Optimal. Leaf size=109 \[ \frac{2 a^3 b \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac{\sec ^3(x) (b-a \sin (x))}{3 \left (a^2-b^2\right )}-\frac{\sec (x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \sin (x)\right )}{3 \left (a^2-b^2\right )^2} \]
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Rubi [A] time = 0.238833, antiderivative size = 109, normalized size of antiderivative = 1., 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 3872
Rule 2866
Rule 12
Rule 2660
Rule 618
Rule 206
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*}
Mathematica [A] time = 0.757917, size = 121, normalized size = 1.11 \[ \frac{\sec ^3(x) \left (-6 a^2 b \cos (2 x)-10 a^2 b+6 a^3 \sin (x)+2 a^3 \sin (3 x)-3 a b^2 \sin (x)+a b^2 \sin (3 x)+4 b^3\right )-\frac{24 a^3 b \tan ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}}{12 (a-b)^2 (a+b)^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.061, size = 200, normalized size = 1.8 \begin{align*} -{\frac{4}{12\,a-12\,b} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+2\,{\frac{1}{ \left ( 4\,a-4\,b \right ) \left ( \tan \left ( x/2 \right ) +1 \right ) ^{2}}}-{\frac{a}{ \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{b}{2\, \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{4}{12\,a+12\,b} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}}-2\,{\frac{1}{ \left ( 4\,a+4\,b \right ) \left ( \tan \left ( x/2 \right ) -1 \right ) ^{2}}}-{\frac{b}{2\, \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{a}{ \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{{a}^{3}b}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}\sqrt{-{a}^{2}+{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,b\tan \left ( x/2 \right ) +2\,a}{\sqrt{-{a}^{2}+{b}^{2}}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.56328, size = 882, normalized size = 8.09 \begin{align*} \left [\frac{3 \, \sqrt{a^{2} - b^{2}} a^{3} b \cos \left (x\right )^{3} \log \left (\frac{{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2} + 2 \,{\left (b \cos \left (x\right ) \sin \left (x\right ) + a \cos \left (x\right )\right )} \sqrt{a^{2} - b^{2}}}{a^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - 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 \left (x\right )^{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 \left (x\right )^{2}\right )} \sin \left (x\right )}{6 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (x\right )^{3}}, \frac{3 \, \sqrt{-a^{2} + b^{2}} a^{3} b \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \sin \left (x\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (x\right )}\right ) \cos \left (x\right )^{3} - a^{4} b + 2 \, a^{2} b^{3} - b^{5} - 3 \,{\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (x\right )^{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 \left (x\right )^{2}\right )} \sin \left (x\right )}{3 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (x\right )^{3}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{4}{\left (x \right )}}{a + b \csc{\left (x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.39852, size = 252, normalized size = 2.31 \begin{align*} -\frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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