Optimal. Leaf size=118 \[ \frac{2 b^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a d \left (a^2-b^2\right )^{3/2}}+\frac{b \sec (c+d x) (b-a \sin (c+d x))}{a d \left (a^2-b^2\right )}+\frac{\sec (c+d x)}{a d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d} \]
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Rubi [A] time = 0.230191, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2898, 2622, 321, 207, 2696, 12, 2660, 618, 204} \[ \frac{2 b^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a d \left (a^2-b^2\right )^{3/2}}+\frac{b \sec (c+d x) (b-a \sin (c+d x))}{a d \left (a^2-b^2\right )}+\frac{\sec (c+d x)}{a d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d} \]
Antiderivative was successfully verified.
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Rule 2898
Rule 2622
Rule 321
Rule 207
Rule 2696
Rule 12
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{\csc (c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\int \left (\frac{\csc (c+d x) \sec ^2(c+d x)}{a}-\frac{b \sec ^2(c+d x)}{a (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac{\int \csc (c+d x) \sec ^2(c+d x) \, dx}{a}-\frac{b \int \frac{\sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{a}\\ &=\frac{b \sec (c+d x) (b-a \sin (c+d x))}{a \left (a^2-b^2\right ) d}+\frac{b \int \frac{b^2}{a+b \sin (c+d x)} \, dx}{a \left (a^2-b^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a d}\\ &=\frac{\sec (c+d x)}{a d}+\frac{b \sec (c+d x) (b-a \sin (c+d x))}{a \left (a^2-b^2\right ) d}+\frac{b^3 \int \frac{1}{a+b \sin (c+d x)} \, dx}{a \left (a^2-b^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{a d}\\ &=-\frac{\tanh ^{-1}(\cos (c+d x))}{a d}+\frac{\sec (c+d x)}{a d}+\frac{b \sec (c+d x) (b-a \sin (c+d x))}{a \left (a^2-b^2\right ) d}+\frac{\left (2 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a \left (a^2-b^2\right ) d}\\ &=-\frac{\tanh ^{-1}(\cos (c+d x))}{a d}+\frac{\sec (c+d x)}{a d}+\frac{b \sec (c+d x) (b-a \sin (c+d x))}{a \left (a^2-b^2\right ) d}-\frac{\left (4 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a \left (a^2-b^2\right ) d}\\ &=\frac{2 b^3 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{3/2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a d}+\frac{\sec (c+d x)}{a d}+\frac{b \sec (c+d x) (b-a \sin (c+d x))}{a \left (a^2-b^2\right ) d}\\ \end{align*}
Mathematica [A] time = 0.362104, size = 191, normalized size = 1.62 \[ \frac{2 b^3 \cos (c+d x) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )+\sqrt{a^2-b^2} \left (a (a-b \sin (c+d x))-\left (a^2-b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{a d (a-b) (a+b) \sqrt{a^2-b^2} \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.101, size = 130, normalized size = 1.1 \begin{align*} 2\,{\frac{{b}^{3}}{d \left ( a-b \right ) \left ( a+b \right ) a\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+{\frac{1}{d \left ( a-b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{1}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{1}{d \left ( a+b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}} \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 = 3.05825, size = 1084, normalized size = 9.19 \begin{align*} \left [\frac{\sqrt{-a^{2} + b^{2}} b^{3} \cos \left (d x + c\right ) \log \left (-\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \,{\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{4} - 2 \, a^{2} b^{2} -{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right ) \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right ) \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 2 \,{\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right )}, -\frac{2 \, \sqrt{a^{2} - b^{2}} b^{3} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - 2 \, a^{4} + 2 \, a^{2} b^{2} +{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right ) \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right ) \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 2 \,{\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right )}\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{\csc{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{a + b \sin{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.18329, size = 182, normalized size = 1.54 \begin{align*} \frac{\frac{2 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )} b^{3}}{{\left (a^{3} - a b^{2}\right )} \sqrt{a^{2} - b^{2}}} + \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} + \frac{2 \,{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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