\(\int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx\) [42]

Optimal result

Integrand size = 24, antiderivative size = 134 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=8 i a^4 x+\frac {4 i a^4 \cot (c+d x)}{d}+\frac {8 a^4 \log (\sin (c+d x))}{d}-\frac {i a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}+\frac {\cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d} \] Output:

8*I*a^4*x+4*I*a^4*cot(d*x+c)/d+8*a^4*ln(sin(d*x+c))/d-1/3*I*a*cot(d*x+c)^3 
*(a+I*a*tan(d*x+c))^3/d-1/4*cot(d*x+c)^4*(a+I*a*tan(d*x+c))^4/d+cot(d*x+c) 
^2*(a^2+I*a^2*tan(d*x+c))^2/d
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.69 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=a^4 \left (\frac {8 i \cot (c+d x)}{d}+\frac {7 \cot ^2(c+d x)}{2 d}-\frac {4 i \cot ^3(c+d x)}{3 d}-\frac {\cot ^4(c+d x)}{4 d}+\frac {8 \log (\tan (c+d x))}{d}-\frac {8 \log (i+\tan (c+d x))}{d}\right ) \] Input:

Integrate[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^4,x]
 

Output:

a^4*(((8*I)*Cot[c + d*x])/d + (7*Cot[c + d*x]^2)/(2*d) - (((4*I)/3)*Cot[c 
+ d*x]^3)/d - Cot[c + d*x]^4/(4*d) + (8*Log[Tan[c + d*x]])/d - (8*Log[I + 
Tan[c + d*x]])/d)
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.13, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {3042, 4031, 3042, 4028, 3042, 4028, 3042, 4025, 27, 3042, 4014, 3042, 25, 3956}

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.

\(\displaystyle \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^4}{\tan (c+d x)^5}dx\)

\(\Big \downarrow \) 4031

\(\displaystyle i \int \cot ^4(c+d x) (i \tan (c+d x) a+a)^4dx-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle i \int \frac {(i \tan (c+d x) a+a)^4}{\tan (c+d x)^4}dx-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}\)

\(\Big \downarrow \) 4028

\(\displaystyle i \left (2 i a \int \cot ^3(c+d x) (i \tan (c+d x) a+a)^3dx-\frac {a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\right )-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle i \left (2 i a \int \frac {(i \tan (c+d x) a+a)^3}{\tan (c+d x)^3}dx-\frac {a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\right )-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}\)

\(\Big \downarrow \) 4028

\(\displaystyle i \left (2 i a \left (2 i a \int \cot ^2(c+d x) (i \tan (c+d x) a+a)^2dx-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\right )-\frac {a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\right )-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle i \left (2 i a \left (2 i a \int \frac {(i \tan (c+d x) a+a)^2}{\tan (c+d x)^2}dx-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\right )-\frac {a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\right )-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}\)

\(\Big \downarrow \) 4025

\(\displaystyle i \left (2 i a \left (2 i a \left (-\frac {a^2 \cot (c+d x)}{d}+\int 2 \cot (c+d x) \left (i a^2-a^2 \tan (c+d x)\right )dx\right )-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\right )-\frac {a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\right )-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle i \left (2 i a \left (2 i a \left (-\frac {a^2 \cot (c+d x)}{d}+2 \int \cot (c+d x) \left (i a^2-a^2 \tan (c+d x)\right )dx\right )-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\right )-\frac {a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\right )-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle i \left (2 i a \left (2 i a \left (-\frac {a^2 \cot (c+d x)}{d}+2 \int \frac {i a^2-a^2 \tan (c+d x)}{\tan (c+d x)}dx\right )-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\right )-\frac {a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\right )-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}\)

\(\Big \downarrow \) 4014

\(\displaystyle i \left (2 i a \left (2 i a \left (-\frac {a^2 \cot (c+d x)}{d}+2 \left (-a^2 x+i a^2 \int \cot (c+d x)dx\right )\right )-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\right )-\frac {a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\right )-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle i \left (2 i a \left (2 i a \left (-\frac {a^2 \cot (c+d x)}{d}+2 \left (-a^2 x+i a^2 \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\right )-\frac {a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\right )-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}\)

\(\Big \downarrow \) 25

\(\displaystyle i \left (2 i a \left (2 i a \left (-\frac {a^2 \cot (c+d x)}{d}+2 \left (a^2 (-x)-i a^2 \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx\right )\right )-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\right )-\frac {a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\right )-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}\)

\(\Big \downarrow \) 3956

\(\displaystyle i \left (2 i a \left (2 i a \left (-\frac {a^2 \cot (c+d x)}{d}+2 \left (-a^2 x+\frac {i a^2 \log (-\sin (c+d x))}{d}\right )\right )-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\right )-\frac {a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\right )-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}\)

Input:

Int[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^4,x]
 

Output:

