\(\int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\) [51]

Optimal result

Integrand size = 34, antiderivative size = 191 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {(7 A+25 i B) x}{8 a^3}-\frac {(i A-3 B) \log (\cos (c+d x))}{a^3 d}+\frac {(7 A+25 i B) \tan (c+d x)}{8 a^3 d}+\frac {(i A-B) \tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(5 A+11 i B) \tan ^3(c+d x)}{24 a d (a+i a \tan (c+d x))^2}-\frac {(i A-3 B) \tan ^2(c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

-1/8*(7*A+25*I*B)*x/a^3-(I*A-3*B)*ln(cos(d*x+c))/a^3/d+1/8*(7*A+25*I*B)*tan(d*x+c)/a^3/d+1/6*(I*A-B)*tan(d*x+c
)^4/d/(a+I*a*tan(d*x+c))^3+1/24*(5*A+11*I*B)*tan(d*x+c)^3/a/d/(a+I*a*tan(d*x+c))^2-1/2*(I*A-3*B)*tan(d*x+c)^2/
d/(a^3+I*a^3*tan(d*x+c))

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {3676, 3606, 3556} \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {(-3 B+i A) \tan ^2(c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(7 A+25 i B) \tan (c+d x)}{8 a^3 d}-\frac {(-3 B+i A) \log (\cos (c+d x))}{a^3 d}-\frac {x (7 A+25 i B)}{8 a^3}+\frac {(-B+i A) \tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(5 A+11 i B) \tan ^3(c+d x)}{24 a d (a+i a \tan (c+d x))^2} \]

[In]

Int[(Tan[c + d*x]^4*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]

[Out]

-1/8*((7*A + (25*I)*B)*x)/a^3 - ((I*A - 3*B)*Log[Cos[c + d*x]])/(a^3*d) + ((7*A + (25*I)*B)*Tan[c + d*x])/(8*a
^3*d) + ((I*A - B)*Tan[c + d*x]^4)/(6*d*(a + I*a*Tan[c + d*x])^3) + ((5*A + (11*I)*B)*Tan[c + d*x]^3)/(24*a*d*
(a + I*a*Tan[c + d*x])^2) - ((I*A - 3*B)*Tan[c + d*x]^2)/(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 3606

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[b*d*(Tan[e + f*x]/f), x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3676

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*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f
*m)), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d
*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, 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 {(i A-B) \tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^3(c+d x) (4 a (i A-B)+a (A+7 i B) \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2} \\ & = \frac {(i A-B) \tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(5 A+11 i B) \tan ^3(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\tan ^2(c+d x) \left (-3 a^2 (5 A+11 i B)+3 a^2 (3 i A-13 B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4} \\ & = \frac {(i A-B) \tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(5 A+11 i B) \tan ^3(c+d x)}{24 a d (a+i a \tan (c+d x))^2}-\frac {(i A-3 B) \tan ^2(c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\int \tan (c+d x) \left (-48 a^3 (i A-3 B)-6 a^3 (7 A+25 i B) \tan (c+d x)\right ) \, dx}{48 a^6} \\ & = -\frac {(7 A+25 i B) x}{8 a^3}+\frac {(7 A+25 i B) \tan (c+d x)}{8 a^3 d}+\frac {(i A-B) \tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(5 A+11 i B) \tan ^3(c+d x)}{24 a d (a+i a \tan (c+d x))^2}-\frac {(i A-3 B) \tan ^2(c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(i A-3 B) \int \tan (c+d x) \, dx}{a^3} \\ & = -\frac {(7 A+25 i B) x}{8 a^3}-\frac {(i A-3 B) \log (\cos (c+d x))}{a^3 d}+\frac {(7 A+25 i B) \tan (c+d x)}{8 a^3 d}+\frac {(i A-B) \tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(5 A+11 i B) \tan ^3(c+d x)}{24 a d (a+i a \tan (c+d x))^2}-\frac {(i A-3 B) \tan ^2(c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.78 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.41 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {-3 ((15 A+49 i B) \log (i-\tan (c+d x))+(A-i B) \log (i+\tan (c+d x)))+3 (14 i A-50 B+3 (-15 i A+49 B) \log (i-\tan (c+d x))+(-3 i A-3 B) \log (i+\tan (c+d x))) \tan (c+d x)+3 (-2 (17 A+63 i B)+3 (15 A+49 i B) \log (i-\tan (c+d x))+3 (A-i B) \log (i+\tan (c+d x))) \tan ^2(c+d x)+(-68 i A+284 B+(45 i A-147 B) \log (i-\tan (c+d x))+3 (i A+B) \log (i+\tan (c+d x))) \tan ^3(c+d x)+48 i B \tan ^4(c+d x)}{48 a^3 d (-i+\tan (c+d x))^3} \]

