\(\int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx\) [43]

Optimal result

Integrand size = 34, antiderivative size = 155 \[ \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {(5 A+3 i B) x}{2 a}+\frac {(5 A+3 i B) \cot (c+d x)}{2 a d}+\frac {(i A-B) \cot ^2(c+d x)}{a d}-\frac {(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac {2 (i A-B) \log (\sin (c+d x))}{a d}+\frac {(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 155, 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} \[ \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac {(-B+i A) \cot ^2(c+d x)}{a d}+\frac {(5 A+3 i B) \cot (c+d x)}{2 a d}+\frac {2 (-B+i A) \log (\sin (c+d x))}{a d}+\frac {(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {x (5 A+3 i B)}{2 a} \]

[In]

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

[Out]

((5*A + (3*I)*B)*x)/(2*a) + ((5*A + (3*I)*B)*Cot[c + d*x])/(2*a*d) + ((I*A - B)*Cot[c + d*x]^2)/(a*d) - ((5*A
+ (3*I)*B)*Cot[c + d*x]^3)/(6*a*d) + (2*(I*A - B)*Log[Sin[c + d*x]])/(a*d) + ((A + I*B)*Cot[c + d*x]^3)/(2*d*(
a + I*a*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 ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \cot ^4(c+d x) (a (5 A+3 i B)-4 a (i A-B) \tan (c+d x)) \, dx}{2 a^2} \\ & = -\frac {(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac {(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \cot ^3(c+d x) (-4 a (i A-B)-a (5 A+3 i B) \tan (c+d x)) \, dx}{2 a^2} \\ & = \frac {(i A-B) \cot ^2(c+d x)}{a d}-\frac {(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac {(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \cot ^2(c+d x) (-a (5 A+3 i B)+4 a (i A-B) \tan (c+d x)) \, dx}{2 a^2} \\ & = \frac {(5 A+3 i B) \cot (c+d x)}{2 a d}+\frac {(i A-B) \cot ^2(c+d x)}{a d}-\frac {(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac {(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \cot (c+d x) (4 a (i A-B)+a (5 A+3 i B) \tan (c+d x)) \, dx}{2 a^2} \\ & = \frac {(5 A+3 i B) x}{2 a}+\frac {(5 A+3 i B) \cot (c+d x)}{2 a d}+\frac {(i A-B) \cot ^2(c+d x)}{a d}-\frac {(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac {(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(2 (i A-B)) \int \cot (c+d x) \, dx}{a} \\ & = \frac {(5 A+3 i B) x}{2 a}+\frac {(5 A+3 i B) \cot (c+d x)}{2 a d}+\frac {(i A-B) \cot ^2(c+d x)}{a d}-\frac {(5 A+3 i B) \cot ^3(c+d x)}{6 a d}+\frac {2 (i A-B) \log (\sin (c+d x))}{a d}+\frac {(A+i B) \cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 1.90 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.74 \[ \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {\frac {3 (A+i B) \cot ^4(c+d x)}{i+\cot (c+d x)}-(5 A+3 i B) \cot ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\tan ^2(c+d x)\right )+6 i (A+i B) \left (\cot ^2(c+d x)+2 (\log (\cos (c+d x))+\log (\tan (c+d x)))\right )}{6 a d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.19

