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

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Rubi [A]  time = 0.0581242, antiderivative size = 45, normalized size of antiderivative = 1., 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 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])

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 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])

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*}

Mathematica [A]  time = 0.197637, 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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Maple [A]  time = 0.019, size = 35, normalized size = 0.8 \begin{align*} 2\,{\it Artanh} \left ( 1/2\,\sqrt{1+\tanh \left ( x \right ) }\sqrt{2} \right ) \sqrt{2}-2\,\sqrt{1+\tanh \left ( x \right ) }-{\frac{2}{5} \left ( 1+\tanh \left ( x \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int{\left (\tanh \left (x\right ) + 1\right )}^{\frac{3}{2}} \tanh \left (x\right )^{2}\,{d x} \end{align*}

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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Fricas [B]  time = 2.34753, size = 1450, normalized size = 32.22 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [A]  time = 17.3534, size = 82, normalized size = 1.82 \begin{align*} - \frac{2 \left (\tanh{\left (x \right )} + 1\right )^{\frac{5}{2}}}{5} - 2 \sqrt{\tanh{\left (x \right )} + 1} - 4 \left (\begin{cases} - \frac{\sqrt{2} \operatorname{acoth}{\left (\frac{\sqrt{2} \sqrt{\tanh{\left (x \right )} + 1}}{2} \right )}}{2} & \text{for}\: \tanh{\left (x \right )} + 1 > 2 \\- \frac{\sqrt{2} \operatorname{atanh}{\left (\frac{\sqrt{2} \sqrt{\tanh{\left (x \right )} + 1}}{2} \right )}}{2} & \text{for}\: \tanh{\left (x \right )} + 1 < 2 \end{cases}\right ) \end{align*}

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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Giac [B]  time = 1.29067, size = 189, normalized size = 4.2 \begin{align*} \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 )} \end{align*}

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))