\(\int \frac {\tan ^{\frac {10}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [241]

Optimal result

Integrand size = 26, antiderivative size = 357 \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {49 \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {49 \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {5 i \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}+\frac {49 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {5 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {49 \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}+\frac {49 \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {5 i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{18 a^2 d}-\frac {49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

49/72*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a^2/d+49/72*arctan(3^(1/2)+2*tan(d*x+c)^(1/3))/a^2/d+49/36*arctan(ta
n(d*x+c)^(1/3))/a^2/d+5/9*I*ln(1+tan(d*x+c)^(2/3))/a^2/d-5/18*I*ln(1-tan(d*x+c)^(2/3)+tan(d*x+c)^(4/3))/a^2/d+
5/9*I*arctan(1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))/a^2/d*3^(1/2)-49/144*ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)
^(2/3))/a^2/d*3^(1/2)+49/144*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/a^2/d*3^(1/2)-49/12*tan(d*x+c)^(1
/3)/a^2/d+5/6*I*tan(d*x+c)^(4/3)/a^2/d/(1+I*tan(d*x+c))-1/4*tan(d*x+c)^(7/3)/d/(a+I*a*tan(d*x+c))^2

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 357, 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 = {3639, 3676, 3609, 3619, 3557, 335, 215, 648, 632, 210, 642, 209, 281, 298, 31} \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {5 i \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}-\frac {49 \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {49 \arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{72 a^2 d}+\frac {49 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac {49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac {5 i \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{9 a^2 d}-\frac {49 \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt {3} a^2 d}+\frac {49 \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt {3} a^2 d}-\frac {5 i \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{18 a^2 d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[In]

Int[Tan[c + d*x]^(10/3)/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(-49*ArcTan[Sqrt[3] - 2*Tan[c + d*x]^(1/3)])/(72*a^2*d) + (49*ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(1/3)])/(72*a^2*
d) + (((5*I)/3)*ArcTan[(1 - 2*Tan[c + d*x]^(2/3))/Sqrt[3]])/(Sqrt[3]*a^2*d) + (49*ArcTan[Tan[c + d*x]^(1/3)])/
(36*a^2*d) + (((5*I)/9)*Log[1 + Tan[c + d*x]^(2/3)])/(a^2*d) - (49*Log[1 - Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c
+ d*x]^(2/3)])/(48*Sqrt[3]*a^2*d) + (49*Log[1 + Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/(48*Sqrt[3]*
a^2*d) - (((5*I)/18)*Log[1 - Tan[c + d*x]^(2/3) + Tan[c + d*x]^(4/3)])/(a^2*d) - (49*Tan[c + d*x]^(1/3))/(12*a
^2*d) + (((5*I)/6)*Tan[c + d*x]^(4/3))/(a^2*d*(1 + I*Tan[c + d*x])) - Tan[c + d*x]^(7/3)/(4*d*(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 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*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] && GtQ[m, 0]

