\(\int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [23]

Optimal result

Integrand size = 34, antiderivative size = 134 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=4 a^3 (A-i B) x+\frac {a^3 (17 A-15 i B) \cot (c+d x)}{6 d}-\frac {4 a^3 (i A+B) \log (\sin (c+d x))}{d}-\frac {a A \cot ^3(c+d x) (a+i a \tan (c+d x))^2}{3 d}-\frac {(5 i A+3 B) \cot ^2(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d} \] Output:

4*a^3*(A-I*B)*x+1/6*a^3*(17*A-15*I*B)*cot(d*x+c)/d-4*a^3*(I*A+B)*ln(sin(d* 
x+c))/d-1/3*a*A*cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2/d-1/6*(5*I*A+3*B)*cot(d* 
x+c)^2*(a^3+I*a^3*tan(d*x+c))/d
 

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.66 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {a^3 \left (6 (4 A-3 i B) \cot (c+d x)+(-9 i A-3 B) \cot ^2(c+d x)-2 A \cot ^3(c+d x)-24 i (A-i B) (\log (\tan (c+d x))-\log (i+\tan (c+d x)))\right )}{6 d} \] Input:

Integrate[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]
 

Output:

(a^3*(6*(4*A - (3*I)*B)*Cot[c + d*x] + ((-9*I)*A - 3*B)*Cot[c + d*x]^2 - 2 
*A*Cot[c + d*x]^3 - (24*I)*(A - I*B)*(Log[Tan[c + d*x]] - Log[I + Tan[c + 
d*x]])))/(6*d)
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.382, Rules used = {3042, 4076, 3042, 4076, 25, 3042, 4074, 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.

\(\displaystyle \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan (c+d x)^4}dx\)

\(\Big \downarrow \) 4076

\(\displaystyle \frac {1}{3} \int \cot ^3(c+d x) (i \tan (c+d x) a+a)^2 (a (5 i A+3 B)-a (A-3 i B) \tan (c+d x))dx-\frac {a A \cot ^3(c+d x) (a+i a \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {(i \tan (c+d x) a+a)^2 (a (5 i A+3 B)-a (A-3 i B) \tan (c+d x))}{\tan (c+d x)^3}dx-\frac {a A \cot ^3(c+d x) (a+i a \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 4076

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int -\cot ^2(c+d x) (i \tan (c+d x) a+a) \left ((17 A-15 i B) a^2+(7 i A+9 B) \tan (c+d x) a^2\right )dx-\frac {(3 B+5 i A) \cot ^2(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+i a \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (-\frac {1}{2} \int \cot ^2(c+d x) (i \tan (c+d x) a+a) \left ((17 A-15 i B) a^2+(7 i A+9 B) \tan (c+d x) a^2\right )dx-\frac {(3 B+5 i A) \cot ^2(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+i a \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (-\frac {1}{2} \int \frac {(i \tan (c+d x) a+a) \left ((17 A-15 i B) a^2+(7 i A+9 B) \tan (c+d x) a^2\right )}{\tan (c+d x)^2}dx-\frac {(3 B+5 i A) \cot ^2(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+i a \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 4074

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {a^3 (17 A-15 i B) \cot (c+d x)}{d}-\int 24 \cot (c+d x) \left (a^3 (i A+B)-a^3 (A-i B) \tan (c+d x)\right )dx\right )-\frac {(3 B+5 i A) \cot ^2(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+i a \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {a^3 (17 A-15 i B) \cot (c+d x)}{d}-24 \int \cot (c+d x) \left (a^3 (i A+B)-a^3 (A-i B) \tan (c+d x)\right )dx\right )-\frac {(3 B+5 i A) \cot ^2(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+i a \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {a^3 (17 A-15 i B) \cot (c+d x)}{d}-24 \int \frac {a^3 (i A+B)-a^3 (A-i B) \tan (c+d x)}{\tan (c+d x)}dx\right )-\frac {(3 B+5 i A) \cot ^2(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+i a \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 4014

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {a^3 (17 A-15 i B) \cot (c+d x)}{d}-24 \left (a^3 (B+i A) \int \cot (c+d x)dx-a^3 x (A-i B)\right )\right )-\frac {(3 B+5 i A) \cot ^2(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+i a \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {a^3 (17 A-15 i B) \cot (c+d x)}{d}-24 \left (a^3 (B+i A) \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-a^3 x (A-i B)\right )\right )-\frac {(3 B+5 i A) \cot ^2(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+i a \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {a^3 (17 A-15 i B) \cot (c+d x)}{d}-24 \left (-a^3 (B+i A) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\left (a^3 x (A-i B)\right )\right )\right )-\frac {(3 B+5 i A) \cot ^2(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+i a \tan (c+d x))^2}{3 d}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {a^3 (17 A-15 i B) \cot (c+d x)}{d}-24 \left (\frac {a^3 (B+i A) \log (-\sin (c+d x))}{d}-a^3 x (A-i B)\right )\right )-\frac {(3 B+5 i A) \cot ^2(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+i a \tan (c+d x))^2}{3 d}\)

