\(\int \sqrt [3]{\tanh (8 x)} \, dx\) [21]

Optimal result

Integrand size = 8, antiderivative size = 69 \[ \int \sqrt [3]{\tanh (8 x)} \, dx=-\frac {1}{16} \sqrt {3} \arctan \left (\frac {1+2 \tanh ^{\frac {2}{3}}(8 x)}{\sqrt {3}}\right )-\frac {1}{16} \log \left (1-\tanh ^{\frac {2}{3}}(8 x)\right )+\frac {1}{32} \log \left (1+\tanh ^{\frac {2}{3}}(8 x)+\tanh ^{\frac {4}{3}}(8 x)\right ) \] Output:

-1/16*3^(1/2)*arctan(1/3*(1+2*tanh(8*x)^(2/3))*3^(1/2))-1/16*ln(1-tanh(8*x 
)^(2/3))+1/32*ln(1+tanh(8*x)^(2/3)+tanh(8*x)^(4/3))
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.30 \[ \int \sqrt [3]{\tanh (8 x)} \, dx=-\frac {\left (\log \left (1-\sqrt [3]{\tanh ^2(8 x)}\right )-\sqrt [3]{-1} \log \left (1+\sqrt [3]{-1} \sqrt [3]{\tanh ^2(8 x)}\right )+(-1)^{2/3} \log \left (1-(-1)^{2/3} \sqrt [3]{\tanh ^2(8 x)}\right )\right ) \tanh ^{\frac {4}{3}}(8 x)}{16 \tanh ^2(8 x)^{2/3}} \] Input:

Integrate[Tanh[8*x]^(1/3),x]
 

Output:

-1/16*((Log[1 - (Tanh[8*x]^2)^(1/3)] - (-1)^(1/3)*Log[1 + (-1)^(1/3)*(Tanh 
[8*x]^2)^(1/3)] + (-1)^(2/3)*Log[1 - (-1)^(2/3)*(Tanh[8*x]^2)^(1/3)])*Tanh 
[8*x]^(4/3))/(Tanh[8*x]^2)^(2/3)
 

Rubi [A] (warning: unable to verify)

Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.375, Rules used = {3042, 3957, 25, 266, 807, 821, 16, 1142, 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.

\(\displaystyle \int \sqrt [3]{\tanh (8 x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt [3]{-i \tan (8 i x)}dx\)

\(\Big \downarrow \) 3957

\(\displaystyle -\frac {1}{8} \int -\frac {\sqrt [3]{\tanh (8 x)}}{1-\tanh ^2(8 x)}d\tanh (8 x)\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{8} \int \frac {\sqrt [3]{\tanh (8 x)}}{1-\tanh ^2(8 x)}d\tanh (8 x)\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {3}{8} \int \frac {\tanh (8 x)}{1-\tanh ^2(8 x)}d\sqrt [3]{\tanh (8 x)}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {3}{16} \int \frac {\tanh ^{\frac {2}{3}}(8 x)}{1-\tanh (8 x)}d\tanh ^{\frac {2}{3}}(8 x)\)

