3.246 \(\int \frac{1}{\tan ^{\frac{5}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=359 \[ \frac{55 i \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac{55 i \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt{3}\right )}{72 a^2 d}+\frac{8 \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{3 \sqrt{3} a^2 d}-\frac{55 i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac{8}{3 a^2 d \tan ^{\frac{2}{3}}(c+d x)}+\frac{11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac{2}{3}}(c+d x)}+\frac{8 \log \left (\tan ^{\frac{2}{3}}(c+d x)+1\right )}{9 a^2 d}+\frac{55 i \log \left (\tan ^{\frac{2}{3}}(c+d x)-\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}-\frac{55 i \log \left (\tan ^{\frac{2}{3}}(c+d x)+\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}-\frac{4 \log \left (\tan ^{\frac{4}{3}}(c+d x)-\tan ^{\frac{2}{3}}(c+d x)+1\right )}{9 a^2 d}+\frac{1}{4 d \tan ^{\frac{2}{3}}(c+d x) (a+i a \tan (c+d x))^2} \]

[Out]

(((55*I)/72)*ArcTan[Sqrt[3] - 2*Tan[c + d*x]^(1/3)])/(a^2*d) - (((55*I)/72)*ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(1
/3)])/(a^2*d) + (8*ArcTan[(1 - 2*Tan[c + d*x]^(2/3))/Sqrt[3]])/(3*Sqrt[3]*a^2*d) - (((55*I)/36)*ArcTan[Tan[c +
d*x]^(1/3)])/(a^2*d) + (8*Log[1 + Tan[c + d*x]^(2/3)])/(9*a^2*d) + (((55*I)/48)*Log[1 - Sqrt[3]*Tan[c + d*x]^
(1/3) + Tan[c + d*x]^(2/3)])/(Sqrt[3]*a^2*d) - (((55*I)/48)*Log[1 + Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^
(2/3)])/(Sqrt[3]*a^2*d) - (4*Log[1 - Tan[c + d*x]^(2/3) + Tan[c + d*x]^(4/3)])/(9*a^2*d) - 8/(3*a^2*d*Tan[c +
d*x]^(2/3)) + 11/(12*a^2*d*(1 + I*Tan[c + d*x])*Tan[c + d*x]^(2/3)) + 1/(4*d*Tan[c + d*x]^(2/3)*(a + I*a*Tan[c
+ d*x])^2)

________________________________________________________________________________________

Rubi [A]  time = 0.564206, antiderivative size = 359, 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 = {3559, 3596, 3529, 3538, 3476, 329, 209, 634, 618, 204, 628, 203, 275, 292, 31} \[ \frac{55 i \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac{55 i \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt{3}\right )}{72 a^2 d}+\frac{8 \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{3 \sqrt{3} a^2 d}-\frac{55 i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac{8}{3 a^2 d \tan ^{\frac{2}{3}}(c+d x)}+\frac{11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac{2}{3}}(c+d x)}+\frac{8 \log \left (\tan ^{\frac{2}{3}}(c+d x)+1\right )}{9 a^2 d}+\frac{55 i \log \left (\tan ^{\frac{2}{3}}(c+d x)-\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}-\frac{55 i \log \left (\tan ^{\frac{2}{3}}(c+d x)+\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt{3} a^2 d}-\frac{4 \log \left (\tan ^{\frac{4}{3}}(c+d x)-\tan ^{\frac{2}{3}}(c+d x)+1\right )}{9 a^2 d}+\frac{1}{4 d \tan ^{\frac{2}{3}}(c+d x) (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(Tan[c + d*x]^(5/3)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(((55*I)/72)*ArcTan[Sqrt[3] - 2*Tan[c + d*x]^(1/3)])/(a^2*d) - (((55*I)/72)*ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(1
/3)])/(a^2*d) + (8*ArcTan[(1 - 2*Tan[c + d*x]^(2/3))/Sqrt[3]])/(3*Sqrt[3]*a^2*d) - (((55*I)/36)*ArcTan[Tan[c +
d*x]^(1/3)])/(a^2*d) + (8*Log[1 + Tan[c + d*x]^(2/3)])/(9*a^2*d) + (((55*I)/48)*Log[1 - Sqrt[3]*Tan[c + d*x]^
(1/3) + Tan[c + d*x]^(2/3)])/(Sqrt[3]*a^2*d) - (((55*I)/48)*Log[1 + Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^
(2/3)])/(Sqrt[3]*a^2*d) - (4*Log[1 - Tan[c + d*x]^(2/3) + Tan[c + d*x]^(4/3)])/(9*a^2*d) - 8/(3*a^2*d*Tan[c +
d*x]^(2/3)) + 11/(12*a^2*d*(1 + I*Tan[c + d*x])*Tan[c + d*x]^(2/3)) + 1/(4*d*Tan[c + d*x]^(2/3)*(a + I*a*Tan[c
+ d*x])^2)

