\(\int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx\) [246]

Optimal result

Integrand size = 26, antiderivative size = 359 \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {55 i \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {55 i \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {8 \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}-\frac {55 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {8 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}+\frac {55 i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {55 i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {4 \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{9 a^2 d}-\frac {8}{3 a^2 d \tan ^{\frac {2}{3}}(c+d x)}+\frac {11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{4 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^2} \]

[Out]

-55/72*I*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a^2/d-55/72*I*arctan(3^(1/2)+2*tan(d*x+c)^(1/3))/a^2/d-55/36*I*ar
ctan(tan(d*x+c)^(1/3))/a^2/d+8/9*ln(1+tan(d*x+c)^(2/3))/a^2/d-4/9*ln(1-tan(d*x+c)^(2/3)+tan(d*x+c)^(4/3))/a^2/
d+8/9*arctan(1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))/a^2/d*3^(1/2)+55/144*I*ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+
c)^(2/3))/a^2/d*3^(1/2)-55/144*I*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/a^2/d*3^(1/2)-8/3/a^2/d/tan(d
*x+c)^(2/3)+11/12/a^2/d/(1+I*tan(d*x+c))/tan(d*x+c)^(2/3)+1/4/d/tan(d*x+c)^(2/3)/(a+I*a*tan(d*x+c))^2

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.577, Rules used = {3640, 3677, 3610, 3619, 3557, 335, 215, 648, 632, 210, 642, 209, 281, 298, 31} \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {8 \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}+\frac {55 i \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {55 i \arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{72 a^2 d}-\frac {55 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {8}{3 a^2 d \tan ^{\frac {2}{3}}(c+d x)}+\frac {11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {2}{3}}(c+d x)}+\frac {8 \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{9 a^2 d}+\frac {55 i \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt {3} a^2 d}-\frac {55 i \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt {3} a^2 d}-\frac {4 \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{9 a^2 d}+\frac {1}{4 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^2} \]

[In]

Int[1/(Tan[c + d*x]^(5/3)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(((55*I)/72)*ArcTan[Sqrt[3] - 2*Tan[c + d*x]^(1/3)])/(a^2*d) - (((55*I)/72)*ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(1
/3)])/(a^2*d) + (8*ArcTan[(1 - 2*Tan[c + d*x]^(2/3))/Sqrt[3]])/(3*Sqrt[3]*a^2*d) - (((55*I)/36)*ArcTan[Tan[c +
d*x]^(1/3)])/(a^2*d) + (8*Log[1 + Tan[c + d*x]^(2/3)])/(9*a^2*d) + (((55*I)/48)*Log[1 - Sqrt[3]*Tan[c + d*x]^
(1/3) + Tan[c + d*x]^(2/3)])/(Sqrt[3]*a^2*d) - (((55*I)/48)*Log[1 + Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^
(2/3)])/(Sqrt[3]*a^2*d) - (4*Log[1 - Tan[c + d*x]^(2/3) + Tan[c + d*x]^(4/3)])/(9*a^2*d) - 8/(3*a^2*d*Tan[c +
d*x]^(2/3)) + 11/(12*a^2*d*(1 + I*Tan[c + d*x])*Tan[c + d*x]^(2/3)) + 1/(4*d*Tan[c + d*x]^(2/3)*(a + I*a*Tan[c
+ d*x])^2)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 215

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[a/b, n]], s = Denominator[Rt[a/b, n]]
, k, u, v}, Simp[u = Int[(r - s*Cos[(2*k - 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x] +
Int[(r + s*Cos[(2*k - 1)*(Pi/n)]*x)/(r^2 + 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x]; 2*(r^2/(a*n))*Int[1/
(r^2 + s^2*x^2), x] + Dist[2*(r/(a*n)), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)
/4, 0] && PosQ[a/b]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

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 3619

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*T
an[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ
[c^2 + d^2, 0] &&  !IntegerQ[2*m]

