Integrand size = 22, antiderivative size = 110 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {i x}{16 a^4}-\frac {1}{8 d (a+i a \tan (c+d x))^4}+\frac {1}{12 a d (a+i a \tan (c+d x))^3}+\frac {1}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {1}{16 d \left (a^4+i a^4 \tan (c+d x)\right )} \] Output:
-1/16*I*x/a^4-1/8/d/(a+I*a*tan(d*x+c))^4+1/12/a/d/(a+I*a*tan(d*x+c))^3+1/1 6/d/(a^2+I*a^2*tan(d*x+c))^2+1/16/d/(a^4+I*a^4*tan(d*x+c))
Time = 0.10 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.85 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sec ^4(c+d x) (16 \cos (2 (c+d x))+(-3-24 i d x) \cos (4 (c+d x))+32 i \sin (2 (c+d x))+3 i \sin (4 (c+d x))+24 d x \sin (4 (c+d x)))}{384 a^4 d (-i+\tan (c+d x))^4} \] Input:
Integrate[Tan[c + d*x]/(a + I*a*Tan[c + d*x])^4,x]
Output:
(Sec[c + d*x]^4*(16*Cos[2*(c + d*x)] + (-3 - (24*I)*d*x)*Cos[4*(c + d*x)] + (32*I)*Sin[2*(c + d*x)] + (3*I)*Sin[4*(c + d*x)] + 24*d*x*Sin[4*(c + d*x )]))/(384*a^4*d*(-I + Tan[c + d*x])^4)
Time = 0.51 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.17, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {3042, 4009, 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.
\(\displaystyle \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4}dx\) |
\(\Big \downarrow \) 4009 |
\(\displaystyle -\frac {i \int \frac {1}{(i \tan (c+d x) a+a)^3}dx}{2 a}-\frac {1}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {i \int \frac {1}{(i \tan (c+d x) a+a)^3}dx}{2 a}-\frac {1}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle -\frac {i \left (\frac {\int \frac {1}{(i \tan (c+d x) a+a)^2}dx}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}\right )}{2 a}-\frac {1}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {i \left (\frac {\int \frac {1}{(i \tan (c+d x) a+a)^2}dx}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}\right )}{2 a}-\frac {1}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle -\frac {i \left (\frac {\frac {\int \frac {1}{i \tan (c+d x) a+a}dx}{2 a}+\frac {i}{4 d (a+i a \tan (c+d x))^2}}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}\right )}{2 a}-\frac {1}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {i \left (\frac {\frac {\int \frac {1}{i \tan (c+d x) a+a}dx}{2 a}+\frac {i}{4 d (a+i a \tan (c+d x))^2}}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}\right )}{2 a}-\frac {1}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle -\frac {i \left (\frac {\frac {\frac {\int 1dx}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))}}{2 a}+\frac {i}{4 d (a+i a \tan (c+d x))^2}}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}\right )}{2 a}-\frac {1}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle -\frac {i \left (\frac {\frac {\frac {x}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))}}{2 a}+\frac {i}{4 d (a+i a \tan (c+d x))^2}}{2 a}+\frac {i}{6 d (a+i a \tan (c+d x))^3}\right )}{2 a}-\frac {1}{8 d (a+i a \tan (c+d x))^4}\) |
Input:
Int[Tan[c + d*x]/(a + I*a*Tan[c + d*x])^4,x]
Output:
-1/8*1/(d*(a + I*a*Tan[c + d*x])^4) - ((I/2)*((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)))/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]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*((a + b*Tan[e + f*x])^m/(2*a *f*m)), x] + Simp[(b*c + a*d)/(2*a*b) Int[(a + b*Tan[e + f*x])^(m + 1), x ], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2 , 0] && LtQ[m, 0]
Time = 0.76 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.55
method | result | size |
risch | \(-\frac {i x}{16 a^{4}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{4} d}-\frac {{\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{4} d}-\frac {{\mathrm e}^{-8 i \left (d x +c \right )}}{128 a^{4} d}\) | \(60\) |
derivativedivides | \(-\frac {i \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}+\frac {i}{12 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {i}{16 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )}-\frac {1}{8 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{4}}-\frac {1}{16 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{2}}\) | \(96\) |