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 3639

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

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 {\tan ^{\frac {7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan ^{\frac {4}{3}}(c+d x) \left (-\frac {7 a}{3}+\frac {13}{3} i a \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{4 a^2} \\ & = \frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \sqrt [3]{\tan (c+d x)} \left (-\frac {80 i a^2}{9}-\frac {98}{9} a^2 \tan (c+d x)\right ) \, dx}{8 a^4} \\ & = -\frac {49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\frac {98 a^2}{9}-\frac {80}{9} i a^2 \tan (c+d x)}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{8 a^4} \\ & = -\frac {49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(10 i) \int \sqrt [3]{\tan (c+d x)} \, dx}{9 a^2}+\frac {49 \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{36 a^2} \\ & = -\frac {49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(10 i) \text {Subst}\left (\int \frac {\sqrt [3]{x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{9 a^2 d}+\frac {49 \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 {49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(10 i) \text {Subst}\left (\int \frac {x^3}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{3 a^2 d}+\frac {49 \text {Subst}\left (\int \frac {1}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{12 a^2 d} \\ & = -\frac {49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(5 i) \text {Subst}\left (\int \frac {x}{1+x^3} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a^2 d}+\frac {49 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {49 \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 {49 \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 {49 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {(5 i) \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {(5 i) \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 {49 \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 {49 \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 {49 \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 {49 \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 {49 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {5 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {49 \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 {49 \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 {49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(5 i) \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{18 a^2 d}-\frac {(5 i) \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a^2 d}-\frac {49 \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 {49 \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 {49 \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {49 \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {49 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {5 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {49 \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 {49 \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 {5 i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{18 a^2 d}-\frac {49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {(5 i) \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 {49 \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {49 \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {5 i \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}+\frac {49 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {5 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {49 \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 {49 \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 {5 i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{18 a^2 d}-\frac {49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.56 (sec) , antiderivative size = 677, normalized size of antiderivative = 1.90 \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\tan ^{\frac {13}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\frac {20 i \log \left (1+\sqrt [3]{\tan ^2(c+d x)}\right ) \sqrt [3]{\tan ^2(c+d x)}}{9 d \tan ^{\frac {2}{3}}(c+d x)}-\frac {20 \sqrt [6]{-1} \log \left (1-e^{-\frac {i \pi }{3}} \sqrt [3]{\tan ^2(c+d x)}\right ) \sqrt [3]{\tan ^2(c+d x)}}{9 d \tan ^{\frac {2}{3}}(c+d x)}-\frac {20 (-1)^{5/6} \log \left (1-e^{\frac {i \pi }{3}} \sqrt [3]{\tan ^2(c+d x)}\right ) \sqrt [3]{\tan ^2(c+d x)}}{9 d \tan ^{\frac {2}{3}}(c+d x)}+\frac {49 i \log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right ) \tan ^2(c+d x)^{5/6}}{18 d \tan ^{\frac {5}{3}}(c+d x)}-\frac {49 i \log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right ) \tan ^2(c+d x)^{5/6}}{18 d \tan ^{\frac {5}{3}}(c+d x)}-\frac {49 \sqrt [6]{-1} \log \left (1-e^{-\frac {i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right ) \tan ^2(c+d x)^{5/6}}{18 d \tan ^{\frac {5}{3}}(c+d x)}+\frac {49 (-1)^{5/6} \log \left (1-e^{\frac {i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right ) \tan ^2(c+d x)^{5/6}}{18 d \tan ^{\frac {5}{3}}(c+d x)}-\frac {49 (-1)^{5/6} \log \left (1-e^{-\frac {5 i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right ) \tan ^2(c+d x)^{5/6}}{18 d \tan ^{\frac {5}{3}}(c+d x)}+\frac {49 \sqrt [6]{-1} \log \left (1-e^{\frac {5 i \pi }{6}} \sqrt [6]{\tan ^2(c+d x)}\right ) \tan ^2(c+d x)^{5/6}}{18 d \tan ^{\frac {5}{3}}(c+d x)}-\frac {49 a \sqrt [3]{\tan (c+d x)}}{3 d (a+i a \tan (c+d x))}-\frac {13 i a \tan ^{\frac {4}{3}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {a \tan ^{\frac {7}{3}}(c+d x)}{d (a+i a \tan (c+d x))}+\frac {i a \tan ^{\frac {10}{3}}(c+d x)}{d (a+i a \tan (c+d x))}}{4 a^2} \]

[In]

Integrate[Tan[c + d*x]^(10/3)/(a + I*a*Tan[c + d*x])^2,x]

[Out]

Tan[c + d*x]^(13/3)/(4*d*(a + I*a*Tan[c + d*x])^2) + ((((20*I)/9)*Log[1 + (Tan[c + d*x]^2)^(1/3)]*(Tan[c + d*x
]^2)^(1/3))/(d*Tan[c + d*x]^(2/3)) - (20*(-1)^(1/6)*Log[1 - (Tan[c + d*x]^2)^(1/3)/E^((I/3)*Pi)]*(Tan[c + d*x]
^2)^(1/3))/(9*d*Tan[c + d*x]^(2/3)) - (20*(-1)^(5/6)*Log[1 - E^((I/3)*Pi)*(Tan[c + d*x]^2)^(1/3)]*(Tan[c + d*x
]^2)^(1/3))/(9*d*Tan[c + d*x]^(2/3)) + (((49*I)/18)*Log[1 - I*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(5/6))/
(d*Tan[c + d*x]^(5/3)) - (((49*I)/18)*Log[1 + I*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(5/6))/(d*Tan[c + d*x
]^(5/3)) - (49*(-1)^(1/6)*Log[1 - (Tan[c + d*x]^2)^(1/6)/E^((I/6)*Pi)]*(Tan[c + d*x]^2)^(5/6))/(18*d*Tan[c + d
*x]^(5/3)) + (49*(-1)^(5/6)*Log[1 - E^((I/6)*Pi)*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(5/6))/(18*d*Tan[c +
d*x]^(5/3)) - (49*(-1)^(5/6)*Log[1 - (Tan[c + d*x]^2)^(1/6)/E^(((5*I)/6)*Pi)]*(Tan[c + d*x]^2)^(5/6))/(18*d*T
an[c + d*x]^(5/3)) + (49*(-1)^(1/6)*Log[1 - E^(((5*I)/6)*Pi)*(Tan[c + d*x]^2)^(1/6)]*(Tan[c + d*x]^2)^(5/6))/(
18*d*Tan[c + d*x]^(5/3)) - (49*a*Tan[c + d*x]^(1/3))/(3*d*(a + I*a*Tan[c + d*x])) - ((13*I)*a*Tan[c + d*x]^(4/
3))/(d*(a + I*a*Tan[c + d*x])) - (a*Tan[c + d*x]^(7/3))/(d*(a + I*a*Tan[c + d*x])) + (I*a*Tan[c + d*x]^(10/3))
/(d*(a + I*a*Tan[c + d*x])))/(4*a^2)