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 {1}{4 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {\frac {14 a}{3}-\frac {8}{3} i a \tan (c+d x)}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))} \, dx}{4 a^2} \\ & = \frac {11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{4 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {\frac {128 a^2}{9}-\frac {110}{9} i a^2 \tan (c+d x)}{\tan ^{\frac {5}{3}}(c+d x)} \, dx}{8 a^4} \\ & = -\frac {8}{3 a^2 d \tan ^{\frac {2}{3}}(c+d x)}+\frac {11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{4 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {-\frac {110 i a^2}{9}-\frac {128}{9} a^2 \tan (c+d x)}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{8 a^4} \\ & = -\frac {8}{3 a^2 d \tan ^{\frac {2}{3}}(c+d x)}+\frac {11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{4 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^2}-\frac {(55 i) \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{36 a^2}-\frac {16 \int \sqrt [3]{\tan (c+d x)} \, dx}{9 a^2} \\ & = -\frac {8}{3 a^2 d \tan ^{\frac {2}{3}}(c+d x)}+\frac {11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{4 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^2}-\frac {(55 i) \text {Subst}\left (\int \frac {1}{x^{2/3} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{36 a^2 d}-\frac {16 \text {Subst}\left (\int \frac {\sqrt [3]{x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{9 a^2 d} \\ & = -\frac {8}{3 a^2 d \tan ^{\frac {2}{3}}(c+d x)}+\frac {11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{4 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^2}-\frac {(55 i) \text {Subst}\left (\int \frac {1}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{12 a^2 d}-\frac {16 \text {Subst}\left (\int \frac {x^3}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{3 a^2 d} \\ & = -\frac {8}{3 a^2 d \tan ^{\frac {2}{3}}(c+d x)}+\frac {11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{4 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^2}-\frac {(55 i) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {(55 i) \text {Subst}\left (\int \frac {1-\frac {\sqrt {3} x}{2}}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {(55 i) \text {Subst}\left (\int \frac {1+\frac {\sqrt {3} x}{2}}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {8 \text {Subst}\left (\int \frac {x}{1+x^3} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a^2 d} \\ & = -\frac {55 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {8}{3 a^2 d \tan ^{\frac {2}{3}}(c+d x)}+\frac {11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{4 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^2}-\frac {(55 i) \text {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{144 a^2 d}-\frac {(55 i) \text {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{144 a^2 d}+\frac {8 \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {8 \text {Subst}\left (\int \frac {1+x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}+\frac {(55 i) \text {Subst}\left (\int \frac {-\sqrt {3}+2 x}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{48 \sqrt {3} a^2 d}-\frac {(55 i) \text {Subst}\left (\int \frac {\sqrt {3}+2 x}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{48 \sqrt {3} a^2 d} \\ & = -\frac {55 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {8 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}+\frac {55 i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {55 i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {8}{3 a^2 d \tan ^{\frac {2}{3}}(c+d x)}+\frac {11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{4 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {(55 i) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {(55 i) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {4 \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {4 \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a^2 d} \\ & = \frac {55 i \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {55 i \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {55 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {8 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}+\frac {55 i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {55 i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {4 \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{9 a^2 d}-\frac {8}{3 a^2 d \tan ^{\frac {2}{3}}(c+d x)}+\frac {11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{4 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {8 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac {2}{3}}(c+d x)\right )}{3 a^2 d} \\ & = \frac {55 i \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {55 i \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {8 \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}-\frac {55 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {8 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}+\frac {55 i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {55 i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {4 \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{9 a^2 d}-\frac {8}{3 a^2 d \tan ^{\frac {2}{3}}(c+d x)}+\frac {11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{4 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.84 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {18 a^2 \sqrt [6]{\tan ^2(c+d x)}+(a+i a \tan (c+d x)) \left (66 a \sqrt [6]{\tan ^2(c+d x)}-(a+i a \tan (c+d x)) \left (-55 \left (\log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right )-\log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right )+\sqrt [3]{-1} \left (\log \left (1-\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )-\log \left (1+\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )+\sqrt [3]{-1} \left (-\log \left (1-(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )+\log \left (1+(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )\right )\right )\right ) \tan (c+d x)-64 \sqrt [6]{\tan ^2(c+d x)} \left (-3+\log \left (1+\sqrt [3]{\tan ^2(c+d x)}\right ) \sqrt [3]{\tan ^2(c+d x)}-\sqrt [3]{-1} \log \left (1-\sqrt [3]{-1} \sqrt [3]{\tan ^2(c+d x)}\right ) \sqrt [3]{\tan ^2(c+d x)}+(-1)^{2/3} \log \left (1+(-1)^{2/3} \sqrt [3]{\tan ^2(c+d x)}\right ) \sqrt [3]{\tan ^2(c+d x)}\right )\right )\right )}{72 a^2 d \tan ^{\frac {2}{3}}(c+d x) \sqrt [6]{\tan ^2(c+d x)} (a+i a \tan (c+d x))^2} \]

