\(\int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\) [57]

Optimal result

Integrand size = 34, antiderivative size = 183 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {(25 A+7 i B) x}{8 a^3}-\frac {(25 A+7 i B) \cot (c+d x)}{8 a^3 d}-\frac {(3 i A-B) \log (\sin (c+d x))}{a^3 d}+\frac {(A+i B) \cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(11 A+5 i B) \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {(3 A+i B) \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

-1/8*(25*A+7*I*B)*x/a^3-1/8*(25*A+7*I*B)*cot(d*x+c)/a^3/d-(3*I*A-B)*ln(sin(d*x+c))/a^3/d+1/6*(A+I*B)*cot(d*x+c
)/d/(a+I*a*tan(d*x+c))^3+1/24*(11*A+5*I*B)*cot(d*x+c)/a/d/(a+I*a*tan(d*x+c))^2+1/2*(3*A+I*B)*cot(d*x+c)/d/(a^3
+I*a^3*tan(d*x+c))

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 183, 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} \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {(25 A+7 i B) \cot (c+d x)}{8 a^3 d}-\frac {(-B+3 i A) \log (\sin (c+d x))}{a^3 d}+\frac {(3 A+i B) \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {x (25 A+7 i B)}{8 a^3}+\frac {(11 A+5 i B) \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {(A+i B) \cot (c+d x)}{6 d (a+i a \tan (c+d x))^3} \]

[In]

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

[Out]

-1/8*((25*A + (7*I)*B)*x)/a^3 - ((25*A + (7*I)*B)*Cot[c + d*x])/(8*a^3*d) - (((3*I)*A - B)*Log[Sin[c + d*x]])/
(a^3*d) + ((A + I*B)*Cot[c + d*x])/(6*d*(a + I*a*Tan[c + d*x])^3) + ((11*A + (5*I)*B)*Cot[c + d*x])/(24*a*d*(a
+ I*a*Tan[c + d*x])^2) + ((3*A + I*B)*Cot[c + d*x])/(2*d*(a^3 + I*a^3*Tan[c + d*x]))

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 (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\cot ^2(c+d x) (a (7 A+i B)-4 a (i A-B) \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2} \\ & = \frac {(A+i B) \cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(11 A+5 i B) \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\cot ^2(c+d x) \left (3 a^2 (13 A+3 i B)-3 a^2 (11 i A-5 B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4} \\ & = \frac {(A+i B) \cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(11 A+5 i B) \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {(3 A+i B) \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \cot ^2(c+d x) \left (6 a^3 (25 A+7 i B)-48 a^3 (3 i A-B) \tan (c+d x)\right ) \, dx}{48 a^6} \\ & = -\frac {(25 A+7 i B) \cot (c+d x)}{8 a^3 d}+\frac {(A+i B) \cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(11 A+5 i B) \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {(3 A+i B) \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \cot (c+d x) \left (-48 a^3 (3 i A-B)-6 a^3 (25 A+7 i B) \tan (c+d x)\right ) \, dx}{48 a^6} \\ & = -\frac {(25 A+7 i B) x}{8 a^3}-\frac {(25 A+7 i B) \cot (c+d x)}{8 a^3 d}+\frac {(A+i B) \cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(11 A+5 i B) \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {(3 A+i B) \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {(3 i A-B) \int \cot (c+d x) \, dx}{a^3} \\ & = -\frac {(25 A+7 i B) x}{8 a^3}-\frac {(25 A+7 i B) \cot (c+d x)}{8 a^3 d}-\frac {(3 i A-B) \log (\sin (c+d x))}{a^3 d}+\frac {(A+i B) \cot (c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(11 A+5 i B) \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {(3 A+i B) \cot (c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 2.58 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.90 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {4 (A+i B) \cot ^4(c+d x)}{(i+\cot (c+d x))^3}+\frac {(11 A+5 i B) \cot ^3(c+d x)}{(i+\cot (c+d x))^2}+\frac {12 (3 A+i B) \cot ^2(c+d x)}{i+\cot (c+d x)}-3 \left ((25 A+7 i B) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )+8 (3 i A-B) (\log (\cos (c+d x))+\log (\tan (c+d x)))\right )}{24 a^3 d} \]

[In]

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

[Out]

