Optimal. Leaf size=183 \[ \frac{a^2 \cos (c+d x)}{b d \left (a^2-b^2\right )}-\frac{b \cos (c+d x)}{d \left (a^2-b^2\right )}-\frac{2 a^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{3/2}}+\frac{a \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac{b \sec (c+d x)}{d \left (a^2-b^2\right )}+\frac{a^3 x}{b^2 \left (a^2-b^2\right )}-\frac{a x}{a^2-b^2} \]
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Rubi [A] time = 0.2734, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 11, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.379, Rules used = {2902, 3473, 8, 2590, 14, 2746, 12, 2735, 2660, 618, 204} \[ \frac{a^2 \cos (c+d x)}{b d \left (a^2-b^2\right )}-\frac{b \cos (c+d x)}{d \left (a^2-b^2\right )}-\frac{2 a^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{3/2}}+\frac{a \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac{b \sec (c+d x)}{d \left (a^2-b^2\right )}+\frac{a^3 x}{b^2 \left (a^2-b^2\right )}-\frac{a x}{a^2-b^2} \]
Antiderivative was successfully verified.
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Rule 2902
Rule 3473
Rule 8
Rule 2590
Rule 14
Rule 2746
Rule 12
Rule 2735
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{a \int \tan ^2(c+d x) \, dx}{a^2-b^2}-\frac{a^2 \int \frac{\sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{a^2-b^2}-\frac{b \int \sin (c+d x) \tan ^2(c+d x) \, dx}{a^2-b^2}\\ &=\frac{a^2 \cos (c+d x)}{b \left (a^2-b^2\right ) d}+\frac{a \tan (c+d x)}{\left (a^2-b^2\right ) d}-\frac{a \int 1 \, dx}{a^2-b^2}+\frac{a^2 \int \frac{a \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )}+\frac{b \operatorname{Subst}\left (\int \frac{1-x^2}{x^2} \, dx,x,\cos (c+d x)\right )}{\left (a^2-b^2\right ) d}\\ &=-\frac{a x}{a^2-b^2}+\frac{a^2 \cos (c+d x)}{b \left (a^2-b^2\right ) d}+\frac{a \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac{a^3 \int \frac{\sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )}+\frac{b \operatorname{Subst}\left (\int \left (-1+\frac{1}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{\left (a^2-b^2\right ) d}\\ &=-\frac{a x}{a^2-b^2}+\frac{a^3 x}{b^2 \left (a^2-b^2\right )}+\frac{a^2 \cos (c+d x)}{b \left (a^2-b^2\right ) d}-\frac{b \cos (c+d x)}{\left (a^2-b^2\right ) d}-\frac{b \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac{a \tan (c+d x)}{\left (a^2-b^2\right ) d}-\frac{a^4 \int \frac{1}{a+b \sin (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=-\frac{a x}{a^2-b^2}+\frac{a^3 x}{b^2 \left (a^2-b^2\right )}+\frac{a^2 \cos (c+d x)}{b \left (a^2-b^2\right ) d}-\frac{b \cos (c+d x)}{\left (a^2-b^2\right ) d}-\frac{b \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac{a \tan (c+d x)}{\left (a^2-b^2\right ) d}-\frac{\left (2 a^4\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 )}{b^2 \left (a^2-b^2\right ) d}\\ &=-\frac{a x}{a^2-b^2}+\frac{a^3 x}{b^2 \left (a^2-b^2\right )}+\frac{a^2 \cos (c+d x)}{b \left (a^2-b^2\right ) d}-\frac{b \cos (c+d x)}{\left (a^2-b^2\right ) d}-\frac{b \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac{a \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac{\left (4 a^4\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 )}{b^2 \left (a^2-b^2\right ) d}\\ &=-\frac{a x}{a^2-b^2}+\frac{a^3 x}{b^2 \left (a^2-b^2\right )}-\frac{2 a^4 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{3/2} d}+\frac{a^2 \cos (c+d x)}{b \left (a^2-b^2\right ) d}-\frac{b \cos (c+d x)}{\left (a^2-b^2\right ) d}-\frac{b \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac{a \tan (c+d x)}{\left (a^2-b^2\right ) d}\\ \end{align*}
Mathematica [A] time = 1.03234, size = 186, normalized size = 1.02 \[ \frac{\frac{a^3 (-(c+d x))+a b^2 (c+d x)+b^3}{b^4-a^2 b^2}-\frac{2 a^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{3/2}}+\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{(a-b) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{\cos (c+d x)}{b}}{d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.102, size = 162, normalized size = 0.9 \begin{align*} -32\,{\frac{1}{d \left ( 32\,a+32\,b \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }}+2\,{\frac{1}{bd \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+2\,{\frac{a\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{b}^{2}}}-2\,{\frac{{a}^{4}}{d \left ( a-b \right ) \left ( a+b \right ){b}^{2}\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 ) }-32\,{\frac{1}{d \left ( 32\,a-32\,b \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \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 = 1.6845, size = 954, normalized size = 5.21 \begin{align*} \left [\frac{\sqrt{-a^{2} + b^{2}} a^{4} \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^{2} b^{3} + 2 \, b^{5} + 2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x \cos \left (d x + c\right ) + 2 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )}, \frac{\sqrt{a^{2} - b^{2}} a^{4} \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 ) - a^{2} b^{3} + b^{5} +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x \cos \left (d x + c\right ) +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} +{\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\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(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.18588, size = 234, normalized size = 1.28 \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 )} a^{4}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt{a^{2} - b^{2}}} - \frac{{\left (d x + c\right )} a}{b^{2}} + \frac{2 \,{\left (a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2} - 2 \, b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 1\right )}{\left (a^{2} b - b^{3}\right )}}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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