Optimal. Leaf size=114 \[ \frac {i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac {\sin (e+f x) \cos ^5(e+f x)}{6 a^3 c^2 f}+\frac {5 \sin (e+f x) \cos ^3(e+f x)}{24 a^3 c^2 f}+\frac {5 \sin (e+f x) \cos (e+f x)}{16 a^3 c^2 f}+\frac {5 x}{16 a^3 c^2} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.13, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3522, 3486, 2635, 8} \[ \frac {i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac {\sin (e+f x) \cos ^5(e+f x)}{6 a^3 c^2 f}+\frac {5 \sin (e+f x) \cos ^3(e+f x)}{24 a^3 c^2 f}+\frac {5 \sin (e+f x) \cos (e+f x)}{16 a^3 c^2 f}+\frac {5 x}{16 a^3 c^2} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 8
Rule 2635
Rule 3486
Rule 3522
Rubi steps
\begin {align*} \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^2} \, dx &=\frac {\int \cos ^6(e+f x) (c-i c \tan (e+f x)) \, dx}{a^3 c^3}\\ &=\frac {i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac {\int \cos ^6(e+f x) \, dx}{a^3 c^2}\\ &=\frac {i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a^3 c^2 f}+\frac {5 \int \cos ^4(e+f x) \, dx}{6 a^3 c^2}\\ &=\frac {i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac {5 \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^2 f}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a^3 c^2 f}+\frac {5 \int \cos ^2(e+f x) \, dx}{8 a^3 c^2}\\ &=\frac {i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac {5 \cos (e+f x) \sin (e+f x)}{16 a^3 c^2 f}+\frac {5 \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^2 f}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a^3 c^2 f}+\frac {5 \int 1 \, dx}{16 a^3 c^2}\\ &=\frac {5 x}{16 a^3 c^2}+\frac {i \cos ^6(e+f x)}{6 a^3 c^2 f}+\frac {5 \cos (e+f x) \sin (e+f x)}{16 a^3 c^2 f}+\frac {5 \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^2 f}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a^3 c^2 f}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 1.35, size = 135, normalized size = 1.18 \[ \frac {\sec ^3(e+f x) (\cos (2 (e+f x))+i \sin (2 (e+f x))) (-120 f x \sin (e+f x)+60 i \sin (e+f x)+45 i \sin (3 (e+f x))+5 i \sin (5 (e+f x))+60 i (2 f x+i) \cos (e+f x)+15 \cos (3 (e+f x))+\cos (5 (e+f x)))}{384 a^3 c^2 f (\tan (e+f x)-i)^3} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.47, size = 79, normalized size = 0.69 \[ \frac {{\left (120 \, f x e^{\left (6 i \, f x + 6 i \, e\right )} - 3 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 30 i \, e^{\left (8 i \, f x + 8 i \, e\right )} + 60 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 15 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{384 \, a^{3} c^{2} f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 1.29, size = 137, normalized size = 1.20 \[ -\frac {-\frac {30 i \, \log \left (\tan \left (f x + e\right ) + i\right )}{a^{3} c^{2}} + \frac {30 i \, \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3} c^{2}} + \frac {3 \, {\left (-15 i \, \tan \left (f x + e\right )^{2} + 38 \, \tan \left (f x + e\right ) + 25 i\right )}}{a^{3} c^{2} {\left (-i \, \tan \left (f x + e\right ) + 1\right )}^{2}} - \frac {55 i \, \tan \left (f x + e\right )^{3} + 201 \, \tan \left (f x + e\right )^{2} - 255 i \, \tan \left (f x + e\right ) - 117}{a^{3} c^{2} {\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{192 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.25, size = 158, normalized size = 1.39 \[ \frac {i}{32 f \,a^{3} c^{2} \left (\tan \left (f x +e \right )+i\right )^{2}}+\frac {5 i \ln \left (\tan \left (f x +e \right )+i\right )}{32 f \,a^{3} c^{2}}+\frac {1}{8 f \,a^{3} c^{2} \left (\tan \left (f x +e \right )+i\right )}-\frac {5 i \ln \left (\tan \left (f x +e \right )-i\right )}{32 f \,a^{3} c^{2}}-\frac {3 i}{32 f \,a^{3} c^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {1}{24 f \,a^{3} c^{2} \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {3}{16 f \,a^{3} c^{2} \left (\tan \left (f x +e \right )-i\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 5.22, size = 88, normalized size = 0.77 \[ \frac {5\,x}{16\,a^3\,c^2}-\frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,5{}\mathrm {i}}{16}+\frac {5\,{\mathrm {tan}\left (e+f\,x\right )}^3}{16}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,25{}\mathrm {i}}{48}+\frac {25\,\mathrm {tan}\left (e+f\,x\right )}{48}+\frac {1}{6}{}\mathrm {i}}{a^3\,c^2\,f\,{\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,{\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}^2} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.58, size = 262, normalized size = 2.30 \[ \begin {cases} - \frac {\left (50331648 i a^{12} c^{8} f^{4} e^{16 i e} e^{4 i f x} + 503316480 i a^{12} c^{8} f^{4} e^{14 i e} e^{2 i f x} - 1006632960 i a^{12} c^{8} f^{4} e^{10 i e} e^{- 2 i f x} - 251658240 i a^{12} c^{8} f^{4} e^{8 i e} e^{- 4 i f x} - 33554432 i a^{12} c^{8} f^{4} e^{6 i e} e^{- 6 i f x}\right ) e^{- 12 i e}}{6442450944 a^{15} c^{10} f^{5}} & \text {for}\: 6442450944 a^{15} c^{10} f^{5} e^{12 i e} \neq 0 \\x \left (\frac {\left (e^{10 i e} + 5 e^{8 i e} + 10 e^{6 i e} + 10 e^{4 i e} + 5 e^{2 i e} + 1\right ) e^{- 6 i e}}{32 a^{3} c^{2}} - \frac {5}{16 a^{3} c^{2}}\right ) & \text {otherwise} \end {cases} + \frac {5 x}{16 a^{3} c^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________