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

Optimal. Leaf size=357 \[ \frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{49 \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{49 \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt{3}\right )}{72 a^2 d}+\frac{5 i \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{3 \sqrt{3} a^2 d}+\frac{49 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \log \left (\tan ^{\frac{2}{3}}(c+d x)+1\right )}{9 a^2 d}-\frac{49 \log \left (\tan ^{\frac{2}{3}}(c+d x)-\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}+\frac{49 \log \left (\tan ^{\frac{2}{3}}(c+d x)+\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}-\frac{5 i \log \left (\tan ^{\frac{4}{3}}(c+d x)-\tan ^{\frac{2}{3}}(c+d x)+1\right )}{18 a^2 d}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.56227, antiderivative size = 357, normalized size of antiderivative = 1., number of steps used = 25, number of rules used = 15, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.577, Rules used = {3558, 3595, 3528, 3538, 3476, 329, 209, 634, 618, 204, 628, 203, 275, 292, 31} \[ \frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{49 \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{49 \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt{3}\right )}{72 a^2 d}+\frac{5 i \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{3 \sqrt{3} a^2 d}+\frac{49 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \log \left (\tan ^{\frac{2}{3}}(c+d x)+1\right )}{9 a^2 d}-\frac{49 \log \left (\tan ^{\frac{2}{3}}(c+d x)-\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}+\frac{49 \log \left (\tan ^{\frac{2}{3}}(c+d x)+\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}-\frac{5 i \log \left (\tan ^{\frac{4}{3}}(c+d x)-\tan ^{\frac{2}{3}}(c+d x)+1\right )}{18 a^2 d}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((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 3595

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]

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 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 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 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 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 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 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 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 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 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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{\tan ^{\frac{10}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{\int \frac{\tan ^{\frac{4}{3}}(c+d x) \left (-\frac{7 a}{3}+\frac{13}{3} i a \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \sqrt [3]{\tan (c+d x)} \left (-\frac{80 i a^2}{9}-\frac{98}{9} a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\frac{98 a^2}{9}-\frac{80}{9} i a^2 \tan (c+d x)}{\tan ^{\frac{2}{3}}(c+d x)} \, dx}{8 a^4}\\ &=-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(10 i) \int \sqrt [3]{\tan (c+d x)} \, dx}{9 a^2}+\frac{49 \int \frac{1}{\tan ^{\frac{2}{3}}(c+d x)} \, dx}{36 a^2}\\ &=-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(10 i) \operatorname{Subst}\left (\int \frac{\sqrt [3]{x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{9 a^2 d}+\frac{49 \operatorname{Subst}\left (\int \frac{1}{x^{2/3} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{36 a^2 d}\\ &=-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(10 i) \operatorname{Subst}\left (\int \frac{x^3}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{3 a^2 d}+\frac{49 \operatorname{Subst}\left (\int \frac{1}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{12 a^2 d}\\ &=-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(5 i) \operatorname{Subst}\left (\int \frac{x}{1+x^3} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{3 a^2 d}+\frac{49 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac{49 \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 )}{36 a^2 d}+\frac{49 \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 )}{36 a^2 d}\\ &=\frac{49 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{(5 i) \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 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 )}{9 a^2 d}+\frac{49 \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{144 a^2 d}+\frac{49 \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{144 a^2 d}-\frac{49 \operatorname{Subst}\left (\int \frac{-\sqrt{3}+2 x}{1-\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{48 \sqrt{3} a^2 d}+\frac{49 \operatorname{Subst}\left (\int \frac{\sqrt{3}+2 x}{1+\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{48 \sqrt{3} a^2 d}\\ &=\frac{49 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac{5 i \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}-\frac{49 \log \left (1-\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}+\frac{49 \log \left (1+\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(5 i) \operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{18 a^2 d}-\frac{(5 i) \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{6 a^2 d}-\frac{49 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,-\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac{49 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}\\ &=-\frac{49 \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{49 \tan ^{-1}\left (\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{49 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac{5 i \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}-\frac{49 \log \left (1-\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}+\frac{49 \log \left (1+\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}-\frac{5 i \log \left (1-\tan ^{\frac{2}{3}}(c+d x)+\tan ^{\frac{4}{3}}(c+d x)\right )}{18 a^2 d}-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{(5 i) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac{2}{3}}(c+d x)\right )}{3 a^2 d}\\ &=-\frac{49 \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{49 \tan ^{-1}\left (\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{5 i \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{3 \sqrt{3} a^2 d}+\frac{49 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac{5 i \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}-\frac{49 \log \left (1-\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}+\frac{49 \log \left (1+\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}-\frac{5 i \log \left (1-\tan ^{\frac{2}{3}}(c+d x)+\tan ^{\frac{4}{3}}(c+d x)\right )}{18 a^2 d}-\frac{49 \sqrt [3]{\tan (c+d x)}}{12 a^2 d}+\frac{5 i \tan ^{\frac{4}{3}}(c+d x)}{6 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^{\frac{7}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 1.62348, size = 192, normalized size = 0.54 \[ -\frac{\sqrt [3]{\tan (c+d x)} \sec ^2(c+d x) \left (9\ 2^{2/3} e^{2 i (c+d x)} \sqrt [3]{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac{1}{3},\frac{1}{3};\frac{4}{3};\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+2 \left (89 \, _2F_1\left (\frac{1}{3},1;\frac{4}{3};-\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-88 i \sin (2 (c+d x))-85 \cos (2 (c+d x))-13\right )\right )}{48 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c + d*x]^2*(9*2^(2/3)*E^((2*I)*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(1/3)*Hypergeometric2F1[1/3, 1/3, 4/
3, (1 - E^((2*I)*(c + d*x)))/2] + 2*(-13 - 85*Cos[2*(c + d*x)] + 89*Hypergeometric2F1[1/3, 1, 4/3, -((-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)]) - (88*I)*Sin[2*(c + d*x)
]))*Tan[c + d*x]^(1/3))/(48*a^2*d*(-I + Tan[c + d*x])^2)

