\(\int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx\) [245]

Optimal result

Integrand size = 26, antiderivative size = 339 \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {7 i \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {7 i \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {2 \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}-\frac {7 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {2 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {7 i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}+\frac {7 i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {\log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{9 a^2 d}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{12 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3640, 3677, 3619, 3557, 335, 281, 206, 31, 648, 632, 210, 642, 301, 209} \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx=-\frac {2 \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}+\frac {7 i \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {7 i \arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{72 a^2 d}-\frac {7 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{12 a^2 d (1+i \tan (c+d x))}+\frac {2 \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{9 a^2 d}-\frac {7 i \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt {3} a^2 d}+\frac {7 i \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt {3} a^2 d}-\frac {\log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{9 a^2 d}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[In]

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

[Out]

(((7*I)/72)*ArcTan[Sqrt[3] - 2*Tan[c + d*x]^(1/3)])/(a^2*d) - (((7*I)/72)*ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(1/3
)])/(a^2*d) - (2*ArcTan[(1 - 2*Tan[c + d*x]^(2/3))/Sqrt[3]])/(3*Sqrt[3]*a^2*d) - (((7*I)/36)*ArcTan[Tan[c + d*
x]^(1/3)])/(a^2*d) + (2*Log[1 + Tan[c + d*x]^(2/3)])/(9*a^2*d) - (((7*I)/48)*Log[1 - Sqrt[3]*Tan[c + d*x]^(1/3
) + Tan[c + d*x]^(2/3)])/(Sqrt[3]*a^2*d) + (((7*I)/48)*Log[1 + Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)
])/(Sqrt[3]*a^2*d) - Log[1 - Tan[c + d*x]^(2/3) + Tan[c + d*x]^(4/3)]/(9*a^2*d) + (7*Tan[c + d*x]^(2/3))/(12*a
^2*d*(1 + I*Tan[c + d*x])) + Tan[c + d*x]^(2/3)/(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 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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 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 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 301

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

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 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 {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\frac {10 a}{3}-\frac {4}{3} i a \tan (c+d x)}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))} \, dx}{4 a^2} \\ & = \frac {7 \tan ^{\frac {2}{3}}(c+d x)}{12 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\frac {32 a^2}{9}-\frac {14}{9} i a^2 \tan (c+d x)}{\sqrt [3]{\tan (c+d x)}} \, dx}{8 a^4} \\ & = \frac {7 \tan ^{\frac {2}{3}}(c+d x)}{12 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(7 i) \int \tan ^{\frac {2}{3}}(c+d x) \, dx}{36 a^2}+\frac {4 \int \frac {1}{\sqrt [3]{\tan (c+d x)}} \, dx}{9 a^2} \\ & = \frac {7 \tan ^{\frac {2}{3}}(c+d x)}{12 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(7 i) \text {Subst}\left (\int \frac {x^{2/3}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{36 a^2 d}+\frac {4 \text {Subst}\left (\int \frac {1}{\sqrt [3]{x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{9 a^2 d} \\ & = \frac {7 \tan ^{\frac {2}{3}}(c+d x)}{12 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(7 i) \text {Subst}\left (\int \frac {x^4}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{12 a^2 d}+\frac {4 \text {Subst}\left (\int \frac {x}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{3 a^2 d} \\ & = \frac {7 \tan ^{\frac {2}{3}}(c+d x)}{12 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(7 i) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {(7 i) \text {Subst}\left (\int \frac {-\frac {1}{2}+\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 {(7 i) \text {Subst}\left (\int \frac {-\frac {1}{2}-\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 {2 \text {Subst}\left (\int \frac {1}{1+x^3} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a^2 d} \\ & = -\frac {7 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{12 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(7 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 {(7 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 {2 \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}+\frac {2 \text {Subst}\left (\int \frac {2-x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {(7 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 {(7 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 {7 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {2 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {7 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 {7 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 {7 \tan ^{\frac {2}{3}}(c+d x)}{12 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {(7 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 {(7 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 {\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 {\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 {7 i \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {7 i \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {7 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {2 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {7 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 {7 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 {\log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{9 a^2 d}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{12 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {2 \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 {7 i \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {7 i \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {2 \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}-\frac {7 i \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {2 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {7 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 {7 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 {\log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{9 a^2 d}+\frac {7 \tan ^{\frac {2}{3}}(c+d x)}{12 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.35 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {18 \tan ^{\frac {2}{3}}(c+d x)+\frac {(a+i a \tan (c+d x)) \left (42 a \tan ^{\frac {2}{3}}(c+d x) \tan ^2(c+d x)^{5/6}-(a+i a \tan (c+d x)) \left (-7 \left (\log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right )-\log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right )+\sqrt [3]{-1} \left (-\sqrt [3]{-1} \log \left (1-\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )+\sqrt [3]{-1} \log \left (1+\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )+\log \left (1-(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )-\log \left (1+(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )\right )\right ) \tan ^{\frac {5}{3}}(c+d x)-8 \left (2 \sqrt {3} \arctan \left (\frac {-1+2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )+2 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )-\log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )\right ) \tan ^2(c+d x)^{5/6}\right )\right )}{a^2 \tan ^2(c+d x)^{5/6}}}{72 d (a+i a \tan (c+d x))^2} \]

