\(\int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx\) [50]

Optimal result

Integrand size = 34, antiderivative size = 170 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {3 (5 i A-3 B) x}{4 a^2}+\frac {3 (5 i A-3 B) \cot (c+d x)}{4 a^2 d}-\frac {(2 A+i B) \cot ^2(c+d x)}{a^2 d}-\frac {2 (2 A+i B) \log (\sin (c+d x))}{a^2 d}+\frac {(5 A+3 i B) \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3677, 3610, 3612, 3556} \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {(2 A+i B) \cot ^2(c+d x)}{a^2 d}+\frac {3 (-3 B+5 i A) \cot (c+d x)}{4 a^2 d}-\frac {2 (2 A+i B) \log (\sin (c+d x))}{a^2 d}+\frac {(5 A+3 i B) \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {3 x (-3 B+5 i A)}{4 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[In]

Int[(Cot[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(3*((5*I)*A - 3*B)*x)/(4*a^2) + (3*((5*I)*A - 3*B)*Cot[c + d*x])/(4*a^2*d) - ((2*A + I*B)*Cot[c + d*x]^2)/(a^2
*d) - (2*(2*A + I*B)*Log[Sin[c + d*x]])/(a^2*d) + ((5*A + (3*I)*B)*Cot[c + d*x]^2)/(4*a^2*d*(1 + I*Tan[c + d*x
])) + ((A + I*B)*Cot[c + d*x]^2)/(4*d*(a + I*a*Tan[c + d*x])^2)

Rule 3556

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

Rule 3610

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] + Dist[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 3612

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] + Dist[(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] && NeQ[a*c + b*d, 0]

Rule 3677

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(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] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[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
] /; FreeQ[{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]

Rubi steps \begin{align*} \text {integral}& = \frac {(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\cot ^3(c+d x) (2 a (3 A+i B)-4 a (i A-B) \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2} \\ & = \frac {(5 A+3 i B) \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \cot ^3(c+d x) \left (16 a^2 (2 A+i B)-6 a^2 (5 i A-3 B) \tan (c+d x)\right ) \, dx}{8 a^4} \\ & = -\frac {(2 A+i B) \cot ^2(c+d x)}{a^2 d}+\frac {(5 A+3 i B) \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \cot ^2(c+d x) \left (-6 a^2 (5 i A-3 B)-16 a^2 (2 A+i B) \tan (c+d x)\right ) \, dx}{8 a^4} \\ & = \frac {3 (5 i A-3 B) \cot (c+d x)}{4 a^2 d}-\frac {(2 A+i B) \cot ^2(c+d x)}{a^2 d}+\frac {(5 A+3 i B) \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \cot (c+d x) \left (-16 a^2 (2 A+i B)+6 a^2 (5 i A-3 B) \tan (c+d x)\right ) \, dx}{8 a^4} \\ & = \frac {3 (5 i A-3 B) x}{4 a^2}+\frac {3 (5 i A-3 B) \cot (c+d x)}{4 a^2 d}-\frac {(2 A+i B) \cot ^2(c+d x)}{a^2 d}+\frac {(5 A+3 i B) \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(2 (2 A+i B)) \int \cot (c+d x) \, dx}{a^2} \\ & = \frac {3 (5 i A-3 B) x}{4 a^2}+\frac {3 (5 i A-3 B) \cot (c+d x)}{4 a^2 d}-\frac {(2 A+i B) \cot ^2(c+d x)}{a^2 d}-\frac {2 (2 A+i B) \log (\sin (c+d x))}{a^2 d}+\frac {(5 A+3 i B) \cot ^2(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \cot ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 2.71 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.84 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {2 (A+i B) \cot ^4(c+d x)}{(i+\cot (c+d x))^2}+\frac {2 (5 A+3 i B) \cot ^3(c+d x)}{i+\cot (c+d x)}+6 (5 i A-3 B) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )-8 (2 A+i B) \left (\cot ^2(c+d x)+2 (\log (\cos (c+d x))+\log (\tan (c+d x)))\right )}{8 a^2 d} \]

[In]

Integrate[(Cot[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^2,x]

[Out]

((2*(A + I*B)*Cot[c + d*x]^4)/(I + Cot[c + d*x])^2 + (2*(5*A + (3*I)*B)*Cot[c + d*x]^3)/(I + Cot[c + d*x]) + 6
*((5*I)*A - 3*B)*Cot[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2] - 8*(2*A + I*B)*(Cot[c + d*x]^2
+ 2*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]])))/(8*a^2*d)

