Optimal. Leaf size=200 \[ -\frac{2 a^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}-\frac{4 a b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}-\frac{a^2 b \cos (c+d x)}{d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac{\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)} \]
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Rubi [A] time = 0.301028, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2731, 2648, 2664, 12, 2660, 618, 204} \[ -\frac{2 a^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}-\frac{4 a b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}-\frac{a^2 b \cos (c+d x)}{d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac{\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)} \]
Antiderivative was successfully verified.
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Rule 2731
Rule 2648
Rule 2664
Rule 12
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{\tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \left (-\frac{1}{2 (a+b)^2 (-1+\sin (c+d x))}+\frac{1}{2 (a-b)^2 (1+\sin (c+d x))}-\frac{a^2}{\left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac{2 a b^2}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac{\int \frac{1}{1+\sin (c+d x)} \, dx}{2 (a-b)^2}-\frac{\int \frac{1}{-1+\sin (c+d x)} \, dx}{2 (a+b)^2}-\frac{\left (2 a b^2\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}-\frac{a^2 \int \frac{1}{(a+b \sin (c+d x))^2} \, dx}{a^2-b^2}\\ &=\frac{\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}-\frac{a^2 b \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac{a^2 \int \frac{a}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}-\frac{\left (4 a b^2\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 )}{\left (a^2-b^2\right )^2 d}\\ &=\frac{\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}-\frac{a^2 b \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac{a^3 \int \frac{1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}+\frac{\left (8 a b^2\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 )}{\left (a^2-b^2\right )^2 d}\\ &=-\frac{4 a b^2 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}+\frac{\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}-\frac{a^2 b \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac{\left (2 a^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 )}{\left (a^2-b^2\right )^2 d}\\ &=-\frac{4 a b^2 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}+\frac{\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}-\frac{a^2 b \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac{\left (4 a^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 )}{\left (a^2-b^2\right )^2 d}\\ &=-\frac{2 a^3 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}-\frac{4 a b^2 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}+\frac{\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}-\frac{a^2 b \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}\\ \end{align*}
Mathematica [A] time = 1.01231, size = 169, normalized size = 0.84 \[ \frac{-\frac{2 a \left (a^2+2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac{a^2 b \cos (c+d x)}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}+\sin \left (\frac{1}{2} (c+d x)\right ) \left (\frac{1}{(a-b)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{1}{(a+b)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}\right )}{d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.088, size = 282, normalized size = 1.4 \begin{align*} -{\frac{1}{d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{a{b}^{2}\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}-2\,{\frac{{a}^{2}b}{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}-2\,{\frac{{a}^{3}}{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{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 ) }-4\,{\frac{a{b}^{2}}{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{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 ) }-{\frac{1}{d \left ( a-b \right ) ^{2}} \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 = 1.95834, size = 1261, normalized size = 6.3 \begin{align*} \left [-\frac{2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} + 2 \,{\left (2 \, a^{4} b - a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} +{\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{4} + 2 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}} \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 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )\right )}}, -\frac{a^{4} b - 2 \, a^{2} b^{3} + b^{5} +{\left (2 \, a^{4} b - a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} -{\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{4} + 2 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) -{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a 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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{2}{\left (c + d x \right )}}{\left (a + b \sin{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 2.15159, size = 339, normalized size = 1.7 \begin{align*} -\frac{2 \,{\left (\frac{{\left (a^{3} + 2 \, a b^{2}\right )}{\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 )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} - b^{2}}} + \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a^{2} b}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}}\right )}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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