[In]

Integrate[1/(Tan[c + d*x]^(5/3)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(18*a^2*(Tan[c + d*x]^2)^(1/6) + (a + I*a*Tan[c + d*x])*(66*a*(Tan[c + d*x]^2)^(1/6) - (a + I*a*Tan[c + d*x])*
(-55*(Log[1 - I*(Tan[c + d*x]^2)^(1/6)] - Log[1 + I*(Tan[c + d*x]^2)^(1/6)] + (-1)^(1/3)*(Log[1 - (-1)^(1/6)*(
Tan[c + d*x]^2)^(1/6)] - Log[1 + (-1)^(1/6)*(Tan[c + d*x]^2)^(1/6)] + (-1)^(1/3)*(-Log[1 - (-1)^(5/6)*(Tan[c +
d*x]^2)^(1/6)] + Log[1 + (-1)^(5/6)*(Tan[c + d*x]^2)^(1/6)])))*Tan[c + d*x] - 64*(Tan[c + d*x]^2)^(1/6)*(-3 +
Log[1 + (Tan[c + d*x]^2)^(1/3)]*(Tan[c + d*x]^2)^(1/3) - (-1)^(1/3)*Log[1 - (-1)^(1/3)*(Tan[c + d*x]^2)^(1/3)
]*(Tan[c + d*x]^2)^(1/3) + (-1)^(2/3)*Log[1 + (-1)^(2/3)*(Tan[c + d*x]^2)^(1/3)]*(Tan[c + d*x]^2)^(1/3)))))/(7
2*a^2*d*Tan[c + d*x]^(2/3)*(Tan[c + d*x]^2)^(1/6)*(a + I*a*Tan[c + d*x])^2)

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}+\frac {30 i \tan \left (d x +c \right )+4 \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )-4 i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+32}{72 {\left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}^{2}}-\frac {119 \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}-\frac {119 i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}-\frac {\ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}+\frac {i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}-\frac {5 i}{12 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {1}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}+\frac {119 \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}-\frac {3}{2 \tan \left (d x +c \right )^{\frac {2}{3}}}}{d \,a^{2}}\)	\(233\)
default	\(\frac {\frac {\ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}+\frac {30 i \tan \left (d x +c \right )+4 \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )-4 i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+32}{72 {\left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}^{2}}-\frac {119 \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}-\frac {119 i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}-\frac {\ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}+\frac {i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}-\frac {5 i}{12 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {1}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}+\frac {119 \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}-\frac {3}{2 \tan \left (d x +c \right )^{\frac {2}{3}}}}{d \,a^{2}}\)	\(233\)

[In]

int(1/tan(d*x+c)^(5/3)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d/a^2*(1/8*ln(tan(d*x+c)^(1/3)-I)+1/72*(30*I*tan(d*x+c)+4*tan(d*x+c)^(2/3)-4*I*tan(d*x+c)^(1/3)+32)/(-I*tan(
d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)^2-119/144*ln(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)-119/72*I*3^(1/2)*arctanh
(1/3*(-I+2*tan(d*x+c)^(1/3))*3^(1/2))-1/16*ln(I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)+1/8*I*3^(1/2)*arctanh(1/3
*(I+2*tan(d*x+c)^(1/3))*3^(1/2))-5/12*I/(tan(d*x+c)^(1/3)+I)+1/36/(tan(d*x+c)^(1/3)+I)^2+119/72*ln(tan(d*x+c)^
(1/3)+I)-3/2/tan(d*x+c)^(2/3))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 681 vs. \(2 (282) = 564\).