[In]

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

[Out]

(18*Tan[c + d*x]^(2/3) + ((a + I*a*Tan[c + d*x])*(42*a*Tan[c + d*x]^(2/3)*(Tan[c + d*x]^2)^(5/6) - (a + I*a*Ta
n[c + d*x])*(-7*(Log[1 - I*(Tan[c + d*x]^2)^(1/6)] - Log[1 + I*(Tan[c + d*x]^2)^(1/6)] + (-1)^(1/3)*(-((-1)^(1
/3)*Log[1 - (-1)^(1/6)*(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)^(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]^(5/3) - 8*(
2*Sqrt[3]*ArcTan[(-1 + 2*Tan[c + d*x]^(2/3))/Sqrt[3]] + 2*Log[1 + Tan[c + d*x]^(2/3)] - Log[1 - Tan[c + d*x]^(
2/3) + Tan[c + d*x]^(4/3)])*(Tan[c + d*x]^2)^(5/6))))/(a^2*(Tan[c + d*x]^2)^(5/6)))/(72*d*(a + I*a*Tan[c + d*x
])^2)

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.66

method	result	size
derivativedivides	\(\frac {-\frac {7 i}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {1}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}+\frac {23 \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}-\frac {\ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}-\frac {i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}+\frac {\ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}+\frac {-28 i \tan \left (d x +c \right )-44 \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )+50 i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+16}{72 {\left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}^{2}}-\frac {23 \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}+\frac {23 i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}}{d \,a^{2}}\)	\(223\)
default	\(\frac {-\frac {7 i}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {1}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}+\frac {23 \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}-\frac {\ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}-\frac {i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}+\frac {\ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}+\frac {-28 i \tan \left (d x +c \right )-44 \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )+50 i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+16}{72 {\left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}^{2}}-\frac {23 \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}+\frac {23 i \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}}{d \,a^{2}}\)	\(223\)

[In]

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

[Out]

1/d/a^2*(-7/36*I/(tan(d*x+c)^(1/3)+I)+1/36/(tan(d*x+c)^(1/3)+I)^2+23/72*ln(tan(d*x+c)^(1/3)+I)-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))+1/8*ln(tan(d*x+c)^(1
/3)-I)+1/72*(-28*I*tan(d*x+c)-44*tan(d*x+c)^(2/3)+50*I*tan(d*x+c)^(1/3)+16)/(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2
/3)-1)^2-23/144*ln(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)+23/72*I*3^(1/2)*arctanh(1/3*(-I+2*tan(d*x+c)^(1/3))
*3^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 515, normalized size of antiderivative = 1.52 \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx=-\frac {{\left (9 \, {\left (i \, \sqrt {3} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 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 (-i \, \sqrt {3} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 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 ) + 23 \, {\left (-3 i \, \sqrt {\frac {1}{3}} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 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 ) + 23 \, {\left (3 i \, \sqrt {\frac {1}{3}} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 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 ) - 46 \, 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 \, 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 {2}{3}} {\left (17 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 20 \, 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(1/tan(d*x+c)^(1/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/144*(9*(I*sqrt(3)*a^2*d*sqrt(1/(a^4*d^2))*e^(4*I*d*x + 4*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*(-I*sqrt(3)*a^
2*d*sqrt(1/(a^4*d^2))*e^(4*I*d*x + 4*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) + 23*(-3*I*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d
^2))*e^(4*I*d*x + 4*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) + 23*(3*I*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d^2))*e^(4*I*d*x +
4*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) - 46*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^(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))^(2/3)*(17*e^(4*I*d*x + 4*I*c) + 20*e^
(2*I*d*x + 2*I*c) + 3))*e^(-4*I*d*x - 4*I*c)/(a^2*d)

Sympy [F]

\[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx=- \frac {\int \frac {1}{\tan ^{\frac {7}{3}}{\left (c + d x \right )} - 2 i \tan ^{\frac {4}{3}}{\left (c + d x \right )} - \sqrt [3]{\tan {\left (c + d x \right )}}}\, dx}{a^{2}} \]

[In]

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

[Out]

-Integral(1/(tan(c + d*x)**(7/3) - 2*I*tan(c + d*x)**(4/3) - tan(c + d*x)**(1/3)), x)/a**2

Maxima [F(-2)]

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

[In]

integrate(1/tan(d*x+c)^(1/3)/(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 = 0.65 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx=-\frac {23 i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right )}{144 \, a^{2} d} + \frac {i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right )}{16 \, a^{2} d} - \frac {\log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{16 \, a^{2} d} - \frac {23 \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{144 \, a^{2} d} + \frac {23 \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right )}{72 \, a^{2} d} + \frac {\log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )}{8 \, a^{2} d} + \frac {-7 i \, \tan \left (d x + c\right )^{\frac {5}{3}} - 10 \, \tan \left (d x + c\right )^{\frac {2}{3}}}{12 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} \]

[In]

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

[Out]

