Integrand size = 15, antiderivative size = 116 \[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=\frac {x}{16 a^4}+\frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {i}{12 a d (a+i a \tan (c+d x))^3}+\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {i}{16 d \left (a^4+i a^4 \tan (c+d x)\right )} \] Output:
1/16*x/a^4+1/8*I/d/(a+I*a*tan(d*x+c))^4+1/12*I/a/d/(a+I*a*tan(d*x+c))^3+1/ 16*I/d/(a^2+I*a^2*tan(d*x+c))^2+1/16*I/d/(a^4+I*a^4*tan(d*x+c))
Time = 0.18 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=-\frac {i a \left (\frac {i \arctan (\tan (c+d x))}{16 a^5}-\frac {1}{8 a (a+i a \tan (c+d x))^4}-\frac {1}{12 a^2 (a+i a \tan (c+d x))^3}-\frac {1}{16 a^3 (a+i a \tan (c+d x))^2}-\frac {1}{16 a^4 (a+i a \tan (c+d x))}\right )}{d} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^(-4),x]
Output:
((-I)*a*(((I/16)*ArcTan[Tan[c + d*x]])/a^5 - 1/(8*a*(a + I*a*Tan[c + d*x]) ^4) - 1/(12*a^2*(a + I*a*Tan[c + d*x])^3) - 1/(16*a^3*(a + I*a*Tan[c + d*x ])^2) - 1/(16*a^4*(a + I*a*Tan[c + d*x]))))/d
Time = 0.47 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 3960, 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.
\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^4}dx\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\int \frac {1}{(i \tan (c+d x) a+a)^3}dx}{2 a}+\frac {i}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {1}{(i \tan (c+d x) a+a)^3}dx}{2 a}+\frac {i}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\int \frac {1}{(i \tan (c+d x) a+a)^2}dx}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}}{2 a}+\frac {i}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {1}{(i \tan (c+d x) a+a)^2}dx}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}}{2 a}+\frac {i}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\frac {\int \frac {1}{i \tan (c+d x) a+a}dx}{2 a}+\frac {i}{4 d (a+i a \tan (c+d x))^2}}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}}{2 a}+\frac {i}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\int \frac {1}{i \tan (c+d x) a+a}dx}{2 a}+\frac {i}{4 d (a+i a \tan (c+d x))^2}}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}}{2 a}+\frac {i}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\frac {\frac {\int 1dx}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))}}{2 a}+\frac {i}{4 d (a+i a \tan (c+d x))^2}}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}}{2 a}+\frac {i}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\frac {\frac {\frac {x}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))}}{2 a}+\frac {i}{4 d (a+i a \tan (c+d x))^2}}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}}{2 a}+\frac {i}{8 d (a+i a \tan (c+d x))^4}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^(-4),x]
Output:
(I/8)/(d*(a + I*a*Tan[c + d*x])^4) + ((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))/(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 = 0.38 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.69
method | result | size |
risch | \(\frac {x}{16 a^{4}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{4} d}+\frac {3 i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{4} d}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{24 a^{4} d}+\frac {i {\mathrm e}^{-8 i \left (d x +c \right )}}{128 a^{4} d}\) | \(80\) |
derivativedivides | \(\frac {i}{8 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{4}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{16 a^{4} d}-\frac {i}{16 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{2}}-\frac {1}{12 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{3}}+\frac {1}{16 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )}\) | \(95\) |
default | \(\frac {i}{8 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{4}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{16 a^{4} d}-\frac {i}{16 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{2}}-\frac {1}{12 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{3}}+\frac {1}{16 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )}\) | \(95\) |