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Maple [A]  time = 0.048, size = 373, normalized size = 1. \begin{align*} -3\,{\frac{\sqrt [3]{\tan \left ( dx+c \right ) }}{{a}^{2}d}}+{\frac{{\frac{i}{18}}}{{a}^{2}d} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}} \left ( -i\sqrt [3]{\tan \left ( dx+c \right ) }+ \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-1 \right ) ^{-2}}+{\frac{1}{18\,{a}^{2}d}\sqrt [3]{\tan \left ( dx+c \right ) } \left ( -i\sqrt [3]{\tan \left ( dx+c \right ) }+ \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-1 \right ) ^{-2}}+{\frac{5\,\tan \left ( dx+c \right ) }{12\,{a}^{2}d} \left ( -i\sqrt [3]{\tan \left ( dx+c \right ) }+ \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-1 \right ) ^{-2}}-{\frac{{\frac{7\,i}{18}}}{{a}^{2}d} \left ( -i\sqrt [3]{\tan \left ( dx+c \right ) }+ \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-1 \right ) ^{-2}}-{\frac{{\frac{89\,i}{144}}}{{a}^{2}d}\ln \left ( -i\sqrt [3]{\tan \left ( dx+c \right ) }+ \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-1 \right ) }+{\frac{89\,\sqrt{3}}{72\,{a}^{2}d}{\it Artanh} \left ({\frac{\sqrt{3}}{3} \left ( -i+2\,\sqrt [3]{\tan \left ( dx+c \right ) } \right ) } \right ) }+{\frac{{\frac{i}{16}}}{{a}^{2}d}\ln \left ( i\sqrt [3]{\tan \left ( dx+c \right ) }+ \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-1 \right ) }+{\frac{\sqrt{3}}{8\,{a}^{2}d}{\it Artanh} \left ({\frac{\sqrt{3}}{3} \left ( i+2\,\sqrt [3]{\tan \left ( dx+c \right ) } \right ) } \right ) }-{\frac{{\frac{i}{8}}}{{a}^{2}d}\ln \left ( \sqrt [3]{\tan \left ( dx+c \right ) }-i \right ) }+{\frac{{\frac{i}{36}}}{{a}^{2}d} \left ( \sqrt [3]{\tan \left ( dx+c \right ) }+i \right ) ^{-2}}+{\frac{{\frac{89\,i}{72}}}{{a}^{2}d}\ln \left ( \sqrt [3]{\tan \left ( dx+c \right ) }+i \right ) }-{\frac{5}{12\,{a}^{2}d} \left ( \sqrt [3]{\tan \left ( dx+c \right ) }+i \right ) ^{-1}} \end{align*}

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))^2,x)

[Out]

