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

Optimal. Leaf size=36 \[ \frac {2 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2} \]

[Out]

-4/3*arctanh(tanh(b*x+a))^(3/2)/b^2+2*x*arctanh(tanh(b*x+a))^(1/2)/b

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Rubi [A]  time = 0.01, antiderivative size = 36, 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} \[ \frac {2 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[ArcTanh[Tanh[a + b*x]]],x]

[Out]

(2*x*Sqrt[ArcTanh[Tanh[a + b*x]]])/b - (4*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*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 \frac {x}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx &=\frac {2 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {2 \int \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac {2 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {2 \operatorname {Subst}\left (\int \sqrt {x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^2}\\ &=\frac {2 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 32, normalized size = 0.89 \[ \frac {2 \left (3 b x-2 \tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 19, normalized size = 0.53 \[ \frac {2 \, \sqrt {b x + a} {\left (b x - 2 \, a\right )}}{3 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(b*x + a)*(b*x - 2*a)/b^2

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giac [A]  time = 0.39, size = 23, normalized size = 0.64 \[ \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )}}{3 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.17, size = 56, normalized size = 1.56 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}-2 a \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}-2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b^{2}} \]

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

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maxima [A]  time = 0.54, size = 30, normalized size = 0.83 \[ \frac {2 \, {\left (b^{2} x^{2} - a b x - 2 \, a^{2}\right )}}{3 \, \sqrt {b x + a} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.21, size = 105, normalized size = 2.92 \[ \frac {2\,\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+3\,b\,x\right )}{3\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/atanh(tanh(a + b*x))^(1/2),x)

[Out]

(2*(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) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 3*b*x))/(3*b^2)

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

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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