3.129 \(\int \tanh ^2(x) (1+\tanh (x))^{3/2} \, dx\)

Optimal. Leaf size=45 \[ 2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )-\frac {2}{5} (\tanh (x)+1)^{5/2}-2 \sqrt {\tanh (x)+1} \]

[Out]

2*arctanh(1/2*(1+tanh(x))^(1/2)*2^(1/2))*2^(1/2)-2*(1+tanh(x))^(1/2)-2/5*(1+tanh(x))^(5/2)

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Rubi [A]  time = 0.06, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3543, 3478, 3480, 206} \[ 2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )-\frac {2}{5} (\tanh (x)+1)^{5/2}-2 \sqrt {\tanh (x)+1} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^2*(1 + Tanh[x])^(3/2),x]

[Out]

2*Sqrt[2]*ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]] - 2*Sqrt[1 + Tanh[x]] - (2*(1 + Tanh[x])^(5/2))/5

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 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[2*a, Int[(a + b*Tan[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && G
tQ[n, 1]

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 3543

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

Rubi steps

\begin {align*} \int \tanh ^2(x) (1+\tanh (x))^{3/2} \, dx &=-\frac {2}{5} (1+\tanh (x))^{5/2}+\int (1+\tanh (x))^{3/2} \, dx\\ &=-2 \sqrt {1+\tanh (x)}-\frac {2}{5} (1+\tanh (x))^{5/2}+2 \int \sqrt {1+\tanh (x)} \, dx\\ &=-2 \sqrt {1+\tanh (x)}-\frac {2}{5} (1+\tanh (x))^{5/2}+4 \operatorname {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\sqrt {1+\tanh (x)}\right )\\ &=2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )-2 \sqrt {1+\tanh (x)}-\frac {2}{5} (1+\tanh (x))^{5/2}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 53, normalized size = 1.18 \[ 2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )-\frac {1}{5} \sqrt {\tanh (x)+1} \text {sech}^2(x) (2 \sinh (2 x)+7 \cosh (2 x)+5) \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^2*(1 + Tanh[x])^(3/2),x]

[Out]

2*Sqrt[2]*ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]] - (Sech[x]^2*(5 + 7*Cosh[2*x] + 2*Sinh[2*x])*Sqrt[1 + Tanh[x]])/5

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fricas [B]  time = 0.69, size = 429, normalized size = 9.53 \[ -\frac {2 \, \sqrt {2} {\left (9 \, \sqrt {2} \cosh \relax (x)^{5} + 45 \, \sqrt {2} \cosh \relax (x) \sinh \relax (x)^{4} + 9 \, \sqrt {2} \sinh \relax (x)^{5} + 10 \, {\left (9 \, \sqrt {2} \cosh \relax (x)^{2} + \sqrt {2}\right )} \sinh \relax (x)^{3} + 10 \, \sqrt {2} \cosh \relax (x)^{3} + 30 \, {\left (3 \, \sqrt {2} \cosh \relax (x)^{3} + \sqrt {2} \cosh \relax (x)\right )} \sinh \relax (x)^{2} + 5 \, {\left (9 \, \sqrt {2} \cosh \relax (x)^{4} + 6 \, \sqrt {2} \cosh \relax (x)^{2} + \sqrt {2}\right )} \sinh \relax (x) + 5 \, \sqrt {2} \cosh \relax (x)\right )} \sqrt {\frac {\cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}} - 5 \, {\left (\sqrt {2} \cosh \relax (x)^{6} + 6 \, \sqrt {2} \cosh \relax (x) \sinh \relax (x)^{5} + \sqrt {2} \sinh \relax (x)^{6} + 3 \, {\left (5 \, \sqrt {2} \cosh \relax (x)^{2} + \sqrt {2}\right )} \sinh \relax (x)^{4} + 3 \, \sqrt {2} \cosh \relax (x)^{4} + 4 \, {\left (5 \, \sqrt {2} \cosh \relax (x)^{3} + 3 \, \sqrt {2} \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 3 \, {\left (5 \, \sqrt {2} \cosh \relax (x)^{4} + 6 \, \sqrt {2} \cosh \relax (x)^{2} + \sqrt {2}\right )} \sinh \relax (x)^{2} + 3 \, \sqrt {2} \cosh \relax (x)^{2} + 6 \, {\left (\sqrt {2} \cosh \relax (x)^{5} + 2 \, \sqrt {2} \cosh \relax (x)^{3} + \sqrt {2} \cosh \relax (x)\right )} \sinh \relax (x) + \sqrt {2}\right )} \log \left (-2 \, \sqrt {2} \sqrt {\frac {\cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} - 2 \, \cosh \relax (x)^{2} - 4 \, \cosh \relax (x) \sinh \relax (x) - 2 \, \sinh \relax (x)^{2} - 1\right )}{5 \, {\left (\cosh \relax (x)^{6} + 6 \, \cosh \relax (x) \sinh \relax (x)^{5} + \sinh \relax (x)^{6} + 3 \, {\left (5 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{4} + 3 \, \cosh \relax (x)^{4} + 4 \, {\left (5 \, \cosh \relax (x)^{3} + 3 \, \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 3 \, {\left (5 \, \cosh \relax (x)^{4} + 6 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{2} + 3 \, \cosh \relax (x)^{2} + 6 \, {\left (\cosh \relax (x)^{5} + 2 \, \cosh \relax (x)^{3} + \cosh \relax (x)\right )} \sinh \relax (x) + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(2*sqrt(2)*(9*sqrt(2)*cosh(x)^5 + 45*sqrt(2)*cosh(x)*sinh(x)^4 + 9*sqrt(2)*sinh(x)^5 + 10*(9*sqrt(2)*cosh
(x)^2 + sqrt(2))*sinh(x)^3 + 10*sqrt(2)*cosh(x)^3 + 30*(3*sqrt(2)*cosh(x)^3 + sqrt(2)*cosh(x))*sinh(x)^2 + 5*(
9*sqrt(2)*cosh(x)^4 + 6*sqrt(2)*cosh(x)^2 + sqrt(2))*sinh(x) + 5*sqrt(2)*cosh(x))*sqrt(cosh(x)/(cosh(x) - sinh
(x))) - 5*(sqrt(2)*cosh(x)^6 + 6*sqrt(2)*cosh(x)*sinh(x)^5 + sqrt(2)*sinh(x)^6 + 3*(5*sqrt(2)*cosh(x)^2 + sqrt
(2))*sinh(x)^4 + 3*sqrt(2)*cosh(x)^4 + 4*(5*sqrt(2)*cosh(x)^3 + 3*sqrt(2)*cosh(x))*sinh(x)^3 + 3*(5*sqrt(2)*co
sh(x)^4 + 6*sqrt(2)*cosh(x)^2 + sqrt(2))*sinh(x)^2 + 3*sqrt(2)*cosh(x)^2 + 6*(sqrt(2)*cosh(x)^5 + 2*sqrt(2)*co
sh(x)^3 + sqrt(2)*cosh(x))*sinh(x) + sqrt(2))*log(-2*sqrt(2)*sqrt(cosh(x)/(cosh(x) - sinh(x)))*(cosh(x) + sinh
(x)) - 2*cosh(x)^2 - 4*cosh(x)*sinh(x) - 2*sinh(x)^2 - 1))/(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + 3*(5
*cosh(x)^2 + 1)*sinh(x)^4 + 3*cosh(x)^4 + 4*(5*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + 3*(5*cosh(x)^4 + 6*cosh(x)^2
+ 1)*sinh(x)^2 + 3*cosh(x)^2 + 6*(cosh(x)^5 + 2*cosh(x)^3 + cosh(x))*sinh(x) + 1)

