3.114 \(\int x \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx\)

Optimal. Leaf size=38 \[ \frac {2 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2} \]

[Out]

2/3*x*arctanh(tanh(b*x+a))^(3/2)/b-4/15*arctanh(tanh(b*x+a))^(5/2)/b^2

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Rubi [A]  time = 0.01, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2168, 2157, 30} \[ \frac {2 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[ArcTanh[Tanh[a + b*x]]],x]

[Out]

(2*x*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b) - (4*ArcTanh[Tanh[a + b*x]]^(5/2))/(15*b^2)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps

\begin {align*} \int x \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac {2 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {2 \int \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{3 b}\\ &=\frac {2 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {2 \operatorname {Subst}\left (\int x^{3/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{3 b^2}\\ &=\frac {2 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 32, normalized size = 0.84 \[ \frac {2 \left (5 b x-2 \tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}{15 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[ArcTanh[Tanh[a + b*x]]],x]

[Out]

(2*(5*b*x - 2*ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2))/(15*b^2)

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fricas [A]  time = 0.51, size = 30, normalized size = 0.79 \[ \frac {2 \, {\left (3 \, b^{2} x^{2} + a b x - 2 \, a^{2}\right )} \sqrt {b x + a}}{15 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*b^2*x^2 + a*b*x - 2*a^2)*sqrt(b*x + a)/b^2

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giac [B]  time = 0.18, size = 75, normalized size = 1.97 \[ \frac {\sqrt {2} {\left (\frac {5 \, \sqrt {2} {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a}{b} + \frac {\sqrt {2} {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )}}{b}\right )}}{15 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*sqrt(2)*(5*sqrt(2)*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a/b + sqrt(2)*(3*(b*x + a)^(5/2) - 10*(b*x + a)^
(3/2)*a + 15*sqrt(b*x + a)*a^2)/b)/b

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maple [A]  time = 0.15, size = 42, normalized size = 1.11 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctanh(tanh(b*x+a))^(1/2),x)

[Out]

2/b^2*(1/5*arctanh(tanh(b*x+a))^(5/2)+1/3*(b*x-arctanh(tanh(b*x+a)))*arctanh(tanh(b*x+a))^(3/2))

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maxima [A]  time = 0.52, size = 30, normalized size = 0.79 \[ \frac {2 \, {\left (3 \, b^{2} x^{2} + a b x - 2 \, a^{2}\right )} \sqrt {b x + a}}{15 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*b^2*x^2 + a*b*x - 2*a^2)*sqrt(b*x + a)/b^2

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mupad [B]  time = 1.04, size = 151, normalized size = 3.97 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+5\,b\,x\right )}{15\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atanh(tanh(a + b*x))^(1/2),x)

[Out]

((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*(log((2
*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)))*(log(2/(exp(2*a)*exp(2*b*
x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 5*b*x))/(15*b^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atanh(tanh(b*x+a))**(1/2),x)

[Out]

Integral(x*sqrt(atanh(tanh(a + b*x))), x)

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