\(\int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx\) [42]

Optimal result

Integrand size = 34, antiderivative size = 131 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {3 (i A-B) x}{2 a}+\frac {3 (i A-B) \cot (c+d x)}{2 a d}-\frac {(2 A+i B) \cot ^2(c+d x)}{2 a d}-\frac {(2 A+i B) \log (\sin (c+d x))}{a d}+\frac {(A+i B) \cot ^2(c+d x)}{2 d (a+i a \tan (c+d x))} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.16 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.85 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {\frac {(A+i B) \cot ^3(c+d x)}{i+\cot (c+d x)}+3 i (A+i B) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )-(2 A+i B) \left (\cot ^2(c+d x)+2 (\log (\cos (c+d x))+\log (\tan (c+d x)))\right )}{2 a d} \] Input:

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

Output:

(((A + I*B)*Cot[c + d*x]^3)/(I + Cot[c + d*x]) + (3*I)*(A + I*B)*Cot[c + d 
*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2] - (2*A + I*B)*(Cot[c 
+ d*x]^2 + 2*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]])))/(2*a*d)
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.97, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3042, 4079, 3042, 4012, 25, 3042, 4012, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4014

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3956

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 4012

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 + 1)/ 
(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 + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a 
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 
]
 

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 4079

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*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( 
b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m 
 + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m 
- b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free 
Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] 
 && LtQ[m, 0] &&  !GtQ[n, 0]
 

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.21

method	result	size
risch	\(-\frac {5 x B}{2 a}+\frac {7 i x A}{2 a}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B}{4 a d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} A}{4 a d}-\frac {2 B c}{a d}+\frac {4 i A c}{a d}-\frac {2 i \left (B \,{\mathrm e}^{2 i \left (d x +c \right )}+i A -B \right )}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{a d}-\frac {2 A \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a d}\)	\(159\)
norman	\(\frac {-\frac {A}{2 a d}-\frac {\left (-i A +B \right ) \tan \left (d x +c \right )}{a d}-\frac {3 \left (-i A +B \right ) \tan \left (d x +c \right )^{3}}{2 a d}-\frac {3 \left (-i A +B \right ) x \tan \left (d x +c \right )^{2}}{2 a}-\frac {3 \left (-i A +B \right ) x \tan \left (d x +c \right )^{4}}{2 a}-\frac {\left (i B +2 A \right ) \tan \left (d x +c \right )^{2}}{2 a d}}{\tan \left (d x +c \right )^{2} \left (1+\tan \left (d x +c \right )^{2}\right )}-\frac {\left (i B +2 A \right ) \ln \left (\tan \left (d x +c \right )\right )}{a d}+\frac {\left (i B +2 A \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 a d}\)	\(189\)
derivativedivides	\(\frac {A \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d a}+\frac {3 i A \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}+\frac {i B \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d a}-\frac {3 B \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}+\frac {i A}{2 d a \left (-i+\tan \left (d x +c \right )\right )}-\frac {B}{2 d a \left (-i+\tan \left (d x +c \right )\right )}-\frac {i B \ln \left (\tan \left (d x +c \right )\right )}{a d}-\frac {2 A \ln \left (\tan \left (d x +c \right )\right )}{d a}+\frac {i A}{a d \tan \left (d x +c \right )}-\frac {B}{a d \tan \left (d x +c \right )}-\frac {A}{2 a d \tan \left (d x +c \right )^{2}}\)	\(201\)
default	\(\frac {A \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d a}+\frac {3 i A \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}+\frac {i B \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d a}-\frac {3 B \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}+\frac {i A}{2 d a \left (-i+\tan \left (d x +c \right )\right )}-\frac {B}{2 d a \left (-i+\tan \left (d x +c \right )\right )}-\frac {i B \ln \left (\tan \left (d x +c \right )\right )}{a d}-\frac {2 A \ln \left (\tan \left (d x +c \right )\right )}{d a}+\frac {i A}{a d \tan \left (d x +c \right )}-\frac {B}{a d \tan \left (d x +c \right )}-\frac {A}{2 a d \tan \left (d x +c \right )^{2}}\)	\(201\)

Input:

int(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

-5/2*x/a*B+7/2*I*x/a*A-1/4*I/a/d*exp(-2*I*(d*x+c))*B-1/4/a/d*exp(-2*I*(d*x 
+c))*A-2/a/d*B*c+4*I/a/d*A*c-2*I*(B*exp(2*I*(d*x+c))+I*A-B)/a/d/(exp(2*I*( 
d*x+c))-1)^2-I/a/d*ln(exp(2*I*(d*x+c))-1)*B-2*A/a/d*ln(exp(2*I*(d*x+c))-1)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.44 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {2 \, {\left (-7 i \, A + 5 \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (4 \, {\left (7 i \, A - 5 \, B\right )} d x + A + 9 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left ({\left (-7 i \, A + 5 \, B\right )} d x - 5 \, A - 5 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, {\left ({\left (2 \, A + i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, {\left (2 \, A + i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (2 \, A + i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + A + i \, B}{4 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \] Input:

