3.3.51 \(\int \cot (c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [251]

3.3.51.1 Optimal result

Integrand size = 29, antiderivative size = 117 \[ \int \cot (c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x-\frac {b \left (3 a A b+3 a^2 B-b^2 B\right ) \log (\cos (c+d x))}{d}+\frac {a^3 A \log (\sin (c+d x))}{d}+\frac {b^2 (A b+2 a B) \tan (c+d x)}{d}+\frac {b B (a+b \tan (c+d x))^2}{2 d} \]

output

(3*A*a^2*b-A*b^3+B*a^3-3*B*a*b^2)*x-b*(3*A*a*b+3*B*a^2-B*b^2)*ln(cos(d*x+c 
))/d+a^3*A*ln(sin(d*x+c))/d+b^2*(A*b+2*B*a)*tan(d*x+c)/d+1/2*b*B*(a+b*tan( 
d*x+c))^2/d
 

3.3.51.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.63 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.98 \[ \int \cot (c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {-(a+i b)^3 (A+i B) \log (i-\tan (c+d x))+2 a^3 A \log (\tan (c+d x))-(a-i b)^3 (A-i B) \log (i+\tan (c+d x))+2 b^2 (A b+2 a B) \tan (c+d x)+b B (a+b \tan (c+d x))^2}{2 d} \]

input

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

output

(-((a + I*b)^3*(A + I*B)*Log[I - Tan[c + d*x]]) + 2*a^3*A*Log[Tan[c + d*x] 
] - (a - I*b)^3*(A - I*B)*Log[I + Tan[c + d*x]] + 2*b^2*(A*b + 2*a*B)*Tan[ 
c + d*x] + b*B*(a + b*Tan[c + d*x])^2)/(2*d)
 

3.3.51.3 Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {3042, 4090, 27, 3042, 4120, 25, 3042, 4107, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4090

\(\displaystyle \frac {1}{2} \int 2 \cot (c+d x) (a+b \tan (c+d x)) \left (A a^2+b (A b+2 a B) \tan ^2(c+d x)+\left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)\right )dx+\frac {b B (a+b \tan (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \int \cot (c+d x) (a+b \tan (c+d x)) \left (A a^2+b (A b+2 a B) \tan ^2(c+d x)+\left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)\right )dx+\frac {b B (a+b \tan (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (c+d x)) \left (A a^2+b (A b+2 a B) \tan (c+d x)^2+\left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)\right )}{\tan (c+d x)}dx+\frac {b B (a+b \tan (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 4120

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A a^3+b \left (3 B a^2+3 A b a-b^2 B\right ) \tan (c+d x)^2+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\tan (c+d x)}dx+\frac {b^2 (2 a B+A b) \tan (c+d x)}{d}+\frac {b B (a+b \tan (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 4107

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

\(\Big \downarrow \) 3042

\(\displaystyle a^3 A \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx+b \left (3 a^2 B+3 a A b-b^2 B\right ) \int \tan (c+d x)dx+x \left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right )+\frac {b^2 (2 a B+A b) \tan (c+d x)}{d}+\frac {b B (a+b \tan (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 25

\(\displaystyle a^3 (-A) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx+b \left (3 a^2 B+3 a A b-b^2 B\right ) \int \tan (c+d x)dx+x \left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right )+\frac {b^2 (2 a B+A b) \tan (c+d x)}{d}+\frac {b B (a+b \tan (c+d x))^2}{2 d}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {a^3 A \log (-\sin (c+d x))}{d}-\frac {b \left (3 a^2 B+3 a A b-b^2 B\right ) \log (\cos (c+d x))}{d}+x \left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right )+\frac {b^2 (2 a B+A b) \tan (c+d x)}{d}+\frac {b B (a+b \tan (c+d x))^2}{2 d}\)

input

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

output

(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*x - (b*(3*a*A*b + 3*a^2*B - b^2*B) 
*Log[Cos[c + d*x]])/d + (a^3*A*Log[-Sin[c + d*x]])/d + (b^2*(A*b + 2*a*B)* 
Tan[c + d*x])/d + (b*B*(a + b*Tan[c + d*x])^2)/(2*d)
 

