3.674 \(\int \frac{1}{\tan ^{\frac{5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx\)

Optimal. Leaf size=525 \[ \frac{\sqrt{3} b^{8/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{a^{5/3} d \left (a^2+b^2\right )}-\frac{3 b^2}{2 a d \left (a^2+b^2\right ) \tan ^{\frac{2}{3}}(c+d x)}+\frac{b \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{2 d \left (a^2+b^2\right )}-\frac{b \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt{3}\right )}{2 d \left (a^2+b^2\right )}-\frac{b \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}+\frac{\sqrt{3} a \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{2 d \left (a^2+b^2\right )}-\frac{3 a}{2 d \left (a^2+b^2\right ) \tan ^{\frac{2}{3}}(c+d x)}+\frac{\sqrt{3} b \log \left (\tan ^{\frac{2}{3}}(c+d x)-\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{4 d \left (a^2+b^2\right )}-\frac{\sqrt{3} b \log \left (\tan ^{\frac{2}{3}}(c+d x)+\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{4 d \left (a^2+b^2\right )}+\frac{a \log \left (\tan ^{\frac{2}{3}}(c+d x)+1\right )}{2 d \left (a^2+b^2\right )}-\frac{a \log \left (\tan ^{\frac{4}{3}}(c+d x)-\tan ^{\frac{2}{3}}(c+d x)+1\right )}{4 d \left (a^2+b^2\right )}-\frac{3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} d \left (a^2+b^2\right )}+\frac{b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} d \left (a^2+b^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.558197, antiderivative size = 525, normalized size of antiderivative = 1., number of steps used = 30, number of rules used = 18, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.783, Rules used = {3574, 3529, 3538, 3476, 329, 209, 634, 618, 204, 628, 203, 275, 292, 31, 3634, 51, 58, 617} \[ \frac{\sqrt{3} b^{8/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{a^{5/3} d \left (a^2+b^2\right )}-\frac{3 b^2}{2 a d \left (a^2+b^2\right ) \tan ^{\frac{2}{3}}(c+d x)}+\frac{b \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{2 d \left (a^2+b^2\right )}-\frac{b \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt{3}\right )}{2 d \left (a^2+b^2\right )}-\frac{b \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}+\frac{\sqrt{3} a \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{2 d \left (a^2+b^2\right )}-\frac{3 a}{2 d \left (a^2+b^2\right ) \tan ^{\frac{2}{3}}(c+d x)}+\frac{\sqrt{3} b \log \left (\tan ^{\frac{2}{3}}(c+d x)-\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{4 d \left (a^2+b^2\right )}-\frac{\sqrt{3} b \log \left (\tan ^{\frac{2}{3}}(c+d x)+\sqrt{3} \sqrt [3]{\tan (c+d x)}+1\right )}{4 d \left (a^2+b^2\right )}+\frac{a \log \left (\tan ^{\frac{2}{3}}(c+d x)+1\right )}{2 d \left (a^2+b^2\right )}-\frac{a \log \left (\tan ^{\frac{4}{3}}(c+d x)-\tan ^{\frac{2}{3}}(c+d x)+1\right )}{4 d \left (a^2+b^2\right )}-\frac{3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} d \left (a^2+b^2\right )}+\frac{b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3574

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

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]

