3.52 \(\int \sec ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx\)

Optimal. Leaf size=163 \[ \frac{7 i a^4 \sec ^3(c+d x)}{8 d}+\frac{21 a^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{3 i \sec ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{10 d}+\frac{21 i \sec ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{40 d}+\frac{21 a^4 \tan (c+d x) \sec (c+d x)}{16 d}+\frac{i a \sec ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d} \]

[Out]

(21*a^4*ArcTanh[Sin[c + d*x]])/(16*d) + (((7*I)/8)*a^4*Sec[c + d*x]^3)/d + (21*a^4*Sec[c + d*x]*Tan[c + d*x])/
(16*d) + ((I/6)*a*Sec[c + d*x]^3*(a + I*a*Tan[c + d*x])^3)/d + (((3*I)/10)*Sec[c + d*x]^3*(a^2 + I*a^2*Tan[c +
d*x])^2)/d + (((21*I)/40)*Sec[c + d*x]^3*(a^4 + I*a^4*Tan[c + d*x]))/d

________________________________________________________________________________________

Rubi [A]  time = 0.161987, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3498, 3486, 3768, 3770} \[ \frac{7 i a^4 \sec ^3(c+d x)}{8 d}+\frac{21 a^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{3 i \sec ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{10 d}+\frac{21 i \sec ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{40 d}+\frac{21 a^4 \tan (c+d x) \sec (c+d x)}{16 d}+\frac{i a \sec ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(21*a^4*ArcTanh[Sin[c + d*x]])/(16*d) + (((7*I)/8)*a^4*Sec[c + d*x]^3)/d + (21*a^4*Sec[c + d*x]*Tan[c + d*x])/
(16*d) + ((I/6)*a*Sec[c + d*x]^3*(a + I*a*Tan[c + d*x])^3)/d + (((3*I)/10)*Sec[c + d*x]^3*(a^2 + I*a^2*Tan[c +
d*x])^2)/d + (((21*I)/40)*Sec[c + d*x]^3*(a^4 + I*a^4*Tan[c + d*x]))/d

Rule 3498

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

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rubi steps

\begin{align*} \int \sec ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx &=\frac{i a \sec ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}+\frac{1}{2} (3 a) \int \sec ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx\\ &=\frac{i a \sec ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}+\frac{3 i \sec ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{10 d}+\frac{1}{10} \left (21 a^2\right ) \int \sec ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=\frac{i a \sec ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}+\frac{3 i \sec ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{10 d}+\frac{21 i \sec ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{40 d}+\frac{1}{8} \left (21 a^3\right ) \int \sec ^3(c+d x) (a+i a \tan (c+d x)) \, dx\\ &=\frac{7 i a^4 \sec ^3(c+d x)}{8 d}+\frac{i a \sec ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}+\frac{3 i \sec ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{10 d}+\frac{21 i \sec ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{40 d}+\frac{1}{8} \left (21 a^4\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{7 i a^4 \sec ^3(c+d x)}{8 d}+\frac{21 a^4 \sec (c+d x) \tan (c+d x)}{16 d}+\frac{i a \sec ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}+\frac{3 i \sec ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{10 d}+\frac{21 i \sec ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{40 d}+\frac{1}{16} \left (21 a^4\right ) \int \sec (c+d x) \, dx\\ &=\frac{21 a^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{7 i a^4 \sec ^3(c+d x)}{8 d}+\frac{21 a^4 \sec (c+d x) \tan (c+d x)}{16 d}+\frac{i a \sec ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}+\frac{3 i \sec ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{10 d}+\frac{21 i \sec ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{40 d}\\ \end{align*}

Mathematica [A]  time = 1.78356, size = 171, normalized size = 1.05 \[ -\frac{a^4 (\cos (4 c)-i \sin (4 c)) (\tan (c+d x)-i)^4 \sec ^2(c+d x) \left (-4608 i \cos (c+d x)+5 (90 \sin (c+d x)+155 \sin (3 (c+d x))-63 \sin (5 (c+d x))-512 i \cos (3 (c+d x)))+5040 \cos ^6(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{3840 d (\cos (d x)+i \sin (d x))^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^4*Sec[c + d*x]^2*(Cos[4*c] - I*Sin[4*c])*((-4608*I)*Cos[c + d*x] + 5040*Cos[c + d*x]^6*(Log[Cos[(c + d*x)/
2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 5*((-512*I)*Cos[3*(c + d*x)] + 90*Sin[c +
d*x] + 155*Sin[3*(c + d*x)] - 63*Sin[5*(c + d*x)]))*(-I + Tan[c + d*x])^4)/(3840*d*(Cos[d*x] + I*Sin[d*x])^4)

________________________________________________________________________________________

Maple [B]  time = 0.064, size = 324, normalized size = 2. \begin{align*}{\frac{{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{6\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{24\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{48\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{48\,d}}-{\frac{13\,{a}^{4}\sin \left ( dx+c \right ) }{16\,d}}+{\frac{21\,{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{{\frac{8\,i}{15}}{a}^{4}\cos \left ( dx+c \right ) }{d}}+{\frac{{\frac{4\,i}{15}}{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d\cos \left ( dx+c \right ) }}-{\frac{{\frac{4\,i}{15}}{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{\frac{4\,i}{15}}{a}^{4}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d}}-{\frac{{\frac{4\,i}{5}}{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}-{\frac{3\,{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{3\,{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{\frac{4\,i}{3}}{a}^{4}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x)

[Out]

