Optimal. Leaf size=157 \[ \frac{\frac{1}{2} b \left (b^2-9 a^2\right ) \left (2 \left (a^2+b^2\right )+3 a b \sin (2 (c+d x))\right )-3 \left (-a^2 b^3+3 a^4 b+b^5\right ) \cos (2 (c+d x))}{6 d \left (a^2+b^2\right )^3 (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{a \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{7/2}} \]
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Rubi [B] time = 1.17309, antiderivative size = 362, normalized size of antiderivative = 2.31, number of steps used = 7, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {1660, 12, 618, 206} \[ -\frac{8 b^3 \left (b \left (3 a^2+4 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )+a \left (a^2+2 b^2\right )\right )}{3 a^5 d \left (a^2+b^2\right ) \left (-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+2 b \tan \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 b^2 \left (a \left (30 a^2 b^2+9 a^4+16 b^4\right ) \tan \left (\frac{1}{2} (c+d x)\right )+b \left (18 a^2 b^2+15 a^4+8 b^4\right )\right )}{3 a^5 d \left (a^2+b^2\right )^2 \left (-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+2 b \tan \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b \left (a b \left (6 a^2 b^2+9 a^4+2 b^4\right ) \tan \left (\frac{1}{2} (c+d x)\right )+9 a^4 b^2+12 a^2 b^4+6 a^6+4 b^6\right )}{a^4 d \left (a^2+b^2\right )^3 \left (-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+2 b \tan \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{a \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{b-a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{7/2}} \]
Antiderivative was successfully verified.
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Rule 1660
Rule 12
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))^4} \, dx &=\frac{2 \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{\left (a+2 b x-a x^2\right )^4} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{d}\\ &=-\frac{8 b^3 \left (a \left (a^2+2 b^2\right )+b \left (3 a^2+4 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right ) 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 )^3}-\frac{\operatorname{Subst}\left (\int \frac{-\frac{4 \left (3 a^6+3 a^4 b^2-12 a^2 b^4-32 b^6\right )}{a^5}+24 b \left (1-\frac{3 b^2}{a^2}-\frac{4 b^4}{a^4}\right ) x+24 \left (a-\frac{b^2}{a}-\frac{2 b^4}{a^3}\right ) x^2-24 b \left (1+\frac{b^2}{a^2}\right ) x^3-12 \left (a+\frac{b^2}{a}\right ) x^4}{\left (a+2 b x-a x^2\right )^3} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{6 \left (a^2+b^2\right ) d}\\ &=-\frac{8 b^3 \left (a \left (a^2+2 b^2\right )+b \left (3 a^2+4 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right ) 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 )^3}+\frac{2 b^2 \left (b \left (15 a^4+18 a^2 b^2+8 b^4\right )+a \left (9 a^4+30 a^2 b^2+16 b^4\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^5 \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 )^2}+\frac{\operatorname{Subst}\left (\int \frac{\frac{96 \left (a^6-a^4 b^2+7 a^2 b^4+4 b^6\right )}{a^4}-\frac{384 b \left (a^2+b^2\right )^2 x}{a^3}-\frac{96 \left (a^2+b^2\right )^2 x^2}{a^2}}{\left (a+2 b x-a x^2\right )^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{48 \left (a^2+b^2\right )^2 d}\\ &=-\frac{8 b^3 \left (a \left (a^2+2 b^2\right )+b \left (3 a^2+4 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right ) 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 )^3}+\frac{2 b^2 \left (b \left (15 a^4+18 a^2 b^2+8 b^4\right )+a \left (9 a^4+30 a^2 b^2+16 b^4\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^5 \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 )^2}-\frac{b \left (6 a^6+9 a^4 b^2+12 a^2 b^4+4 b^6+a b \left (9 a^4+6 a^2 b^2+2 b^4\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 \left (a^2+b^2\right )^3 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{\operatorname{Subst}\left (\int -\frac{192 a \left (2 a^2-3 b^2\right )}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{192 \left (a^2+b^2\right )^3 d}\\ &=-\frac{8 b^3 \left (a \left (a^2+2 b^2\right )+b \left (3 a^2+4 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right ) 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 )^3}+\frac{2 b^2 \left (b \left (15 a^4+18 a^2 b^2+8 b^4\right )+a \left (9 a^4+30 a^2 b^2+16 b^4\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^5 \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 )^2}-\frac{b \left (6 a^6+9 a^4 b^2+12 a^2 b^4+4 b^6+a b \left (9 a^4+6 a^2 b^2+2 b^4\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 \left (a^2+b^2\right )^3 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 (a \left (2 a^2-3 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 )}{\left (a^2+b^2\right )^3 d}\\ &=-\frac{8 b^3 \left (a \left (a^2+2 b^2\right )+b \left (3 a^2+4 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right ) 