Rule 3559

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d))
, Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan
[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2
+ d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3596

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
0]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

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{1}{\tan ^{\frac{5}{3}}(c+d x) (a+i a \tan (c+d x))^2} \, dx &=\frac{1}{4 d \tan ^{\frac{2}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac{\int \frac{\frac{14 a}{3}-\frac{8}{3} i a \tan (c+d x)}{\tan ^{\frac{5}{3}}(c+d x) (a+i a \tan (c+d x))} \, dx}{4 a^2}\\ &=\frac{11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac{2}{3}}(c+d x)}+\frac{1}{4 d \tan ^{\frac{2}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac{\int \frac{\frac{128 a^2}{9}-\frac{110}{9} i a^2 \tan (c+d x)}{\tan ^{\frac{5}{3}}(c+d x)} \, dx}{8 a^4}\\ &=-\frac{8}{3 a^2 d \tan ^{\frac{2}{3}}(c+d x)}+\frac{11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac{2}{3}}(c+d x)}+\frac{1}{4 d \tan ^{\frac{2}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac{\int \frac{-\frac{110 i a^2}{9}-\frac{128}{9} a^2 \tan (c+d x)}{\tan ^{\frac{2}{3}}(c+d x)} \, dx}{8 a^4}\\ &=-\frac{8}{3 a^2 d \tan ^{\frac{2}{3}}(c+d x)}+\frac{11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac{2}{3}}(c+d x)}+\frac{1}{4 d \tan ^{\frac{2}{3}}(c+d x) (a+i a \tan (c+d x))^2}-\frac{(55 i) \int \frac{1}{\tan ^{\frac{2}{3}}(c+d x)} \, dx}{36 a^2}-\frac{16 \int \sqrt [3]{\tan (c+d x)} \, dx}{9 a^2}\\ &=-\frac{8}{3 a^2 d \tan ^{\frac{2}{3}}(c+d x)}+\frac{11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac{2}{3}}(c+d x)}+\frac{1}{4 d \tan ^{\frac{2}{3}}(c+d x) (a+i a \tan (c+d x))^2}-\frac{(55 i) \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{16 \operatorname{Subst}\left (\int \frac{\sqrt [3]{x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{9 a^2 d}\\ &=-\frac{8}{3 a^2 d \tan ^{\frac{2}{3}}(c+d x)}+\frac{11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac{2}{3}}(c+d x)}+\frac{1}{4 d \tan ^{\frac{2}{3}}(c+d x) (a+i a \tan (c+d x))^2}-\frac{(55 i) \operatorname{Subst}\left (\int \frac{1}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{12 a^2 d}-\frac{16 \operatorname{Subst}\left (\int \frac{x^3}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{3 a^2 d}\\ &=-\frac{8}{3 a^2 d \tan ^{\frac{2}{3}}(c+d x)}+\frac{11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac{2}{3}}(c+d x)}+\frac{1}{4 d \tan ^{\frac{2}{3}}(c+d x) (a+i a \tan (c+d x))^2}-\frac{(55 i) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac{(55 i) \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{(55 i) \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{8 \operatorname{Subst}\left (\int \frac{x}{1+x^3} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{3 a^2 d}\\ &=-\frac{55 i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac{8}{3 a^2 d \tan ^{\frac{2}{3}}(c+d x)}+\frac{11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac{2}{3}}(c+d x)}+\frac{1}{4 d \tan ^{\frac{2}{3}}(c+d x) (a+i a \tan (c+d x))^2}-\frac{(55 i) \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{(55 i) \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{8 \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}-\frac{8 \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{(55 i) \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{(55 i) \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{55 i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac{8 \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}+\frac{55 i \log \left (1-\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}-\frac{55 i \log \left (1+\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}-\frac{8}{3 a^2 d \tan ^{\frac{2}{3}}(c+d x)}+\frac{11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac{2}{3}}(c+d x)}+\frac{1}{4 d \tan ^{\frac{2}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac{(55 i) \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{(55 i) \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{4 \operatorname{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{4 \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{3 a^2 d}\\ &=\frac{55 i \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac{55 i \tan ^{-1}\left (\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac{55 i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac{8 \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}+\frac{55 i \log \left (1-\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}-\frac{55 i \log \left (1+\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}-\frac{4 \log \left (1-\tan ^{\frac{2}{3}}(c+d x)+\tan ^{\frac{4}{3}}(c+d x)\right )}{9 a^2 d}-\frac{8}{3 a^2 d \tan ^{\frac{2}{3}}(c+d x)}+\frac{11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac{2}{3}}(c+d x)}+\frac{1}{4 d \tan ^{\frac{2}{3}}(c+d x) (a+i a \tan (c+d x))^2}+\frac{8 \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{55 i \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac{55 i \tan ^{-1}\left (\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac{8 \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{3 \sqrt{3} a^2 d}-\frac{55 i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac{8 \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{9 a^2 d}+\frac{55 i \log \left (1-\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}-\frac{55 i \log \left (1+\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{48 \sqrt{3} a^2 d}-\frac{4 \log \left (1-\tan ^{\frac{2}{3}}(c+d x)+\tan ^{\frac{4}{3}}(c+d x)\right )}{9 a^2 d}-\frac{8}{3 a^2 d \tan ^{\frac{2}{3}}(c+d x)}+\frac{11}{12 a^2 d (1+i \tan (c+d x)) \tan ^{\frac{2}{3}}(c+d x)}+\frac{1}{4 d \tan ^{\frac{2}{3}}(c+d x) (a+i a \tan (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 3.65144, size = 205, normalized size = 0.57 \[ \frac{\sqrt [3]{\tan (c+d x)} \sec (c+d x) \left (-\frac{36 i 2^{2/3} e^{3 i (c+d x)} \, _2F_1\left (\frac{1}{3},\frac{1}{3};\frac{4}{3};\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )}{\left (1+e^{2 i (c+d x)}\right )^{2/3}}+476 i \, _2F_1\left (\frac{1}{3},1;\frac{4}{3};-\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (\cos (2 (c+d x)) \sec (c+d x)+2 i \sin (c+d x))+4 \csc (c+d x) (53 i \sin (2 (c+d x))+50 \cos (2 (c+d x))-14)\right )}{96 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Tan[c + d*x]^(5/3)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(Sec[c + d*x]*(((-36*I)*2^(2/3)*E^((3*I)*(c + d*x))*Hypergeometric2F1[1/3, 1/3, 4/3, (1 - E^((2*I)*(c + d*x)))
/2])/(1 + E^((2*I)*(c + d*x)))^(2/3) + (476*I)*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)]*Sec[c + d*x] + (2*I)*Sin[c + d*x]) + 4*Csc[c + d*x]*(-14 + 50*Cos[2
*(c + d*x)] + (53*I)*Sin[2*(c + d*x)]))*Tan[c + d*x]^(1/3))/(96*a^2*d*(-I + Tan[c + d*x])^2)

