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

Optimal. Leaf size=69 \[ -\frac{1}{16} \log \left (1-\tanh ^{\frac{2}{3}}(8 x)\right )+\frac{1}{32} \log \left (\tanh ^{\frac{4}{3}}(8 x)+\tanh ^{\frac{2}{3}}(8 x)+1\right )-\frac{1}{16} \sqrt{3} \tan ^{-1}\left (\frac{2 \tanh ^{\frac{2}{3}}(8 x)+1}{\sqrt{3}}\right ) \]

[Out]

-(Sqrt[3]*ArcTan[(1 + 2*Tanh[8*x]^(2/3))/Sqrt[3]])/16 - Log[1 - Tanh[8*x]^(2/3)]/16 + Log[1 + Tanh[8*x]^(2/3)
+ Tanh[8*x]^(4/3)]/32

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Rubi [A]  time = 0.0597553, antiderivative size = 69, normalized size of antiderivative = 1., 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, 292, 31, 634, 618, 204, 628} \[ -\frac{1}{16} \log \left (1-\tanh ^{\frac{2}{3}}(8 x)\right )+\frac{1}{32} \log \left (\tanh ^{\frac{4}{3}}(8 x)+\tanh ^{\frac{2}{3}}(8 x)+1\right )-\frac{1}{16} \sqrt{3} \tan ^{-1}\left (\frac{2 \tanh ^{\frac{2}{3}}(8 x)+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[3]*ArcTan[(1 + 2*Tanh[8*x]^(2/3))/Sqrt[3]])/16 - Log[1 - Tanh[8*x]^(2/3)]/16 + Log[1 + Tanh[8*x]^(2/3)
+ Tanh[8*x]^(4/3)]/32

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 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 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 \sqrt [3]{\tanh (8 x)} \, dx &=-\left (\frac{1}{8} \operatorname{Subst}\left (\int \frac{\sqrt [3]{x}}{-1+x^2} \, dx,x,\tanh (8 x)\right )\right )\\ &=-\left (\frac{3}{8} \operatorname{Subst}\left (\int \frac{x^3}{-1+x^6} \, dx,x,\sqrt [3]{\tanh (8 x)}\right )\right )\\ &=-\left (\frac{3}{16} \operatorname{Subst}\left (\int \frac{x}{-1+x^3} \, dx,x,\tanh ^{\frac{2}{3}}(8 x)\right )\right )\\ &=-\left (\frac{1}{16} \operatorname{Subst}\left (\int \frac{1}{-1+x} \, dx,x,\tanh ^{\frac{2}{3}}(8 x)\right )\right )+\frac{1}{16} \operatorname{Subst}\left (\int \frac{-1+x}{1+x+x^2} \, dx,x,\tanh ^{\frac{2}{3}}(8 x)\right )\\ &=-\frac{1}{16} \log \left (1-\tanh ^{\frac{2}{3}}(8 x)\right )+\frac{1}{32} \operatorname{Subst}\left (\int \frac{1+2 x}{1+x+x^2} \, dx,x,\tanh ^{\frac{2}{3}}(8 x)\right )-\frac{3}{32} \operatorname{Subst}\left (\int \frac{1}{1+x+x^2} \, dx,x,\tanh ^{\frac{2}{3}}(8 x)\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 )+\frac{3}{16} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 \tanh ^{\frac{2}{3}}(8 x)\right )\\ &=-\frac{1}{16} \sqrt{3} \tan ^{-1}\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 )\\ \end{align*}

Mathematica [C]  time = 0.0258012, size = 26, normalized size = 0.38 \[ \frac{3}{32} \tanh ^{\frac{4}{3}}(8 x) \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\tanh ^2(8 x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Hypergeometric2F1[2/3, 1, 5/3, Tanh[8*x]^2]*Tanh[8*x]^(4/3))/32

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Maple [A]  time = 0.023, size = 102, normalized size = 1.5 \begin{align*} -{\frac{1}{16}\ln \left ( \sqrt [3]{\tanh \left ( 8\,x \right ) }-1 \right ) }+{\frac{1}{32}\ln \left ( \left ( \tanh \left ( 8\,x \right ) \right ) ^{{\frac{2}{3}}}+\sqrt [3]{\tanh \left ( 8\,x \right ) }+1 \right ) }+{\frac{\sqrt{3}}{16}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,\sqrt [3]{\tanh \left ( 8\,x \right ) }+1 \right ) } \right ) }-{\frac{1}{16}\ln \left ( \sqrt [3]{\tanh \left ( 8\,x \right ) }+1 \right ) }+{\frac{1}{32}\ln \left ( \left ( \tanh \left ( 8\,x \right ) \right ) ^{{\frac{2}{3}}}-\sqrt [3]{\tanh \left ( 8\,x \right ) }+1 \right ) }-{\frac{\sqrt{3}}{16}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,\sqrt [3]{\tanh \left ( 8\,x \right ) }-1 \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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))-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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \tanh \left (8 \, x\right )^{\frac{1}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tanh(8*x)^(1/3), x)

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Fricas [B]  time = 2.21668, size = 563, normalized size = 8.16 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)*(sin
h(8*x)/cosh(8*x))^(1/3) + 1)/(cosh(8*x)^2 + 2*cosh(8*x)*sinh(8*x) + sinh(8*x)^2 + 1))

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Sympy [A]  time = 2.94131, size = 63, normalized size = 0.91 \begin{align*} - \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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

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Giac [B]  time = 1.24303, size = 258, normalized size = 3.74 \begin{align*} \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{1}{3}} + 1\right )}\right ) - \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{1}{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}} + \left (\frac{e^{\left (16 \, x\right )} - 1}{e^{\left (16 \, x\right )} + 1}\right )^{\frac{1}{3}} + 1\right ) + \frac{1}{32} \, \log \left (\left (\frac{e^{\left (16 \, x\right )} - 1}{e^{\left (16 \, x\right )} + 1}\right )^{\frac{2}{3}} - \left (\frac{e^{\left (16 \, x\right )} - 1}{e^{\left (16 \, x\right )} + 1}\right )^{\frac{1}{3}} + 1\right ) - \frac{1}{16} \, \log \left ({\left | \left (\frac{e^{\left (16 \, x\right )} - 1}{e^{\left (16 \, x\right )} + 1}\right )^{\frac{1}{3}} + 1 \right |}\right ) - \frac{1}{16} \, \log \left ({\left | \left (\frac{e^{\left (16 \, x\right )} - 1}{e^{\left (16 \, x\right )} + 1}\right )^{\frac{1}{3}} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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