\(\int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx\) [234]

Optimal result

Integrand size = 26, antiderivative size = 319 \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {5 \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {5 \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {2 i \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}+\frac {5 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {2 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}+\frac {5 \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {5 \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{3 a d}-\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{a d}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))} \]

[Out]

5/12*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a/d+5/12*arctan(3^(1/2)+2*tan(d*x+c)^(1/3))/a/d+5/6*arctan(tan(d*x+c)
^(1/3))/a/d+2/3*I*ln(1+tan(d*x+c)^(2/3))/a/d-1/3*I*ln(1-tan(d*x+c)^(2/3)+tan(d*x+c)^(4/3))/a/d-2/3*I*arctan(1/
3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))/a/d*3^(1/2)+5/24*ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/a/d*3^(1/2)
-5/24*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/a/d*3^(1/2)-2*I*tan(d*x+c)^(2/3)/a/d-1/2*tan(d*x+c)^(5/3
)/d/(a+I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 319, 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 = {3631, 3609, 3619, 3557, 335, 281, 206, 31, 648, 632, 210, 642, 301, 209} \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {2 i \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}-\frac {5 \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {5 \arctan \left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{12 a d}+\frac {5 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{a d}+\frac {2 i \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{3 a d}+\frac {5 \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}-\frac {5 \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}-\frac {i \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{3 a d} \]

[In]

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

[Out]

(-5*ArcTan[Sqrt[3] - 2*Tan[c + d*x]^(1/3)])/(12*a*d) + (5*ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(1/3)])/(12*a*d) - (
(2*I)*ArcTan[(1 - 2*Tan[c + d*x]^(2/3))/Sqrt[3]])/(Sqrt[3]*a*d) + (5*ArcTan[Tan[c + d*x]^(1/3)])/(6*a*d) + (((
2*I)/3)*Log[1 + Tan[c + d*x]^(2/3)])/(a*d) + (5*Log[1 - Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/(8*S
qrt[3]*a*d) - (5*Log[1 + Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/(8*Sqrt[3]*a*d) - ((I/3)*Log[1 - Ta
n[c + d*x]^(2/3) + Tan[c + d*x]^(4/3)])/(a*d) - ((2*I)*Tan[c + d*x]^(2/3))/(a*d) - Tan[c + d*x]^(5/3)/(2*d*(a
+ I*a*Tan[c + d*x]))

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 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 3631

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*
c - a*d)*((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*(a + b*Tan[e + f*x]))), x] + Dist[1/(2*a^2), Int[(c + d*Tan[e +
f*x])^(n - 2)*Simp[a*c^2 + a*d^2*(n - 1) - b*c*d*n - d*(a*c*(n - 2) + b*d*n)*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] && GtQ[n, 1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \tan ^{\frac {2}{3}}(c+d x) \left (\frac {5 a}{3}-\frac {8}{3} i a \tan (c+d x)\right ) \, dx}{2 a^2} \\ & = -\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{a d}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \frac {\frac {8 i a}{3}+\frac {5}{3} a \tan (c+d x)}{\sqrt [3]{\tan (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{a d}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(4 i) \int \frac {1}{\sqrt [3]{\tan (c+d x)}} \, dx}{3 a}+\frac {5 \int \tan ^{\frac {2}{3}}(c+d x) \, dx}{6 a} \\ & = -\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{a d}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(4 i) \text {Subst}\left (\int \frac {1}{\sqrt [3]{x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{3 a d}+\frac {5 \text {Subst}\left (\int \frac {x^{2/3}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{6 a d} \\ & = -\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{a d}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(4 i) \text {Subst}\left (\int \frac {x}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{a d}+\frac {5 \text {Subst}\left (\int \frac {x^4}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 a d} \\ & = -\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{a d}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(2 i) \text {Subst}\left (\int \frac {1}{1+x^3} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{a d}+\frac {5 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {5 \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 )}{6 a d}+\frac {5 \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 )}{6 a d} \\ & = \frac {5 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{a d}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(2 i) \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a 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 )}{3 a d}+\frac {5 \text {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{24 a d}+\frac {5 \text {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{24 a d}+\frac {5 \text {Subst}\left (\int \frac {-\sqrt {3}+2 x}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{8 \sqrt {3} a d}-\frac {5 \text {Subst}\left (\int \frac {\sqrt {3}+2 x}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{8 \sqrt {3} a d} \\ & = \frac {5 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {2 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}+\frac {5 \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {5 \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{a d}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {i \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}+\frac {i \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{a d}-\frac {5 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {5 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d} \\ & = -\frac {5 \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {5 \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {5 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {2 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}+\frac {5 \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {5 \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{3 a d}-\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{a d}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {(2 i) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac {2}{3}}(c+d x)\right )}{a d} \\ & = -\frac {5 \arctan \left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {5 \arctan \left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {2 i \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}+\frac {5 \arctan \left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {2 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}+\frac {5 \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {5 \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{3 a d}-\frac {2 i \tan ^{\frac {2}{3}}(c+d x)}{a d}-\frac {\tan ^{\frac {5}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.79 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {\frac {6 \tan ^{\frac {11}{3}}(c+d x)}{a+i a \tan (c+d x)}+\frac {2 i \left (4 \sqrt {3} \arctan \left (\frac {-1+2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )+4 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )-2 \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )-12 \tan ^{\frac {2}{3}}(c+d x)+3 \tan ^{\frac {8}{3}}(c+d x)\right )}{a}+\frac {\tan ^{\frac {5}{3}}(c+d x) \left (5 i \log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right )-5 i \log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right )+5 \sqrt [6]{-1} \log \left (1-\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )-5 \sqrt [6]{-1} \log \left (1+\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )+5 (-1)^{5/6} \log \left (1-(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )-5 (-1)^{5/6} \log \left (1+(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )-6 \tan ^2(c+d x)^{5/6}\right )}{a \tan ^2(c+d x)^{5/6}}}{12 d} \]

