\(\int \frac {1}{\sqrt [3]{\tan (5 x)}} \, dx\) [398]

Optimal result

Integrand size = 8, antiderivative size = 57 \[ \int \frac {1}{\sqrt [3]{\tan (5 x)}} \, dx=-\frac {1}{10} \sqrt {3} \arctan \left (\frac {1-2 \tan ^{\frac {2}{3}}(5 x)}{\sqrt {3}}\right )+\frac {3}{20} \log \left (1+\tan ^{\frac {2}{3}}(5 x)\right )-\frac {1}{20} \log \left (1+\tan ^2(5 x)\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.21 \[ \int \frac {1}{\sqrt [3]{\tan (5 x)}} \, dx=\frac {1}{10} \sqrt {3} \arctan \left (\frac {-1+2 \tan ^{\frac {2}{3}}(5 x)}{\sqrt {3}}\right )+\frac {1}{10} \log \left (1+\tan ^{\frac {2}{3}}(5 x)\right )-\frac {1}{20} \log \left (1-\tan ^{\frac {2}{3}}(5 x)+\tan ^{\frac {4}{3}}(5 x)\right ) \] Input:

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

Output:

(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) + Tan[5*x]^(4/3)]/20
 

Rubi [A] (warning: unable to verify)

Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.79, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.375, Rules used = {3042, 3957, 266, 807, 750, 16, 1142, 25, 1083, 217, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

\(\displaystyle \frac {1}{5} \int \frac {1}{\sqrt [3]{\tan (5 x)} \left (\tan ^2(5 x)+1\right )}d\tan (5 x)\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {3}{5} \int \frac {\sqrt [3]{\tan (5 x)}}{\tan ^2(5 x)+1}d\sqrt [3]{\tan (5 x)}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {3}{10} \int \frac {1}{\tan (5 x)+1}d\tan ^{\frac {2}{3}}(5 x)\)

\(\Big \downarrow \) 750

\(\displaystyle \frac {3}{10} \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(5 x)\right )d\tan ^{\frac {2}{3}}(5 x)+\frac {1}{3} \int \frac {1}{\tan ^{\frac {2}{3}}(5 x)+1}d\tan ^{\frac {2}{3}}(5 x)\right )\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {3}{10} \left (\frac {1}{3} \int \left (2-\tan ^{\frac {2}{3}}(5 x)\right )d\tan ^{\frac {2}{3}}(5 x)+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(5 x)+1\right )\right )\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {3}{10} \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(5 x)-\frac {1}{2} \int \left (2 \tan ^{\frac {2}{3}}(5 x)-1\right )d\tan ^{\frac {2}{3}}(5 x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(5 x)+1\right )\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {3}{10} \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\tan ^{\frac {2}{3}}(5 x)+\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(5 x)\right )d\tan ^{\frac {2}{3}}(5 x)\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(5 x)+1\right )\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {3}{10} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(5 x)\right )d\tan ^{\frac {2}{3}}(5 x)-3 \int \frac {1}{-2 \tan ^{\frac {2}{3}}(5 x)-2}d\left (2 \tan ^{\frac {2}{3}}(5 x)-1\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(5 x)+1\right )\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {3}{10} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 \tan ^{\frac {2}{3}}(5 x)\right )d\tan ^{\frac {2}{3}}(5 x)+\sqrt {3} \arctan \left (\frac {2 \tan ^{\frac {2}{3}}(5 x)-1}{\sqrt {3}}\right )\right )+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(5 x)+1\right )\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {3}{10} \left (\frac {\arctan \left (\frac {2 \tan ^{\frac {2}{3}}(5 x)-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (\tan ^{\frac {2}{3}}(5 x)+1\right )\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 16

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

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 217

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

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 750

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2)   Int[1/ 
(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[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 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[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 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[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]
 