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.66

method	result	size
derivativedivides	\(\frac {-3 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\frac {i}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}+\frac {89 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}-\frac {5}{12 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}-\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}-\frac {-30 \tan \left (d x +c \right )-4 i \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )-4 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+28 i}{72 {\left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}^{2}}-\frac {89 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}+\frac {89 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}}{d \,a^{2}}\)	\(235\)
default	\(\frac {-3 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\frac {i}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}+\frac {89 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}-\frac {5}{12 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}-\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}-\frac {-30 \tan \left (d x +c \right )-4 i \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )-4 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+28 i}{72 {\left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}^{2}}-\frac {89 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}+\frac {89 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}}{d \,a^{2}}\)	\(235\)

[In]

int(tan(d*x+c)^(10/3)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d/a^2*(-3*tan(d*x+c)^(1/3)+1/36*I/(tan(d*x+c)^(1/3)+I)^2+89/72*I*ln(tan(d*x+c)^(1/3)+I)-5/12/(tan(d*x+c)^(1/
3)+I)+1/16*I*ln(I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)+1/8*3^(1/2)*arctanh(1/3*(I+2*tan(d*x+c)^(1/3))*3^(1/2))
-1/8*I*ln(tan(d*x+c)^(1/3)-I)-1/72*(-30*tan(d*x+c)-4*I*tan(d*x+c)^(2/3)-4*tan(d*x+c)^(1/3)+28*I)/(-I*tan(d*x+c
)^(1/3)+tan(d*x+c)^(2/3)-1)^2-89/144*I*ln(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)+89/72*3^(1/2)*arctanh(1/3*(-
I+2*tan(d*x+c)^(1/3))*3^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 521, normalized size of antiderivative = 1.46 \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (9 \, {\left (\sqrt {3} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} + i \, 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} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} - i \, 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 ) + 89 \, {\left (3 \, \sqrt {\frac {1}{3}} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} - i \, 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 ) - 89 \, {\left (3 \, \sqrt {\frac {1}{3}} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} + i \, 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 ) + 178 i \, e^{\left (4 i \, d x + 4 i \, c\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 i \, e^{\left (4 i \, d x + 4 i \, c\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 (173 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 26 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{144 \, a^{2} d} \]

[In]

integrate(tan(d*x+c)^(10/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/144*(9*(sqrt(3)*a^2*d*sqrt(1/(a^4*d^2))*e^(4*I*d*x + 4*I*c) + I*e^(4*I*d*x + 4*I*c))*log(1/2*sqrt(3)*a^2*d*s
qrt(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)*a^2*d*
sqrt(1/(a^4*d^2))*e^(4*I*d*x + 4*I*c) - I*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) + 89*(3*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d^2))*
e^(4*I*d*x + 4*I*c) - I*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) - 89*(3*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d^2))*e^(4*I*d*x + 4*I*
c) + I*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) + 178*I*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*I*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)*(173*e^(4*I*d*x + 4*I*c) + 26
*e^(2*I*d*x + 2*I*c) - 3))*e^(-4*I*d*x - 4*I*c)/(a^2*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(tan(d*x+c)**(10/3)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(tan(d*x+c)^(10/3)/(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 = 0.62 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.68 \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {89 \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right )}{144 \, a^{2} d} - \frac {\sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right )}{16 \, a^{2} d} + \frac {i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{16 \, a^{2} d} - \frac {89 i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{144 \, a^{2} d} + \frac {89 i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right )}{72 \, a^{2} d} - \frac {i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )}{8 \, a^{2} d} - \frac {3 \, \tan \left (d x + c\right )^{\frac {1}{3}}}{a^{2} d} - \frac {-16 i \, \tan \left (d x + c\right )^{\frac {4}{3}} - 13 \, \tan \left (d x + c\right )^{\frac {1}{3}}}{12 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} \]

