Optimal. Leaf size=138 \[ \frac{\left (a^2-b^2\right ) \sin (c+d x)}{d \left (a^2+b^2\right )^2}+\frac{2 a b \cos (c+d x)}{d \left (a^2+b^2\right )^2}-\frac{b^3}{d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))}-\frac{3 a b^2 \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}} \]
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Rubi [A] time = 1.04833, antiderivative size = 231, normalized size of antiderivative = 1.67, number of steps used = 11, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {6742, 639, 203, 638, 618, 206} \[ -\frac{2 b^3 \left (a+b \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a d \left (a^2+b^2\right )^2 \left (-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+2 b \tan \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{2 \left (\left (a^2-b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )+2 a b\right )}{d \left (a^2+b^2\right )^2 \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )}+\frac{2 b^4 \tanh ^{-1}\left (\frac{b-a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a d \left (a^2+b^2\right )^{5/2}}-\frac{2 b^2 \left (3 a^2+b^2\right ) \tanh ^{-1}\left (\frac{b-a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a d \left (a^2+b^2\right )^{5/2}} \]
Antiderivative was successfully verified.
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Rule 6742
Rule 639
Rule 203
Rule 638
Rule 618
Rule 206
Rubi steps
\begin{align*} \int \frac{\cos ^3(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx &=\frac{2 \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{\left (1+x^2\right )^2 \left (a+2 b x-a x^2\right )^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{d}\\ &=\frac{2 \operatorname{Subst}\left (\int \left (\frac{2 \left (a^2-b^2-2 a b x\right )}{\left (a^2+b^2\right )^2 \left (1+x^2\right )^2}+\frac{-a^2+b^2}{\left (a^2+b^2\right )^2 \left (1+x^2\right )}-\frac{2 b^3 x}{a \left (a^2+b^2\right ) \left (-a-2 b x+a x^2\right )^2}-\frac{b^2 \left (3 a^2+b^2\right )}{a \left (a^2+b^2\right )^2 \left (-a-2 b x+a x^2\right )}\right ) \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{d}\\ &=\frac{4 \operatorname{Subst}\left (\int \frac{a^2-b^2-2 a b x}{\left (1+x^2\right )^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^2 d}-\frac{\left (2 \left (a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^2 d}-\frac{\left (4 b^3\right ) \operatorname{Subst}\left (\int \frac{x}{\left (-a-2 b x+a x^2\right )^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a \left (a^2+b^2\right ) d}-\frac{\left (2 b^2 \left (3 a^2+b^2\right )\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 )}{a \left (a^2+b^2\right )^2 d}\\ &=-\frac{\left (a^2-b^2\right ) x}{\left (a^2+b^2\right )^2}+\frac{2 \left (2 a b+\left (a^2-b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^2 d \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}-\frac{2 b^3 \left (a+b \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac{1}{2} (c+d x)\right )-a \tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}+\frac{\left (2 b^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 )}{a \left (a^2+b^2\right )^2 d}+\frac{\left (2 \left (a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^2 d}+\frac{\left (4 b^2 \left (3 a^2+b^2\right )\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 )}{a \left (a^2+b^2\right )^2 d}\\ &=-\frac{2 b^2 \left (3 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 )}{a \left (a^2+b^2\right )^{5/2} d}+\frac{2 \left (2 a b+\left (a^2-b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^2 d \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}-\frac{2 b^3 \left (a+b \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac{1}{2} (c+d x)\right )-a \tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}-\frac{\left (4 b^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 )}{a \left (a^2+b^2\right )^2 d}\\ &=\frac{2 b^4 \tanh ^{-1}\left (\frac{b-a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a \left (a^2+b^2\right )^{5/2} d}-\frac{2 b^2 \left (3 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 )}{a \left (a^2+b^2\right )^{5/2} d}+\frac{2 \left (2 a b+\left (a^2-b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2+b^2\right )^2 d \left (1+\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}-\frac{2 b^3 \left (a+b \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a \left (a^2+b^2\right )^2 d \left (a+2 b \tan \left (\frac{1}{2} (c+d x)\right )-a \tan ^2\left (\frac{1}{2} (c+d x)\right )\right )}\\ \end{align*}
Mathematica [A] time = 0.774981, size = 130, normalized size = 0.94 \[ \frac{\frac{a \left (a^2+b^2\right ) \sin (2 (c+d x))+b \left (a^2+b^2\right ) \cos (2 (c+d x))+3 b \left (a^2-b^2\right )}{\left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))}+\frac{12 a b^2 \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}}{2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.185, size = 172, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}} \left ({\frac{1}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a} \left ( -{\frac{{b}^{2}\tan \left ( 1/2\,dx+c/2 \right ) }{a}}-b \right ) }-3\,{\frac{a}{\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \right ) }-2\,{\frac{ \left ( -{a}^{2}+{b}^{2} \right ) \tan \left ( 1/2\,dx+c/2 \right ) -2\,ab}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \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 = 0.553538, size = 697, normalized size = 5.05 \begin{align*} \frac{2 \, a^{4} b - 2 \, a^{2} b^{3} - 4 \, b^{5} + 2 \,{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \,{\left (a^{2} b^{2} \cos \left (d x + c\right ) + a b^{3} \sin \left (d x + c\right )\right )} \sqrt{a^{2} + b^{2}} \log \left (-\frac{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\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 \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(-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 [B] time = 1.28279, size = 386, normalized size = 2.8 \begin{align*} -\frac{\frac{3 \, a b^{2} \log \left (\frac{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} + b^{2}}} - \frac{2 \,{\left (a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a^{3} b + a b^{3}\right )}}{{\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 )^{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 )}}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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