[In]

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

[Out]

((6*Tan[c + d*x]^(11/3))/(a + I*a*Tan[c + d*x]) + ((2*I)*(4*Sqrt[3]*ArcTan[(-1 + 2*Tan[c + d*x]^(2/3))/Sqrt[3]
] + 4*Log[1 + Tan[c + d*x]^(2/3)] - 2*Log[1 - Tan[c + d*x]^(2/3) + Tan[c + d*x]^(4/3)] - 12*Tan[c + d*x]^(2/3)
+ 3*Tan[c + d*x]^(8/3)))/a + (Tan[c + d*x]^(5/3)*((5*I)*Log[1 - I*(Tan[c + d*x]^2)^(1/6)] - (5*I)*Log[1 + I*(
Tan[c + d*x]^2)^(1/6)] + 5*(-1)^(1/6)*Log[1 - (-1)^(1/6)*(Tan[c + d*x]^2)^(1/6)] - 5*(-1)^(1/6)*Log[1 + (-1)^(
1/6)*(Tan[c + d*x]^2)^(1/6)] + 5*(-1)^(5/6)*Log[1 - (-1)^(5/6)*(Tan[c + d*x]^2)^(1/6)] - 5*(-1)^(5/6)*Log[1 +
(-1)^(5/6)*(Tan[c + d*x]^2)^(1/6)] - 6*(Tan[c + d*x]^2)^(5/6)))/(a*(Tan[c + d*x]^2)^(5/6)))/(12*d)

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.63

method	result	size
derivativedivides	\(\frac {-\frac {3 i \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{2}+\frac {13 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{12}-\frac {1}{6 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{4}-\frac {4 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )-2 i}{12 \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}-\frac {13 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{24}-\frac {13 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{12}-\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{8}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{4}}{d a}\)	\(201\)
default	\(\frac {-\frac {3 i \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{2}+\frac {13 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{12}-\frac {1}{6 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{4}-\frac {4 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )-2 i}{12 \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}-\frac {13 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{24}-\frac {13 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{12}-\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{8}+\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{4}}{d a}\)	\(201\)

[In]

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

[Out]

