3.3.42 \(\int \frac {\tan ^{\frac {8}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [242]

Optimal. Leaf size=337 \[ \frac {25 \text {ArcTan}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {25 \text {ArcTan}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {2 i \text {ArcTan}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}-\frac {25 \text {ArcTan}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {2 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {25 \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}+\frac {25 \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}+\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{9 a^2 d}+\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

-25/72*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a^2/d-25/72*arctan(3^(1/2)+2*tan(d*x+c)^(1/3))/a^2/d-25/36*arctan(t
an(d*x+c)^(1/3))/a^2/d-2/9*I*ln(1+tan(d*x+c)^(2/3))/a^2/d+1/9*I*ln(1-tan(d*x+c)^(2/3)+tan(d*x+c)^(4/3))/a^2/d+
2/9*I*arctan(1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))/a^2/d*3^(1/2)-25/144*ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)
^(2/3))/a^2/d*3^(1/2)+25/144*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/a^2/d*3^(1/2)+2/3*I*tan(d*x+c)^(2
/3)/a^2/d/(1+I*tan(d*x+c))-1/4*tan(d*x+c)^(5/3)/d/(a+I*a*tan(d*x+c))^2

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Rubi [A]
time = 0.40, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 14, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3639, 3676, 3619, 3557, 335, 281, 206, 31, 648, 632, 210, 642, 301, 209} \begin {gather*} \frac {2 i \text {ArcTan}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}+\frac {25 \text {ArcTan}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {25 \text {ArcTan}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{72 a^2 d}-\frac {25 \text {ArcTan}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}-\frac {2 i \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{9 a^2 d}-\frac {25 \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 {25 \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 {i \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 {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(8/3)/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(25*ArcTan[Sqrt[3] - 2*Tan[c + d*x]^(1/3)])/(72*a^2*d) - (25*ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(1/3)])/(72*a^2*d
) + (((2*I)/3)*ArcTan[(1 - 2*Tan[c + d*x]^(2/3))/Sqrt[3]])/(Sqrt[3]*a^2*d) - (25*ArcTan[Tan[c + d*x]^(1/3)])/(
36*a^2*d) - (((2*I)/9)*Log[1 + Tan[c + d*x]^(2/3)])/(a^2*d) - (25*Log[1 - Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c +
d*x]^(2/3)])/(48*Sqrt[3]*a^2*d) + (25*Log[1 + Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/(48*Sqrt[3]*a
^2*d) + ((I/9)*Log[1 - Tan[c + d*x]^(2/3) + Tan[c + d*x]^(4/3)])/(a^2*d) + (((2*I)/3)*Tan[c + d*x]^(2/3))/(a^2
*d*(1 + I*Tan[c + d*x])) - Tan[c + d*x]^(5/3)/(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 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*} \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan ^{\frac {2}{3}}(c+d x) \left (-\frac {5 a}{3}+\frac {11}{3} i a \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {-\frac {32 i a^2}{9}-\frac {50}{9} a^2 \tan (c+d x)}{\sqrt [3]{\tan (c+d x)}} \, dx}{8 a^4}\\ &=\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(4 i) \int \frac {1}{\sqrt [3]{\tan (c+d x)}} \, dx}{9 a^2}-\frac {25 \int \tan ^{\frac {2}{3}}(c+d x) \, dx}{36 a^2}\\ &=\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(4 i) \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 {25 \text {Subst}\left (\int \frac {x^{2/3}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{36 a^2 d}\\ &=\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(4 i) \text {Subst}\left (\int \frac {x}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{3 a^2 d}-\frac {25 \text {Subst}\left (\int \frac {x^4}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{12 a^2 d}\\ &=\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(2 i) \text {Subst}\left (\int \frac {1}{1+x^3} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a^2 d}-\frac {25 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {25 \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 {25 \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 {25 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(2 i) \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {(2 i) \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 {25 \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 {25 \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 {25 \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 {25 \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 {25 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {2 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {25 \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 {25 \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 {2 i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {i \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 {i \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 {25 \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 {25 \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 {25 \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {25 \tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {25 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {2 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {25 \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 {25 \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 {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{9 a^2 d}+\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {(2 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 {25 \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {25 \tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {2 i \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}-\frac {25 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {2 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {25 \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 {25 \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 {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{9 a^2 d}+\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 1.76, size = 194, normalized size = 0.58 \begin {gather*} -\frac {i \sec ^2(c+d x) \left (9 \sqrt [3]{2} e^{2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )-82 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+4 (8+8 \cos (2 (c+d x))+11 i \sin (2 (c+d x)))\right ) \tan ^{\frac {2}{3}}(c+d x)}{96 a^2 d (-i+\tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^(8/3)/(a + I*a*Tan[c + d*x])^2,x]