-1/4*(Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^4)/d + I*(-1/3*(a*Cot[c + d*x] 
^3*(a + I*a*Tan[c + d*x])^3)/d + (2*I)*a*((2*I)*a*(-((a^2*Cot[c + d*x])/d) 
 + 2*(-(a^2*x) + (I*a^2*Log[-Sin[c + d*x]])/d)) - (a*Cot[c + d*x]^2*(a + I 
*a*Tan[c + d*x])^2)/(2*d)))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3956

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d 
*x], x]]/d, x] /; FreeQ[{c, d}, x]
 

rule 4014

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a 
*d)/(a^2 + b^2)   Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] /; 
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && N 
eQ[a*c + b*d, 0]
 

rule 4025

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*c - a*d)^2*((a + b*Tan[e + f*x])^(m + 
 1)/(b*f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2)   Int[(a + b*Tan[e 
+ f*x])^(m + 1)*Simp[a*c^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Ta 
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] 
 && LtQ[m, -1] && NeQ[a^2 + b^2, 0]
 

rule 4028

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*b*(a + b*Tan[e + f*x])^(m - 1)*((c + 
 d*Tan[e + f*x])^(n + 1)/(f*(m - 1)*(a*c - b*d))), x] + Simp[2*(a^2/(a*c - 
b*d))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), x], 
x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0 
] && NeQ[c^2 + d^2, 0] && EqQ[m + n, 0] && GtQ[m, 1/2]
 

rule 4031

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-d)*(a + b*Tan[e + f*x])^m*((c + d*Ta 
n[e + f*x])^(n + 1)/(f*m*(c^2 + d^2))), x] + Simp[a/(a*c - b*d)   Int[(a + 
b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d 
, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^ 
2, 0] && EqQ[m + n + 1, 0] &&  !LtQ[m, -1]
 

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.56

method	result	size
parallelrisch	\(-\frac {a^{4} \left (3 \cot \left (d x +c \right )^{4}+16 i \cot \left (d x +c \right )^{3}-96 i d x -42 \cot \left (d x +c \right )^{2}-96 i \cot \left (d x +c \right )-96 \ln \left (\tan \left (d x +c \right )\right )+48 \ln \left (\sec \left (d x +c \right )^{2}\right )\right )}{12 d}\)	\(75\)
derivativedivides	\(\frac {a^{4} \left (8 i \cot \left (d x +c \right )-\frac {\cot \left (d x +c \right )^{4}}{4}-\frac {4 i \cot \left (d x +c \right )^{3}}{3}+\frac {7 \cot \left (d x +c \right )^{2}}{2}-4 \ln \left (\cot \left (d x +c \right )^{2}+1\right )-8 i \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )\right )}{d}\)	\(78\)
default	\(\frac {a^{4} \left (8 i \cot \left (d x +c \right )-\frac {\cot \left (d x +c \right )^{4}}{4}-\frac {4 i \cot \left (d x +c \right )^{3}}{3}+\frac {7 \cot \left (d x +c \right )^{2}}{2}-4 \ln \left (\cot \left (d x +c \right )^{2}+1\right )-8 i \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )\right )}{d}\)	\(78\)
risch	\(-\frac {16 i a^{4} c}{d}-\frac {4 a^{4} \left (30 \,{\mathrm e}^{6 i \left (d x +c \right )}-63 \,{\mathrm e}^{4 i \left (d x +c \right )}+50 \,{\mathrm e}^{2 i \left (d x +c \right )}-14\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {8 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(88\)
norman	\(\frac {-\frac {a^{4}}{4 d}+\frac {7 a^{4} \tan \left (d x +c \right )^{2}}{2 d}-\frac {4 i a^{4} \tan \left (d x +c \right )}{3 d}+\frac {8 i a^{4} \tan \left (d x +c \right )^{3}}{d}+8 i a^{4} x \tan \left (d x +c \right )^{4}}{\tan \left (d x +c \right )^{4}}+\frac {8 a^{4} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {4 a^{4} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}\)	\(117\)

Input:

int(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
 

Output:

-1/12*a^4*(3*cot(d*x+c)^4+16*I*cot(d*x+c)^3-96*I*d*x-42*cot(d*x+c)^2-96*I* 
cot(d*x+c)-96*ln(tan(d*x+c))+48*ln(sec(d*x+c)^2))/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.30 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {4 \, {\left (30 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 63 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 50 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 14 \, a^{4} - 6 \, {\left (a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
 

Output:

-4/3*(30*a^4*e^(6*I*d*x + 6*I*c) - 63*a^4*e^(4*I*d*x + 4*I*c) + 50*a^4*e^( 
2*I*d*x + 2*I*c) - 14*a^4 - 6*(a^4*e^(8*I*d*x + 8*I*c) - 4*a^4*e^(6*I*d*x 
+ 6*I*c) + 6*a^4*e^(4*I*d*x + 4*I*c) - 4*a^4*e^(2*I*d*x + 2*I*c) + a^4)*lo 
g(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I* 
c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)
 