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

Output:

-1/4*(2*(-7*I*A + 5*B)*d*x*e^(6*I*d*x + 6*I*c) + (4*(7*I*A - 5*B)*d*x + A 
+ 9*I*B)*e^(4*I*d*x + 4*I*c) + 2*((-7*I*A + 5*B)*d*x - 5*A - 5*I*B)*e^(2*I 
*d*x + 2*I*c) + 4*((2*A + I*B)*e^(6*I*d*x + 6*I*c) - 2*(2*A + I*B)*e^(4*I* 
d*x + 4*I*c) + (2*A + I*B)*e^(2*I*d*x + 2*I*c))*log(e^(2*I*d*x + 2*I*c) - 
1) + A + I*B)/(a*d*e^(6*I*d*x + 6*I*c) - 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e 
^(2*I*d*x + 2*I*c))
 

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.52 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {2 A - 2 i B e^{2 i c} e^{2 i d x} + 2 i B}{a d e^{4 i c} e^{4 i d x} - 2 a d e^{2 i c} e^{2 i d x} + a d} + \begin {cases} \frac {\left (- A - i B\right ) e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: a d e^{2 i c} \neq 0 \\x \left (- \frac {7 i A - 5 B}{2 a} + \frac {\left (7 i A e^{2 i c} + i A - 5 B e^{2 i c} - B\right ) e^{- 2 i c}}{2 a}\right ) & \text {otherwise} \end {cases} + \frac {x \left (7 i A - 5 B\right )}{2 a} - \frac {\left (2 A + i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a d} \] Input:

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

Output:

(2*A - 2*I*B*exp(2*I*c)*exp(2*I*d*x) + 2*I*B)/(a*d*exp(4*I*c)*exp(4*I*d*x) 
 - 2*a*d*exp(2*I*c)*exp(2*I*d*x) + a*d) + Piecewise(((-A - I*B)*exp(-2*I*c 
)*exp(-2*I*d*x)/(4*a*d), Ne(a*d*exp(2*I*c), 0)), (x*(-(7*I*A - 5*B)/(2*a) 
+ (7*I*A*exp(2*I*c) + I*A - 5*B*exp(2*I*c) - B)*exp(-2*I*c)/(2*a)), True)) 
 + x*(7*I*A - 5*B)/(2*a) - (2*A + I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/(a* 
d)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {{\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{4 \, a d} + \frac {{\left (7 \, A + 5 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{4 \, a d} - \frac {{\left (2 \, A + i \, B\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a d} - \frac {3 \, {\left (-i \, A + B\right )} \tan \left (d x + c\right )^{2} - {\left (A + 2 i \, B\right )} \tan \left (d x + c\right ) - i \, A}{2 \, a d {\left (\tan \left (d x + c\right ) - i\right )} \tan \left (d x + c\right )^{2}} \] Input:

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

Output:

1/4*(A - I*B)*log(tan(d*x + c) + I)/(a*d) + 1/4*(7*A + 5*I*B)*log(tan(d*x 
+ c) - I)/(a*d) - (2*A + I*B)*log(abs(tan(d*x + c)))/(a*d) - 1/2*(3*(-I*A 
+ B)*tan(d*x + c)^2 - (A + 2*I*B)*tan(d*x + c) - I*A)/(a*d*(tan(d*x + c) - 
 I)*tan(d*x + c)^2)
 

Mupad [B] (verification not implemented)

Time = 3.70 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.17 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {3\,A}{2\,a}+\frac {B\,3{}\mathrm {i}}{2\,a}\right )+\frac {A}{2\,a}-\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {B}{a}+\frac {A\,1{}\mathrm {i}}{2\,a}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}+{\mathrm {tan}\left (c+d\,x\right )}^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (2\,A+B\,1{}\mathrm {i}\right )}{a\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{4\,a\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (7\,A+B\,5{}\mathrm {i}\right )}{4\,a\,d} \] Input:

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

Output:

(log(tan(c + d*x) + 1i)*(A - B*1i))/(4*a*d) - (log(tan(c + d*x))*(2*A + B* 
1i))/(a*d) - (tan(c + d*x)^2*((3*A)/(2*a) + (B*3i)/(2*a)) + A/(2*a) - tan( 
c + d*x)*((A*1i)/(2*a) - B/a))/(d*(tan(c + d*x)^2 + tan(c + d*x)^3*1i)) + 
(log(tan(c + d*x) - 1i)*(7*A + B*5i))/(4*a*d)
 

Reduce [F]

\[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {\left (\int \frac {\cot \left (d x +c \right )^{3}}{\tan \left (d x +c \right ) i +1}d x \right ) a +\left (\int \frac {\cot \left (d x +c \right )^{3} \tan \left (d x +c \right )}{\tan \left (d x +c \right ) i +1}d x \right ) b}{a} \] Input:

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

Output:

(int(cot(c + d*x)**3/(tan(c + d*x)*i + 1),x)*a + int((cot(c + d*x)**3*tan( 
c + d*x))/(tan(c + d*x)*i + 1),x)*b)/a