[In]

integrate(tan(d*x+c)^(10/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-89/144*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
*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*log(
tan(d*x + c)^(2/3) + I*tan(d*x + c)^(1/3) - 1)/(a^2*d) - 89/144*I*log(tan(d*x + c)^(2/3) - I*tan(d*x + c)^(1/3
) - 1)/(a^2*d) + 89/72*I*log(tan(d*x + c)^(1/3) + I)/(a^2*d) - 1/8*I*log(tan(d*x + c)^(1/3) - I)/(a^2*d) - 3*t
an(d*x + c)^(1/3)/(a^2*d) - 1/12*(-16*I*tan(d*x + c)^(4/3) - 13*tan(d*x + c)^(1/3))/(a^2*d*(tan(d*x + c) - I)^
2)

Mupad [B] (verification not implemented)

Time = 6.26 (sec) , antiderivative size = 668, normalized size of antiderivative = 1.87 \[ \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

int(tan(c + d*x)^(10/3)/(a + a*tan(c + d*x)*1i)^2,x)

[Out]

log(((a^6*d^3*90329344i)/3 - 17694720*a^10*d^5*tan(c + d*x)^(1/3)*(1i/(512*a^6*d^3))^(2/3))*(1i/(512*a^6*d^3))
^(1/3) + (a^4*d^2*tan(c + d*x)^(1/3)*11394848i)/3)*(1i/(512*a^6*d^3))^(1/3) + log(((a^6*d^3*90329344i)/3 - 176
94720*a^10*d^5*tan(c + d*x)^(1/3)*(-704969i/(373248*a^6*d^3))^(2/3))*(-704969i/(373248*a^6*d^3))^(1/3) + (a^4*
d^2*tan(c + d*x)^(1/3)*11394848i)/3)*(-704969i/(373248*a^6*d^3))^(1/3) + ((tan(c + d*x)^(1/3)*13i)/(12*a^2*d)
- (4*tan(c + d*x)^(4/3))/(3*a^2*d))/(2*tan(c + d*x) + tan(c + d*x)^2*1i - 1i) - (3*tan(c + d*x)^(1/3))/(a^2*d)
+ (log(((3^(1/2)*1i - 1)*((a^6*d^3*90329344i)/3 - 4423680*a^10*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)^2*(1i/
(512*a^6*d^3))^(2/3))*(1i/(512*a^6*d^3))^(1/3))/2 + (a^4*d^2*tan(c + d*x)^(1/3)*11394848i)/3)*(3^(1/2)*1i - 1)
*(1i/(512*a^6*d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)*((a^6*d^3*90329344i)/3 - 4423680*a^10*d^5*tan(c + d*x)^(
1/3)*(3^(1/2)*1i + 1)^2*(1i/(512*a^6*d^3))^(2/3))*(1i/(512*a^6*d^3))^(1/3))/2 - (a^4*d^2*tan(c + d*x)^(1/3)*11
394848i)/3)*(3^(1/2)*1i + 1)*(1i/(512*a^6*d^3))^(1/3))/2 + (log(((3^(1/2)*1i - 1)*((a^6*d^3*90329344i)/3 - 442
3680*a^10*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)^2*(-704969i/(373248*a^6*d^3))^(2/3))*(-704969i/(373248*a^6*d
^3))^(1/3))/2 + (a^4*d^2*tan(c + d*x)^(1/3)*11394848i)/3)*(3^(1/2)*1i - 1)*(-704969i/(373248*a^6*d^3))^(1/3))/
2 - (log(((3^(1/2)*1i + 1)*((a^6*d^3*90329344i)/3 - 4423680*a^10*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)^2*(-7
04969i/(373248*a^6*d^3))^(2/3))*(-704969i/(373248*a^6*d^3))^(1/3))/2 - (a^4*d^2*tan(c + d*x)^(1/3)*11394848i)/
3)*(3^(1/2)*1i + 1)*(-704969i/(373248*a^6*d^3))^(1/3))/2