Sympy [A] (verification not implemented)

Time = 1.50 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.25 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {8 a^{4} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 120 a^{4} e^{6 i c} e^{6 i d x} + 252 a^{4} e^{4 i c} e^{4 i d x} - 200 a^{4} e^{2 i c} e^{2 i d x} + 56 a^{4}}{3 d e^{8 i c} e^{8 i d x} - 12 d e^{6 i c} e^{6 i d x} + 18 d e^{4 i c} e^{4 i d x} - 12 d e^{2 i c} e^{2 i d x} + 3 d} \] Input:

integrate(cot(d*x+c)**5*(a+I*a*tan(d*x+c))**4,x)
 

Output:

8*a**4*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-120*a**4*exp(6*I*c)*exp(6*I*d 
*x) + 252*a**4*exp(4*I*c)*exp(4*I*d*x) - 200*a**4*exp(2*I*c)*exp(2*I*d*x) 
+ 56*a**4)/(3*d*exp(8*I*c)*exp(8*I*d*x) - 12*d*exp(6*I*c)*exp(6*I*d*x) + 1 
8*d*exp(4*I*c)*exp(4*I*d*x) - 12*d*exp(2*I*c)*exp(2*I*d*x) + 3*d)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.72 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {-96 i \, {\left (d x + c\right )} a^{4} + 48 \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 96 \, a^{4} \log \left (\tan \left (d x + c\right )\right ) + \frac {-96 i \, a^{4} \tan \left (d x + c\right )^{3} - 42 \, a^{4} \tan \left (d x + c\right )^{2} + 16 i \, a^{4} \tan \left (d x + c\right ) + 3 \, a^{4}}{\tan \left (d x + c\right )^{4}}}{12 \, d} \] Input:

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")
 

Output:

-1/12*(-96*I*(d*x + c)*a^4 + 48*a^4*log(tan(d*x + c)^2 + 1) - 96*a^4*log(t 
an(d*x + c)) + (-96*I*a^4*tan(d*x + c)^3 - 42*a^4*tan(d*x + c)^2 + 16*I*a^ 
4*tan(d*x + c) + 3*a^4)/tan(d*x + c)^4)/d
 

Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.67 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {8 \, a^{4} \log \left (\tan \left (d x + c\right ) + i\right )}{d} + \frac {8 \, a^{4} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {-96 i \, a^{4} \tan \left (d x + c\right )^{3} - 42 \, a^{4} \tan \left (d x + c\right )^{2} + 16 i \, a^{4} \tan \left (d x + c\right ) + 3 \, a^{4}}{12 \, d \tan \left (d x + c\right )^{4}} \] Input:

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
 

Output:

-8*a^4*log(tan(d*x + c) + I)/d + 8*a^4*log(abs(tan(d*x + c)))/d - 1/12*(-9 
6*I*a^4*tan(d*x + c)^3 - 42*a^4*tan(d*x + c)^2 + 16*I*a^4*tan(d*x + c) + 3 
*a^4)/(d*tan(d*x + c)^4)
 

Mupad [B] (verification not implemented)

Time = 1.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.60 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^4\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,16{}\mathrm {i}}{d}-\frac {-a^4\,{\mathrm {tan}\left (c+d\,x\right )}^3\,8{}\mathrm {i}-\frac {7\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {a^4\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{3}+\frac {a^4}{4}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^4} \] Input:

int(cot(c + d*x)^5*(a + a*tan(c + d*x)*1i)^4,x)
 

Output:

(a^4*atan(2*tan(c + d*x) + 1i)*16i)/d - ((a^4*tan(c + d*x)*4i)/3 + a^4/4 - 
 (7*a^4*tan(c + d*x)^2)/2 - a^4*tan(c + d*x)^3*8i)/(d*tan(c + d*x)^4)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.95 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^{4} \left (896 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} i -128 \cos \left (d x +c \right ) \sin \left (d x +c \right ) i -768 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{4}+768 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4}+768 \sin \left (d x +c \right )^{4} d i x -183 \sin \left (d x +c \right )^{4}+384 \sin \left (d x +c \right )^{2}-24\right )}{96 \sin \left (d x +c \right )^{4} d} \] Input:

int(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4,x)
 

Output:

(a**4*(896*cos(c + d*x)*sin(c + d*x)**3*i - 128*cos(c + d*x)*sin(c + d*x)* 
i - 768*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4 + 768*log(tan((c + d* 
x)/2))*sin(c + d*x)**4 + 768*sin(c + d*x)**4*d*i*x - 183*sin(c + d*x)**4 + 
 384*sin(c + d*x)**2 - 24))/(96*sin(c + d*x)**4*d)