\(\int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) [83]

Optimal result

Integrand size = 24, antiderivative size = 159 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {65 x}{16 a^4}-\frac {65 \cot (c+d x)}{16 a^4 d}-\frac {4 i \log (\sin (c+d x))}{a^4 d}+\frac {31 \cot (c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {2 \cot (c+d x)}{a^4 d (1+i \tan (c+d x))}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^3} \]

[Out]

-65/16*x/a^4-65/16*cot(d*x+c)/a^4/d-4*I*ln(sin(d*x+c))/a^4/d+31/48*cot(d*x+c)/a^4/d/(1+I*tan(d*x+c))^2+2*cot(d
*x+c)/a^4/d/(1+I*tan(d*x+c))+1/8*cot(d*x+c)/d/(a+I*a*tan(d*x+c))^4+7/24*cot(d*x+c)/a/d/(a+I*a*tan(d*x+c))^3

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3640, 3677, 3610, 3612, 3556} \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {65 \cot (c+d x)}{16 a^4 d}-\frac {4 i \log (\sin (c+d x))}{a^4 d}+\frac {2 \cot (c+d x)}{a^4 d (1+i \tan (c+d x))}+\frac {31 \cot (c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac {65 x}{16 a^4}+\frac {7 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4} \]

[In]

Int[Cot[c + d*x]^2/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-65*x)/(16*a^4) - (65*Cot[c + d*x])/(16*a^4*d) - ((4*I)*Log[Sin[c + d*x]])/(a^4*d) + (31*Cot[c + d*x])/(48*a^
4*d*(1 + I*Tan[c + d*x])^2) + (2*Cot[c + d*x])/(a^4*d*(1 + I*Tan[c + d*x])) + Cot[c + d*x]/(8*d*(a + I*a*Tan[c
+ d*x])^4) + (7*Cot[c + d*x])/(24*a*d*(a + I*a*Tan[c + d*x])^3)

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 3640

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

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 {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {\int \frac {\cot ^2(c+d x) (9 a-5 i a \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx}{8 a^2} \\ & = \frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\cot ^2(c+d x) \left (68 a^2-56 i a^2 \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{48 a^4} \\ & = \frac {31 \cot (c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\cot ^2(c+d x) \left (396 a^3-372 i a^3 \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{192 a^6} \\ & = \frac {31 \cot (c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac {2 \cot (c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\int \cot ^2(c+d x) \left (1560 a^4-1536 i a^4 \tan (c+d x)\right ) \, dx}{384 a^8} \\ & = -\frac {65 \cot (c+d x)}{16 a^4 d}+\frac {31 \cot (c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac {2 \cot (c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\int \cot (c+d x) \left (-1536 i a^4-1560 a^4 \tan (c+d x)\right ) \, dx}{384 a^8} \\ & = -\frac {65 x}{16 a^4}-\frac {65 \cot (c+d x)}{16 a^4 d}+\frac {31 \cot (c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac {2 \cot (c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {(4 i) \int \cot (c+d x) \, dx}{a^4} \\ & = -\frac {65 x}{16 a^4}-\frac {65 \cot (c+d x)}{16 a^4 d}-\frac {4 i \log (\sin (c+d x))}{a^4 d}+\frac {31 \cot (c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac {2 \cot (c+d x)}{d \left (a^4+i a^4 \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 = 1.76 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\frac {14 \cot ^4(c+d x)}{a^4 (i+\cot (c+d x))^3}+\frac {\frac {31 \cot ^3(c+d x)}{(i+\cot (c+d x))^2}+\frac {96 \cot ^2(c+d x)}{i+\cot (c+d x)}-195 \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )-192 i (\log (\cos (c+d x))+\log (\tan (c+d x)))}{a^4}+\frac {6 \cot (c+d x)}{(a+i a \tan (c+d x))^4}}{48 d} \]

[In]

Integrate[Cot[c + d*x]^2/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((14*Cot[c + d*x]^4)/(a^4*(I + Cot[c + d*x])^3) + ((31*Cot[c + d*x]^3)/(I + Cot[c + d*x])^2 + (96*Cot[c + d*x]
^2)/(I + Cot[c + d*x]) - 195*Cot[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2] - (192*I)*(Log[Cos[
c + d*x]] + Log[Tan[c + d*x]]))/a^4 + (6*Cot[c + d*x])/(a + I*a*Tan[c + d*x])^4)/(48*d)

