3.233 \(\int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=343 \[ \frac {7 \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {7 \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{12 a d}-\frac {7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {5 i \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \sqrt {3} a d}+\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{6 a d}+\frac {7 \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}-\frac {7 \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}+\frac {5 i \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{12 a d} \]

[Out]

-7/12*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/a/d-7/12*arctan(3^(1/2)+2*tan(d*x+c)^(1/3))/a/d-7/6*arctan(tan(d*x+c
)^(1/3))/a/d-5/6*I*ln(1+tan(d*x+c)^(2/3))/a/d+5/12*I*ln(1-tan(d*x+c)^(2/3)+tan(d*x+c)^(4/3))/a/d-5/6*I*arctan(
1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))/a/d*3^(1/2)+7/24*ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/a/d*3^(1/
2)-7/24*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))/a/d*3^(1/2)+7/2*tan(d*x+c)^(1/3)/a/d-5/4*I*tan(d*x+c)^
(4/3)/a/d-1/2*tan(d*x+c)^(7/3)/d/(a+I*a*tan(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.46, antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 14, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3550, 3528, 3538, 3476, 329, 209, 634, 618, 204, 628, 203, 275, 292, 31} \[ -\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}+\frac {7 \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {7 \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{12 a d}-\frac {5 i \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \sqrt {3} a d}-\frac {7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{6 a d}+\frac {7 \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}-\frac {7 \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}+\frac {5 i \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{12 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*ArcTan[Sqrt[3] - 2*Tan[c + d*x]^(1/3)])/(12*a*d) - (7*ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(1/3)])/(12*a*d) - ((
(5*I)/2)*ArcTan[(1 - 2*Tan[c + d*x]^(2/3))/Sqrt[3]])/(Sqrt[3]*a*d) - (7*ArcTan[Tan[c + d*x]^(1/3)])/(6*a*d) -
(((5*I)/6)*Log[1 + Tan[c + d*x]^(2/3)])/(a*d) + (7*Log[1 - Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/(
8*Sqrt[3]*a*d) - (7*Log[1 + Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/(8*Sqrt[3]*a*d) + (((5*I)/12)*Lo
g[1 - Tan[c + d*x]^(2/3) + Tan[c + d*x]^(4/3)])/(a*d) + (7*Tan[c + d*x]^(1/3))/(2*a*d) - (((5*I)/4)*Tan[c + d*
x]^(4/3))/(a*d) - Tan[c + d*x]^(7/3)/(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 203

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

Rule 204

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

Rule 209

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*Int[1/(r^2 +
s^2*x^2), x])/(a*n) + 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 275

