3.398 \(\int \frac{1}{\sqrt [3]{\tan (5 x)}} \, dx\)

Optimal. Leaf size=57 \[ -\frac{1}{10} \sqrt{3} \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(5 x)}{\sqrt{3}}\right )+\frac{3}{20} \log \left (\tan ^{\frac{2}{3}}(5 x)+1\right )-\frac{1}{20} \log \left (\tan ^2(5 x)+1\right ) \]

[Out]

-(Sqrt[3]*ArcTan[(1 - 2*Tan[5*x]^(2/3))/Sqrt[3]])/10 + (3*Log[1 + Tan[5*x]^(2/3)])/20 - Log[1 + Tan[5*x]^2]/20

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Rubi [A]  time = 0.0584444, antiderivative size = 69, normalized size of antiderivative = 1.21, number of steps used = 9, number of rules used = 9, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.125, Rules used = {3476, 329, 275, 200, 31, 634, 618, 204, 628} \[ -\frac{1}{10} \sqrt{3} \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(5 x)}{\sqrt{3}}\right )+\frac{1}{10} \log \left (\tan ^{\frac{2}{3}}(5 x)+1\right )-\frac{1}{20} \log \left (\tan ^{\frac{4}{3}}(5 x)-\tan ^{\frac{2}{3}}(5 x)+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[Tan[5*x]^(-1/3),x]

[Out]

-(Sqrt[3]*ArcTan[(1 - 2*Tan[5*x]^(2/3))/Sqrt[3]])/10 + Log[1 + Tan[5*x]^(2/3)]/10 - Log[1 - Tan[5*x]^(2/3) + T
an[5*x]^(4/3)]/20

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 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 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 31

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

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]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt [3]{\tan (5 x)}} \, dx &=\frac{1}{5} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{x} \left (1+x^2\right )} \, dx,x,\tan (5 x)\right )\\ &=\frac{3}{5} \operatorname{Subst}\left (\int \frac{x}{1+x^6} \, dx,x,\sqrt [3]{\tan (5 x)}\right )\\ &=\frac{3}{10} \operatorname{Subst}\left (\int \frac{1}{1+x^3} \, dx,x,\tan ^{\frac{2}{3}}(5 x)\right )\\ &=\frac{1}{10} \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,\tan ^{\frac{2}{3}}(5 x)\right )+\frac{1}{10} \operatorname{Subst}\left (\int \frac{2-x}{1-x+x^2} \, dx,x,\tan ^{\frac{2}{3}}(5 x)\right )\\ &=\frac{1}{10} \log \left (1+\tan ^{\frac{2}{3}}(5 x)\right )-\frac{1}{20} \operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac{2}{3}}(5 x)\right )+\frac{3}{20} \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,\tan ^{\frac{2}{3}}(5 x)\right )\\ &=\frac{1}{10} \log \left (1+\tan ^{\frac{2}{3}}(5 x)\right )-\frac{1}{20} \log \left (1-\tan ^{\frac{2}{3}}(5 x)+\tan ^{\frac{4}{3}}(5 x)\right )-\frac{3}{10} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac{2}{3}}(5 x)\right )\\ &=-\frac{1}{10} \sqrt{3} \tan ^{-1}\left (\frac{1-2 \tan ^{\frac{2}{3}}(5 x)}{\sqrt{3}}\right )+\frac{1}{10} \log \left (1+\tan ^{\frac{2}{3}}(5 x)\right )-\frac{1}{20} \log \left (1-\tan ^{\frac{2}{3}}(5 x)+\tan ^{\frac{4}{3}}(5 x)\right )\\ \end{align*}

Mathematica [A]  time = 0.050121, size = 69, normalized size = 1.21 \[ \frac{1}{10} \sqrt{3} \tan ^{-1}\left (\frac{2 \tan ^{\frac{2}{3}}(5 x)-1}{\sqrt{3}}\right )+\frac{1}{10} \log \left (\tan ^{\frac{2}{3}}(5 x)+1\right )-\frac{1}{20} \log \left (\tan ^{\frac{4}{3}}(5 x)-\tan ^{\frac{2}{3}}(5 x)+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[5*x]^(-1/3),x]