((4*(A + I*B)*Cot[c + d*x]^4)/(I + Cot[c + d*x])^3 + ((11*A + (5*I)*B)*Cot[c + d*x]^3)/(I + Cot[c + d*x])^2 +
(12*(3*A + I*B)*Cot[c + d*x]^2)/(I + Cot[c + d*x]) - 3*((25*A + (7*I)*B)*Cot[c + d*x]*Hypergeometric2F1[-1/2,
1, 1/2, -Tan[c + d*x]^2] + 8*((3*I)*A - B)*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]])))/(24*a^3*d)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.17

method	result	size
risch	\(-\frac {15 i x B}{8 a^{3}}-\frac {49 x A}{8 a^{3}}+\frac {11 \,{\mathrm e}^{-2 i \left (d x +c \right )} B}{16 a^{3} d}-\frac {23 i {\mathrm e}^{-2 i \left (d x +c \right )} A}{16 a^{3} d}+\frac {5 \,{\mathrm e}^{-4 i \left (d x +c \right )} B}{32 a^{3} d}-\frac {7 i {\mathrm e}^{-4 i \left (d x +c \right )} A}{32 a^{3} d}+\frac {{\mathrm e}^{-6 i \left (d x +c \right )} B}{48 a^{3} d}-\frac {i {\mathrm e}^{-6 i \left (d x +c \right )} A}{48 a^{3} d}-\frac {2 i B c}{a^{3} d}-\frac {6 A c}{a^{3} d}-\frac {2 i A}{a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{a^{3} d}-\frac {3 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A}{a^{3} d}\)	\(214\)
derivativedivides	\(-\frac {A \cot \left (d x +c \right )}{a^{3} d}+\frac {3 i A \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2 a^{3} d}+\frac {25 A \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{8 a^{3} d}-\frac {B \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2 a^{3} d}+\frac {7 i B \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{8 a^{3} d}-\frac {31 A}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )}-\frac {17 i B}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )}+\frac {9 i A}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{2}}-\frac {7 B}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{2}}+\frac {A}{6 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{3}}+\frac {i B}{6 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{3}}\)	\(226\)
default	\(-\frac {A \cot \left (d x +c \right )}{a^{3} d}+\frac {3 i A \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2 a^{3} d}+\frac {25 A \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{8 a^{3} d}-\frac {B \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2 a^{3} d}+\frac {7 i B \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{8 a^{3} d}-\frac {31 A}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )}-\frac {17 i B}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )}+\frac {9 i A}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{2}}-\frac {7 B}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{2}}+\frac {A}{6 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{3}}+\frac {i B}{6 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{3}}\)	\(226\)

[In]

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

[Out]

-15/8*I*x/a^3*B-49/8*x/a^3*A+11/16/a^3/d*exp(-2*I*(d*x+c))*B-23/16*I/a^3/d*exp(-2*I*(d*x+c))*A+5/32/a^3/d*exp(
-4*I*(d*x+c))*B-7/32*I/a^3/d*exp(-4*I*(d*x+c))*A+1/48/a^3/d*exp(-6*I*(d*x+c))*B-1/48*I/a^3/d*exp(-6*I*(d*x+c))
*A-2*I/a^3/d*B*c-6/a^3/d*A*c-2*I*A/a^3/d/(exp(2*I*(d*x+c))-1)+1/a^3/d*ln(exp(2*I*(d*x+c))-1)*B-3*I/a^3/d*ln(ex
p(2*I*(d*x+c))-1)*A

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.95 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {12 \, {\left (49 \, A + 15 i \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} - 6 \, {\left (2 \, {\left (49 \, A + 15 i \, B\right )} d x - 55 i \, A + 11 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (-39 i \, A + 17 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (19 i \, A - 13 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 96 \, {\left ({\left (3 i \, A - B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (-3 i \, A + B\right )} e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 2 i \, A + 2 \, B}{96 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} - a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \]

[In]

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

[Out]

-1/96*(12*(49*A + 15*I*B)*d*x*e^(8*I*d*x + 8*I*c) - 6*(2*(49*A + 15*I*B)*d*x - 55*I*A + 11*B)*e^(6*I*d*x + 6*I
*c) + 3*(-39*I*A + 17*B)*e^(4*I*d*x + 4*I*c) - (19*I*A - 13*B)*e^(2*I*d*x + 2*I*c) + 96*((3*I*A - B)*e^(8*I*d*
x + 8*I*c) + (-3*I*A + B)*e^(6*I*d*x + 6*I*c))*log(e^(2*I*d*x + 2*I*c) - 1) - 2*I*A + 2*B)/(a^3*d*e^(8*I*d*x +
8*I*c) - a^3*d*e^(6*I*d*x + 6*I*c))

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.86 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=- \frac {2 i A}{a^{3} d e^{2 i c} e^{2 i d x} - a^{3} d} + \begin {cases} \frac {\left (\left (- 512 i A a^{6} d^{2} e^{6 i c} + 512 B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (- 5376 i A a^{6} d^{2} e^{8 i c} + 3840 B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (- 35328 i A a^{6} d^{2} e^{10 i c} + 16896 B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {- 49 A - 15 i B}{8 a^{3}} + \frac {\left (- 49 A e^{6 i c} - 23 A e^{4 i c} - 7 A e^{2 i c} - A - 15 i B e^{6 i c} - 11 i B e^{4 i c} - 5 i B e^{2 i c} - i B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (- 49 A - 15 i B\right )}{8 a^{3}} - \frac {i \left (3 A + i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{3} d} \]