________________________________________________________________________________________

Maple [A]  time = 0.05, size = 371, normalized size = 1. \begin{align*}{\frac{1}{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{{\frac{i}{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{{\frac{5\,i}{12}}\tan \left ( dx+c \right ) }{{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{4}{9\,{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{119}{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{{\frac{119\,i}{72}}\sqrt{3}}{{a}^{2}d}{\it Artanh} \left ({\frac{\sqrt{3}}{3} \left ( -i+2\,\sqrt [3]{\tan \left ( dx+c \right ) } \right ) } \right ) }-{\frac{1}{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{{\frac{i}{8}}\sqrt{3}}{{a}^{2}d}{\it Artanh} \left ({\frac{\sqrt{3}}{3} \left ( i+2\,\sqrt [3]{\tan \left ( dx+c \right ) } \right ) } \right ) }+{\frac{1}{8\,{a}^{2}d}\ln \left ( \sqrt [3]{\tan \left ( dx+c \right ) }-i \right ) }-{\frac{{\frac{5\,i}{12}}}{{a}^{2}d} \left ( \sqrt [3]{\tan \left ( dx+c \right ) }+i \right ) ^{-1}}+{\frac{1}{36\,{a}^{2}d} \left ( \sqrt [3]{\tan \left ( dx+c \right ) }+i \right ) ^{-2}}+{\frac{119}{72\,{a}^{2}d}\ln \left ( \sqrt [3]{\tan \left ( dx+c \right ) }+i \right ) }-{\frac{3}{2\,{a}^{2}d} \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/tan(d*x+c)^(5/3)/(a+I*a*tan(d*x+c))^2,x)

[Out]

1/18/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*I/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*I/d/a^2/(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)^2*tan(d*x+c)+4/9/d/a^2
/(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)^2-119/144/d/a^2*ln(-I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1)-119/72*I/
d/a^2*3^(1/2)*arctanh(1/3*(-I+2*tan(d*x+c)^(1/3))*3^(1/2))-1/16/d/a^2*ln(I*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3)-1
)+1/8*I/d/a^2*3^(1/2)*arctanh(1/3*(I+2*tan(d*x+c)^(1/3))*3^(1/2))+1/8/d/a^2*ln(tan(d*x+c)^(1/3)-I)-5/12*I/d/a^
2/(tan(d*x+c)^(1/3)+I)+1/36/d/a^2/(tan(d*x+c)^(1/3)+I)^2+119/72/d/a^2*ln(tan(d*x+c)^(1/3)+I)-3/2/a^2/d/tan(d*x
+c)^(2/3)