\(\Big \downarrow \) 821

\(\displaystyle \frac {3}{16} \left (\frac {1}{3} \int \frac {1}{1-\tanh ^{\frac {2}{3}}(8 x)}d\tanh ^{\frac {2}{3}}(8 x)-\frac {1}{3} \int \frac {1-\tanh ^{\frac {2}{3}}(8 x)}{2 \tanh ^{\frac {2}{3}}(8 x)+1}d\tanh ^{\frac {2}{3}}(8 x)\right )\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {3}{16} \left (-\frac {1}{3} \int \frac {1-\tanh ^{\frac {2}{3}}(8 x)}{2 \tanh ^{\frac {2}{3}}(8 x)+1}d\tanh ^{\frac {2}{3}}(8 x)-\frac {1}{3} \log \left (1-\tanh ^{\frac {2}{3}}(8 x)\right )\right )\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {3}{16} \left (\frac {1}{3} \left (\frac {1}{2} \int 1d\tanh ^{\frac {2}{3}}(8 x)-\frac {3}{2} \int \frac {1}{2 \tanh ^{\frac {2}{3}}(8 x)+1}d\tanh ^{\frac {2}{3}}(8 x)\right )-\frac {1}{3} \log \left (1-\tanh ^{\frac {2}{3}}(8 x)\right )\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {3}{16} \left (\frac {1}{3} \left (\frac {1}{2} \int 1d\tanh ^{\frac {2}{3}}(8 x)+3 \int \frac {1}{-2 \tanh ^{\frac {2}{3}}(8 x)-4}d\left (2 \tanh ^{\frac {2}{3}}(8 x)+1\right )\right )-\frac {1}{3} \log \left (1-\tanh ^{\frac {2}{3}}(8 x)\right )\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {3}{16} \left (\frac {1}{3} \left (\frac {1}{2} \int 1d\tanh ^{\frac {2}{3}}(8 x)-\sqrt {3} \arctan \left (\frac {2 \tanh ^{\frac {2}{3}}(8 x)+1}{\sqrt {3}}\right )\right )-\frac {1}{3} \log \left (1-\tanh ^{\frac {2}{3}}(8 x)\right )\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {3}{16} \left (\frac {1}{3} \left (\frac {1}{2} \log \left (2 \tanh ^{\frac {2}{3}}(8 x)+1\right )-\sqrt {3} \arctan \left (\frac {2 \tanh ^{\frac {2}{3}}(8 x)+1}{\sqrt {3}}\right )\right )-\frac {1}{3} \log \left (1-\tanh ^{\frac {2}{3}}(8 x)\right )\right )\)

Input:

Int[Tanh[8*x]^(1/3),x]
 

Output:

(3*(-1/3*Log[1 - Tanh[8*x]^(2/3)] + (-(Sqrt[3]*ArcTan[(1 + 2*Tanh[8*x]^(2/ 
3))/Sqrt[3]]) + Log[1 + 2*Tanh[8*x]^(2/3)]/2)/3))/16
 

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 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 821

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 
1)   Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[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 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 = 0.16 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.48

method	result	size
derivativedivides	\(-\frac {\ln \left (\tanh \left (8 x \right )^{\frac {1}{3}}-1\right )}{16}+\frac {\ln \left (\tanh \left (8 x \right )^{\frac {2}{3}}+\tanh \left (8 x \right )^{\frac {1}{3}}+1\right )}{32}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \tanh \left (8 x \right )^{\frac {1}{3}}+1\right ) \sqrt {3}}{3}\right )}{16}-\frac {\ln \left (\tanh \left (8 x \right )^{\frac {1}{3}}+1\right )}{16}+\frac {\ln \left (\tanh \left (8 x \right )^{\frac {2}{3}}-\tanh \left (8 x \right )^{\frac {1}{3}}+1\right )}{32}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \tanh \left (8 x \right )^{\frac {1}{3}}-1\right ) \sqrt {3}}{3}\right )}{16}\)	\(102\)
default	\(-\frac {\ln \left (\tanh \left (8 x \right )^{\frac {1}{3}}-1\right )}{16}+\frac {\ln \left (\tanh \left (8 x \right )^{\frac {2}{3}}+\tanh \left (8 x \right )^{\frac {1}{3}}+1\right )}{32}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \tanh \left (8 x \right )^{\frac {1}{3}}+1\right ) \sqrt {3}}{3}\right )}{16}-\frac {\ln \left (\tanh \left (8 x \right )^{\frac {1}{3}}+1\right )}{16}+\frac {\ln \left (\tanh \left (8 x \right )^{\frac {2}{3}}-\tanh \left (8 x \right )^{\frac {1}{3}}+1\right )}{32}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \tanh \left (8 x \right )^{\frac {1}{3}}-1\right ) \sqrt {3}}{3}\right )}{16}\)	\(102\)