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 292

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 329

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 618

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 628

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 634

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 3476

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 3528

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 3538

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 3550

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*} \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx &=-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \tan ^{\frac {4}{3}}(c+d x) \left (\frac {7 a}{3}-\frac {10}{3} i a \tan (c+d x)\right ) \, dx}{2 a^2}\\ &=-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \sqrt [3]{\tan (c+d x)} \left (\frac {10 i a}{3}+\frac {7}{3} a \tan (c+d x)\right ) \, dx}{2 a^2}\\ &=\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \frac {-\frac {7 a}{3}+\frac {10}{3} i a \tan (c+d x)}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{2 a^2}\\ &=\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(5 i) \int \sqrt [3]{\tan (c+d x)} \, dx}{3 a}-\frac {7 \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{6 a}\\ &=\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {\sqrt [3]{x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{3 a d}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{x^{2/3} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{6 a d}\\ &=\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {x^3}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{a d}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 a d}\\ &=\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {x}{1+x^3} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{2 a d}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {7 \operatorname {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 )}{6 a d}-\frac {7 \operatorname {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 )}{6 a d}\\ &=-\frac {7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {(5 i) \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {1+x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{24 a d}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{24 a d}+\frac {7 \operatorname {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 {7 \operatorname {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 {7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {5 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {7 \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 {7 \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 {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{12 a d}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{4 a d}+\frac {7 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {7 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}\\ &=\frac {7 \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {7 \tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {5 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {7 \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 {7 \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 i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{12 a d}+\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {(5 i) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac {2}{3}}(c+d x)\right )}{2 a d}\\ &=\frac {7 \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {7 \tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {5 i \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \sqrt {3} a d}-\frac {7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {5 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {7 \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 {7 \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 i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{12 a d}+\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 1.68, size = 181, normalized size = 0.53 \[ -\frac {i \sqrt [3]{\tan (c+d x)} \sec ^2(c+d x) \left (3\ 2^{2/3} \left (1+e^{2 i (c+d x)}\right )^{4/3} \, _2F_1\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )-34 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};-\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (i \sin (2 (c+d x))+\cos (2 (c+d x))+1)+18 i \sin (2 (c+d x))+22 \cos (2 (c+d x))+34\right )}{16 a d (\tan (c+d x)-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/16*I)*Sec[c + d*x]^2*(34 + 22*Cos[2*(c + d*x)] + 3*2^(2/3)*(1 + E^((2*I)*(c + d*x)))^(4/3)*Hypergeometric
2F1[1/3, 1/3, 4/3, (1 - E^((2*I)*(c + d*x)))/2] - 34*Hypergeometric2F1[1/3, 1, 4/3, -((-1 + E^((2*I)*(c + d*x)
))/(1 + E^((2*I)*(c + d*x))))]*(1 + Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) + (18*I)*Sin[2*(c + d*x)])*Tan[c +
d*x]^(1/3))/(a*d*(-I + Tan[c + d*x]))

