Optimal. Leaf size=203 \[ \frac{a b^2 \sin (c+d x)}{d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}-\frac{4 a^2 b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{2 b^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)} \]
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Rubi [A] time = 0.399241, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {4397, 2731, 2648, 2664, 12, 2659, 208} \[ \frac{a b^2 \sin (c+d x)}{d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}-\frac{4 a^2 b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{2 b^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)} \]
Antiderivative was successfully verified.
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Rule 4397
Rule 2731
Rule 2648
Rule 2664
Rule 12
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{1}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx &=\int \frac{\cot ^2(c+d x)}{(b+a \cos (c+d x))^2} \, dx\\ &=\int \left (-\frac{1}{2 (a+b)^2 (-1+\cos (c+d x))}+\frac{1}{2 (a-b)^2 (1+\cos (c+d x))}-\frac{b^2}{\left (-a^2+b^2\right ) (b+a \cos (c+d x))^2}-\frac{2 a^2 b}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))}\right ) \, dx\\ &=\frac{\int \frac{1}{1+\cos (c+d x)} \, dx}{2 (a-b)^2}-\frac{\int \frac{1}{-1+\cos (c+d x)} \, dx}{2 (a+b)^2}-\frac{\left (2 a^2 b\right ) \int \frac{1}{b+a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^2}+\frac{b^2 \int \frac{1}{(b+a \cos (c+d x))^2} \, dx}{a^2-b^2}\\ &=-\frac{\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}+\frac{a b^2 \sin (c+d x)}{\left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac{b^2 \int \frac{b}{b+a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^2}-\frac{\left (4 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(-a+b) 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^2 b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac{\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}+\frac{a b^2 \sin (c+d x)}{\left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac{b^3 \int \frac{1}{b+a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac{4 a^2 b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac{\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}+\frac{a b^2 \sin (c+d x)}{\left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac{\left (2 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(-a+b) 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^2 b \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac{2 b^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac{\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}+\frac{a b^2 \sin (c+d x)}{\left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 1.24578, size = 128, normalized size = 0.63 \[ \frac{\frac{4 b \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{\frac{2 a b^2 \sin (c+d x)}{(a+b)^2 (a \cos (c+d x)+b)}+\tan \left (\frac{1}{2} (c+d x)\right )}{(a-b)^2}-\frac{\cot \left (\frac{1}{2} (c+d x)\right )}{(a+b)^2}}{2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.148, size = 162, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{1}{2\,{a}^{2}-4\,ab+2\,{b}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{b}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}} \left ( -{\frac{ab\tan \left ( 1/2\,dx+c/2 \right ) }{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b}}-{\frac{2\,{a}^{2}+{b}^{2}}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \right ) }-{\frac{1}{2\, \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \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 = 0.590717, size = 1164, normalized size = 5.73 \begin{align*} \left [\frac{6 \, a^{3} b^{2} - 6 \, a b^{4} +{\left (2 \, a^{2} b^{2} + b^{4} +{\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 2 \,{\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )}{2 \,{\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d\right )} \sin \left (d x + c\right )}, \frac{3 \, a^{3} b^{2} - 3 \, a b^{4} -{\left (2 \, a^{2} b^{2} + b^{4} +{\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) -{\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \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 )}{{\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d\right )} \sin \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{1}{\left (a \sin{\left (c + d x \right )} + b \tan{\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 = 1.12958, size = 390, normalized size = 1.92 \begin{align*} \frac{\frac{4 \,{\left (2 \, a^{2} b + b^{3}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2} - 2 \, a b + b^{2}} - \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 7 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{3} + a^{2} b + a b^{2} - b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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