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(-\frac {\ln \left (\tan \left (5 x \right )^{\frac {4}{3}}-\tan \left (5 x \right )^{\frac {2}{3}}+1\right )}{20}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \tan \left (5 x \right )^{\frac {2}{3}}-1\right ) \sqrt {3}}{3}\right )}{10}+\frac {\ln \left (1+\tan \left (5 x \right )^{\frac {2}{3}}\right )}{10}\)	\(53\)
default	\(-\frac {\ln \left (\tan \left (5 x \right )^{\frac {4}{3}}-\tan \left (5 x \right )^{\frac {2}{3}}+1\right )}{20}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \tan \left (5 x \right )^{\frac {2}{3}}-1\right ) \sqrt {3}}{3}\right )}{10}+\frac {\ln \left (1+\tan \left (5 x \right )^{\frac {2}{3}}\right )}{10}\)	\(53\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\sqrt [3]{\tan (5 x)}} \, dx=\frac {1}{10} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} \tan \left (5 \, x\right )^{\frac {2}{3}} - \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{20} \, \log \left (\tan \left (5 \, x\right )^{\frac {4}{3}} - \tan \left (5 \, x\right )^{\frac {2}{3}} + 1\right ) + \frac {1}{10} \, \log \left (\tan \left (5 \, x\right )^{\frac {2}{3}} + 1\right ) \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {1}{\sqrt [3]{\tan (5 x)}} \, dx=\int \frac {1}{\sqrt [3]{\tan {\left (5 x \right )}}}\, dx \] Input:

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

Output:

Integral(tan(5*x)**(-1/3), x)
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\sqrt [3]{\tan (5 x)}} \, dx=\frac {1}{10} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \tan \left (5 \, x\right )^{\frac {2}{3}} - 1\right )}\right ) - \frac {1}{20} \, \log \left (\tan \left (5 \, x\right )^{\frac {4}{3}} - \tan \left (5 \, x\right )^{\frac {2}{3}} + 1\right ) + \frac {1}{10} \, \log \left (\tan \left (5 \, x\right )^{\frac {2}{3}} + 1\right ) \] Input:

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

Output:

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)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\sqrt [3]{\tan (5 x)}} \, dx=\frac {1}{10} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \tan \left (5 \, x\right )^{\frac {2}{3}} - 1\right )}\right ) - \frac {1}{20} \, \log \left (\tan \left (5 \, x\right )^{\frac {4}{3}} - \tan \left (5 \, x\right )^{\frac {2}{3}} + 1\right ) + \frac {1}{10} \, \log \left (\tan \left (5 \, x\right )^{\frac {2}{3}} + 1\right ) \] Input:

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

Output:

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)
 

Mupad [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\sqrt [3]{\tan (5 x)}} \, dx=\frac {\ln \left (81\,{\mathrm {tan}\left (5\,x\right )}^{2/3}+81\right )}{10}-\ln \left (81-162\,{\mathrm {tan}\left (5\,x\right )}^{2/3}+\sqrt {3}\,81{}\mathrm {i}\right )\,\left (\frac {1}{20}+\frac {\sqrt {3}\,1{}\mathrm {i}}{20}\right )+\ln \left (162\,{\mathrm {tan}\left (5\,x\right )}^{2/3}-81+\sqrt {3}\,81{}\mathrm {i}\right )\,\left (-\frac {1}{20}+\frac {\sqrt {3}\,1{}\mathrm {i}}{20}\right ) \] Input:

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

Output:

log(81*tan(5*x)^(2/3) + 81)/10 - log(3^(1/2)*81i - 162*tan(5*x)^(2/3) + 81 
)*((3^(1/2)*1i)/20 + 1/20) + log(3^(1/2)*81i + 162*tan(5*x)^(2/3) - 81)*(( 
3^(1/2)*1i)/20 - 1/20)
 

Reduce [F]

\[ \int \frac {1}{\sqrt [3]{\tan (5 x)}} \, dx=\int \frac {1}{\tan \left (5 x \right )^{\frac {1}{3}}}d x \] Input:

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

Output:

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