Optimal. Leaf size=90 \[ \frac{4 \sin ^5(c+d x)}{5 a^3 d}-\frac{5 \sin ^3(c+d x)}{3 a^3 d}+\frac{\sin (c+d x)}{a^3 d}+\frac{4 i \cos ^5(c+d x)}{5 a^3 d}-\frac{i \cos ^3(c+d x)}{3 a^3 d} \]
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Rubi [A] time = 0.220501, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {3092, 3090, 2633, 2565, 30, 2564, 14} \[ \frac{4 \sin ^5(c+d x)}{5 a^3 d}-\frac{5 \sin ^3(c+d x)}{3 a^3 d}+\frac{\sin (c+d x)}{a^3 d}+\frac{4 i \cos ^5(c+d x)}{5 a^3 d}-\frac{i \cos ^3(c+d x)}{3 a^3 d} \]
Antiderivative was successfully verified.
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Rule 3092
Rule 3090
Rule 2633
Rule 2565
Rule 30
Rule 2564
Rule 14
Rubi steps
\begin{align*} \int \frac{\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx &=\frac{i \int \cos ^2(c+d x) (i a \cos (c+d x)+a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac{i \int \left (-i a^3 \cos ^5(c+d x)-3 a^3 \cos ^4(c+d x) \sin (c+d x)+3 i a^3 \cos ^3(c+d x) \sin ^2(c+d x)+a^3 \cos ^2(c+d x) \sin ^3(c+d x)\right ) \, dx}{a^6}\\ &=\frac{i \int \cos ^2(c+d x) \sin ^3(c+d x) \, dx}{a^3}-\frac{(3 i) \int \cos ^4(c+d x) \sin (c+d x) \, dx}{a^3}+\frac{\int \cos ^5(c+d x) \, dx}{a^3}-\frac{3 \int \cos ^3(c+d x) \sin ^2(c+d x) \, dx}{a^3}\\ &=-\frac{i \operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^3 d}+\frac{(3 i) \operatorname{Subst}\left (\int x^4 \, dx,x,\cos (c+d x)\right )}{a^3 d}-\frac{\operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{a^3 d}-\frac{3 \operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\sin (c+d x)\right )}{a^3 d}\\ &=\frac{3 i \cos ^5(c+d x)}{5 a^3 d}+\frac{\sin (c+d x)}{a^3 d}-\frac{2 \sin ^3(c+d x)}{3 a^3 d}+\frac{\sin ^5(c+d x)}{5 a^3 d}-\frac{i \operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (c+d x)\right )}{a^3 d}-\frac{3 \operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\sin (c+d x)\right )}{a^3 d}\\ &=-\frac{i \cos ^3(c+d x)}{3 a^3 d}+\frac{4 i \cos ^5(c+d x)}{5 a^3 d}+\frac{\sin (c+d x)}{a^3 d}-\frac{5 \sin ^3(c+d x)}{3 a^3 d}+\frac{4 \sin ^5(c+d x)}{5 a^3 d}\\ \end{align*}
Mathematica [A] time = 0.0787693, size = 111, normalized size = 1.23 \[ \frac{\sin (c+d x)}{4 a^3 d}+\frac{\sin (3 (c+d x))}{6 a^3 d}+\frac{\sin (5 (c+d x))}{20 a^3 d}+\frac{i \cos (c+d x)}{4 a^3 d}+\frac{i \cos (3 (c+d x))}{6 a^3 d}+\frac{i \cos (5 (c+d x))}{20 a^3 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.137, size = 90, normalized size = 1. \begin{align*} 2\,{\frac{1}{d{a}^{3}} \left ({\frac{-2\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{4}}}+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-1}+4/5\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-5}+{\frac{2\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{2}}}-8/3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-3} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.13629, size = 93, normalized size = 1.03 \begin{align*} \frac{3 i \, \cos \left (5 \, d x + 5 \, c\right ) + 10 i \, \cos \left (3 \, d x + 3 \, c\right ) + 15 i \, \cos \left (d x + c\right ) + 3 \, \sin \left (5 \, d x + 5 \, c\right ) + 10 \, \sin \left (3 \, d x + 3 \, c\right ) + 15 \, \sin \left (d x + c\right )}{60 \, a^{3} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.461456, size = 128, normalized size = 1.42 \begin{align*} \frac{{\left (15 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 10 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{60 \, a^{3} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 0.64325, size = 133, normalized size = 1.48 \begin{align*} \begin{cases} \frac{\left (120 i a^{6} d^{2} e^{8 i c} e^{- i d x} + 80 i a^{6} d^{2} e^{6 i c} e^{- 3 i d x} + 24 i a^{6} d^{2} e^{4 i c} e^{- 5 i d x}\right ) e^{- 9 i c}}{480 a^{9} d^{3}} & \text{for}\: 480 a^{9} d^{3} e^{9 i c} \neq 0 \\\frac{x \left (e^{4 i c} + 2 e^{2 i c} + 1\right ) e^{- 5 i c}}{4 a^{3}} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.19289, size = 99, normalized size = 1.1 \begin{align*} \frac{2 \,{\left (15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 30 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 20 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 7\right )}}{15 \, a^{3} d{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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