method	result	size
risch	\(\frac {7 i x B}{2 a}+\frac {9 x A}{2 a}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} B}{4 a d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} A}{4 a d}+\frac {4 i B c}{a d}+\frac {4 A c}{a d}+\frac {4 i A \,{\mathrm e}^{4 i \left (d x +c \right )}-6 i A \,{\mathrm e}^{2 i \left (d x +c \right )}+2 B \,{\mathrm e}^{2 i \left (d x +c \right )}+\frac {14 i A}{3}-2 B}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{a d}+\frac {2 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A}{a d}\)	\(185\)
norman	\(\frac {-\frac {A}{3 a d}-\frac {\left (-i A +B \right ) \tan \left (d x +c \right )}{2 a d}+\frac {\left (3 i B +5 A \right ) \left (\tan ^{2}\left (d x +c \right )\right )}{3 a d}+\frac {\left (3 i B +5 A \right ) \left (\tan ^{4}\left (d x +c \right )\right )}{2 a d}+\frac {\left (3 i B +5 A \right ) x \left (\tan ^{3}\left (d x +c \right )\right )}{2 a}+\frac {\left (3 i B +5 A \right ) x \left (\tan ^{5}\left (d x +c \right )\right )}{2 a}-\frac {\left (-i A +B \right ) \left (\tan ^{3}\left (d x +c \right )\right )}{a d}}{\tan \left (d x +c \right )^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )}+\frac {\left (-i A +B \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{a d}-\frac {2 \left (-i A +B \right ) \ln \left (\tan \left (d x +c \right )\right )}{a d}\)	\(212\)
derivativedivides	\(\frac {i B}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {5 A \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}+\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d a}+\frac {i B}{a d \tan \left (d x +c \right )}+\frac {A}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {3 i B \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}+\frac {2 i A \ln \left (\tan \left (d x +c \right )\right )}{a d}+\frac {i A}{2 a d \tan \left (d x +c \right )^{2}}-\frac {A}{3 a d \tan \left (d x +c \right )^{3}}-\frac {i A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d a}+\frac {2 A}{a d \tan \left (d x +c \right )}-\frac {B}{2 a d \tan \left (d x +c \right )^{2}}-\frac {2 B \ln \left (\tan \left (d x +c \right )\right )}{a d}\)	\(236\)
default	\(\frac {i B}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {5 A \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}+\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d a}+\frac {i B}{a d \tan \left (d x +c \right )}+\frac {A}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {3 i B \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}+\frac {2 i A \ln \left (\tan \left (d x +c \right )\right )}{a d}+\frac {i A}{2 a d \tan \left (d x +c \right )^{2}}-\frac {A}{3 a d \tan \left (d x +c \right )^{3}}-\frac {i A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d a}+\frac {2 A}{a d \tan \left (d x +c \right )}-\frac {B}{2 a d \tan \left (d x +c \right )^{2}}-\frac {2 B \ln \left (\tan \left (d x +c \right )\right )}{a d}\)	\(236\)

[In]

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

[Out]

7/2*I*x/a*B+9/2*x/a*A-1/4/a/d*exp(-2*I*(d*x+c))*B+1/4*I/a/d*exp(-2*I*(d*x+c))*A+4*I/a/d*B*c+4/a/d*A*c+2/3*(6*I
*A*exp(4*I*(d*x+c))-9*I*A*exp(2*I*(d*x+c))+3*B*exp(2*I*(d*x+c))+7*I*A-3*B)/a/d/(exp(2*I*(d*x+c))-1)^3-2/a/d*ln
(exp(2*I*(d*x+c))-1)*B+2*I/a/d*ln(exp(2*I*(d*x+c))-1)*A

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.61 \[ \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {6 \, {\left (9 \, A + 7 i \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} - 3 \, {\left (6 \, {\left (9 \, A + 7 i \, B\right )} d x - 17 i \, A + B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (6 \, {\left (9 \, A + 7 i \, B\right )} d x - 27 i \, A + 11 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (6 \, {\left (9 \, A + 7 i \, B\right )} d x - 65 i \, A + 33 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 24 \, {\left ({\left (-i \, A + B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 3 \, {\left (i \, A - B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (-i \, A + B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (i \, A - B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 3 i \, A + 3 \, B}{12 \, {\left (a d e^{\left (8 i \, d x + 8 i \, c\right )} - 3 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]

[In]

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

[Out]

1/12*(6*(9*A + 7*I*B)*d*x*e^(8*I*d*x + 8*I*c) - 3*(6*(9*A + 7*I*B)*d*x - 17*I*A + B)*e^(6*I*d*x + 6*I*c) + 3*(
6*(9*A + 7*I*B)*d*x - 27*I*A + 11*B)*e^(4*I*d*x + 4*I*c) - (6*(9*A + 7*I*B)*d*x - 65*I*A + 33*B)*e^(2*I*d*x +
2*I*c) - 24*((-I*A + B)*e^(8*I*d*x + 8*I*c) + 3*(I*A - B)*e^(6*I*d*x + 6*I*c) + 3*(-I*A + B)*e^(4*I*d*x + 4*I*
c) + (I*A - B)*e^(2*I*d*x + 2*I*c))*log(e^(2*I*d*x + 2*I*c) - 1) - 3*I*A + 3*B)/(a*d*e^(8*I*d*x + 8*I*c) - 3*a
*d*e^(6*I*d*x + 6*I*c) + 3*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 = 0.40 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.63 \[ \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {12 i A e^{4 i c} e^{4 i d x} + 14 i A - 6 B + \left (- 18 i A e^{2 i c} + 6 B e^{2 i c}\right ) e^{2 i d x}}{3 a d e^{6 i c} e^{6 i d x} - 9 a d e^{4 i c} e^{4 i d x} + 9 a d e^{2 i c} e^{2 i d x} - 3 a d} + \begin {cases} \frac {\left (i A - B\right ) e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: a d e^{2 i c} \neq 0 \\x \left (- \frac {9 A + 7 i B}{2 a} + \frac {\left (9 A e^{2 i c} + A + 7 i B e^{2 i c} + i B\right ) e^{- 2 i c}}{2 a}\right ) & \text {otherwise} \end {cases} + \frac {x \left (9 A + 7 i B\right )}{2 a} + \frac {2 i \left (A + i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a d} \]