Input:

int(tanh(8*x)^(1/3),x,method=_RETURNVERBOSE)
 

Output:

-1/16*ln(tanh(8*x)^(1/3)-1)+1/32*ln(tanh(8*x)^(2/3)+tanh(8*x)^(1/3)+1)+1/1 
6*3^(1/2)*arctan(1/3*(2*tanh(8*x)^(1/3)+1)*3^(1/2))-1/16*ln(tanh(8*x)^(1/3 
)+1)+1/32*ln(tanh(8*x)^(2/3)-tanh(8*x)^(1/3)+1)-1/16*3^(1/2)*arctan(1/3*(2 
*tanh(8*x)^(1/3)-1)*3^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (52) = 104\).

Time = 0.08 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.59 \[ \int \sqrt [3]{\tanh (8 x)} \, dx=-\frac {1}{16} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} \left (\frac {\sinh \left (8 \, x\right )}{\cosh \left (8 \, x\right )}\right )^{\frac {2}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{16} \, \log \left (\left (\frac {\sinh \left (8 \, x\right )}{\cosh \left (8 \, x\right )}\right )^{\frac {2}{3}} - 1\right ) + \frac {1}{32} \, \log \left (\frac {\cosh \left (8 \, x\right )^{2} + 2 \, \cosh \left (8 \, x\right ) \sinh \left (8 \, x\right ) + \sinh \left (8 \, x\right )^{2} + {\left (\cosh \left (8 \, x\right )^{2} + 2 \, \cosh \left (8 \, x\right ) \sinh \left (8 \, x\right ) + \sinh \left (8 \, x\right )^{2} + 1\right )} \left (\frac {\sinh \left (8 \, x\right )}{\cosh \left (8 \, x\right )}\right )^{\frac {2}{3}} + {\left (\cosh \left (8 \, x\right )^{2} + 2 \, \cosh \left (8 \, x\right ) \sinh \left (8 \, x\right ) + \sinh \left (8 \, x\right )^{2} - 1\right )} \left (\frac {\sinh \left (8 \, x\right )}{\cosh \left (8 \, x\right )}\right )^{\frac {1}{3}} + 1}{\cosh \left (8 \, x\right )^{2} + 2 \, \cosh \left (8 \, x\right ) \sinh \left (8 \, x\right ) + \sinh \left (8 \, x\right )^{2} + 1}\right ) \] Input:

integrate(tanh(8*x)^(1/3),x, algorithm="fricas")
 

Output:

-1/16*sqrt(3)*arctan(2/3*sqrt(3)*(sinh(8*x)/cosh(8*x))^(2/3) + 1/3*sqrt(3) 
) - 1/16*log((sinh(8*x)/cosh(8*x))^(2/3) - 1) + 1/32*log((cosh(8*x)^2 + 2* 
cosh(8*x)*sinh(8*x) + sinh(8*x)^2 + (cosh(8*x)^2 + 2*cosh(8*x)*sinh(8*x) + 
 sinh(8*x)^2 + 1)*(sinh(8*x)/cosh(8*x))^(2/3) + (cosh(8*x)^2 + 2*cosh(8*x) 
*sinh(8*x) + sinh(8*x)^2 - 1)*(sinh(8*x)/cosh(8*x))^(1/3) + 1)/(cosh(8*x)^ 
2 + 2*cosh(8*x)*sinh(8*x) + sinh(8*x)^2 + 1))
 

Sympy [A] (verification not implemented)

Time = 1.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91 \[ \int \sqrt [3]{\tanh (8 x)} \, dx=- \frac {\log {\left (\tanh ^{\frac {2}{3}}{\left (8 x \right )} - 1 \right )}}{16} + \frac {\log {\left (\tanh ^{\frac {4}{3}}{\left (8 x \right )} + \tanh ^{\frac {2}{3}}{\left (8 x \right )} + 1 \right )}}{32} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} \left (\tanh ^{\frac {2}{3}}{\left (8 x \right )} + \frac {1}{2}\right )}{3} \right )}}{16} \] Input:

integrate(tanh(8*x)**(1/3),x)
 