default | \(-\frac {i \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}+\frac {i}{12 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {i}{16 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )}-\frac {1}{8 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{4}}-\frac {1}{16 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{2}}\) | \(96\) |
norman | \(\frac {-\frac {i x}{16 a}+\frac {1}{12 a d}-\frac {\tan \left (d x +c \right )^{4}}{4 a d}+\frac {i \tan \left (d x +c \right )}{16 d a}-\frac {53 i \tan \left (d x +c \right )^{3}}{48 d a}-\frac {11 i \tan \left (d x +c \right )^{5}}{48 a d}-\frac {i \tan \left (d x +c \right )^{7}}{16 a d}-\frac {i x \tan \left (d x +c \right )^{2}}{4 a}-\frac {3 i x \tan \left (d x +c \right )^{4}}{8 a}-\frac {i x \tan \left (d x +c \right )^{6}}{4 a}-\frac {i x \tan \left (d x +c \right )^{8}}{16 a}+\frac {5 \tan \left (d x +c \right )^{2}}{6 a d}}{\left (1+\tan \left (d x +c \right )^{2}\right )^{4} a^{3}}\) | \(191\) |
Input:
int(tan(d*x+c)/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
-1/16*I*x/a^4+1/16/a^4/d*exp(-2*I*(d*x+c))-1/48/a^4/d*exp(-6*I*(d*x+c))-1/ 128/a^4/d*exp(-8*I*(d*x+c))
Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.49 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (-24 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} + 24 \, e^{\left (6 i \, d x + 6 i \, c\right )} - 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{384 \, a^{4} d} \] Input:
integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
Output:
1/384*(-24*I*d*x*e^(8*I*d*x + 8*I*c) + 24*e^(6*I*d*x + 6*I*c) - 8*e^(2*I*d *x + 2*I*c) - 3)*e^(-8*I*d*x - 8*I*c)/(a^4*d)
Time = 0.27 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.42 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {\left (6144 a^{8} d^{2} e^{14 i c} e^{- 2 i d x} - 2048 a^{8} d^{2} e^{10 i c} e^{- 6 i d x} - 768 a^{8} d^{2} e^{8 i c} e^{- 8 i d x}\right ) e^{- 16 i c}}{98304 a^{12} d^{3}} & \text {for}\: a^{12} d^{3} e^{16 i c} \neq 0 \\x \left (\frac {\left (- i e^{8 i c} - 2 i e^{6 i c} + 2 i e^{2 i c} + i\right ) e^{- 8 i c}}{16 a^{4}} + \frac {i}{16 a^{4}}\right ) & \text {otherwise} \end {cases} - \frac {i x}{16 a^{4}} \] Input:
integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))**4,x)
Output:
Piecewise(((6144*a**8*d**2*exp(14*I*c)*exp(-2*I*d*x) - 2048*a**8*d**2*exp( 10*I*c)*exp(-6*I*d*x) - 768*a**8*d**2*exp(8*I*c)*exp(-8*I*d*x))*exp(-16*I* c)/(98304*a**12*d**3), Ne(a**12*d**3*exp(16*I*c), 0)), (x*((-I*exp(8*I*c) - 2*I*exp(6*I*c) + 2*I*exp(2*I*c) + I)*exp(-8*I*c)/(16*a**4) + I/(16*a**4) ), True)) - I*x/(16*a**4)
Exception generated. \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(tan(d*x+c)/(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 = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.75 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\log \left (\tan \left (d x + c\right ) + i\right )}{32 \, a^{4} d} - \frac {\log \left (\tan \left (d x + c\right ) - i\right )}{32 \, a^{4} d} - \frac {3 i \, \tan \left (d x + c\right )^{3} + 12 \, \tan \left (d x + c\right )^{2} - 19 i \, \tan \left (d x + c\right ) - 4}{48 \, a^{4} d {\left (\tan \left (d x + c\right ) - i\right )}^{4}} \] Input:
integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
Output:
1/32*log(tan(d*x + c) + I)/(a^4*d) - 1/32*log(tan(d*x + c) - I)/(a^4*d) - 1/48*(3*I*tan(d*x + c)^3 + 12*tan(d*x + c)^2 - 19*I*tan(d*x + c) - 4)/(a^4 *d*(tan(d*x + c) - I)^4)
Time = 1.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.55 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {x\,1{}\mathrm {i}}{16\,a^4}+\frac {-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}}{16}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{4}+\frac {\mathrm {tan}\left (c+d\,x\right )\,19{}\mathrm {i}}{48}+\frac {1}{12}}{a^4\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \] Input:
int(tan(c + d*x)/(a + a*tan(c + d*x)*1i)^4,x)
Output:
((tan(c + d*x)*19i)/48 - tan(c + d*x)^2/4 - (tan(c + d*x)^3*1i)/16 + 1/12) /(a^4*d*(tan(c + d*x)*1i + 1)^4) - (x*1i)/(16*a^4)
\[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {\tan \left (d x +c \right )}{\tan \left (d x +c \right )^{4}-4 \tan \left (d x +c \right )^{3} i -6 \tan \left (d x +c \right )^{2}+4 \tan \left (d x +c \right ) i +1}d x}{a^{4}} \] Input:
int(tan(d*x+c)/(a+I*a*tan(d*x+c))^4,x)
Output:
int(tan(c + d*x)/(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