3.33 \(\int \sqrt {\tanh ^3(x)} \, dx\)

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

[Out]

-2*coth(x)*(tanh(x)^3)^(1/2)+arctan(tanh(x)^(1/2))*(tanh(x)^3)^(1/2)/tanh(x)^(3/2)+arctanh(tanh(x)^(1/2))*(tan
h(x)^3)^(1/2)/tanh(x)^(3/2)

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Rubi [A]  time = 0.04, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {3658, 3473, 3476, 329, 212, 206, 203} \[ \frac {\tanh ^{-1}\left (\sqrt {\tanh (x)}\right ) \sqrt {\tanh ^3(x)}}{\tanh ^{\frac {3}{2}}(x)}+\frac {\sqrt {\tanh ^3(x)} \tan ^{-1}\left (\sqrt {\tanh (x)}\right )}{\tanh ^{\frac {3}{2}}(x)}-2 \sqrt {\tanh ^3(x)} \coth (x) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Tanh[x]^3],x]

[Out]

-2*Coth[x]*Sqrt[Tanh[x]^3] + (ArcTan[Sqrt[Tanh[x]]]*Sqrt[Tanh[x]^3])/Tanh[x]^(3/2) + (ArcTanh[Sqrt[Tanh[x]]]*S
qrt[Tanh[x]^3])/Tanh[x]^(3/2)

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 206

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

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

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 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

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 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps

\begin {align*} \int \sqrt {\tanh ^3(x)} \, dx &=\frac {\sqrt {\tanh ^3(x)} \int \tanh ^{\frac {3}{2}}(x) \, dx}{\tanh ^{\frac {3}{2}}(x)}\\ &=-2 \coth (x) \sqrt {\tanh ^3(x)}+\frac {\sqrt {\tanh ^3(x)} \int \frac {1}{\sqrt {\tanh (x)}} \, dx}{\tanh ^{\frac {3}{2}}(x)}\\ &=-2 \coth (x) \sqrt {\tanh ^3(x)}-\frac {\sqrt {\tanh ^3(x)} \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (-1+x^2\right )} \, dx,x,\tanh (x)\right )}{\tanh ^{\frac {3}{2}}(x)}\\ &=-2 \coth (x) \sqrt {\tanh ^3(x)}-\frac {\left (2 \sqrt {\tanh ^3(x)}\right ) \operatorname {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\sqrt {\tanh (x)}\right )}{\tanh ^{\frac {3}{2}}(x)}\\ &=-2 \coth (x) \sqrt {\tanh ^3(x)}+\frac {\sqrt {\tanh ^3(x)} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {\tanh (x)}\right )}{\tanh ^{\frac {3}{2}}(x)}+\frac {\sqrt {\tanh ^3(x)} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {\tanh (x)}\right )}{\tanh ^{\frac {3}{2}}(x)}\\ &=-2 \coth (x) \sqrt {\tanh ^3(x)}+\frac {\tan ^{-1}\left (\sqrt {\tanh (x)}\right ) \sqrt {\tanh ^3(x)}}{\tanh ^{\frac {3}{2}}(x)}+\frac {\tanh ^{-1}\left (\sqrt {\tanh (x)}\right ) \sqrt {\tanh ^3(x)}}{\tanh ^{\frac {3}{2}}(x)}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 38, normalized size = 0.67 \[ \frac {\sqrt {\tanh ^3(x)} \left (\tanh ^{-1}\left (\sqrt {\tanh (x)}\right )-2 \sqrt {\tanh (x)}+\tan ^{-1}\left (\sqrt {\tanh (x)}\right )\right )}{\tanh ^{\frac {3}{2}}(x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Tanh[x]^3],x]

[Out]

((ArcTan[Sqrt[Tanh[x]]] + ArcTanh[Sqrt[Tanh[x]]] - 2*Sqrt[Tanh[x]])*Sqrt[Tanh[x]^3])/Tanh[x]^(3/2)

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fricas [B]  time = 0.51, size = 106, normalized size = 1.86 \[ -2 \, \sqrt {\frac {\sinh \relax (x)}{\cosh \relax (x)}} + \arctan \left (-\cosh \relax (x)^{2} - 2 \, \cosh \relax (x) \sinh \relax (x) - \sinh \relax (x)^{2} + {\left (\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} + 1\right )} \sqrt {\frac {\sinh \relax (x)}{\cosh \relax (x)}}\right ) - \frac {1}{2} \, \log \left (-\cosh \relax (x)^{2} - 2 \, \cosh \relax (x) \sinh \relax (x) - \sinh \relax (x)^{2} + {\left (\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} + 1\right )} \sqrt {\frac {\sinh \relax (x)}{\cosh \relax (x)}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((tanh(x)^3)^(1/2),x, algorithm="fricas")

[Out]

-2*sqrt(sinh(x)/cosh(x)) + arctan(-cosh(x)^2 - 2*cosh(x)*sinh(x) - sinh(x)^2 + (cosh(x)^2 + 2*cosh(x)*sinh(x)
+ sinh(x)^2 + 1)*sqrt(sinh(x)/cosh(x))) - 1/2*log(-cosh(x)^2 - 2*cosh(x)*sinh(x) - sinh(x)^2 + (cosh(x)^2 + 2*
cosh(x)*sinh(x) + sinh(x)^2 + 1)*sqrt(sinh(x)/cosh(x)))

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giac [A]  time = 0.15, size = 55, normalized size = 0.96 \[ \frac {4}{\sqrt {e^{\left (4 \, x\right )} - 1} - e^{\left (2 \, x\right )} - 1} + \arctan \left (\sqrt {e^{\left (4 \, x\right )} - 1} - e^{\left (2 \, x\right )}\right ) - \frac {1}{2} \, \log \left (-\sqrt {e^{\left (4 \, x\right )} - 1} + e^{\left (2 \, x\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((tanh(x)^3)^(1/2),x, algorithm="giac")

[Out]

4/(sqrt(e^(4*x) - 1) - e^(2*x) - 1) + arctan(sqrt(e^(4*x) - 1) - e^(2*x)) - 1/2*log(-sqrt(e^(4*x) - 1) + e^(2*
x))

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maple [A]  time = 0.10, size = 43, normalized size = 0.75 \[ -\frac {\sqrt {\tanh ^{3}\relax (x )}\, \left (4 \left (\sqrt {\tanh }\relax (x )\right )+\ln \left (\sqrt {\tanh }\relax (x )-1\right )-\ln \left (\sqrt {\tanh }\relax (x )+1\right )-2 \arctan \left (\sqrt {\tanh }\relax (x )\right )\right )}{2 \tanh \relax (x )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tanh(x)^3)^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\tanh \relax (x)^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((tanh(x)^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(tanh(x)^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \sqrt {{\mathrm {tanh}\relax (x)}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tanh(x)^3)^(1/2),x)

[Out]

int((tanh(x)^3)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\tanh ^{3}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((tanh(x)**3)**(1/2),x)

[Out]

Integral(sqrt(tanh(x)**3), x)

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