1/d/a*(-3/2*I*tan(d*x+c)^(2/3)+13/12*I*ln(tan(d*x+c)^(1/3)+I)-1/6/(tan(d*x+c)^(1/3)+I)+1/4*I*ln(tan(d*x+c)^(1/
3)-I)-1/12*(4*tan(d*x+c)^(1/3)-2*I)/(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)-13/24*I*ln(-I*tan(d*x+c)^(1/3)+ta
n(d*x+c)^(2/3)-1)-13/12*3^(1/2)*arctanh(1/3*(-I+2*tan(d*x+c)^(1/3))*3^(1/2))-1/8*I*ln(I*tan(d*x+c)^(1/3)+tan(d
*x+c)^(2/3)-1)+1/4*3^(1/2)*arctanh(1/3*(I+2*tan(d*x+c)^(1/3))*3^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 494, normalized size of antiderivative = 1.55 \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {{\left (3 \, {\left (\sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (\frac {1}{2} \, \sqrt {3} a d \sqrt {\frac {1}{a^{2} 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 ) - 3 \, {\left (\sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (-\frac {1}{2} \, \sqrt {3} a d \sqrt {\frac {1}{a^{2} 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 ) - 13 \, {\left (3 \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (\frac {3}{2} \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} 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 ) + 13 \, {\left (3 \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (-\frac {3}{2} \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} 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 ) + 26 i \, e^{\left (2 i \, d x + 2 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 ) + 6 i \, e^{\left (2 i \, d x + 2 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 ) - 6 \, \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 (7 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{24 \, a d} \]

[In]

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

[Out]

1/24*(3*(sqrt(3)*a*d*sqrt(1/(a^2*d^2))*e^(2*I*d*x + 2*I*c) - I*e^(2*I*d*x + 2*I*c))*log(1/2*sqrt(3)*a*d*sqrt(1
/(a^2*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) - 3*(sqrt(3)*a*d*sqrt(1/
(a^2*d^2))*e^(2*I*d*x + 2*I*c) + I*e^(2*I*d*x + 2*I*c))*log(-1/2*sqrt(3)*a*d*sqrt(1/(a^2*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) - 13*(3*sqrt(1/3)*a*d*sqrt(1/(a^2*d^2))*e^(2*I*d*x
+ 2*I*c) + I*e^(2*I*d*x + 2*I*c))*log(3/2*sqrt(1/3)*a*d*sqrt(1/(a^2*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) + 13*(3*sqrt(1/3)*a*d*sqrt(1/(a^2*d^2))*e^(2*I*d*x + 2*I*c) - I*e^(2*I*d
*x + 2*I*c))*log(-3/2*sqrt(1/3)*a*d*sqrt(1/(a^2*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) + 26*I*e^(2*I*d*x + 2*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) + 6*I*e^(2*I*d*x + 2*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) - 6*(
(-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(7*I*e^(2*I*d*x + 2*I*c) + I))*e^(-2*I*d*x - 2*I
*c)/(a*d)

Sympy [F]

\[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=- \frac {i \int \frac {\tan ^{\frac {8}{3}}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \]

[In]

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

[Out]

-I*Integral(tan(c + d*x)**(8/3)/(tan(c + d*x) - I), x)/a

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.58 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.72 \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {13 \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right )}{24 \, a d} - \frac {\sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right )}{8 \, a d} - \frac {i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{8 \, a d} - \frac {13 i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{24 \, a d} + \frac {13 i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right )}{12 \, a d} + \frac {i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )}{4 \, a d} - \frac {3 i \, \tan \left (d x + c\right )^{\frac {2}{3}}}{2 \, a d} - \frac {\tan \left (d x + c\right )^{\frac {2}{3}}}{2 \, a d {\left (\tan \left (d x + c\right ) - i\right )}} \]

[In]

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

[Out]

13/24*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) + I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) - I))/(a*d) - 1/8*sqrt
(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) - I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) + I))/(a*d) - 1/8*I*log(tan(d*x
+ c)^(2/3) + I*tan(d*x + c)^(1/3) - 1)/(a*d) - 13/24*I*log(tan(d*x + c)^(2/3) - I*tan(d*x + c)^(1/3) - 1)/(a*d
) + 13/12*I*log(tan(d*x + c)^(1/3) + I)/(a*d) + 1/4*I*log(tan(d*x + c)^(1/3) - I)/(a*d) - 3/2*I*tan(d*x + c)^(
2/3)/(a*d) - 1/2*tan(d*x + c)^(2/3)/(a*d*(tan(d*x + c) - I))