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giac [B]  time = 0.16, size = 140, normalized size = 3.11 \[ \frac {1}{5} \, \sqrt {2} {\left (\frac {2 \, {\left (25 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{4} - 60 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3} + 70 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{2} - 40 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 40 \, e^{\left (2 \, x\right )} + 9\right )}}{{\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )} - 1\right )}^{5}} - 5 \, \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*sqrt(2)*(2*(25*(sqrt(e^(4*x) + e^(2*x)) - e^(2*x))^4 - 60*(sqrt(e^(4*x) + e^(2*x)) - e^(2*x))^3 + 70*(sqrt
(e^(4*x) + e^(2*x)) - e^(2*x))^2 - 40*sqrt(e^(4*x) + e^(2*x)) + 40*e^(2*x) + 9)/(sqrt(e^(4*x) + e^(2*x)) - e^(
2*x) - 1)^5 - 5*log(-2*sqrt(e^(4*x) + e^(2*x)) + 2*e^(2*x) + 1))

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maple [A]  time = 0.06, size = 35, normalized size = 0.78 \[ 2 \arctanh \left (\frac {\sqrt {1+\tanh \relax (x )}\, \sqrt {2}}{2}\right ) \sqrt {2}-2 \sqrt {1+\tanh \relax (x )}-\frac {2 \left (1+\tanh \relax (x )\right )^{\frac {5}{2}}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctanh(1/2*(1+tanh(x))^(1/2)*2^(1/2))*2^(1/2)-2*(1+tanh(x))^(1/2)-2/5*(1+tanh(x))^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (\tanh \relax (x) + 1\right )}^{\frac {3}{2}} \tanh \relax (x)^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.11, size = 34, normalized size = 0.76 \[ 2\,\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\relax (x)+1}}{2}\right )-2\,\sqrt {\mathrm {tanh}\relax (x)+1}-\frac {2\,{\left (\mathrm {tanh}\relax (x)+1\right )}^{5/2}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^2*(tanh(x) + 1)^(3/2),x)

[Out]

2*2^(1/2)*atanh((2^(1/2)*(tanh(x) + 1)^(1/2))/2) - 2*(tanh(x) + 1)^(1/2) - (2*(tanh(x) + 1)^(5/2))/5

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sympy [A]  time = 16.13, size = 82, normalized size = 1.82 \[ - \frac {2 \left (\tanh {\relax (x )} + 1\right )^{\frac {5}{2}}}{5} - 2 \sqrt {\tanh {\relax (x )} + 1} - 4 \left (\begin {cases} - \frac {\sqrt {2} \operatorname {acoth}{\left (\frac {\sqrt {2} \sqrt {\tanh {\relax (x )} + 1}}{2} \right )}}{2} & \text {for}\: \tanh {\relax (x )} + 1 > 2 \\- \frac {\sqrt {2} \operatorname {atanh}{\left (\frac {\sqrt {2} \sqrt {\tanh {\relax (x )} + 1}}{2} \right )}}{2} & \text {for}\: \tanh {\relax (x )} + 1 < 2 \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(tanh(x) + 1)**(5/2)/5 - 2*sqrt(tanh(x) + 1) - 4*Piecewise((-sqrt(2)*acoth(sqrt(2)*sqrt(tanh(x) + 1)/2)/2,
tanh(x) + 1 > 2), (-sqrt(2)*atanh(sqrt(2)*sqrt(tanh(x) + 1)/2)/2, tanh(x) + 1 < 2))

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