[In]

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

[Out]

(12*I*A*exp(4*I*c)*exp(4*I*d*x) + 14*I*A - 6*B + (-18*I*A*exp(2*I*c) + 6*B*exp(2*I*c))*exp(2*I*d*x))/(3*a*d*ex
p(6*I*c)*exp(6*I*d*x) - 9*a*d*exp(4*I*c)*exp(4*I*d*x) + 9*a*d*exp(2*I*c)*exp(2*I*d*x) - 3*a*d) + Piecewise(((I
*A - B)*exp(-2*I*c)*exp(-2*I*d*x)/(4*a*d), Ne(a*d*exp(2*I*c), 0)), (x*(-(9*A + 7*I*B)/(2*a) + (9*A*exp(2*I*c)
+ A + 7*I*B*exp(2*I*c) + I*B)*exp(-2*I*c)/(2*a)), True)) + x*(9*A + 7*I*B)/(2*a) + 2*I*(A + I*B)*log(exp(2*I*d
*x) - exp(-2*I*c))/(a*d)

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.89 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.19 \[ \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {\frac {3 \, {\left (-i \, A - B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a} + \frac {3 \, {\left (9 i \, A - 7 \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a} + \frac {24 \, {\left (-i \, A + B\right )} \log \left (\tan \left (d x + c\right )\right )}{a} + \frac {3 \, {\left (-9 i \, A \tan \left (d x + c\right ) + 7 \, B \tan \left (d x + c\right ) - 11 \, A - 9 i \, B\right )}}{a {\left (\tan \left (d x + c\right ) - i\right )}} + \frac {2 i \, {\left (22 \, A \tan \left (d x + c\right )^{3} + 22 i \, B \tan \left (d x + c\right )^{3} + 12 i \, A \tan \left (d x + c\right )^{2} - 6 \, B \tan \left (d x + c\right )^{2} - 3 \, A \tan \left (d x + c\right ) - 3 i \, B \tan \left (d x + c\right ) - 2 i \, A\right )}}{a \tan \left (d x + c\right )^{3}}}{12 \, d} \]

[In]

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

[Out]

-1/12*(3*(-I*A - B)*log(tan(d*x + c) + I)/a + 3*(9*I*A - 7*B)*log(tan(d*x + c) - I)/a + 24*(-I*A + B)*log(tan(
d*x + c))/a + 3*(-9*I*A*tan(d*x + c) + 7*B*tan(d*x + c) - 11*A - 9*I*B)/(a*(tan(d*x + c) - I)) + 2*I*(22*A*tan
(d*x + c)^3 + 22*I*B*tan(d*x + c)^3 + 12*I*A*tan(d*x + c)^2 - 6*B*tan(d*x + c)^2 - 3*A*tan(d*x + c) - 3*I*B*ta
n(d*x + c) - 2*I*A)/(a*tan(d*x + c)^3))/d

Mupad [B] (verification not implemented)

Time = 7.95 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.12 \[ \int \frac {\cot ^4(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {3\,A}{2\,a}+\frac {B\,1{}\mathrm {i}}{2\,a}\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (-\frac {3\,B}{2\,a}+\frac {A\,5{}\mathrm {i}}{2\,a}\right )-\frac {A}{3\,a}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {B}{2\,a}+\frac {A\,1{}\mathrm {i}}{6\,a}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}+{\mathrm {tan}\left (c+d\,x\right )}^3\right )}+\frac {2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{a\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{4\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-7\,B+A\,9{}\mathrm {i}\right )}{4\,a\,d} \]

[In]

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

[Out]

(tan(c + d*x)^2*((3*A)/(2*a) + (B*1i)/(2*a)) + tan(c + d*x)^3*((A*5i)/(2*a) - (3*B)/(2*a)) - A/(3*a) + tan(c +
d*x)*((A*1i)/(6*a) - B/(2*a)))/(d*(tan(c + d*x)^3 + tan(c + d*x)^4*1i)) + (2*log(tan(c + d*x))*(A*1i - B))/(a
*d) + (log(tan(c + d*x) + 1i)*(A*1i + B))/(4*a*d) - (log(tan(c + d*x) - 1i)*(A*9i - 7*B))/(4*a*d)