Time = 0.27 (sec) , antiderivative size = 681, normalized size of antiderivative = 1.90 \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=-\frac {9 \, {\left (\sqrt {3} {\left (-i \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + i \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {1}{a^{4} d^{2}}} + e^{\left (6 i \, d x + 6 i \, c\right )} - e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (\frac {1}{2} \, \sqrt {3} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) + 9 \, {\left (\sqrt {3} {\left (i \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - i \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {1}{a^{4} d^{2}}} + e^{\left (6 i \, d x + 6 i \, c\right )} - e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (-\frac {1}{2} \, \sqrt {3} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) + 119 \, {\left (3 \, \sqrt {\frac {1}{3}} {\left (i \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - i \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {1}{a^{4} d^{2}}} + e^{\left (6 i \, d x + 6 i \, c\right )} - e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (\frac {3}{2} \, \sqrt {\frac {1}{3}} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) + 119 \, {\left (3 \, \sqrt {\frac {1}{3}} {\left (-i \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + i \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {1}{a^{4} d^{2}}} + e^{\left (6 i \, d x + 6 i \, c\right )} - e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (-\frac {3}{2} \, \sqrt {\frac {1}{3}} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) - 238 \, {\left (e^{\left (6 i \, d x + 6 i \, c\right )} - e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (\left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + i\right ) - 18 \, {\left (e^{\left (6 i \, d x + 6 i \, c\right )} - e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (\left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - i\right ) + 3 \, \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} {\left (103 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 75 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 31 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i\right )}}{144 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \]

[In]

integrate(1/tan(d*x+c)^(5/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/144*(9*(sqrt(3)*(-I*a^2*d*e^(6*I*d*x + 6*I*c) + I*a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a^4*d^2)) + e^(6*I*d*x
+ 6*I*c) - e^(4*I*d*x + 4*I*c))*log(1/2*sqrt(3)*a^2*d*sqrt(1/(a^4*d^2)) + ((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2
*I*d*x + 2*I*c) + 1))^(1/3) + 1/2*I) + 9*(sqrt(3)*(I*a^2*d*e^(6*I*d*x + 6*I*c) - I*a^2*d*e^(4*I*d*x + 4*I*c))*
sqrt(1/(a^4*d^2)) + e^(6*I*d*x + 6*I*c) - e^(4*I*d*x + 4*I*c))*log(-1/2*sqrt(3)*a^2*d*sqrt(1/(a^4*d^2)) + ((-I
*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) + 1/2*I) + 119*(3*sqrt(1/3)*(I*a^2*d*e^(6*I*d*x + 6
*I*c) - I*a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a^4*d^2)) + e^(6*I*d*x + 6*I*c) - e^(4*I*d*x + 4*I*c))*log(3/2*sq
rt(1/3)*a^2*d*sqrt(1/(a^4*d^2)) + ((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) - 1/2*I) + 11
9*(3*sqrt(1/3)*(-I*a^2*d*e^(6*I*d*x + 6*I*c) + I*a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a^4*d^2)) + e^(6*I*d*x + 6
*I*c) - e^(4*I*d*x + 4*I*c))*log(-3/2*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d^2)) + ((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*
I*d*x + 2*I*c) + 1))^(1/3) - 1/2*I) - 238*(e^(6*I*d*x + 6*I*c) - e^(4*I*d*x + 4*I*c))*log(((-I*e^(2*I*d*x + 2*
I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) + I) - 18*(e^(6*I*d*x + 6*I*c) - e^(4*I*d*x + 4*I*c))*log(((-I*e^(2
*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) - I) + 3*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I
*c) + 1))^(1/3)*(103*I*e^(6*I*d*x + 6*I*c) + 75*I*e^(4*I*d*x + 4*I*c) - 31*I*e^(2*I*d*x + 2*I*c) - 3*I))/(a^2*
d*e^(6*I*d*x + 6*I*c) - a^2*d*e^(4*I*d*x + 4*I*c))

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(1/tan(d*x+c)**(5/3)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/tan(d*x+c)^(5/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 2.00 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {119 i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right )}{144 \, a^{2} d} - \frac {i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right )}{16 \, a^{2} d} - \frac {\log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{16 \, a^{2} d} - \frac {119 \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{144 \, a^{2} d} + \frac {119 \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right )}{72 \, a^{2} d} + \frac {\log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )}{8 \, a^{2} d} - \frac {32 \, \tan \left (d x + c\right )^{2} - 53 i \, \tan \left (d x + c\right ) - 18}{12 \, {\left (\tan \left (d x + c\right )^{\frac {4}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}}\right )}^{2} a^{2} d} \]