3.3.51.3.1 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 4090

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

rule 4107

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2 
)/tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[B*x, x] + (Simp[A   Int[1/Tan[ 
e + f*x], x], x] + Simp[C   Int[Tan[e + f*x], x], x]) /; FreeQ[{e, f, A, B, 
 C}, x] && NeQ[A, C]
 

rule 4120

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

3.3.51.4 Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.03

method	result	size
parallelrisch	\(\frac {\left (-A \,a^{3}+3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \ln \left (\sec ^{2}\left (d x +c \right )\right )+2 A \,a^{3} \ln \left (\tan \left (d x +c \right )\right )+B \,b^{3} \left (\tan ^{2}\left (d x +c \right )\right )+\left (2 A \,b^{3}+6 B a \,b^{2}\right ) \tan \left (d x +c \right )+6 \left (A \,a^{2} b -\frac {1}{3} A \,b^{3}+\frac {1}{3} B \,a^{3}-B a \,b^{2}\right ) d x}{2 d}\)	\(121\)
norman	\(\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) x +\frac {b^{2} \left (A b +3 B a \right ) \tan \left (d x +c \right )}{d}+\frac {B \,b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {A \,a^{3} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(124\)
derivativedivides	\(\frac {\frac {B \,b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2}+A \,b^{3} \tan \left (d x +c \right )+3 B a \,b^{2} \tan \left (d x +c \right )+\frac {\left (-A \,a^{3}+3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )+A \,a^{3} \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(130\)
default	\(\frac {\frac {B \,b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{2}+A \,b^{3} \tan \left (d x +c \right )+3 B a \,b^{2} \tan \left (d x +c \right )+\frac {\left (-A \,a^{3}+3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )+A \,a^{3} \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(130\)
risch	\(\frac {6 i B \,a^{2} b c}{d}+3 i B \,a^{2} b x +\frac {2 i b^{2} \left (-i B b \,{\mathrm e}^{2 i \left (d x +c \right )}+A b \,{\mathrm e}^{2 i \left (d x +c \right )}+3 B a \,{\mathrm e}^{2 i \left (d x +c \right )}+A b +3 B a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {2 i B \,b^{3} c}{d}+3 A \,a^{2} b x -A \,b^{3} x +B \,a^{3} x -3 B a \,b^{2} x -i A \,a^{3} x +3 i A a \,b^{2} x -i B \,b^{3} x +\frac {6 i A a \,b^{2} c}{d}-\frac {2 i a^{3} A c}{d}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A a \,b^{2}}{d}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B \,a^{2} b}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B \,b^{3}}{d}+\frac {A \,a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(264\)

input

int(cot(d*x+c)*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE 
)
 

output

1/2*((-A*a^3+3*A*a*b^2+3*B*a^2*b-B*b^3)*ln(sec(d*x+c)^2)+2*A*a^3*ln(tan(d* 
x+c))+B*b^3*tan(d*x+c)^2+(2*A*b^3+6*B*a*b^2)*tan(d*x+c)+6*(A*a^2*b-1/3*A*b 
^3+1/3*B*a^3-B*a*b^2)*d*x)/d
 

3.3.51.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.14 \[ \int \cot (c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {B b^{3} \tan \left (d x + c\right )^{2} + A a^{3} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + 2 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} d x - {\left (3 \, B a^{2} b + 3 \, A a b^{2} - B b^{3}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \tan \left (d x + c\right )}{2 \, d} \]

input

integrate(cot(d*x+c)*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fri 
cas")
 

output

1/2*(B*b^3*tan(d*x + c)^2 + A*a^3*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1)) 
 + 2*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*d*x - (3*B*a^2*b + 3*A*a*b^2 
- B*b^3)*log(1/(tan(d*x + c)^2 + 1)) + 2*(3*B*a*b^2 + A*b^3)*tan(d*x + c)) 
/d
 