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.23

method	result	size
risch	\(-\frac {17 x B}{4 a^{2}}+\frac {31 i x A}{4 a^{2}}-\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )} B}{4 a^{2} d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} A}{a^{2} d}-\frac {i {\mathrm e}^{-4 i \left (d x +c \right )} B}{16 a^{2} d}-\frac {{\mathrm e}^{-4 i \left (d x +c \right )} A}{16 a^{2} d}-\frac {4 B c}{a^{2} d}+\frac {8 i A c}{a^{2} d}-\frac {2 i \left (-i A \,{\mathrm e}^{2 i \left (d x +c \right )}+B \,{\mathrm e}^{2 i \left (d x +c \right )}+2 i A -B \right )}{a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {2 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{a^{2} d}-\frac {4 A \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\)	\(209\)
derivativedivides	\(\frac {A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {5 B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}+\frac {15 i A \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {2 A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}}+\frac {i B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}}+\frac {7 i A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {9 B \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {2 i B \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}+\frac {2 i A}{a^{2} d \tan \left (d x +c \right )}-\frac {A}{2 a^{2} d \tan \left (d x +c \right )^{2}}-\frac {4 A \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}-\frac {B}{a^{2} d \tan \left (d x +c \right )}\)	\(243\)
default	\(\frac {A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {5 B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}+\frac {15 i A \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {2 A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}}+\frac {i B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{2}}+\frac {7 i A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {9 B \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {2 i B \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}+\frac {2 i A}{a^{2} d \tan \left (d x +c \right )}-\frac {A}{2 a^{2} d \tan \left (d x +c \right )^{2}}-\frac {4 A \ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}-\frac {B}{a^{2} d \tan \left (d x +c \right )}\)	\(243\)

[In]

int(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-17/4*x/a^2*B+31/4*I*x/a^2*A-3/4*I/a^2/d*exp(-2*I*(d*x+c))*B-1/a^2/d*exp(-2*I*(d*x+c))*A-1/16*I/a^2/d*exp(-4*I
*(d*x+c))*B-1/16/a^2/d*exp(-4*I*(d*x+c))*A-4/a^2/d*B*c+8*I/a^2/d*A*c-2*I*(-I*A*exp(2*I*(d*x+c))+B*exp(2*I*(d*x
+c))+2*I*A-B)/a^2/d/(exp(2*I*(d*x+c))-1)^2-2*I/a^2/d*ln(exp(2*I*(d*x+c))-1)*B-4/a^2*A/d*ln(exp(2*I*(d*x+c))-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.26 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {4 \, {\left (-31 i \, A + 17 \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, {\left (2 \, {\left (31 i \, A - 17 \, B\right )} d x + 12 \, A + 11 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (4 \, {\left (-31 i \, A + 17 \, B\right )} d x - 95 \, A - 55 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (7 \, A + 5 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 32 \, {\left ({\left (2 \, A + i \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} - 2 \, {\left (2 \, A + i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (2 \, A + i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + A + i \, B}{16 \, {\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} - 2 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \]

[In]

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/16*(4*(-31*I*A + 17*B)*d*x*e^(8*I*d*x + 8*I*c) + 4*(2*(31*I*A - 17*B)*d*x + 12*A + 11*I*B)*e^(6*I*d*x + 6*I
*c) + (4*(-31*I*A + 17*B)*d*x - 95*A - 55*I*B)*e^(4*I*d*x + 4*I*c) + 2*(7*A + 5*I*B)*e^(2*I*d*x + 2*I*c) + 32*
((2*A + I*B)*e^(8*I*d*x + 8*I*c) - 2*(2*A + I*B)*e^(6*I*d*x + 6*I*c) + (2*A + I*B)*e^(4*I*d*x + 4*I*c))*log(e^
(2*I*d*x + 2*I*c) - 1) + A + I*B)/(a^2*d*e^(8*I*d*x + 8*I*c) - 2*a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4*I*d*x
+ 4*I*c))

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.90 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {4 A + 2 i B + \left (- 2 A e^{2 i c} - 2 i B e^{2 i c}\right ) e^{2 i d x}}{a^{2} d e^{4 i c} e^{4 i d x} - 2 a^{2} d e^{2 i c} e^{2 i d x} + a^{2} d} + \begin {cases} \frac {\left (\left (- 4 A a^{2} d e^{2 i c} - 4 i B a^{2} d e^{2 i c}\right ) e^{- 4 i d x} + \left (- 64 A a^{2} d e^{4 i c} - 48 i B a^{2} d e^{4 i c}\right ) e^{- 2 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (- \frac {31 i A - 17 B}{4 a^{2}} + \frac {\left (31 i A e^{4 i c} + 8 i A e^{2 i c} + i A - 17 B e^{4 i c} - 6 B e^{2 i c} - B\right ) e^{- 4 i c}}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (31 i A - 17 B\right )}{4 a^{2}} - \frac {2 \cdot \left (2 A + i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{2} d} \]