-23/144*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) - 23/144*log(tan(d*x + c)^(2/3) - I*tan(d*x + c)^(1/3
) - 1)/(a^2*d) + 23/72*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*(-
7*I*tan(d*x + c)^(5/3) - 10*tan(d*x + c)^(2/3))/(a^2*d*(tan(d*x + c) - I)^2)

Mupad [B] (verification not implemented)

Time = 5.81 (sec) , antiderivative size = 630, normalized size of antiderivative = 1.86 \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {23\,\ln \left (\frac {529\,\left (\frac {1795840\,a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (\frac {1}{a^6\,d^3}\right )}^{1/3}}{3}+\frac {a^6\,d^3\,1464064{}\mathrm {i}}{3}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{2/3}}{5184}-\frac {33856\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{1/3}}{72}+\ln \left (\left (1873920\,a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{1/3}+\frac {a^6\,d^3\,1464064{}\mathrm {i}}{3}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{2/3}-\frac {33856\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{1/3}-\frac {-\frac {7\,{\mathrm {tan}\left (c+d\,x\right )}^{5/3}}{12\,a^2\,d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^{2/3}\,5{}\mathrm {i}}{6\,a^2\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}}+\frac {23\,\ln \left (-\frac {33856\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}+\frac {529\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (\frac {a^6\,d^3\,1464064{}\mathrm {i}}{3}+\frac {897920\,a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{1/3}}{3}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{2/3}}{20736}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{1/3}}{144}-\frac {23\,\ln \left (-\frac {33856\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}+\frac {529\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (\frac {a^6\,d^3\,1464064{}\mathrm {i}}{3}-\frac {897920\,a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{1/3}}{3}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{2/3}}{20736}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{a^6\,d^3}\right )}^{1/3}}{144}+\ln \left (-\frac {33856\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}+{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,\left (\frac {a^6\,d^3\,1464064{}\mathrm {i}}{3}+1873920\,a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{1/3}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{2/3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{1/3}-\ln \left (-\frac {33856\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}+{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,\left (\frac {a^6\,d^3\,1464064{}\mathrm {i}}{3}-1873920\,a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{1/3}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{2/3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (\frac {1}{512\,a^6\,d^3}\right )}^{1/3} \]

[In]

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

[Out]

(23*log((529*((a^6*d^3*1464064i)/3 + (1795840*a^8*d^4*tan(c + d*x)^(1/3)*(1/(a^6*d^3))^(1/3))/3)*(1/(a^6*d^3))
^(2/3))/5184 - (33856*a^2*d*tan(c + d*x)^(1/3))/3)*(1/(a^6*d^3))^(1/3))/72 + log(((a^6*d^3*1464064i)/3 + 18739
20*a^8*d^4*tan(c + d*x)^(1/3)*(1/(512*a^6*d^3))^(1/3))*(1/(512*a^6*d^3))^(2/3) - (33856*a^2*d*tan(c + d*x)^(1/
3))/3)*(1/(512*a^6*d^3))^(1/3) - ((tan(c + d*x)^(2/3)*5i)/(6*a^2*d) - (7*tan(c + d*x)^(5/3))/(12*a^2*d))/(2*ta
n(c + d*x) + tan(c + d*x)^2*1i - 1i) + (23*log((529*(3^(1/2)*1i - 1)^2*((a^6*d^3*1464064i)/3 + (897920*a^8*d^4
*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)*(1/(a^6*d^3))^(1/3))/3)*(1/(a^6*d^3))^(2/3))/20736 - (33856*a^2*d*tan(c +
d*x)^(1/3))/3)*(3^(1/2)*1i - 1)*(1/(a^6*d^3))^(1/3))/144 - (23*log((529*(3^(1/2)*1i + 1)^2*((a^6*d^3*1464064i
)/3 - (897920*a^8*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)*(1/(a^6*d^3))^(1/3))/3)*(1/(a^6*d^3))^(2/3))/20736 -
(33856*a^2*d*tan(c + d*x)^(1/3))/3)*(3^(1/2)*1i + 1)*(1/(a^6*d^3))^(1/3))/144 + log(((3^(1/2)*1i)/2 - 1/2)^2*
((a^6*d^3*1464064i)/3 + 1873920*a^8*d^4*tan(c + d*x)^(1/3)*((3^(1/2)*1i)/2 - 1/2)*(1/(512*a^6*d^3))^(1/3))*(1/
(512*a^6*d^3))^(2/3) - (33856*a^2*d*tan(c + d*x)^(1/3))/3)*((3^(1/2)*1i)/2 - 1/2)*(1/(512*a^6*d^3))^(1/3) - lo
g(((3^(1/2)*1i)/2 + 1/2)^2*((a^6*d^3*1464064i)/3 - 1873920*a^8*d^4*tan(c + d*x)^(1/3)*((3^(1/2)*1i)/2 + 1/2)*(
1/(512*a^6*d^3))^(1/3))*(1/(512*a^6*d^3))^(2/3) - (33856*a^2*d*tan(c + d*x)^(1/3))/3)*((3^(1/2)*1i)/2 + 1/2)*(
1/(512*a^6*d^3))^(1/3)