________________________________________________________________________________________

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(1/tan(d*x+c)^(5/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [B]  time = 2.85816, size = 2078, normalized size = 5.79 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(5/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3456*(3*(sqrt(3)*(72*I*a^2*d*e^(6*I*d*x + 6*I*c) - 72*I*a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a^4*d^2)) - 72*e^
(6*I*d*x + 6*I*c) + 72*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) + 3*(sqrt(3)*(-72*I*a^2*d*e^(6*I*d*x + 6*I*c) + 72*I*a^2*d*e^(
4*I*d*x + 4*I*c))*sqrt(1/(a^4*d^2)) - 72*e^(6*I*d*x + 6*I*c) + 72*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) + (sqrt(1/3)*(-856
8*I*a^2*d*e^(6*I*d*x + 6*I*c) + 8568*I*a^2*d*e^(4*I*d*x + 4*I*c))*sqrt(1/(a^4*d^2)) - 2856*e^(6*I*d*x + 6*I*c)
+ 2856*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) + (sqrt(1/3)*(8568*I*a^2*d*e^(6*I*d*x + 6*I*c) - 8568*I*a^2*d*e^(4*I*d*x +
4*I*c))*sqrt(1/(a^4*d^2)) - 2856*e^(6*I*d*x + 6*I*c) + 2856*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) + 5712*(e^(6*I*d*x + 6
*I*c) - 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) + 432*(e^
(6*I*d*x + 6*I*c) - 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) + 24*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*(-309*I*e^(6*I*d*x + 6*I*c) - 225*I*e^(
4*I*d*x + 4*I*c) + 93*I*e^(2*I*d*x + 2*I*c) + 9*I))/(a^2*d*e^(6*I*d*x + 6*I*c) - a^2*d*e^(4*I*d*x + 4*I*c))

________________________________________________________________________________________

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(1/tan(d*x+c)**(5/3)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.21927, size = 321, normalized size = 0.89 \begin{align*} \frac{119 i \, \sqrt{3} \log \left (-\frac{\sqrt{3} - 2 \, \tan \left (d x + c\right )^{\frac{1}{3}} + i}{\sqrt{3} + 2 \, \tan \left (d x + c\right )^{\frac{1}{3}} - i}\right )}{144 \, a^{2} d} - \frac{i \, \sqrt{3} \log \left (-\frac{\sqrt{3} - 2 \, \tan \left (d x + c\right )^{\frac{1}{3}} - i}{\sqrt{3} + 2 \, \tan \left (d x + c\right )^{\frac{1}{3}} + i}\right )}{16 \, a^{2} d} - \frac{\log \left (\tan \left (d x + c\right )^{\frac{2}{3}} + i \, \tan \left (d x + c\right )^{\frac{1}{3}} - 1\right )}{16 \, a^{2} d} - \frac{119 \, \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{119 \, \log \left (\tan \left (d x + c\right )^{\frac{1}{3}} + i\right )}{72 \, a^{2} d} + \frac{\log \left (\tan \left (d x + c\right )^{\frac{1}{3}} - i\right )}{8 \, a^{2} d} - \frac{32 \, \tan \left (d x + c\right )^{2} - 53 i \, \tan \left (d x + c\right ) - 18}{12 \,{\left (\tan \left (d x + c\right )^{\frac{4}{3}} - i \, \tan \left (d x + c\right )^{\frac{1}{3}}\right )}^{2} a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(5/3)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

119/144*I*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) + I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) - I))/(a^2*d) - 1/
16*I*sqrt(3)*log(-(sqrt(3) - 2*tan(d*x + c)^(1/3) - I)/(sqrt(3) + 2*tan(d*x + c)^(1/3) + I))/(a^2*d) - 1/16*lo
g(tan(d*x + c)^(2/3) + I*tan(d*x + c)^(1/3) - 1)/(a^2*d) - 119/144*log(tan(d*x + c)^(2/3) - I*tan(d*x + c)^(1/
3) - 1)/(a^2*d) + 119/72*log(tan(d*x + c)^(1/3) + I)/(a^2*d) + 1/8*log(tan(d*x + c)^(1/3) - I)/(a^2*d) - 1/12*
(32*tan(d*x + c)^2 - 53*I*tan(d*x + c) - 18)/((tan(d*x + c)^(4/3) - I*tan(d*x + c)^(1/3))^2*a^2*d)