norman | \(\frac {\frac {x}{16 a}+\frac {15 \tan \left (d x +c \right )}{16 a d}+\frac {5 \tan \left (d x +c \right )^{3}}{48 a d}+\frac {11 \tan \left (d x +c \right )^{5}}{48 a d}+\frac {\tan \left (d x +c \right )^{7}}{16 a d}+\frac {x \tan \left (d x +c \right )^{2}}{4 a}+\frac {3 x \tan \left (d x +c \right )^{4}}{8 a}+\frac {x \tan \left (d x +c \right )^{6}}{4 a}+\frac {x \tan \left (d x +c \right )^{8}}{16 a}+\frac {i}{3 a d}-\frac {2 i \tan \left (d x +c \right )^{2}}{3 a d}}{a^{3} \left (1+\tan \left (d x +c \right )^{2}\right )^{4}}\) | \(168\) |
Input:
int(1/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/16*x/a^4+1/8*I/a^4/d*exp(-2*I*(d*x+c))+3/32*I/a^4/d*exp(-4*I*(d*x+c))+1/ 24*I/a^4/d*exp(-6*I*(d*x+c))+1/128*I/a^4/d*exp(-8*I*(d*x+c))
Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.56 \[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (24 \, d x e^{\left (8 i \, d x + 8 i \, c\right )} + 48 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 36 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{384 \, a^{4} d} \] Input:
integrate(1/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
Output:
1/384*(24*d*x*e^(8*I*d*x + 8*I*c) + 48*I*e^(6*I*d*x + 6*I*c) + 36*I*e^(4*I *d*x + 4*I*c) + 16*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(-8*I*d*x - 8*I*c)/(a^4* d)
Time = 0.24 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.63 \[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {\left (98304 i a^{12} d^{3} e^{18 i c} e^{- 2 i d x} + 73728 i a^{12} d^{3} e^{16 i c} e^{- 4 i d x} + 32768 i a^{12} d^{3} e^{14 i c} e^{- 6 i d x} + 6144 i a^{12} d^{3} e^{12 i c} e^{- 8 i d x}\right ) e^{- 20 i c}}{786432 a^{16} d^{4}} & \text {for}\: a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (\frac {\left (e^{8 i c} + 4 e^{6 i c} + 6 e^{4 i c} + 4 e^{2 i c} + 1\right ) e^{- 8 i c}}{16 a^{4}} - \frac {1}{16 a^{4}}\right ) & \text {otherwise} \end {cases} + \frac {x}{16 a^{4}} \] Input:
integrate(1/(a+I*a*tan(d*x+c))**4,x)
Output:
Piecewise(((98304*I*a**12*d**3*exp(18*I*c)*exp(-2*I*d*x) + 73728*I*a**12*d **3*exp(16*I*c)*exp(-4*I*d*x) + 32768*I*a**12*d**3*exp(14*I*c)*exp(-6*I*d* x) + 6144*I*a**12*d**3*exp(12*I*c)*exp(-8*I*d*x))*exp(-20*I*c)/(786432*a** 16*d**4), Ne(a**16*d**4*exp(20*I*c), 0)), (x*((exp(8*I*c) + 4*exp(6*I*c) + 6*exp(4*I*c) + 4*exp(2*I*c) + 1)*exp(-8*I*c)/(16*a**4) - 1/(16*a**4)), Tr ue)) + x/(16*a**4)
Exception generated. \[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(1/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.72 \[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=\frac {i \, \log \left (\tan \left (d x + c\right ) + i\right )}{32 \, a^{4} d} - \frac {i \, \log \left (\tan \left (d x + c\right ) - i\right )}{32 \, a^{4} d} + \frac {3 \, \tan \left (d x + c\right )^{3} - 12 i \, \tan \left (d x + c\right )^{2} - 19 \, \tan \left (d x + c\right ) + 16 i}{48 \, a^{4} d {\left (\tan \left (d x + c\right ) - i\right )}^{4}} \] Input:
integrate(1/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
Output:
1/32*I*log(tan(d*x + c) + I)/(a^4*d) - 1/32*I*log(tan(d*x + c) - I)/(a^4*d ) + 1/48*(3*tan(d*x + c)^3 - 12*I*tan(d*x + c)^2 - 19*tan(d*x + c) + 16*I) /(a^4*d*(tan(d*x + c) - I)^4)
Time = 0.62 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.52 \[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=\frac {x}{16\,a^4}-\frac {-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{16}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{4}+\frac {19\,\mathrm {tan}\left (c+d\,x\right )}{48}-\frac {1}{3}{}\mathrm {i}}{a^4\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \] Input:
int(1/(a + a*tan(c + d*x)*1i)^4,x)
Output:
x/(16*a^4) - ((19*tan(c + d*x))/48 + (tan(c + d*x)^2*1i)/4 - tan(c + d*x)^ 3/16 - 1i/3)/(a^4*d*(tan(c + d*x)*1i + 1)^4)
\[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {1}{\tan \left (d x +c \right )^{4}-4 \tan \left (d x +c \right )^{3} i -6 \tan \left (d x +c \right )^{2}+4 \tan \left (d x +c \right ) i +1}d x}{a^{4}} \] Input:
int(1/(a+I*a*tan(d*x+c))^4,x)
Output:
int(1/(tan(c + d*x)**4 - 4*tan(c + d*x)**3*i - 6*tan(c + d*x)**2 + 4*tan(c + d*x)*i + 1),x)/a**4