-3*tan(d*x+c)^(1/3)/a^2/d+1/18*I/d/a^2/(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)^2*tan(d*x+c)^(2/3)+1/18/d/a^2/
(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)^2*tan(d*x+c)^(1/3)+5/12/d/a^2/(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1
)^2*tan(d*x+c)-7/18*I/d/a^2/(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)^2-89/144*I/d/a^2*ln(-I*tan(d*x+c)^(1/3)+t
an(d*x+c)^(2/3)-1)+89/72/d/a^2*3^(1/2)*arctanh(1/3*(-I+2*tan(d*x+c)^(1/3))*3^(1/2))+1/16*I/d/a^2*ln(I*tan(d*x+
c)^(1/3)+tan(d*x+c)^(2/3)-1)+1/8/d/a^2*3^(1/2)*arctanh(1/3*(I+2*tan(d*x+c)^(1/3))*3^(1/2))-1/8*I/d/a^2*ln(tan(
d*x+c)^(1/3)-I)+1/36*I/d/a^2/(tan(d*x+c)^(1/3)+I)^2+89/72*I/d/a^2*ln(tan(d*x+c)^(1/3)+I)-5/12/d/a^2/(tan(d*x+c
)^(1/3)+I)

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

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))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.7649, size = 1573, normalized size = 4.41 \begin{align*} \text{result too large to display} \end{align*}

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))^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) + 9*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) - (9*sqrt(3)*a^2*
d*sqrt(1/(a^4*d^2))*e^(4*I*d*x + 4*I*c) - 9*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) + (267*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d^2
))*e^(4*I*d*x + 4*I*c) - 89*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) - (267*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d^2))*e^(4*I*d*x +
4*I*c) + 89*I*e^(4*I*d*x + 4*I*c))*log(-3/2*sqrt(1/3)*a^2*d*sqrt(1/(a^4*d^2)) + ((-I*e^(2*I*d*x + 2*I*c) + I)
/(e^(2*I*d*x + 2*I*c) + 1))^(1/3) - 1/2*I) + 178*I*e^(4*I*d*x + 4*I*c)*log(((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2
*I*d*x + 2*I*c) + 1))^(1/3) + I) - 18*I*e^(4*I*d*x + 4*I*c)*log(((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*
I*c) + 1))^(1/3) - I) - 3*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*(173*e^(4*I*d*x + 4*I
*c) + 26*e^(2*I*d*x + 2*I*c) - 3))*e^(-4*I*d*x - 4*I*c)/(a^2*d)

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

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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.18894, size = 329, normalized size = 0.92 \begin{align*} -\frac{89 \, \sqrt{3} \log \left (-\frac{\sqrt{3} - 2 \, \tan \left (d x + c\right )^{\frac{1}{3}} + i}{\sqrt{3} + 2 \, \tan \left (d x + c\right )^{\frac{1}{3}} - i}\right )}{144 \, a^{2} d} - \frac{\sqrt{3} \log \left (-\frac{\sqrt{3} - 2 \, \tan \left (d x + c\right )^{\frac{1}{3}} - i}{\sqrt{3} + 2 \, \tan \left (d x + c\right )^{\frac{1}{3}} + i}\right )}{16 \, a^{2} d} + \frac{i \, \log \left (\tan \left (d x + c\right )^{\frac{2}{3}} + i \, \tan \left (d x + c\right )^{\frac{1}{3}} - 1\right )}{16 \, a^{2} d} - \frac{89 i \, \log \left (\tan \left (d x + c\right )^{\frac{2}{3}} - i \, \tan \left (d x + c\right )^{\frac{1}{3}} - 1\right )}{144 \, a^{2} d} + \frac{89 i \, \log \left (\tan \left (d x + c\right )^{\frac{1}{3}} + i\right )}{72 \, a^{2} d} - \frac{i \, \log \left (\tan \left (d x + c\right )^{\frac{1}{3}} - i\right )}{8 \, a^{2} d} - \frac{3 \, \tan \left (d x + c\right )^{\frac{1}{3}}}{a^{2} d} - \frac{-16 i \, \tan \left (d x + c\right )^{\frac{4}{3}} - 13 \, \tan \left (d x + c\right )^{\frac{1}{3}}}{12 \, a^{2} d{\left (\tan \left (d x + c\right ) - i\right )}^{2}} \end{align*}

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))^2,x, algorithm="giac")

[Out]

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