Optimal. Leaf size=227 \[ \frac{b^4 \sin (c+d x)}{a d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}-\frac{2 b^5 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{4 b^3 \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{x}{a^2}-\frac{\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)} \]
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Rubi [A] time = 0.491171, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4397, 2897, 2648, 2664, 12, 2659, 208} \[ \frac{b^4 \sin (c+d x)}{a d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}-\frac{2 b^5 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{4 b^3 \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{x}{a^2}-\frac{\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)} \]
Antiderivative was successfully verified.
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Rule 4397
Rule 2897
Rule 2648
Rule 2664
Rule 12
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\cos ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx &=\int \frac{\cos ^2(c+d x) \cot ^2(c+d x)}{(b+a \cos (c+d x))^2} \, dx\\ &=\int \left (-\frac{1}{a^2}-\frac{1}{2 (a-b)^2 (-1-\cos (c+d x))}+\frac{1}{2 (a+b)^2 (1-\cos (c+d x))}+\frac{b^4}{a^2 \left (a^2-b^2\right ) (-b-a \cos (c+d x))^2}+\frac{2 \left (2 a^2 b^3-b^5\right )}{a^2 \left (a^2-b^2\right )^2 (-b-a \cos (c+d x))}\right ) \, dx\\ &=-\frac{x}{a^2}-\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{b^4 \int \frac{1}{(-b-a \cos (c+d x))^2} \, dx}{a^2 \left (a^2-b^2\right )}+\frac{\left (2 b^3 \left (2 a^2-b^2\right )\right ) \int \frac{1}{-b-a \cos (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )^2}\\ &=-\frac{x}{a^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{b^4 \sin (c+d x)}{a \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac{b^4 \int \frac{b}{-b-a \cos (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )^2}+\frac{\left (4 b^3 \left (2 a^2-b^2\right )\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 )}{a^2 \left (a^2-b^2\right )^2 d}\\ &=-\frac{x}{a^2}-\frac{4 b^3 \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 (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{b^4 \sin (c+d x)}{a \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac{b^5 \int \frac{1}{-b-a \cos (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )^2}\\ &=-\frac{x}{a^2}-\frac{4 b^3 \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 (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{b^4 \sin (c+d x)}{a \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac{\left (2 b^5\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 )}{a^2 \left (a^2-b^2\right )^2 d}\\ &=-\frac{x}{a^2}-\frac{2 b^5 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 (a-b)^{5/2} (a+b)^{5/2} d}-\frac{4 b^3 \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 (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{b^4 \sin (c+d x)}{a \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 2.16558, size = 151, normalized size = 0.67 \[ \frac{-\frac{4 b^3 \left (b^2-4 a^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{5/2}}-\frac{2 (c+d x)}{a^2}+\frac{2 b^4 \sin (c+d x)}{a (a-b)^2 (a+b)^2 (a \cos (c+d x)+b)}+\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{(a-b)^2}-\frac{\cot \left (\frac{1}{2} (c+d x)\right )}{(a+b)^2}}{2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.171, size = 255, normalized size = 1.1 \begin{align*}{\frac{1}{2\,d \left ({a}^{2}-2\,ab+{b}^{2} \right ) }\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{{a}^{2}d}}-2\,{\frac{{b}^{4}\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}a \left ( \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 \right ) }}-8\,{\frac{{b}^{3}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{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 ) }+2\,{\frac{{b}^{5}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}{a}^{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 ) }-{\frac{1}{2\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \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 = 0.634744, size = 1507, normalized size = 6.64 \begin{align*} \text{result too large to display} \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{\cos ^{2}{\left (c + d x \right )}}{\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 = 1.22763, size = 447, normalized size = 1.97 \begin{align*} \frac{\frac{4 \,{\left (4 \, a^{2} b^{3} - b^{5}\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^{6} - 2 \, a^{4} b^{2} + a^{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^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 4 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{4} + a^{3} b + a^{2} b^{2} - a b^{3}}{{\left (a^{5} - 2 \, a^{3} b^{2} + a 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 )}} - \frac{2 \,{\left (d x + c\right )}}{a^{2}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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