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.83

method	result	size
risch	\(-\frac {129 x}{16 a^{4}}-\frac {9 i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{4} d}-\frac {15 i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{4} d}-\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{12 a^{4} d}-\frac {i {\mathrm e}^{-8 i \left (d x +c \right )}}{128 a^{4} d}-\frac {8 c}{a^{4} d}-\frac {2 i}{d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {4 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{4} d}\)	\(132\)
derivativedivides	\(\frac {2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{4}}-\frac {65 \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {i}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {17 i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {5}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {49}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {4 i \ln \left (\tan \left (d x +c \right )\right )}{a^{4} d}-\frac {1}{a^{4} d \tan \left (d x +c \right )}\)	\(147\)
default	\(\frac {2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{4}}-\frac {65 \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {i}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {17 i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {5}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {49}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {4 i \ln \left (\tan \left (d x +c \right )\right )}{a^{4} d}-\frac {1}{a^{4} d \tan \left (d x +c \right )}\)	\(147\)
norman	\(\frac {-\frac {7 i \left (\tan ^{5}\left (d x +c \right )\right )}{a d}-\frac {1}{a d}-\frac {949 \left (\tan ^{4}\left (d x +c \right )\right )}{48 a d}-\frac {715 \left (\tan ^{6}\left (d x +c \right )\right )}{48 a d}-\frac {65 \left (\tan ^{8}\left (d x +c \right )\right )}{16 a d}-\frac {65 x \tan \left (d x +c \right )}{16 a}-\frac {65 x \left (\tan ^{3}\left (d x +c \right )\right )}{4 a}-\frac {195 x \left (\tan ^{5}\left (d x +c \right )\right )}{8 a}-\frac {65 x \left (\tan ^{7}\left (d x +c \right )\right )}{4 a}-\frac {65 x \left (\tan ^{9}\left (d x +c \right )\right )}{16 a}-\frac {175 \left (\tan ^{2}\left (d x +c \right )\right )}{16 a d}-\frac {2 i \left (\tan ^{7}\left (d x +c \right )\right )}{a d}-\frac {14 i \tan \left (d x +c \right )}{3 d a}-\frac {26 i \left (\tan ^{3}\left (d x +c \right )\right )}{3 d a}}{\tan \left (d x +c \right ) a^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{4}}-\frac {4 i \ln \left (\tan \left (d x +c \right )\right )}{a^{4} d}+\frac {2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{4}}\)	\(269\)

[In]

int(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

-129/16*x/a^4-9/4*I/a^4/d*exp(-2*I*(d*x+c))-15/32*I/a^4/d*exp(-4*I*(d*x+c))-1/12*I/a^4/d*exp(-6*I*(d*x+c))-1/1
28*I/a^4/d*exp(-8*I*(d*x+c))-8/a^4/d*c-2*I/d/a^4/(exp(2*I*(d*x+c))-1)-4*I/a^4/d*ln(exp(2*I*(d*x+c))-1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.86 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {3096 \, d x e^{\left (10 i \, d x + 10 i \, c\right )} - 24 \, {\left (129 \, d x - 68 i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 1536 \, {\left (i \, e^{\left (10 i \, d x + 10 i \, c\right )} - i \, e^{\left (8 i \, d x + 8 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 684 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 148 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 29 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i}{384 \, {\left (a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} - a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )}\right )}} \]

[In]

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/384*(3096*d*x*e^(10*I*d*x + 10*I*c) - 24*(129*d*x - 68*I)*e^(8*I*d*x + 8*I*c) + 1536*(I*e^(10*I*d*x + 10*I*
c) - I*e^(8*I*d*x + 8*I*c))*log(e^(2*I*d*x + 2*I*c) - 1) - 684*I*e^(6*I*d*x + 6*I*c) - 148*I*e^(4*I*d*x + 4*I*
c) - 29*I*e^(2*I*d*x + 2*I*c) - 3*I)/(a^4*d*e^(10*I*d*x + 10*I*c) - a^4*d*e^(8*I*d*x + 8*I*c))