[Out]

(Sqrt[3]*ArcTan[(-1 + 2*Tan[5*x]^(2/3))/Sqrt[3]])/10 + Log[1 + Tan[5*x]^(2/3)]/10 - Log[1 - Tan[5*x]^(2/3) + T
an[5*x]^(4/3)]/20

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Maple [A]  time = 0.012, size = 53, normalized size = 0.9 \begin{align*}{\frac{1}{10}\ln \left ( 1+ \left ( \tan \left ( 5\,x \right ) \right ) ^{{\frac{2}{3}}} \right ) }-{\frac{1}{20}\ln \left ( 1- \left ( \tan \left ( 5\,x \right ) \right ) ^{{\frac{2}{3}}}+ \left ( \tan \left ( 5\,x \right ) \right ) ^{{\frac{4}{3}}} \right ) }+{\frac{\sqrt{3}}{10}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\, \left ( \tan \left ( 5\,x \right ) \right ) ^{2/3}-1 \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/tan(5*x)^(1/3),x)

[Out]

1/10*ln(1+tan(5*x)^(2/3))-1/20*ln(1-tan(5*x)^(2/3)+tan(5*x)^(4/3))+1/10*3^(1/2)*arctan(1/3*(2*tan(5*x)^(2/3)-1
)*3^(1/2))

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Maxima [A]  time = 1.4146, size = 70, normalized size = 1.23 \begin{align*} \frac{1}{10} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, \tan \left (5 \, x\right )^{\frac{2}{3}} - 1\right )}\right ) - \frac{1}{20} \, \log \left (\tan \left (5 \, x\right )^{\frac{4}{3}} - \tan \left (5 \, x\right )^{\frac{2}{3}} + 1\right ) + \frac{1}{10} \, \log \left (\tan \left (5 \, x\right )^{\frac{2}{3}} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(5*x)^(1/3),x, algorithm="maxima")

[Out]

1/10*sqrt(3)*arctan(1/3*sqrt(3)*(2*tan(5*x)^(2/3) - 1)) - 1/20*log(tan(5*x)^(4/3) - tan(5*x)^(2/3) + 1) + 1/10
*log(tan(5*x)^(2/3) + 1)

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Fricas [A]  time = 1.86973, size = 192, normalized size = 3.37 \begin{align*} \frac{1}{10} \, \sqrt{3} \arctan \left (\frac{2}{3} \, \sqrt{3} \tan \left (5 \, x\right )^{\frac{2}{3}} - \frac{1}{3} \, \sqrt{3}\right ) - \frac{1}{20} \, \log \left (\tan \left (5 \, x\right )^{\frac{4}{3}} - \tan \left (5 \, x\right )^{\frac{2}{3}} + 1\right ) + \frac{1}{10} \, \log \left (\tan \left (5 \, x\right )^{\frac{2}{3}} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(5*x)^(1/3),x, algorithm="fricas")

[Out]

1/10*sqrt(3)*arctan(2/3*sqrt(3)*tan(5*x)^(2/3) - 1/3*sqrt(3)) - 1/20*log(tan(5*x)^(4/3) - tan(5*x)^(2/3) + 1)
+ 1/10*log(tan(5*x)^(2/3) + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(5*x)**(1/3),x)

[Out]

Integral(tan(5*x)**(-1/3), x)

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Giac [A]  time = 1.16806, size = 70, normalized size = 1.23 \begin{align*} \frac{1}{10} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, \tan \left (5 \, x\right )^{\frac{2}{3}} - 1\right )}\right ) - \frac{1}{20} \, \log \left (\tan \left (5 \, x\right )^{\frac{4}{3}} - \tan \left (5 \, x\right )^{\frac{2}{3}} + 1\right ) + \frac{1}{10} \, \log \left (\tan \left (5 \, x\right )^{\frac{2}{3}} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(5*x)^(1/3),x, algorithm="giac")

[Out]

1/10*sqrt(3)*arctan(1/3*sqrt(3)*(2*tan(5*x)^(2/3) - 1)) - 1/20*log(tan(5*x)^(4/3) - tan(5*x)^(2/3) + 1) + 1/10
*log(tan(5*x)^(2/3) + 1)