3.3.51.6 Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.74 \[ \int \cot (c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\begin {cases} - \frac {A a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {A a^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 3 A a^{2} b x + \frac {3 A a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - A b^{3} x + \frac {A b^{3} \tan {\left (c + d x \right )}}{d} + B a^{3} x + \frac {3 B a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - 3 B a b^{2} x + \frac {3 B a b^{2} \tan {\left (c + d x \right )}}{d} - \frac {B b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b^{3} \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \tan {\left (c \right )}\right ) \left (a + b \tan {\left (c \right )}\right )^{3} \cot {\left (c \right )} & \text {otherwise} \end {cases} \]

input

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

output

Piecewise((-A*a**3*log(tan(c + d*x)**2 + 1)/(2*d) + A*a**3*log(tan(c + d*x 
))/d + 3*A*a**2*b*x + 3*A*a*b**2*log(tan(c + d*x)**2 + 1)/(2*d) - A*b**3*x 
 + A*b**3*tan(c + d*x)/d + B*a**3*x + 3*B*a**2*b*log(tan(c + d*x)**2 + 1)/ 
(2*d) - 3*B*a*b**2*x + 3*B*a*b**2*tan(c + d*x)/d - B*b**3*log(tan(c + d*x) 
**2 + 1)/(2*d) + B*b**3*tan(c + d*x)**2/(2*d), Ne(d, 0)), (x*(A + B*tan(c) 
)*(a + b*tan(c))**3*cot(c), True))
 

3.3.51.7 Maxima [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.06 \[ \int \cot (c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {B b^{3} \tan \left (d x + c\right )^{2} + 2 \, A a^{3} \log \left (\tan \left (d x + c\right )\right ) + 2 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} {\left (d x + c\right )} - {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \tan \left (d x + c\right )}{2 \, d} \]

input

integrate(cot(d*x+c)*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="max 
ima")
 

output

1/2*(B*b^3*tan(d*x + c)^2 + 2*A*a^3*log(tan(d*x + c)) + 2*(B*a^3 + 3*A*a^2 
*b - 3*B*a*b^2 - A*b^3)*(d*x + c) - (A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3 
)*log(tan(d*x + c)^2 + 1) + 2*(3*B*a*b^2 + A*b^3)*tan(d*x + c))/d
 

3.3.51.8 Giac [A] (verification not implemented)

Time = 1.13 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.10 \[ \int \cot (c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {B b^{3} \tan \left (d x + c\right )^{2} + 2 \, A a^{3} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) + 6 \, B a b^{2} \tan \left (d x + c\right ) + 2 \, A b^{3} \tan \left (d x + c\right ) + 2 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} {\left (d x + c\right )} - {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \]

input

integrate(cot(d*x+c)*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="gia 
c")
 

output

1/2*(B*b^3*tan(d*x + c)^2 + 2*A*a^3*log(abs(tan(d*x + c))) + 6*B*a*b^2*tan 
(d*x + c) + 2*A*b^3*tan(d*x + c) + 2*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^ 
3)*(d*x + c) - (A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*log(tan(d*x + c)^2 
+ 1))/d
 

3.3.51.9 Mupad [B] (verification not implemented)

Time = 8.48 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.01 \[ \int \cot (c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A\,b^3+3\,B\,a\,b^2\right )}{d}+\frac {A\,a^3\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}+\frac {B\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \]

input

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

output

(tan(c + d*x)*(A*b^3 + 3*B*a*b^2))/d + (A*a^3*log(tan(c + d*x)))/d - (log( 
tan(c + d*x) + 1i)*(A - B*1i)*(a*1i + b)^3*1i)/(2*d) - (log(tan(c + d*x) - 
 1i)*(A + B*1i)*(a*1i - b)^3*1i)/(2*d) + (B*b^3*tan(c + d*x)^2)/(2*d)