Output:

-log(tanh(8*x)**(2/3) - 1)/16 + log(tanh(8*x)**(4/3) + tanh(8*x)**(2/3) + 
1)/32 - sqrt(3)*atan(2*sqrt(3)*(tanh(8*x)**(2/3) + 1/2)/3)/16
 

Maxima [F]

\[ \int \sqrt [3]{\tanh (8 x)} \, dx=\int { \tanh \left (8 \, x\right )^{\frac {1}{3}} \,d x } \] Input:

integrate(tanh(8*x)^(1/3),x, algorithm="maxima")
 

Output:

integrate(tanh(8*x)^(1/3), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (52) = 104\).

Time = 0.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.59 \[ \int \sqrt [3]{\tanh (8 x)} \, dx=-\frac {1}{16} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {e^{\left (16 \, x\right )} - 1}{e^{\left (16 \, x\right )} + 1}\right )^{\frac {2}{3}} + 1\right )}\right ) + \frac {1}{32} \, \log \left (\left (\frac {e^{\left (16 \, x\right )} - 1}{e^{\left (16 \, x\right )} + 1}\right )^{\frac {2}{3}} + \frac {\left (\frac {e^{\left (16 \, x\right )} - 1}{e^{\left (16 \, x\right )} + 1}\right )^{\frac {1}{3}} {\left (e^{\left (16 \, x\right )} - 1\right )}}{e^{\left (16 \, x\right )} + 1} + 1\right ) - \frac {1}{16} \, \log \left ({\left | \left (\frac {e^{\left (16 \, x\right )} - 1}{e^{\left (16 \, x\right )} + 1}\right )^{\frac {2}{3}} - 1 \right |}\right ) \] Input:

integrate(tanh(8*x)^(1/3),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

-1/16*sqrt(3)*arctan(1/3*sqrt(3)*(2*((e^(16*x) - 1)/(e^(16*x) + 1))^(2/3) 
+ 1)) + 1/32*log(((e^(16*x) - 1)/(e^(16*x) + 1))^(2/3) + ((e^(16*x) - 1)/( 
e^(16*x) + 1))^(1/3)*(e^(16*x) - 1)/(e^(16*x) + 1) + 1) - 1/16*log(abs(((e 
^(16*x) - 1)/(e^(16*x) + 1))^(2/3) - 1))
 

Mupad [B] (verification not implemented)

Time = 2.60 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03 \[ \int \sqrt [3]{\tanh (8 x)} \, dx=-\frac {\ln \left (81\,{\mathrm {tanh}\left (8\,x\right )}^{2/3}-81\right )}{16}-\ln \left (162\,{\mathrm {tanh}\left (8\,x\right )}^{2/3}\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )-81\right )\,\left (-\frac {1}{32}+\frac {\sqrt {3}\,1{}\mathrm {i}}{32}\right )+\ln \left (-162\,{\mathrm {tanh}\left (8\,x\right )}^{2/3}\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )-81\right )\,\left (\frac {1}{32}+\frac {\sqrt {3}\,1{}\mathrm {i}}{32}\right ) \] Input:

int(tanh(8*x)^(1/3),x)
 

Output:

log(- 162*tanh(8*x)^(2/3)*((3^(1/2)*1i)/4 + 1/4) - 81)*((3^(1/2)*1i)/32 + 
1/32) - log(162*tanh(8*x)^(2/3)*((3^(1/2)*1i)/4 - 1/4) - 81)*((3^(1/2)*1i) 
/32 - 1/32) - log(81*tanh(8*x)^(2/3) - 81)/16
 

Reduce [F]

\[ \int \sqrt [3]{\tanh (8 x)} \, dx=\int \tanh \left (8 x \right )^{\frac {1}{3}}d x \] Input:

int(tanh(8*x)^(1/3),x)
 

Output:

int(tanh(8*x)**(1/3),x)