1/6/d*a^4*sin(d*x+c)^5/cos(d*x+c)^6+1/24/d*a^4*sin(d*x+c)^5/cos(d*x+c)^4-1/48/d*a^4*sin(d*x+c)^5/cos(d*x+c)^2-
1/48*a^4*sin(d*x+c)^3/d-13/16*a^4*sin(d*x+c)/d+21/16/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+8/15*I/d*a^4*cos(d*x+c)+4
/15*I/d*a^4*sin(d*x+c)^4/cos(d*x+c)-4/15*I/d*a^4*sin(d*x+c)^4/cos(d*x+c)^3+4/15*I/d*a^4*cos(d*x+c)*sin(d*x+c)^
2-4/5*I/d*a^4*sin(d*x+c)^4/cos(d*x+c)^5-3/2/d*a^4*sin(d*x+c)^3/cos(d*x+c)^4-3/4/d*a^4*sin(d*x+c)^3/cos(d*x+c)^
2+4/3*I/d*a^4/cos(d*x+c)^3+1/2*a^4*sec(d*x+c)*tan(d*x+c)/d

________________________________________________________________________________________

Maxima [A]  time = 1.10252, size = 332, normalized size = 2.04 \begin{align*} -\frac{5 \, a^{4}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{5} + 8 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 180 \, a^{4}{\left (\frac{2 \,{\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, a^{4}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac{640 i \, a^{4}}{\cos \left (d x + c\right )^{3}} - \frac{128 i \,{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} a^{4}}{\cos \left (d x + c\right )^{5}}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/480*(5*a^4*(2*(3*sin(d*x + c)^5 + 8*sin(d*x + c)^3 - 3*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3
*sin(d*x + c)^2 - 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) + 180*a^4*(2*(sin(d*x + c)^3 + sin(d
*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 120*a^4*(2
*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 640*I*a^4/cos(d*x + c)^3
- 128*I*(5*cos(d*x + c)^2 - 3)*a^4/cos(d*x + c)^5)/d

________________________________________________________________________________________

Fricas [B]  time = 1.21524, size = 1064, normalized size = 6.53 \begin{align*} \frac{-630 i \, a^{4} e^{\left (11 i \, d x + 11 i \, c\right )} + 6670 i \, a^{4} e^{\left (9 i \, d x + 9 i \, c\right )} + 10116 i \, a^{4} e^{\left (7 i \, d x + 7 i \, c\right )} + 8316 i \, a^{4} e^{\left (5 i \, d x + 5 i \, c\right )} + 3570 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} + 630 i \, a^{4} e^{\left (i \, d x + i \, c\right )} + 315 \,{\left (a^{4} e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 315 \,{\left (a^{4} e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{240 \,{\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(-630*I*a^4*e^(11*I*d*x + 11*I*c) + 6670*I*a^4*e^(9*I*d*x + 9*I*c) + 10116*I*a^4*e^(7*I*d*x + 7*I*c) + 8
316*I*a^4*e^(5*I*d*x + 5*I*c) + 3570*I*a^4*e^(3*I*d*x + 3*I*c) + 630*I*a^4*e^(I*d*x + I*c) + 315*(a^4*e^(12*I*
d*x + 12*I*c) + 6*a^4*e^(10*I*d*x + 10*I*c) + 15*a^4*e^(8*I*d*x + 8*I*c) + 20*a^4*e^(6*I*d*x + 6*I*c) + 15*a^4
*e^(4*I*d*x + 4*I*c) + 6*a^4*e^(2*I*d*x + 2*I*c) + a^4)*log(e^(I*d*x + I*c) + I) - 315*(a^4*e^(12*I*d*x + 12*I
*c) + 6*a^4*e^(10*I*d*x + 10*I*c) + 15*a^4*e^(8*I*d*x + 8*I*c) + 20*a^4*e^(6*I*d*x + 6*I*c) + 15*a^4*e^(4*I*d*
x + 4*I*c) + 6*a^4*e^(2*I*d*x + 2*I*c) + a^4)*log(e^(I*d*x + I*c) - I))/(d*e^(12*I*d*x + 12*I*c) + 6*d*e^(10*I
*d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8*I*c) + 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) + 6*d*e^(2*I*d
*x + 2*I*c) + d)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{4} \left (\int - 6 \tan ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \tan ^{4}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 4 i \tan{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int - 4 i \tan ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \sec ^{3}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+I*a*tan(d*x+c))**4,x)

[Out]

a**4*(Integral(-6*tan(c + d*x)**2*sec(c + d*x)**3, x) + Integral(tan(c + d*x)**4*sec(c + d*x)**3, x) + Integra
l(4*I*tan(c + d*x)*sec(c + d*x)**3, x) + Integral(-4*I*tan(c + d*x)**3*sec(c + d*x)**3, x) + Integral(sec(c +
d*x)**3, x))

________________________________________________________________________________________

Giac [A]  time = 1.35276, size = 323, normalized size = 1.98 \begin{align*} \frac{315 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 315 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (75 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 960 i \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} + 1175 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 4800 i \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 1890 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 4480 i \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 1890 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 1920 i \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 1175 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1728 i \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 75 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 448 i \, a^{4}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/240*(315*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 315*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(75*a^4*tan
(1/2*d*x + 1/2*c)^11 + 960*I*a^4*tan(1/2*d*x + 1/2*c)^10 + 1175*a^4*tan(1/2*d*x + 1/2*c)^9 - 4800*I*a^4*tan(1/
2*d*x + 1/2*c)^8 - 1890*a^4*tan(1/2*d*x + 1/2*c)^7 + 4480*I*a^4*tan(1/2*d*x + 1/2*c)^6 - 1890*a^4*tan(1/2*d*x
+ 1/2*c)^5 - 1920*I*a^4*tan(1/2*d*x + 1/2*c)^4 + 1175*a^4*tan(1/2*d*x + 1/2*c)^3 + 1728*I*a^4*tan(1/2*d*x + 1/
2*c)^2 + 75*a^4*tan(1/2*d*x + 1/2*c) - 448*I*a^4)/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d