Mupad [B] (verification not implemented)

Time = 7.32 (sec) , antiderivative size = 616, normalized size of antiderivative = 1.93 \[ \int \frac {\tan ^{\frac {8}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx=\ln \left (\left (a^3\,d^3\,312480{}\mathrm {i}-a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}\,307584{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{2/3}+24336\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}+\ln \left (\left (a^3\,d^3\,312480{}\mathrm {i}-a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-\frac {2197{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}\,307584{}\mathrm {i}\right )\,{\left (-\frac {2197{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{2/3}+24336\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,{\left (-\frac {2197{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^{2/3}\,3{}\mathrm {i}}{2\,a\,d}+\frac {\ln \left (\frac {{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (a^3\,d^3\,312480{}\mathrm {i}-a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}\,153792{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{2/3}}{4}+24336\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}}{2}-\frac {\ln \left (\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (a^3\,d^3\,312480{}\mathrm {i}+a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}\,153792{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{2/3}}{4}+24336\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}}{2}+\frac {\ln \left (\frac {{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (a^3\,d^3\,312480{}\mathrm {i}-a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {2197{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}\,153792{}\mathrm {i}\right )\,{\left (-\frac {2197{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{2/3}}{4}+24336\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {2197{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}}{2}-\frac {\ln \left (\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (a^3\,d^3\,312480{}\mathrm {i}+a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {2197{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}\,153792{}\mathrm {i}\right )\,{\left (-\frac {2197{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{2/3}}{4}+24336\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {2197{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}}{2}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^{2/3}\,1{}\mathrm {i}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]

[In]

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

[Out]

log((a^3*d^3*312480i - a^4*d^4*tan(c + d*x)^(1/3)*(-1i/(64*a^3*d^3))^(1/3)*307584i)*(-1i/(64*a^3*d^3))^(2/3) +
24336*a*d*tan(c + d*x)^(1/3))*(-1i/(64*a^3*d^3))^(1/3) + log((a^3*d^3*312480i - a^4*d^4*tan(c + d*x)^(1/3)*(-
2197i/(1728*a^3*d^3))^(1/3)*307584i)*(-2197i/(1728*a^3*d^3))^(2/3) + 24336*a*d*tan(c + d*x)^(1/3))*(-2197i/(17
28*a^3*d^3))^(1/3) - (tan(c + d*x)^(2/3)*3i)/(2*a*d) + (log(((3^(1/2)*1i - 1)^2*(a^3*d^3*312480i - a^4*d^4*tan
(c + d*x)^(1/3)*(3^(1/2)*1i - 1)*(-1i/(64*a^3*d^3))^(1/3)*153792i)*(-1i/(64*a^3*d^3))^(2/3))/4 + 24336*a*d*tan
(c + d*x)^(1/3))*(3^(1/2)*1i - 1)*(-1i/(64*a^3*d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)^2*(a^3*d^3*312480i + a^
4*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)*(-1i/(64*a^3*d^3))^(1/3)*153792i)*(-1i/(64*a^3*d^3))^(2/3))/4 + 2433
6*a*d*tan(c + d*x)^(1/3))*(3^(1/2)*1i + 1)*(-1i/(64*a^3*d^3))^(1/3))/2 + (log(((3^(1/2)*1i - 1)^2*(a^3*d^3*312
480i - a^4*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)*(-2197i/(1728*a^3*d^3))^(1/3)*153792i)*(-2197i/(1728*a^3*d^
3))^(2/3))/4 + 24336*a*d*tan(c + d*x)^(1/3))*(3^(1/2)*1i - 1)*(-2197i/(1728*a^3*d^3))^(1/3))/2 - (log(((3^(1/2
)*1i + 1)^2*(a^3*d^3*312480i + a^4*d^4*tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)*(-2197i/(1728*a^3*d^3))^(1/3)*15379
2i)*(-2197i/(1728*a^3*d^3))^(2/3))/4 + 24336*a*d*tan(c + d*x)^(1/3))*(3^(1/2)*1i + 1)*(-2197i/(1728*a^3*d^3))^
(1/3))/2 - (tan(c + d*x)^(2/3)*1i)/(2*a*d*(tan(c + d*x)*1i + 1))