[Out]

((-1/96*I)*Sec[c + d*x]^2*(9*2^(1/3)*E^((2*I)*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(2/3)*Hypergeometric2F1[2/3
, 2/3, 5/3, (1 - E^((2*I)*(c + d*x)))/2] - 82*Hypergeometric2F1[2/3, 1, 5/3, -((-1 + E^((2*I)*(c + d*x)))/(1 +
E^((2*I)*(c + d*x))))]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) + 4*(8 + 8*Cos[2*(c + d*x)] + (11*I)*Sin[2*(c
+ d*x)]))*Tan[c + d*x]^(2/3))/(a^2*d*(-I + Tan[c + d*x])^2)

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Maple [A]
time = 0.20, size = 225, normalized size = 0.67

method	result	size
derivativedivides	\(\frac {-\frac {41 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}-\frac {i}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}+\frac {11}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}-\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}+\frac {\sqrt {3}\, \arctanh \left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}+\frac {44 \tan \left (d x +c \right )-64 i \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )-58 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+20 i}{72 \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )^{2}}+\frac {41 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}+\frac {41 \sqrt {3}\, \arctanh \left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}+\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}}{d \,a^{2}}\)	\(225\)
default	\(\frac {-\frac {41 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}-\frac {i}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}+\frac {11}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}-\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}+\frac {\sqrt {3}\, \arctanh \left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}+\frac {44 \tan \left (d x +c \right )-64 i \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )-58 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+20 i}{72 \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )^{2}}+\frac {41 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}+\frac {41 \sqrt {3}\, \arctanh \left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}+\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}}{d \,a^{2}}\)	\(225\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(8/3)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d/a^2*(-41/72*I*ln(tan(d*x+c)^(1/3)+I)-1/36*I/(tan(d*x+c)^(1/3)+I)^2+11/36/(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))+1/72*(44*tan(d*x+
c)-64*I*tan(d*x+c)^(2/3)-58*tan(d*x+c)^(1/3)+20*I)/(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)^2+41/144*I*ln(-I*t
an(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)+41/72*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))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(8/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [A]
time = 0.72, size = 521, normalized size = 1.55 \begin {gather*} \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 ) + 41 \, {\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 ) - 41 \, {\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 ) - 82 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 {2}{3}} {\left (-19 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 16 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{144 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(8/3)/(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) + 41*(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) - 41*(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) - 82*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))^(2/3)*(-19*I*e^(4*I*d*x + 4*I*c) - 1
6*I*e^(2*I*d*x + 2*I*c) + 3*I))*e^(-4*I*d*x - 4*I*c)/(a^2*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**(8/3)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]
time = 0.94, size = 228, normalized size = 0.68 \begin {gather*} -\frac {41 \, \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 {41 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 {41 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 {11 \, \tan \left (d x + c\right )^{\frac {5}{3}} - 8 i \, \tan \left (d x + c\right )^{\frac {2}{3}}}{12 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(8/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-41/144*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) + 41/144*I*log(tan(d*x + c)^(2/3) - I*tan(d*x + c)^(1/3
) - 1)/(a^2*d) - 41/72*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) + 1/1
2*(11*tan(d*x + c)^(5/3) - 8*I*tan(d*x + c)^(2/3))/(a^2*d*(tan(d*x + c) - I)^2)