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 51

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

Rule 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, -Sim
p[Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (Dist[3/(2*b*q), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d
*x)^(1/3)], x] + Dist[3/(2*b*q^2), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x
] && NegQ[(b*c - a*d)/b]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int \frac{1}{\tan ^{\frac{5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx &=\frac{\int \frac{a-b \tan (c+d x)}{\tan ^{\frac{5}{3}}(c+d x)} \, dx}{a^2+b^2}+\frac{b^2 \int \frac{1+\tan ^2(c+d x)}{\tan ^{\frac{5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx}{a^2+b^2}\\ &=-\frac{3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}+\frac{\int \frac{-b-a \tan (c+d x)}{\tan ^{\frac{2}{3}}(c+d x)} \, dx}{a^2+b^2}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{x^{5/3} (a+b x)} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right ) d}\\ &=-\frac{3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}-\frac{3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}-\frac{a \int \sqrt [3]{\tan (c+d x)} \, dx}{a^2+b^2}-\frac{b \int \frac{1}{\tan ^{\frac{2}{3}}(c+d x)} \, dx}{a^2+b^2}-\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{x^{2/3} (a+b x)} \, dx,x,\tan (c+d x)\right )}{a \left (a^2+b^2\right ) d}\\ &=\frac{b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac{3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}-\frac{3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}-\frac{a \operatorname{Subst}\left (\int \frac{\sqrt [3]{x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right ) d}-\frac{b \operatorname{Subst}\left (\int \frac{1}{x^{2/3} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right ) d}-\frac{\left (3 b^{7/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{a^{2/3}}{b^{2/3}}-\frac{\sqrt [3]{a} x}{\sqrt [3]{b}}+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 a^{4/3} \left (a^2+b^2\right ) d}-\frac{\left (3 b^{8/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt [3]{a}}{\sqrt [3]{b}}+x} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}\\ &=-\frac{3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}+\frac{b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac{3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}-\frac{3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}-\frac{(3 a) \operatorname{Subst}\left (\int \frac{x^3}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{\left (3 b^{8/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt [3]{a}}\right )}{a^{5/3} \left (a^2+b^2\right ) d}\\ &=\frac{\sqrt{3} b^{8/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{a^{5/3} \left (a^2+b^2\right ) d}-\frac{3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}+\frac{b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac{3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}-\frac{3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}-\frac{(3 a) \operatorname{Subst}\left (\int \frac{x}{1+x^3} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}-\frac{b \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{b \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 )}{\left (a^2+b^2\right ) d}-\frac{b \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 )}{\left (a^2+b^2\right ) d}\\ &=\frac{\sqrt{3} b^{8/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{a^{5/3} \left (a^2+b^2\right ) d}-\frac{b \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}+\frac{b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac{3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}-\frac{3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}+\frac{a \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}-\frac{a \operatorname{Subst}\left (\int \frac{1+x}{1-x+x^2} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}-\frac{b \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{4 \left (a^2+b^2\right ) d}-\frac{b \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{4 \left (a^2+b^2\right ) d}+\frac{\left (\sqrt{3} b\right ) \operatorname{Subst}\left (\int \frac{-\sqrt{3}+2 x}{1-\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{4 \left (a^2+b^2\right ) d}-\frac{\left (\sqrt{3} b\right ) \operatorname{Subst}\left (\int \frac{\sqrt{3}+2 x}{1+\sqrt{3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{4 \left (a^2+b^2\right ) d}\\ &=\frac{\sqrt{3} b^{8/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{a^{5/3} \left (a^2+b^2\right ) d}-\frac{b \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}+\frac{a \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}+\frac{\sqrt{3} b \log \left (1-\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac{\sqrt{3} b \log \left (1+\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac{b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac{3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}-\frac{3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}-\frac{a \operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac{(3 a) \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,\tan ^{\frac{2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac{b \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,-\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac{b \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac{b \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}-\frac{b \tan ^{-1}\left (\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac{\sqrt{3} b^{8/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{a^{5/3} \left (a^2+b^2\right ) d}-\frac{b \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}+\frac{a \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}+\frac{\sqrt{3} b \log \left (1-\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac{\sqrt{3} b \log \left (1+\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac{b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac{a \log \left (1-\tan ^{\frac{2}{3}}(c+d x)+\tan ^{\frac{4}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac{3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}-\frac{3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}+\frac{(3 a) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac{2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac{b \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}-\frac{b \tan ^{-1}\left (\sqrt{3}+2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac{\sqrt{3} b^{8/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{a^{5/3} \left (a^2+b^2\right ) d}+\frac{\sqrt{3} a \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(c+d x)}{\sqrt{3}}\right )}{2 \left (a^2+b^2\right ) d}-\frac{b \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}+\frac{a \log \left (1+\tan ^{\frac{2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}+\frac{\sqrt{3} b \log \left (1-\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac{\sqrt{3} b \log \left (1+\sqrt{3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac{2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac{b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac{a \log \left (1-\tan ^{\frac{2}{3}}(c+d x)+\tan ^{\frac{4}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac{3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}-\frac{3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac{2}{3}}(c+d x)}\\ \end{align*}

