \[ -\frac {2 a^{3} b \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {-a^{2}+b^{2}}}-\frac {4}{3 \left (\tan \left (\frac {x}{2}\right )-1\right )^{3} \left (4 a +4 b \right )}-\frac {2}{\left (4 a +4 b \right ) \left (\tan \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {a}{\left (a +b \right )^{2} \left (\tan \left (\frac {x}{2}\right )-1\right )}-\frac {b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {x}{2}\right )-1\right )}-\frac {4}{3 \left (\tan \left (\frac {x}{2}\right )+1\right )^{3} \left (4 a -4 b \right )}+\frac {2}{\left (4 a -4 b \right ) \left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {a}{\left (a -b \right )^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )}+\frac {b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.65, size = 298, normalized size = 2.73 \[ \frac {\frac {2\,\left (4\,a^2\,b-b^3\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (2\,a^2\,b-b^3\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,a^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,a^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{a^4-2\,a^2\,b^2+b^4}-\frac {4\,a^2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (a^2+2\,b^2\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2-1}+\frac {2\,a^3\,b\,\mathrm {atanh}\left (\frac {2\,a\,b^4+2\,a^5-4\,a^3\,b^2+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{2\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]
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 \[ \int \frac {\sec ^{4}{\relax (x )}}{a + b \csc {\relax (x )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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