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Mupad [B]
time = 5.26, size = 642, normalized size = 1.91 \begin {gather*} \ln \left (-\left (\frac {a^6\,d^3\,8915200{}\mathrm {i}}{3}+a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{1/3}\,5412864{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{2/3}-\frac {107584\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}\right )\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{1/3}+\ln \left (-\left (\frac {a^6\,d^3\,8915200{}\mathrm {i}}{3}+a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (\frac {68921{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{1/3}\,5412864{}\mathrm {i}\right )\,{\left (\frac {68921{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{2/3}-\frac {107584\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}\right )\,{\left (\frac {68921{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{1/3}+\frac {\frac {2\,{\mathrm {tan}\left (c+d\,x\right )}^{2/3}}{3\,a^2\,d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^{5/3}\,11{}\mathrm {i}}{12\,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 {\ln \left (-\frac {107584\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}-\frac {{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (\frac {a^6\,d^3\,8915200{}\mathrm {i}}{3}+a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{1/3}\,2706432{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{2/3}}{4}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{1/3}}{2}-\frac {\ln \left (-\frac {107584\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}-\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (\frac {a^6\,d^3\,8915200{}\mathrm {i}}{3}-a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{1/3}\,2706432{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{2/3}}{4}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{1/3}}{2}+\frac {\ln \left (-\frac {107584\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}-\frac {{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (\frac {a^6\,d^3\,8915200{}\mathrm {i}}{3}+a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {68921{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{1/3}\,2706432{}\mathrm {i}\right )\,{\left (\frac {68921{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{2/3}}{4}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {68921{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{1/3}}{2}-\frac {\ln \left (-\frac {107584\,a^2\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{3}-\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (\frac {a^6\,d^3\,8915200{}\mathrm {i}}{3}-a^8\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {68921{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{1/3}\,2706432{}\mathrm {i}\right )\,{\left (\frac {68921{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{2/3}}{4}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {68921{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{1/3}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(8/3)/(a + a*tan(c + d*x)*1i)^2,x)

[Out]

log(- ((a^6*d^3*8915200i)/3 + a^8*d^4*tan(c + d*x)^(1/3)*(-1i/(512*a^6*d^3))^(1/3)*5412864i)*(-1i/(512*a^6*d^3
))^(2/3) - (107584*a^2*d*tan(c + d*x)^(1/3))/3)*(-1i/(512*a^6*d^3))^(1/3) + log(- ((a^6*d^3*8915200i)/3 + a^8*
d^4*tan(c + d*x)^(1/3)*(68921i/(373248*a^6*d^3))^(1/3)*5412864i)*(68921i/(373248*a^6*d^3))^(2/3) - (107584*a^2
*d*tan(c + d*x)^(1/3))/3)*(68921i/(373248*a^6*d^3))^(1/3) + ((2*tan(c + d*x)^(2/3))/(3*a^2*d) + (tan(c + d*x)^
(5/3)*11i)/(12*a^2*d))/(2*tan(c + d*x) + tan(c + d*x)^2*1i - 1i) + (log(- (107584*a^2*d*tan(c + d*x)^(1/3))/3
- ((3^(1/2)*1i - 1)^2*((a^6*d^3*8915200i)/3 + a^8*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)*(-1i/(512*a^6*d^3))^
(1/3)*2706432i)*(-1i/(512*a^6*d^3))^(2/3))/4)*(3^(1/2)*1i - 1)*(-1i/(512*a^6*d^3))^(1/3))/2 - (log(- (107584*a
^2*d*tan(c + d*x)^(1/3))/3 - ((3^(1/2)*1i + 1)^2*((a^6*d^3*8915200i)/3 - a^8*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1
i + 1)*(-1i/(512*a^6*d^3))^(1/3)*2706432i)*(-1i/(512*a^6*d^3))^(2/3))/4)*(3^(1/2)*1i + 1)*(-1i/(512*a^6*d^3))^
(1/3))/2 + (log(- (107584*a^2*d*tan(c + d*x)^(1/3))/3 - ((3^(1/2)*1i - 1)^2*((a^6*d^3*8915200i)/3 + a^8*d^4*ta
n(c + d*x)^(1/3)*(3^(1/2)*1i - 1)*(68921i/(373248*a^6*d^3))^(1/3)*2706432i)*(68921i/(373248*a^6*d^3))^(2/3))/4
)*(3^(1/2)*1i - 1)*(68921i/(373248*a^6*d^3))^(1/3))/2 - (log(- (107584*a^2*d*tan(c + d*x)^(1/3))/3 - ((3^(1/2)
*1i + 1)^2*((a^6*d^3*8915200i)/3 - a^8*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)*(68921i/(373248*a^6*d^3))^(1/3)
*2706432i)*(68921i/(373248*a^6*d^3))^(2/3))/4)*(3^(1/2)*1i + 1)*(68921i/(373248*a^6*d^3))^(1/3))/2

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