________________________________________________________________________________________

fricas [B]  time = 0.64, size = 643, normalized size = 1.87 \[ \frac {{\left (3 \, \sqrt {3} {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {1}{a^{2} d^{2}}} + 3 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 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 ) - {\left (3 \, \sqrt {3} {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {1}{a^{2} d^{2}}} - 3 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 3 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 ) - {\left (51 \, \sqrt {\frac {1}{3}} {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {1}{a^{2} d^{2}}} - 17 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 17 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 ) + {\left (51 \, \sqrt {\frac {1}{3}} {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {1}{a^{2} d^{2}}} + 17 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 17 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 ) + {\left (-34 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 34 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (\left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + i\right ) + {\left (-6 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (\left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - i\right ) + 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 {1}{3}} {\left (10 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 17 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}{24 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((3*sqrt(3)*(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt(1/(a^2*d^2)) + 3*I*e^(4*I*d*x + 4*I*
c) + 3*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*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt(1/(a^2*d
^2)) - 3*I*e^(4*I*d*x + 4*I*c) - 3*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) - (51*sqrt(1/3)*(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^
(2*I*d*x + 2*I*c))*sqrt(1/(a^2*d^2)) - 17*I*e^(4*I*d*x + 4*I*c) - 17*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) + (51*sqrt(1/3
)*(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt(1/(a^2*d^2)) + 17*I*e^(4*I*d*x + 4*I*c) + 17*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) + (-34*I*e^(4*I*d*x + 4*I*c) - 34*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^(4*I*d*x + 4*I*c) - 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))^(1/3)*(10*e^(4*I*d*x + 4*I*c) + 17*e^(2*I*d*x + 2*I*c) + 1))/(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*
x + 2*I*c))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (d x + c\right )^{\frac {10}{3}}}{i \, a \tan \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.32, size = 291, normalized size = 0.85 \[ \frac {3 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )}{d a}-\frac {3 i \left (\tan ^{\frac {4}{3}}\left (d x +c \right )\right )}{4 d a}-\frac {17 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{12 d a}+\frac {1}{6 d a \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{8 d a}+\frac {\sqrt {3}\, \arctanh \left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{4 d a}-\frac {\tan ^{\frac {1}{3}}\left (d x +c \right )}{6 d a \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}-\frac {i}{6 d a \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}+\frac {17 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{24 d a}-\frac {17 \sqrt {3}\, \arctanh \left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{12 d a}-\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{4 d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*tan(d*x+c)^(1/3)/d/a-3/4*I/d/a*tan(d*x+c)^(4/3)-17/12*I/d/a*ln(tan(d*x+c)^(1/3)+I)+1/6/d/a/(tan(d*x+c)^(1/3)
+I)+1/8*I/d/a*ln(I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)+1/4/d/a*3^(1/2)*arctanh(1/3*(I+2*tan(d*x+c)^(1/3))*3^(
1/2))-1/6/d/a/(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)*tan(d*x+c)^(1/3)-1/6*I/d/a/(-I*tan(d*x+c)^(1/3)+tan(d*x
+c)^(2/3)-1)+17/24*I/d/a*ln(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)-17/12/d/a*3^(1/2)*arctanh(1/3*(-I+2*tan(d*
x+c)^(1/3))*3^(1/2))-1/4*I/d/a*ln(tan(d*x+c)^(1/3)-I)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 5.72, size = 655, normalized size = 1.91 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((a^3*d^3*703584i - 414720*a^5*d^5*tan(c + d*x)^(1/3)*(1i/(64*a^3*d^3))^(2/3))*(1i/(64*a^3*d^3))^(1/3) + a^
2*d^2*tan(c + d*x)^(1/3)*182376i)*(1i/(64*a^3*d^3))^(1/3) + log((a^3*d^3*703584i - 414720*a^5*d^5*tan(c + d*x)
^(1/3)*(4913i/(1728*a^3*d^3))^(2/3))*(4913i/(1728*a^3*d^3))^(1/3) + a^2*d^2*tan(c + d*x)^(1/3)*182376i)*(4913i
/(1728*a^3*d^3))^(1/3) + (3*tan(c + d*x)^(1/3))/(a*d) - (tan(c + d*x)^(4/3)*3i)/(4*a*d) + (log(((3^(1/2)*1i -
1)*(a^3*d^3*703584i - 103680*a^5*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)^2*(1i/(64*a^3*d^3))^(2/3))*(1i/(64*a^
3*d^3))^(1/3))/2 + a^2*d^2*tan(c + d*x)^(1/3)*182376i)*(3^(1/2)*1i - 1)*(1i/(64*a^3*d^3))^(1/3))/2 - (log(((3^
(1/2)*1i + 1)*(a^3*d^3*703584i - 103680*a^5*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1i + 1)^2*(1i/(64*a^3*d^3))^(2/3))
*(1i/(64*a^3*d^3))^(1/3))/2 - a^2*d^2*tan(c + d*x)^(1/3)*182376i)*(3^(1/2)*1i + 1)*(1i/(64*a^3*d^3))^(1/3))/2
+ (log(((3^(1/2)*1i - 1)*(a^3*d^3*703584i - 103680*a^5*d^5*tan(c + d*x)^(1/3)*(3^(1/2)*1i - 1)^2*(4913i/(1728*
a^3*d^3))^(2/3))*(4913i/(1728*a^3*d^3))^(1/3))/2 + a^2*d^2*tan(c + d*x)^(1/3)*182376i)*(3^(1/2)*1i - 1)*(4913i
/(1728*a^3*d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)*(a^3*d^3*703584i - 103680*a^5*d^5*tan(c + d*x)^(1/3)*(3^(1/
2)*1i + 1)^2*(4913i/(1728*a^3*d^3))^(2/3))*(4913i/(1728*a^3*d^3))^(1/3))/2 - a^2*d^2*tan(c + d*x)^(1/3)*182376
i)*(3^(1/2)*1i + 1)*(4913i/(1728*a^3*d^3))^(1/3))/2 + tan(c + d*x)^(1/3)/(2*a*d*(tan(c + d*x)*1i + 1))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________