[In]

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

[Out]

-2*I*A/(a**3*d*exp(2*I*c)*exp(2*I*d*x) - a**3*d) + Piecewise((((-512*I*A*a**6*d**2*exp(6*I*c) + 512*B*a**6*d**
2*exp(6*I*c))*exp(-6*I*d*x) + (-5376*I*A*a**6*d**2*exp(8*I*c) + 3840*B*a**6*d**2*exp(8*I*c))*exp(-4*I*d*x) + (
-35328*I*A*a**6*d**2*exp(10*I*c) + 16896*B*a**6*d**2*exp(10*I*c))*exp(-2*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3
), Ne(a**9*d**3*exp(12*I*c), 0)), (x*(-(-49*A - 15*I*B)/(8*a**3) + (-49*A*exp(6*I*c) - 23*A*exp(4*I*c) - 7*A*e
xp(2*I*c) - A - 15*I*B*exp(6*I*c) - 11*I*B*exp(4*I*c) - 5*I*B*exp(2*I*c) - I*B)*exp(-6*I*c)/(8*a**3)), True))
+ x*(-49*A - 15*I*B)/(8*a**3) - I*(3*A + I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/(a**3*d)

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.91 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.99 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\frac {6 \, {\left (i \, A + B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} + \frac {6 \, {\left (-49 i \, A + 15 \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {96 \, {\left (3 i \, A - B\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{3}} + \frac {96 \, {\left (-3 i \, A \tan \left (d x + c\right ) + B \tan \left (d x + c\right ) + A\right )}}{a^{3} \tan \left (d x + c\right )} + \frac {539 \, A \tan \left (d x + c\right )^{3} + 165 i \, B \tan \left (d x + c\right )^{3} - 1821 i \, A \tan \left (d x + c\right )^{2} + 579 \, B \tan \left (d x + c\right )^{2} - 2085 \, A \tan \left (d x + c\right ) - 699 i \, B \tan \left (d x + c\right ) + 819 i \, A - 301 \, B}{a^{3} {\left (i \, \tan \left (d x + c\right ) + 1\right )}^{3}}}{96 \, d} \]

[In]

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

[Out]

-1/96*(6*(I*A + B)*log(tan(d*x + c) + I)/a^3 + 6*(-49*I*A + 15*B)*log(tan(d*x + c) - I)/a^3 + 96*(3*I*A - B)*l
og(tan(d*x + c))/a^3 + 96*(-3*I*A*tan(d*x + c) + B*tan(d*x + c) + A)/(a^3*tan(d*x + c)) + (539*A*tan(d*x + c)^
3 + 165*I*B*tan(d*x + c)^3 - 1821*I*A*tan(d*x + c)^2 + 579*B*tan(d*x + c)^2 - 2085*A*tan(d*x + c) - 699*I*B*ta
n(d*x + c) + 819*I*A - 301*B)/(a^3*(I*tan(d*x + c) + 1)^3))/d

Mupad [B] (verification not implemented)

Time = 7.32 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.08 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {25\,A}{8\,a^3}+\frac {B\,7{}\mathrm {i}}{8\,a^3}\right )-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-\frac {17\,B}{8\,a^3}+\frac {A\,63{}\mathrm {i}}{8\,a^3}\right )+\frac {A\,1{}\mathrm {i}}{a^3}-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {71\,A}{12\,a^3}+\frac {B\,17{}\mathrm {i}}{12\,a^3}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,3{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-B+A\,3{}\mathrm {i}\right )}{a^3\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{16\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-15\,B+A\,49{}\mathrm {i}\right )}{16\,a^3\,d} \]

[In]

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

[Out]

(log(tan(c + d*x) - 1i)*(A*49i - 15*B))/(16*a^3*d) - (log(tan(c + d*x))*(A*3i - B))/(a^3*d) - (log(tan(c + d*x
) + 1i)*(A*1i + B))/(16*a^3*d) - (tan(c + d*x)^3*((25*A)/(8*a^3) + (B*7i)/(8*a^3)) - tan(c + d*x)^2*((A*63i)/(
8*a^3) - (17*B)/(8*a^3)) + (A*1i)/a^3 - tan(c + d*x)*((71*A)/(12*a^3) + (B*17i)/(12*a^3)))/(d*(tan(c + d*x)*1i
- 3*tan(c + d*x)^2 - tan(c + d*x)^3*3i + tan(c + d*x)^4))