[In]

Integrate[(Tan[c + d*x]^4*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(-3*((15*A + (49*I)*B)*Log[I - Tan[c + d*x]] + (A - I*B)*Log[I + Tan[c + d*x]]) + 3*((14*I)*A - 50*B + 3*((-15
*I)*A + 49*B)*Log[I - Tan[c + d*x]] + ((-3*I)*A - 3*B)*Log[I + Tan[c + d*x]])*Tan[c + d*x] + 3*(-2*(17*A + (63
*I)*B) + 3*(15*A + (49*I)*B)*Log[I - Tan[c + d*x]] + 3*(A - I*B)*Log[I + Tan[c + d*x]])*Tan[c + d*x]^2 + ((-68
*I)*A + 284*B + ((45*I)*A - 147*B)*Log[I - Tan[c + d*x]] + 3*(I*A + B)*Log[I + Tan[c + d*x]])*Tan[c + d*x]^3 +
(48*I)*B*Tan[c + d*x]^4)/(48*a^3*d*(-I + Tan[c + d*x])^3)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.12

method	result	size
risch	\(-\frac {49 i x B}{8 a^{3}}-\frac {15 x A}{8 a^{3}}-\frac {23 \,{\mathrm e}^{-2 i \left (d x +c \right )} B}{16 a^{3} d}+\frac {11 i {\mathrm e}^{-2 i \left (d x +c \right )} A}{16 a^{3} d}+\frac {7 \,{\mathrm e}^{-4 i \left (d x +c \right )} B}{32 a^{3} d}-\frac {5 i {\mathrm e}^{-4 i \left (d x +c \right )} A}{32 a^{3} d}-\frac {{\mathrm e}^{-6 i \left (d x +c \right )} B}{48 a^{3} d}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )} A}{48 a^{3} d}-\frac {6 i B c}{a^{3} d}-\frac {2 A c}{a^{3} d}-\frac {2 B}{d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{a^{3} d}-\frac {i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A}{a^{3} d}\)	\(214\)
derivativedivides	\(\frac {i B \tan \left (d x +c \right )}{d \,a^{3}}-\frac {A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {3 B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{3}}-\frac {25 i B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {i A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{3}}-\frac {7 A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {17 A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}+\frac {31 i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}+\frac {7 i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {9 B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}\)	\(215\)
default	\(\frac {i B \tan \left (d x +c \right )}{d \,a^{3}}-\frac {A}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i B}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {3 B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{3}}-\frac {25 i B \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {i A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{3}}-\frac {7 A \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {17 A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}+\frac {31 i B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}+\frac {7 i A}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {9 B}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}\)	\(215\)

[In]

int(tan(d*x+c)^4*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {12 \, {\left (15 \, A + 49 i \, B\right )} d x e^{\left (8 i \, d x + 8 i \, c\right )} + 6 \, {\left (2 \, {\left (15 \, A + 49 i \, B\right )} d x - 11 i \, A + 55 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (-17 i \, A + 39 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (-13 i \, A + 19 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 96 \, {\left ({\left (i \, A - 3 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (i \, A - 3 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 2 i \, A + 2 \, B}{96 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \]

[In]

integrate(tan(d*x+c)^4*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/96*(12*(15*A + 49*I*B)*d*x*e^(8*I*d*x + 8*I*c) + 6*(2*(15*A + 49*I*B)*d*x - 11*I*A + 55*B)*e^(6*I*d*x + 6*I
*c) + 3*(-17*I*A + 39*B)*e^(4*I*d*x + 4*I*c) - (-13*I*A + 19*B)*e^(2*I*d*x + 2*I*c) + 96*((I*A - 3*B)*e^(8*I*d
*x + 8*I*c) + (I*A - 3*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 = 0.60 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.76 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=- \frac {2 B}{a^{3} d e^{2 i c} e^{2 i d x} + a^{3} d} + \begin {cases} \frac {\left (\left (512 i A a^{6} d^{2} e^{6 i c} - 512 B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (- 3840 i A a^{6} d^{2} e^{8 i c} + 5376 B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (16896 i A a^{6} d^{2} e^{10 i c} - 35328 B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {- 15 A - 49 i B}{8 a^{3}} + \frac {\left (- 15 A e^{6 i c} + 11 A e^{4 i c} - 5 A e^{2 i c} + A - 49 i B e^{6 i c} + 23 i B e^{4 i c} - 7 i B e^{2 i c} + i B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (- 15 A - 49 i B\right )}{8 a^{3}} - \frac {i \left (A + 3 i B\right ) \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{3} d} \]

