Integrand size = 15, antiderivative size = 88 \[ \int \frac {1}{(a+i a \tan (c+d x))^3} \, dx=\frac {x}{8 a^3}+\frac {i}{6 d (a+i a \tan (c+d x))^3}+\frac {i}{8 a d (a+i a \tan (c+d x))^2}+\frac {i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \] Output:
1/8*x/a^3+1/6*I/d/(a+I*a*tan(d*x+c))^3+1/8*I/a/d/(a+I*a*tan(d*x+c))^2+1/8* I/d/(a^3+I*a^3*tan(d*x+c))
Time = 0.15 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(a+i a \tan (c+d x))^3} \, dx=\frac {-10-9 i \tan (c+d x)+3 \tan ^2(c+d x)+3 \arctan (\tan (c+d x)) (-i+\tan (c+d x))^3}{24 a^3 d (-i+\tan (c+d x))^3} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^(-3),x]
Output:
(-10 - (9*I)*Tan[c + d*x] + 3*Tan[c + d*x]^2 + 3*ArcTan[Tan[c + d*x]]*(-I + Tan[c + d*x])^3)/(24*a^3*d*(-I + Tan[c + d*x])^3)
Time = 0.36 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 3960, 3042, 3960, 3042, 3960, 24}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^3}dx\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\int \frac {1}{(i \tan (c+d x) a+a)^2}dx}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {1}{(i \tan (c+d x) a+a)^2}dx}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\int \frac {1}{i \tan (c+d x) a+a}dx}{2 a}+\frac {i}{4 d (a+i a \tan (c+d x))^2}}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {1}{i \tan (c+d x) a+a}dx}{2 a}+\frac {i}{4 d (a+i a \tan (c+d x))^2}}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\frac {\int 1dx}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))}}{2 a}+\frac {i}{4 d (a+i a \tan (c+d x))^2}}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\frac {\frac {x}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))}}{2 a}+\frac {i}{4 d (a+i a \tan (c+d x))^2}}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^(-3),x]
Output:
(I/6)/(d*(a + I*a*Tan[c + d*x])^3) + ((I/4)/(d*(a + I*a*Tan[c + d*x])^2) + (x/(2*a) + (I/2)/(d*(a + I*a*Tan[c + d*x])))/(2*a))/(2*a)
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] + Simp[1/(2*a) Int[(a + b*Tan[c + d*x])^ (n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, 0]
Time = 0.33 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.70
method | result | size |
risch | \(\frac {x}{8 a^{3}}+\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{3} d}+\frac {3 i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{3} d}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{3} d}\) | \(62\) |
derivativedivides | \(\frac {\arctan \left (\tan \left (d x +c \right )\right )}{8 a^{3} d}-\frac {i}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}-\frac {1}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}+\frac {1}{8 a^{3} d \left (-i+\tan \left (d x +c \right )\right )}\) | \(75\) |
default | \(\frac {\arctan \left (\tan \left (d x +c \right )\right )}{8 a^{3} d}-\frac {i}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}-\frac {1}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}+\frac {1}{8 a^{3} d \left (-i+\tan \left (d x +c \right )\right )}\) | \(75\) |
norman | \(\frac {\frac {x}{8 a}+\frac {7 \tan \left (d x +c \right )}{8 a d}+\frac {\tan \left (d x +c \right )^{3}}{3 a d}+\frac {\tan \left (d x +c \right )^{5}}{8 a d}+\frac {3 x \tan \left (d x +c \right )^{2}}{8 a}+\frac {3 x \tan \left (d x +c \right )^{4}}{8 a}+\frac {x \tan \left (d x +c \right )^{6}}{8 a}+\frac {5 i}{12 a d}-\frac {i \tan \left (d x +c \right )^{2}}{4 a d}}{a^{2} \left (1+\tan \left (d x +c \right )^{2}\right )^{3}}\) | \(138\) |
Input:
int(1/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/8*x/a^3+3/16*I/a^3/d*exp(-2*I*(d*x+c))+3/32*I/a^3/d*exp(-4*I*(d*x+c))+1/ 48*I/a^3/d*exp(-6*I*(d*x+c))
Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (12 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 18 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 9 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \] Input:
integrate(1/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/96*(12*d*x*e^(6*I*d*x + 6*I*c) + 18*I*e^(4*I*d*x + 4*I*c) + 9*I*e^(2*I*d *x + 2*I*c) + 2*I)*e^(-6*I*d*x - 6*I*c)/(a^3*d)
Time = 0.20 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.76 \[ \int \frac {1}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (4608 i a^{6} d^{2} e^{10 i c} e^{- 2 i d x} + 2304 i a^{6} d^{2} e^{8 i c} e^{- 4 i d x} + 512 i a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (e^{6 i c} + 3 e^{4 i c} + 3 e^{2 i c} + 1\right ) e^{- 6 i c}}{8 a^{3}} - \frac {1}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x}{8 a^{3}} \] Input:
integrate(1/(a+I*a*tan(d*x+c))**3,x)
Output:
Piecewise(((4608*I*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) + 2304*I*a**6*d**2* exp(8*I*c)*exp(-4*I*d*x) + 512*I*a**6*d**2*exp(6*I*c)*exp(-6*I*d*x))*exp(- 12*I*c)/(24576*a**9*d**3), Ne(a**9*d**3*exp(12*I*c), 0)), (x*((exp(6*I*c) + 3*exp(4*I*c) + 3*exp(2*I*c) + 1)*exp(-6*I*c)/(8*a**3) - 1/(8*a**3)), Tru e)) + x/(8*a**3)
Exception generated. \[ \int \frac {1}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(1/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \, \log \left (\tan \left (d x + c\right ) + i\right )}{16 \, a^{3} d} - \frac {i \, \log \left (\tan \left (d x + c\right ) - i\right )}{16 \, a^{3} d} - \frac {i \, {\left (3 i \, \tan \left (d x + c\right )^{2} + 9 \, \tan \left (d x + c\right ) - 10 i\right )}}{24 \, a^{3} d {\left (\tan \left (d x + c\right ) - i\right )}^{3}} \] Input:
integrate(1/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/16*I*log(tan(d*x + c) + I)/(a^3*d) - 1/16*I*log(tan(d*x + c) - I)/(a^3*d ) - 1/24*I*(3*I*tan(d*x + c)^2 + 9*tan(d*x + c) - 10*I)/(a^3*d*(tan(d*x + c) - I)^3)
Time = 0.53 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.57 \[ \int \frac {1}{(a+i a \tan (c+d x))^3} \, dx=\frac {x}{8\,a^3}-\frac {\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{8}+\frac {3\,\mathrm {tan}\left (c+d\,x\right )}{8}-\frac {5}{12}{}\mathrm {i}}{a^3\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \] Input:
int(1/(a + a*tan(c + d*x)*1i)^3,x)
Output:
x/(8*a^3) - ((3*tan(c + d*x))/8 + (tan(c + d*x)^2*1i)/8 - 5i/12)/(a^3*d*(t an(c + d*x)*1i + 1)^3)
\[ \int \frac {1}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\int \frac {1}{\tan \left (d x +c \right )^{3} i +3 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x}{a^{3}} \] Input:
int(1/(a+I*a*tan(d*x+c))^3,x)
Output:
( - int(1/(tan(c + d*x)**3*i + 3*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x ))/a**3