Input:

Int[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]
 

Output:

-1/3*(a*A*Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^2)/d + (((a^3*(17*A - (15* 
I)*B)*Cot[c + d*x])/d - 24*(-(a^3*(A - I*B)*x) + (a^3*(I*A + B)*Log[-Sin[c 
 + d*x]])/d))/2 - (((5*I)*A + 3*B)*Cot[c + d*x]^2*(a^3 + I*a^3*Tan[c + d*x 
]))/(2*d))/3
 

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 4074

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

rule 4076

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

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.68

method	result	size
parallelrisch	\(\frac {4 \left (\frac {\left (i A +B \right ) \ln \left (\sec \left (d x +c \right )^{2}\right )}{2}-\left (i A +B \right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {A \cot \left (d x +c \right )^{3}}{12}-\frac {\cot \left (d x +c \right )^{2} \left (3 i A +B \right )}{8}+\left (-\frac {3 i B}{4}+A \right ) \cot \left (d x +c \right )+\left (-i B +A \right ) x d \right ) a^{3}}{d}\)	\(91\)
derivativedivides	\(\frac {a^{3} \left (-\frac {3 i A \cot \left (d x +c \right )^{2}}{2}-\frac {A \cot \left (d x +c \right )^{3}}{3}-3 i B \cot \left (d x +c \right )-\frac {B \cot \left (d x +c \right )^{2}}{2}+4 A \cot \left (d x +c \right )+\frac {\left (4 i A +4 B \right ) \ln \left (\cot \left (d x +c \right )^{2}+1\right )}{2}+\left (4 i B -4 A \right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )\right )}{d}\)	\(105\)
default	\(\frac {a^{3} \left (-\frac {3 i A \cot \left (d x +c \right )^{2}}{2}-\frac {A \cot \left (d x +c \right )^{3}}{3}-3 i B \cot \left (d x +c \right )-\frac {B \cot \left (d x +c \right )^{2}}{2}+4 A \cot \left (d x +c \right )+\frac {\left (4 i A +4 B \right ) \ln \left (\cot \left (d x +c \right )^{2}+1\right )}{2}+\left (4 i B -4 A \right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )\right )}{d}\)	\(105\)
risch	\(\frac {8 i a^{3} B c}{d}-\frac {8 a^{3} A c}{d}+\frac {2 a^{3} \left (24 i A \,{\mathrm e}^{4 i \left (d x +c \right )}+12 B \,{\mathrm e}^{4 i \left (d x +c \right )}-33 i A \,{\mathrm e}^{2 i \left (d x +c \right )}-21 B \,{\mathrm e}^{2 i \left (d x +c \right )}+13 i A +9 B \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {4 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{d}-\frac {4 i a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A}{d}\)	\(145\)
norman	\(\frac {\frac {\left (-3 i B \,a^{3}+4 A \,a^{3}\right ) \tan \left (d x +c \right )^{2}}{d}+\left (-4 i B \,a^{3}+4 A \,a^{3}\right ) x \tan \left (d x +c \right )^{3}-\frac {A \,a^{3}}{3 d}-\frac {\left (3 i A \,a^{3}+B \,a^{3}\right ) \tan \left (d x +c \right )}{2 d}}{\tan \left (d x +c \right )^{3}}-\frac {4 \left (i A \,a^{3}+B \,a^{3}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {2 \left (i A \,a^{3}+B \,a^{3}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}\)	\(149\)

Input:

int(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x,method=_RETURNVER 
BOSE)
 

Output:

4*(1/2*(I*A+B)*ln(sec(d*x+c)^2)-(I*A+B)*ln(tan(d*x+c))-1/12*A*cot(d*x+c)^3 
-1/8*cot(d*x+c)^2*(3*I*A+B)+(-3/4*I*B+A)*cot(d*x+c)+(A-I*B)*x*d)*a^3/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.35 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (12 \, {\left (-2 i \, A - B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (11 i \, A + 7 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-13 i \, A - 9 \, B\right )} a^{3} + 6 \, {\left ({\left (i \, A + B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (-i \, A - B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (i \, A + B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, A - B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \] Input:

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm= 
"fricas")
 

Output:

-2/3*(12*(-2*I*A - B)*a^3*e^(4*I*d*x + 4*I*c) + 3*(11*I*A + 7*B)*a^3*e^(2* 
I*d*x + 2*I*c) + (-13*I*A - 9*B)*a^3 + 6*((I*A + B)*a^3*e^(6*I*d*x + 6*I*c 
) + 3*(-I*A - B)*a^3*e^(4*I*d*x + 4*I*c) + 3*(I*A + B)*a^3*e^(2*I*d*x + 2* 
I*c) + (-I*A - B)*a^3)*log(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(6*I*d*x + 6*I*c 
) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)
 