[In]

integrate(cot(d*x+c)**3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**2,x)

[Out]

(4*A + 2*I*B + (-2*A*exp(2*I*c) - 2*I*B*exp(2*I*c))*exp(2*I*d*x))/(a**2*d*exp(4*I*c)*exp(4*I*d*x) - 2*a**2*d*e
xp(2*I*c)*exp(2*I*d*x) + a**2*d) + Piecewise((((-4*A*a**2*d*exp(2*I*c) - 4*I*B*a**2*d*exp(2*I*c))*exp(-4*I*d*x
) + (-64*A*a**2*d*exp(4*I*c) - 48*I*B*a**2*d*exp(4*I*c))*exp(-2*I*d*x))*exp(-6*I*c)/(64*a**4*d**2), Ne(a**4*d*
*2*exp(6*I*c), 0)), (x*(-(31*I*A - 17*B)/(4*a**2) + (31*I*A*exp(4*I*c) + 8*I*A*exp(2*I*c) + I*A - 17*B*exp(4*I
*c) - 6*B*exp(2*I*c) - B)*exp(-4*I*c)/(4*a**2)), True)) + x*(31*I*A - 17*B)/(4*a**2) - 2*(2*A + I*B)*log(exp(2
*I*d*x) - exp(-2*I*c))/(a**2*d)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 1.13 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.04 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {4 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {4 \, {\left (31 \, A + 17 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac {64 \, {\left (2 \, A + i \, B\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{2}} + \frac {3 \, A \tan \left (d x + c\right )^{4} - 3 i \, B \tan \left (d x + c\right )^{4} + 114 i \, A \tan \left (d x + c\right )^{3} - 78 \, B \tan \left (d x + c\right )^{3} + 173 \, A \tan \left (d x + c\right )^{2} + 115 i \, B \tan \left (d x + c\right )^{2} - 32 i \, A \tan \left (d x + c\right ) + 32 \, B \tan \left (d x + c\right ) + 16 \, A}{{\left (\tan \left (d x + c\right )^{2} - i \, \tan \left (d x + c\right )\right )}^{2} a^{2}}}{32 \, d} \]

[In]

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/32*(4*(A - I*B)*log(tan(d*x + c) + I)/a^2 + 4*(31*A + 17*I*B)*log(tan(d*x + c) - I)/a^2 - 64*(2*A + I*B)*log
(tan(d*x + c))/a^2 + (3*A*tan(d*x + c)^4 - 3*I*B*tan(d*x + c)^4 + 114*I*A*tan(d*x + c)^3 - 78*B*tan(d*x + c)^3
+ 173*A*tan(d*x + c)^2 + 115*I*B*tan(d*x + c)^2 - 32*I*A*tan(d*x + c) + 32*B*tan(d*x + c) + 16*A)/((tan(d*x +
c)^2 - I*tan(d*x + c))^2*a^2))/d

Mupad [B] (verification not implemented)

Time = 7.91 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.11 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-\frac {7\,B}{2\,a^2}+\frac {A\,11{}\mathrm {i}}{2\,a^2}\right )-{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {15\,A}{4\,a^2}+\frac {B\,9{}\mathrm {i}}{4\,a^2}\right )+\frac {A\,1{}\mathrm {i}}{2\,a^2}+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {A}{a^2}+\frac {B\,1{}\mathrm {i}}{a^2}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}+2\,{\mathrm {tan}\left (c+d\,x\right )}^3-{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}\right )}-\frac {2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (2\,A+B\,1{}\mathrm {i}\right )}{a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{8\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (31\,A+B\,17{}\mathrm {i}\right )}{8\,a^2\,d} \]

[In]

int((cot(c + d*x)^3*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^2,x)

[Out]

(tan(c + d*x)^2*((A*11i)/(2*a^2) - (7*B)/(2*a^2)) - tan(c + d*x)^3*((15*A)/(4*a^2) + (B*9i)/(4*a^2)) + (A*1i)/
(2*a^2) + tan(c + d*x)*(A/a^2 + (B*1i)/a^2))/(d*(2*tan(c + d*x)^3 - tan(c + d*x)^2*1i + tan(c + d*x)^4*1i)) -
(2*log(tan(c + d*x))*(2*A + B*1i))/(a^2*d) + (log(tan(c + d*x) + 1i)*(A - B*1i))/(8*a^2*d) + (log(tan(c + d*x)
- 1i)*(31*A + B*17i))/(8*a^2*d)