Optimal. Leaf size=114 \[ -\frac{b (4 a-b) \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} d (a-b)^{5/2}}+\frac{b^2 \sin (c+d x)}{2 a d (a-b)^2 \left (a-(a-b) \sin ^2(c+d x)\right )}+\frac{\sin (c+d x)}{d (a-b)^2} \]
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Rubi [A] time = 0.181111, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3676, 390, 385, 208} \[ -\frac{b (4 a-b) \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} d (a-b)^{5/2}}+\frac{b^2 \sin (c+d x)}{2 a d (a-b)^2 \left (a-(a-b) \sin ^2(c+d x)\right )}+\frac{\sin (c+d x)}{d (a-b)^2} \]
Antiderivative was successfully verified.
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Rule 3676
Rule 390
Rule 385
Rule 208
Rubi steps
\begin{align*} \int \frac{\cos (c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{\left (a-(a-b) x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{(a-b)^2}-\frac{(2 a-b) b-2 (a-b) b x^2}{(a-b)^2 \left (a+(-a+b) x^2\right )^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{\sin (c+d x)}{(a-b)^2 d}-\frac{\operatorname{Subst}\left (\int \frac{(2 a-b) b-2 (a-b) b x^2}{\left (a+(-a+b) x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{(a-b)^2 d}\\ &=\frac{\sin (c+d x)}{(a-b)^2 d}+\frac{b^2 \sin (c+d x)}{2 a (a-b)^2 d \left (a-(a-b) \sin ^2(c+d x)\right )}-\frac{((4 a-b) b) \operatorname{Subst}\left (\int \frac{1}{a+(-a+b) x^2} \, dx,x,\sin (c+d x)\right )}{2 a (a-b)^2 d}\\ &=-\frac{(4 a-b) b \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} (a-b)^{5/2} d}+\frac{\sin (c+d x)}{(a-b)^2 d}+\frac{b^2 \sin (c+d x)}{2 a (a-b)^2 d \left (a-(a-b) \sin ^2(c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 0.52163, size = 119, normalized size = 1.04 \[ \frac{-\frac{\sqrt{a} \sin (c+d x) \left (a^2+a (a-b) \cos (2 (c+d x))+a b+b^2\right )}{(a-b)^2 \left ((a-b) \sin ^2(c+d x)-a\right )}-\frac{b (4 a-b) \tanh ^{-1}\left (\frac{\sqrt{a-b} \sin (c+d x)}{\sqrt{a}}\right )}{(a-b)^{5/2}}}{2 a^{3/2} d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.085, size = 118, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({\frac{\sin \left ( dx+c \right ) }{{a}^{2}-2\,ab+{b}^{2}}}+{\frac{b}{ \left ( a-b \right ) ^{2}} \left ( -{\frac{\sin \left ( dx+c \right ) b}{2\,a \left ( a \left ( \sin \left ( dx+c \right ) \right ) ^{2}-b \left ( \sin \left ( dx+c \right ) \right ) ^{2}-a \right ) }}-{\frac{4\,a-b}{2\,a}{\it Artanh} \left ({ \left ( a-b \right ) \sin \left ( dx+c \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a-b \right ) }}}} \right ) } \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 [B] time = 1.8103, size = 973, normalized size = 8.54 \begin{align*} \left [-\frac{{\left (4 \, a b^{2} - b^{3} +{\left (4 \, a^{2} b - 5 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{a^{2} - a b} \log \left (-\frac{{\left (a - b\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{a^{2} - a b} \sin \left (d x + c\right ) - 2 \, a + b}{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + b}\right ) - 2 \,{\left (2 \, a^{3} b - a^{2} b^{2} - a b^{3} + 2 \,{\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \,{\left ({\left (a^{6} - 4 \, a^{5} b + 6 \, a^{4} b^{2} - 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{5} b - 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} - a^{2} b^{4}\right )} d\right )}}, \frac{{\left (4 \, a b^{2} - b^{3} +{\left (4 \, a^{2} b - 5 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{-a^{2} + a b} \arctan \left (\frac{\sqrt{-a^{2} + a b} \sin \left (d x + c\right )}{a}\right ) +{\left (2 \, a^{3} b - a^{2} b^{2} - a b^{3} + 2 \,{\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{6} - 4 \, a^{5} b + 6 \, a^{4} b^{2} - 4 \, a^{3} b^{3} + a^{2} b^{4}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{5} b - 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} - a^{2} b^{4}\right )} d\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 = 1.72388, size = 205, normalized size = 1.8 \begin{align*} -\frac{\frac{b^{2} \sin \left (d x + c\right )}{{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )}{\left (a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right )^{2} - a\right )}} + \frac{{\left (4 \, a b - b^{2}\right )} \arctan \left (-\frac{a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{\sqrt{-a^{2} + a b}}\right )}{{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \sqrt{-a^{2} + a b}} - \frac{2 \, \sin \left (d x + c\right )}{a^{2} - 2 \, a b + b^{2}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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