[In]

integrate(1/tan(d*x+c)^(5/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

119/144*I*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) + I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) - I))/(a^2*d) - 1/
16*I*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) - I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) + I))/(a^2*d) - 1/16*lo
g(tan(d*x + c)^(2/3) + I*tan(d*x + c)^(1/3) - 1)/(a^2*d) - 119/144*log(tan(d*x + c)^(2/3) - I*tan(d*x + c)^(1/
3) - 1)/(a^2*d) + 119/72*log(tan(d*x + c)^(1/3) + I)/(a^2*d) + 1/8*log(tan(d*x + c)^(1/3) - I)/(a^2*d) - 1/12*
(32*tan(d*x + c)^2 - 53*I*tan(d*x + c) - 18)/((tan(d*x + c)^(4/3) - I*tan(d*x + c)^(1/3))^2*a^2*d)

Mupad [B] (verification not implemented)

Time = 6.10 (sec) , antiderivative size = 660, normalized size of antiderivative = 1.84 \[ \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

int(1/(tan(c + d*x)^(5/3)*(a + a*tan(c + d*x)*1i)^2),x)

[Out]

log(((a^6*d^3*215607040i)/3 + 28311552*a^10*d^5*tan(c + d*x)^(1/3)*(1/(512*a^6*d^3))^(2/3))*(1/(512*a^6*d^3))^
(1/3) - (27116768*a^4*d^2*tan(c + d*x)^(1/3))/3)*(1/(512*a^6*d^3))^(1/3) + (119*log((119*((a^6*d^3*215607040i)
/3 + (232013824*a^10*d^5*tan(c + d*x)^(1/3)*(1/(a^6*d^3))^(2/3))/3)*(1/(a^6*d^3))^(1/3))/72 - (27116768*a^4*d^
2*tan(c + d*x)^(1/3))/3)*(1/(a^6*d^3))^(1/3))/72 - ((53*tan(c + d*x))/(12*a^2*d) - 3i/(2*a^2*d) + (tan(c + d*x
)^2*8i)/(3*a^2*d))/(2*tan(c + d*x)^(5/3) - tan(c + d*x)^(2/3)*1i + tan(c + d*x)^(8/3)*1i) + log(((3^(1/2)*1i)/
2 - 1/2)*((a^6*d^3*215607040i)/3 + 28311552*a^10*d^5*tan(c + d*x)^(1/3)*((3^(1/2)*1i)/2 - 1/2)^2*(1/(512*a^6*d
^3))^(2/3))*(1/(512*a^6*d^3))^(1/3) - (27116768*a^4*d^2*tan(c + d*x)^(1/3))/3)*((3^(1/2)*1i)/2 - 1/2)*(1/(512*
a^6*d^3))^(1/3) - log(((3^(1/2)*1i)/2 + 1/2)*((a^6*d^3*215607040i)/3 + 28311552*a^10*d^5*tan(c + d*x)^(1/3)*((
3^(1/2)*1i)/2 + 1/2)^2*(1/(512*a^6*d^3))^(2/3))*(1/(512*a^6*d^3))^(1/3) + (27116768*a^4*d^2*tan(c + d*x)^(1/3)
)/3)*((3^(1/2)*1i)/2 + 1/2)*(1/(512*a^6*d^3))^(1/3) + (119*log((119*(3^(1/2)*1i - 1)*((a^6*d^3*215607040i)/3 +
(58003456*a^10*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)^2*(1/(a^6*d^3))^(2/3))/3)*(1/(a^6*d^3))^(1/3))/144 - (
27116768*a^4*d^2*tan(c + d*x)^(1/3))/3)*(3^(1/2)*1i - 1)*(1/(a^6*d^3))^(1/3))/144 - (119*log((119*(3^(1/2)*1i
+ 1)*((a^6*d^3*215607040i)/3 + (58003456*a^10*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)^2*(1/(a^6*d^3))^(2/3))/3
)*(1/(a^6*d^3))^(1/3))/144 + (27116768*a^4*d^2*tan(c + d*x)^(1/3))/3)*(3^(1/2)*1i + 1)*(1/(a^6*d^3))^(1/3))/14
4