[In]

integrate(tan(d*x+c)**4*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**3,x)

[Out]

-2*B/(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*e
xp(6*I*c))*exp(-6*I*d*x) + (-3840*I*A*a**6*d**2*exp(8*I*c) + 5376*B*a**6*d**2*exp(8*I*c))*exp(-4*I*d*x) + (168
96*I*A*a**6*d**2*exp(10*I*c) - 35328*B*a**6*d**2*exp(10*I*c))*exp(-2*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), N
e(a**9*d**3*exp(12*I*c), 0)), (x*(-(-15*A - 49*I*B)/(8*a**3) + (-15*A*exp(6*I*c) + 11*A*exp(4*I*c) - 5*A*exp(2
*I*c) + A - 49*I*B*exp(6*I*c) + 23*I*B*exp(4*I*c) - 7*I*B*exp(2*I*c) + I*B)*exp(-6*I*c)/(8*a**3)), True)) + x*
(-15*A - 49*I*B)/(8*a**3) - I*(A + 3*I*B)*log(exp(2*I*d*x) + exp(-2*I*c))/(a**3*d)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(tan(d*x+c)^4*(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 = 1.08 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.74 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {6 \, {\left (i \, A + B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} - \frac {6 \, {\left (-15 i \, A + 49 \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {96 i \, B \tan \left (d x + c\right )}{a^{3}} - \frac {165 i \, A \tan \left (d x + c\right )^{3} - 539 \, B \tan \left (d x + c\right )^{3} + 291 \, A \tan \left (d x + c\right )^{2} + 1245 i \, B \tan \left (d x + c\right )^{2} - 171 i \, A \tan \left (d x + c\right ) + 981 \, B \tan \left (d x + c\right ) - 29 \, A - 259 i \, B}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]

[In]

integrate(tan(d*x+c)^4*(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*(-15*I*A + 49*B)*log(tan(d*x + c) - I)/a^3 + 96*I*B*tan(d*x +
c)/a^3 - (165*I*A*tan(d*x + c)^3 - 539*B*tan(d*x + c)^3 + 291*A*tan(d*x + c)^2 + 1245*I*B*tan(d*x + c)^2 - 171
*I*A*tan(d*x + c) + 981*B*tan(d*x + c) - 29*A - 259*I*B)/(a^3*(tan(d*x + c) - I)^3))/d

Mupad [B] (verification not implemented)

Time = 8.03 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.96 \[ \int \frac {\tan ^4(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,7{}\mathrm {i}}{2\,a^3}+\frac {\left (-3\,B+A\,1{}\mathrm {i}\right )\,27{}\mathrm {i}}{8\,a^3}\right )+\frac {4\,B}{3\,a^3}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {5\,B}{2\,a^3}+\frac {17\,\left (-3\,B+A\,1{}\mathrm {i}\right )}{8\,a^3}\right )+\frac {17\,\left (-3\,B+A\,1{}\mathrm {i}\right )}{12\,a^3}}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{16\,a^3\,d}+\frac {B\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-49\,B+A\,15{}\mathrm {i}\right )}{16\,a^3\,d} \]

[In]

int((tan(c + d*x)^4*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

(tan(c + d*x)*((B*7i)/(2*a^3) + ((A*1i - 3*B)*27i)/(8*a^3)) + (4*B)/(3*a^3) - tan(c + d*x)^2*((5*B)/(2*a^3) +
(17*(A*1i - 3*B))/(8*a^3)) + (17*(A*1i - 3*B))/(12*a^3))/(d*(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)
^3*1i + 1)) + (log(tan(c + d*x) + 1i)*(A*1i + B))/(16*a^3*d) + (B*tan(c + d*x)*1i)/(a^3*d) + (log(tan(c + d*x)
- 1i)*(A*15i - 49*B))/(16*a^3*d)