Optimal. Leaf size=68 \[ \frac {2 \sin ^5(c+d x)}{5 a^2 d}-\frac {\sin ^3(c+d x)}{a^2 d}+\frac {\sin (c+d x)}{a^2 d}+\frac {2 i \cos ^5(c+d x)}{5 a^2 d} \]
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Rubi [A] time = 0.18, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ \frac {2 \sin ^5(c+d x)}{5 a^2 d}-\frac {\sin ^3(c+d x)}{a^2 d}+\frac {\sin (c+d x)}{a^2 d}+\frac {2 i \cos ^5(c+d x)}{5 a^2 d} \]
Antiderivative was successfully verified.
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Rule 14
Rule 30
Rule 2564
Rule 2565
Rule 2633
Rule 3090
Rule 3092
Rubi steps
\begin {align*} \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx &=-\frac {\int \cos ^3(c+d x) (i a \cos (c+d x)+a \sin (c+d x))^2 \, dx}{a^4}\\ &=-\frac {\int \left (-a^2 \cos ^5(c+d x)+2 i a^2 \cos ^4(c+d x) \sin (c+d x)+a^2 \cos ^3(c+d x) \sin ^2(c+d x)\right ) \, dx}{a^4}\\ &=-\frac {(2 i) \int \cos ^4(c+d x) \sin (c+d x) \, dx}{a^2}+\frac {\int \cos ^5(c+d x) \, dx}{a^2}-\frac {\int \cos ^3(c+d x) \sin ^2(c+d x) \, dx}{a^2}\\ &=\frac {(2 i) \operatorname {Subst}\left (\int x^4 \, dx,x,\cos (c+d x)\right )}{a^2 d}-\frac {\operatorname {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\sin (c+d x)\right )}{a^2 d}-\frac {\operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d}\\ &=\frac {2 i \cos ^5(c+d x)}{5 a^2 d}+\frac {\sin (c+d x)}{a^2 d}-\frac {2 \sin ^3(c+d x)}{3 a^2 d}+\frac {\sin ^5(c+d x)}{5 a^2 d}-\frac {\operatorname {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\sin (c+d x)\right )}{a^2 d}\\ &=\frac {2 i \cos ^5(c+d x)}{5 a^2 d}+\frac {\sin (c+d x)}{a^2 d}-\frac {\sin ^3(c+d x)}{a^2 d}+\frac {2 \sin ^5(c+d x)}{5 a^2 d}\\ \end {align*}
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Mathematica [A] time = 0.08, size = 111, normalized size = 1.63 \[ \frac {\sin (c+d x)}{2 a^2 d}+\frac {\sin (3 (c+d x))}{8 a^2 d}+\frac {\sin (5 (c+d x))}{40 a^2 d}+\frac {i \cos (c+d x)}{4 a^2 d}+\frac {i \cos (3 (c+d x))}{8 a^2 d}+\frac {i \cos (5 (c+d x))}{40 a^2 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.42, size = 52, normalized size = 0.76 \[ \frac {{\left (-5 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 15 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{40 \, a^{2} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 93, normalized size = 1.37 \[ \frac {\frac {5}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right )}} + \frac {35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 90 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 70 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{5}}}{20 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 108, normalized size = 1.59 \[ \frac {\frac {2}{8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 i}-\frac {2 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{4}}+\frac {5 i}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}+\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{5}}-\frac {3}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}+\frac {7}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}}{d \,a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.02, size = 90, normalized size = 1.32 \[ -\frac {2\,\left (-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,10{}\mathrm {i}+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2{}\mathrm {i}\right )}{5\,a^2\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )}^5\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.41, size = 165, normalized size = 2.43 \[ \begin {cases} \frac {\left (- 2560 i a^{6} d^{3} e^{10 i c} e^{i d x} + 7680 i a^{6} d^{3} e^{8 i c} e^{- i d x} + 2560 i a^{6} d^{3} e^{6 i c} e^{- 3 i d x} + 512 i a^{6} d^{3} e^{4 i c} e^{- 5 i d x}\right ) e^{- 9 i c}}{20480 a^{8} d^{4}} & \text {for}\: 20480 a^{8} d^{4} e^{9 i c} \neq 0 \\\frac {x \left (e^{6 i c} + 3 e^{4 i c} + 3 e^{2 i c} + 1\right ) e^{- 5 i c}}{8 a^{2}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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