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.58 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {\left (- 442368 i a^{12} d^{3} e^{18 i c} e^{- 2 i d x} - 92160 i a^{12} d^{3} e^{16 i c} e^{- 4 i d x} - 16384 i a^{12} d^{3} e^{14 i c} e^{- 6 i d x} - 1536 i a^{12} d^{3} e^{12 i c} e^{- 8 i d x}\right ) e^{- 20 i c}}{196608 a^{16} d^{4}} & \text {for}\: a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (\frac {\left (- 129 e^{8 i c} - 72 e^{6 i c} - 30 e^{4 i c} - 8 e^{2 i c} - 1\right ) e^{- 8 i c}}{16 a^{4}} + \frac {129}{16 a^{4}}\right ) & \text {otherwise} \end {cases} - \frac {2 i}{a^{4} d e^{2 i c} e^{2 i d x} - a^{4} d} - \frac {129 x}{16 a^{4}} - \frac {4 i \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{4} d} \]

[In]

integrate(cot(d*x+c)**2/(a+I*a*tan(d*x+c))**4,x)

[Out]

Piecewise(((-442368*I*a**12*d**3*exp(18*I*c)*exp(-2*I*d*x) - 92160*I*a**12*d**3*exp(16*I*c)*exp(-4*I*d*x) - 16
384*I*a**12*d**3*exp(14*I*c)*exp(-6*I*d*x) - 1536*I*a**12*d**3*exp(12*I*c)*exp(-8*I*d*x))*exp(-20*I*c)/(196608
*a**16*d**4), Ne(a**16*d**4*exp(20*I*c), 0)), (x*((-129*exp(8*I*c) - 72*exp(6*I*c) - 30*exp(4*I*c) - 8*exp(2*I
*c) - 1)*exp(-8*I*c)/(16*a**4) + 129/(16*a**4)), True)) - 2*I/(a**4*d*exp(2*I*c)*exp(2*I*d*x) - a**4*d) - 129*
x/(16*a**4) - 4*I*log(exp(2*I*d*x) - exp(-2*I*c))/(a**4*d)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 1.15 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.77 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\frac {12 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} - \frac {1548 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} + \frac {1536 i \, \log \left (\tan \left (d x + c\right )\right )}{a^{4}} + \frac {384 \, {\left (-4 i \, \tan \left (d x + c\right ) + 1\right )}}{a^{4} \tan \left (d x + c\right )} + \frac {3225 i \, \tan \left (d x + c\right )^{4} + 14076 \, \tan \left (d x + c\right )^{3} - 23286 i \, \tan \left (d x + c\right )^{2} - 17404 \, \tan \left (d x + c\right ) + 5017 i}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]

[In]

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

-1/384*(12*I*log(tan(d*x + c) + I)/a^4 - 1548*I*log(tan(d*x + c) - I)/a^4 + 1536*I*log(tan(d*x + c))/a^4 + 384
*(-4*I*tan(d*x + c) + 1)/(a^4*tan(d*x + c)) + (3225*I*tan(d*x + c)^4 + 14076*tan(d*x + c)^3 - 23286*I*tan(d*x
+ c)^2 - 17404*tan(d*x + c) + 5017*I)/(a^4*(tan(d*x + c) - I)^4))/d

Mupad [B] (verification not implemented)

Time = 4.62 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.04 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,129{}\mathrm {i}}{32\,a^4\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,a^4\,d}-\frac {\frac {1}{a^4}-\frac {851\,{\mathrm {tan}\left (c+d\,x\right )}^2}{48\,a^4}+\frac {65\,{\mathrm {tan}\left (c+d\,x\right )}^4}{16\,a^4}+\frac {\mathrm {tan}\left (c+d\,x\right )\,26{}\mathrm {i}}{3\,a^4}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,57{}\mathrm {i}}{4\,a^4}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^5-{\mathrm {tan}\left (c+d\,x\right )}^4\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^3+{\mathrm {tan}\left (c+d\,x\right )}^2\,4{}\mathrm {i}+\mathrm {tan}\left (c+d\,x\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,4{}\mathrm {i}}{a^4\,d} \]

[In]

int(cot(c + d*x)^2/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

(log(tan(c + d*x) - 1i)*129i)/(32*a^4*d) - (log(tan(c + d*x) + 1i)*1i)/(32*a^4*d) - ((tan(c + d*x)*26i)/(3*a^4
) + 1/a^4 - (851*tan(c + d*x)^2)/(48*a^4) - (tan(c + d*x)^3*57i)/(4*a^4) + (65*tan(c + d*x)^4)/(16*a^4))/(d*(t
an(c + d*x) + tan(c + d*x)^2*4i - 6*tan(c + d*x)^3 - tan(c + d*x)^4*4i + tan(c + d*x)^5)) - (log(tan(c + d*x))
*4i)/(a^4*d)