Sympy [A] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.36 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=- \frac {4 i a^{3} \left (A - i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {26 i A a^{3} + 18 B a^{3} + \left (- 66 i A a^{3} e^{2 i c} - 42 B a^{3} e^{2 i c}\right ) e^{2 i d x} + \left (48 i A a^{3} e^{4 i c} + 24 B a^{3} e^{4 i c}\right ) e^{4 i d x}}{3 d e^{6 i c} e^{6 i d x} - 9 d e^{4 i c} e^{4 i d x} + 9 d e^{2 i c} e^{2 i d x} - 3 d} \] Input:

integrate(cot(d*x+c)**4*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)
 

Output:

-4*I*a**3*(A - I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (26*I*A*a**3 + 18* 
B*a**3 + (-66*I*A*a**3*exp(2*I*c) - 42*B*a**3*exp(2*I*c))*exp(2*I*d*x) + ( 
48*I*A*a**3*exp(4*I*c) + 24*B*a**3*exp(4*I*c))*exp(4*I*d*x))/(3*d*exp(6*I* 
c)*exp(6*I*d*x) - 9*d*exp(4*I*c)*exp(4*I*d*x) + 9*d*exp(2*I*c)*exp(2*I*d*x 
) - 3*d)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.86 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {24 \, {\left (d x + c\right )} {\left (A - i \, B\right )} a^{3} - 12 \, {\left (-i \, A - B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 24 \, {\left (i \, A + B\right )} a^{3} \log \left (\tan \left (d x + c\right )\right ) + \frac {6 \, {\left (4 \, A - 3 i \, B\right )} a^{3} \tan \left (d x + c\right )^{2} + 3 \, {\left (-3 i \, A - B\right )} a^{3} \tan \left (d x + c\right ) - 2 \, A a^{3}}{\tan \left (d x + c\right )^{3}}}{6 \, d} \] Input:

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm= 
"maxima")
 

Output:

1/6*(24*(d*x + c)*(A - I*B)*a^3 - 12*(-I*A - B)*a^3*log(tan(d*x + c)^2 + 1 
) - 24*(I*A + B)*a^3*log(tan(d*x + c)) + (6*(4*A - 3*I*B)*a^3*tan(d*x + c) 
^2 + 3*(-3*I*A - B)*a^3*tan(d*x + c) - 2*A*a^3)/tan(d*x + c)^3)/d
 

Giac [A] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.87 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {4 \, {\left (-i \, A a^{3} - B a^{3}\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{d} - \frac {4 \, {\left (i \, A a^{3} + B a^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {2 \, A a^{3} - 6 \, {\left (4 \, A a^{3} - 3 i \, B a^{3}\right )} \tan \left (d x + c\right )^{2} + 3 \, {\left (3 i \, A a^{3} + B a^{3}\right )} \tan \left (d x + c\right )}{6 \, d \tan \left (d x + c\right )^{3}} \] Input:

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm= 
"giac")
 

Output:

-4*(-I*A*a^3 - B*a^3)*log(tan(d*x + c) + I)/d - 4*(I*A*a^3 + B*a^3)*log(ab 
s(tan(d*x + c)))/d - 1/6*(2*A*a^3 - 6*(4*A*a^3 - 3*I*B*a^3)*tan(d*x + c)^2 
 + 3*(3*I*A*a^3 + B*a^3)*tan(d*x + c))/(d*tan(d*x + c)^3)
 

Mupad [B] (verification not implemented)

Time = 3.40 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.69 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {\frac {A\,a^3}{3}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (4\,A\,a^3-B\,a^3\,3{}\mathrm {i}\right )+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^3}{2}+\frac {A\,a^3\,3{}\mathrm {i}}{2}\right )}{d\,{\mathrm {tan}\left (c+d\,x\right )}^3}-\frac {a^3\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{d} \] Input:

int(cot(c + d*x)^4*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.67 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {a^{3} \left (52 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a -36 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b i -4 \cos \left (d x +c \right ) a +48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} a i +48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} b -48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a i -48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} b +48 \sin \left (d x +c \right )^{3} a d x +9 \sin \left (d x +c \right )^{3} a i -48 \sin \left (d x +c \right )^{3} b d i x +3 \sin \left (d x +c \right )^{3} b -18 \sin \left (d x +c \right ) a i -6 \sin \left (d x +c \right ) b \right )}{12 \sin \left (d x +c \right )^{3} d} \] Input:

int(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)
 

Output:

(a**3*(52*cos(c + d*x)*sin(c + d*x)**2*a - 36*cos(c + d*x)*sin(c + d*x)**2 
*b*i - 4*cos(c + d*x)*a + 48*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**3* 
a*i + 48*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**3*b - 48*log(tan((c + 
d*x)/2))*sin(c + d*x)**3*a*i - 48*log(tan((c + d*x)/2))*sin(c + d*x)**3*b 
+ 48*sin(c + d*x)**3*a*d*x + 9*sin(c + d*x)**3*a*i - 48*sin(c + d*x)**3*b* 
d*i*x + 3*sin(c + d*x)**3*b - 18*sin(c + d*x)*a*i - 6*sin(c + d*x)*b))/(12 
*sin(c + d*x)**3*d)