Mathematica [C]  time = 0.21842, size = 104, normalized size = 0.2 \[ -\frac{3 \left (b^2 \, _2F_1\left (-\frac{2}{3},1;\frac{1}{3};-\frac{b \tan (c+d x)}{a}\right )+a \left (a \, _2F_1\left (-\frac{1}{3},1;\frac{2}{3};-\tan ^2(c+d x)\right )+2 b \tan (c+d x) \, _2F_1\left (\frac{1}{6},1;\frac{7}{6};-\tan ^2(c+d x)\right )\right )\right )}{2 a d \left (a^2+b^2\right ) \tan ^{\frac{2}{3}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b^2*Hypergeometric2F1[-2/3, 1, 1/3, -((b*Tan[c + d*x])/a)] + a*(a*Hypergeometric2F1[-1/3, 1, 2/3, -Tan[c
+ d*x]^2] + 2*b*Hypergeometric2F1[1/6, 1, 7/6, -Tan[c + d*x]^2]*Tan[c + d*x])))/(2*a*(a^2 + b^2)*d*Tan[c + d*x
]^(2/3))

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Maple [A]  time = 0.037, size = 558, normalized size = 1.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d/a*b^2/(a^2+b^2)/(1/b*a)^(2/3)*ln(tan(d*x+c)^(1/3)+(1/b*a)^(1/3))+1/2/d/a*b^2/(a^2+b^2)/(1/b*a)^(2/3)*ln(t
an(d*x+c)^(2/3)-(1/b*a)^(1/3)*tan(d*x+c)^(1/3)+(1/b*a)^(2/3))-1/d/a*b^2/(a^2+b^2)/(1/b*a)^(2/3)*3^(1/2)*arctan
(1/3*3^(1/2)*(2/(1/b*a)^(1/3)*tan(d*x+c)^(1/3)-1))-3/2/a/d/tan(d*x+c)^(2/3)+3/2/d/(3*a^2+3*b^2)*a*ln(1+tan(d*x
+c)^(2/3))-3/d/(3*a^2+3*b^2)*b*arctan(tan(d*x+c)^(1/3))+3/4/d/(3*a^2+3*b^2)*ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(
d*x+c)^(2/3))*3^(1/2)*b-3/4/d/(3*a^2+3*b^2)*ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*a-3/2/d/(3*a^2+3*b
^2)*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))*b-3/2/d/(3*a^2+3*b^2)*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))*3^(1/2)*a-3/
4/d/(3*a^2+3*b^2)*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(1/2)*b-3/4/d/(3*a^2+3*b^2)*ln(1+3^(1/2)*t
an(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*a+3/2/d/(3*a^2+3*b^2)*arctan(3^(1/2)+2*tan(d*x+c)^(1/3))*3^(1/2)*a-3/2/d/(3*
a^2+3*b^2)*arctan(3^(1/2)+2*tan(d*x+c)^(1/3))*b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [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+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \tan{\left (c + d x \right )}\right ) \tan ^{\frac{5}{3}}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)**(5/3)/(a+b*tan(d*x+c)),x)

[Out]

Integral(1/((a + b*tan(c + d*x))*tan(c + d*x)**(5/3)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \tan \left (d x + c\right ) + a\right )} \tan \left (d x + c\right )^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*tan(d*x + c) + a)*tan(d*x + c)^(5/3)), x)