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 )^3}+\frac{2 b^2 \left (b \left (15 a^4+18 a^2 b^2+8 b^4\right )+a \left (9 a^4+30 a^2 b^2+16 b^4\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^5 \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 )^2}-\frac{b \left (6 a^6+9 a^4 b^2+12 a^2 b^4+4 b^6+a b \left (9 a^4+6 a^2 b^2+2 b^4\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 \left (a^2+b^2\right )^3 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 a \left (2 a^2-3 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 )}{\left (a^2+b^2\right )^3 d}\\ &=-\frac{a \left (2 a^2-3 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 )^{7/2} d}-\frac{8 b^3 \left (a \left (a^2+2 b^2\right )+b \left (3 a^2+4 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^5 \left (a^2+b^2\right ) 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 )^3}+\frac{2 b^2 \left (b \left (15 a^4+18 a^2 b^2+8 b^4\right )+a \left (9 a^4+30 a^2 b^2+16 b^4\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^5 \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 )^2}-\frac{b \left (6 a^6+9 a^4 b^2+12 a^2 b^4+4 b^6+a b \left (9 a^4+6 a^2 b^2+2 b^4\right ) \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 \left (a^2+b^2\right )^3 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 [C] time = 1.05464, size = 165, normalized size = 1.05 \[ \frac{\frac{6 a \left (2 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}}+\frac{\frac{1}{2} b \left (b^2-9 a^2\right ) \left (2 \left (a^2+b^2\right )+3 a b \sin (2 (c+d x))\right )-3 \left (-a^2 b^3+3 a^4 b+b^5\right ) \cos (2 (c+d x))}{(a-i b)^3 (a+i b)^3 (a \cos (c+d x)+b \sin (c+d x))^3}}{6 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.234, size = 494, normalized size = 3.2 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{ \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 ) ^{3}} \left ( -1/2\,{\frac{{b}^{2} \left ( 9\,{a}^{4}+6\,{a}^{2}{b}^{2}+2\,{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{a \left ({a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6} \right ) }}-1/2\,{\frac{b \left ( 6\,{a}^{6}-27\,{a}^{4}{b}^{2}-12\,{a}^{2}{b}^{4}-4\,{b}^{6} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{{a}^{2} \left ({a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6} \right ) }}+1/3\,{\frac{{b}^{2} \left ( 54\,{a}^{6}-21\,{a}^{4}{b}^{2}-4\,{a}^{2}{b}^{4}-4\,{b}^{6} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{ \left ({a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6} \right ){a}^{3}}}+{\frac{b \left ( 6\,{a}^{6}-20\,{a}^{4}{b}^{2}-3\,{a}^{2}{b}^{4}-2\,{b}^{6} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{{a}^{2} \left ({a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6} \right ) }}-1/2\,{\frac{{b}^{2} \left ( 27\,{a}^{4}+4\,{a}^{2}{b}^{2}+2\,{b}^{4} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{a \left ({a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6} \right ) }}-1/6\,{\frac{b \left ( 18\,{a}^{4}+5\,{a}^{2}{b}^{2}+2\,{b}^{4} \right ) }{{a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6}}} \right ) }+{\frac{a \left ( 2\,{a}^{2}-3\,{b}^{2} \right ) }{{a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6}}{\it Artanh} \left ({\frac{1}{2} \left ( 2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \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 = 0.595758, size = 1173, normalized size = 7.47 \begin{align*} -\frac{22 \, a^{4} b^{3} + 14 \, a^{2} b^{5} - 8 \, b^{7} + 12 \,{\left (3 \, a^{6} b + 2 \, a^{4} b^{3} + b^{7}\right )} \cos \left (d x + c\right )^{2} + 6 \,{\left (9 \, a^{5} b^{2} + 8 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \,{\left ({\left (2 \, a^{6} - 9 \, a^{4} b^{2} + 9 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (2 \, a^{4} b^{2} - 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right ) +{\left (2 \, a^{3} b^{3} - 3 \, a b^{5} +{\left (6 \, a^{5} b - 11 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \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 )}{12 \,{\left ({\left (a^{11} + a^{9} b^{2} - 6 \, a^{7} b^{4} - 14 \, a^{5} b^{6} - 11 \, a^{3} b^{8} - 3 \, a b^{10}\right )} d \cos \left (d x + c\right )^{3} + 3 \,{\left (a^{9} b^{2} + 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} + 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right ) +{\left ({\left (3 \, a^{10} b + 11 \, a^{8} b^{3} + 14 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - a^{2} b^{9} - b^{11}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{8} b^{3} + 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} + 4 \, a^{2} b^{9} + b^{11}\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(-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 = 1.27717, size = 707, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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