3.160 \(\int \frac{x^2}{\coth ^{-1}(\tanh (a+b x))} \, dx\)

Optimal. Leaf size=56 \[ \frac{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^2}+\frac{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac{x^2}{2 b} \]

[Out]

x^2/(2*b) + (x*(b*x - ArcCoth[Tanh[a + b*x]]))/b^2 + ((b*x - ArcCoth[Tanh[a + b*x]])^2*Log[ArcCoth[Tanh[a + b*
x]]])/b^3

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Rubi [A]  time = 0.0346625, antiderivative size = 56, 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 = {2159, 2158, 2157, 29} \[ \frac{x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^2}+\frac{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac{x^2}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2/ArcCoth[Tanh[a + b*x]],x]

[Out]

x^2/(2*b) + (x*(b*x - ArcCoth[Tanh[a + b*x]]))/b^2 + ((b*x - ArcCoth[Tanh[a + b*x]])^2*Log[ArcCoth[Tanh[a + b*
x]]])/b^3

Rule 2159

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^n/(a*n), x] - Dis
t[(b*u - a*v)/a, Int[v^(n - 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && Ne
Q[n, 1]

Rule 2158

Int[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(b*x)/a, x] - Dist[(b*u
- a*v)/a, Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]

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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rubi steps

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

Mathematica [A]  time = 0.0383789, size = 55, normalized size = 0.98 \[ -\frac{x \left (\coth ^{-1}(\tanh (a+b x))-b x\right )}{b^2}+\frac{\left (\coth ^{-1}(\tanh (a+b x))-b x\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac{x^2}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/ArcCoth[Tanh[a + b*x]],x]

[Out]

x^2/(2*b) - (x*(-(b*x) + ArcCoth[Tanh[a + b*x]]))/b^2 + ((-(b*x) + ArcCoth[Tanh[a + b*x]])^2*Log[ArcCoth[Tanh[
a + b*x]]])/b^3

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Maple [C]  time = 1.134, size = 28786, normalized size = 514. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [C]  time = 1.82435, size = 69, normalized size = 1.23 \begin{align*} \frac{b x^{2} +{\left (i \, \pi - 2 \, a\right )} x}{2 \, b^{2}} - \frac{{\left (\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}\right )} \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{4 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2 + (I*pi - 2*a)*x)/b^2 - 1/4*(pi^2 + 4*I*pi*a - 4*a^2)*log(-I*pi + 2*b*x + 2*a)/b^3

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Fricas [A]  time = 1.66004, size = 225, normalized size = 4.02 \begin{align*} \frac{4 \, b^{2} x^{2} - 8 \, a b x - 16 \, \pi a \arctan \left (-\frac{2 \, b x + 2 \, a - \sqrt{4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right ) -{\left (\pi ^{2} - 4 \, a^{2}\right )} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right )}{8 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*b^2*x^2 - 8*a*b*x - 16*pi*a*arctan(-(2*b*x + 2*a - sqrt(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2))/pi) - (pi^
2 - 4*a^2)*log(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a^2))/b^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\operatorname{acoth}{\left (\tanh{\left (a + b x \right )} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/acoth(tanh(a + b*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\operatorname{arcoth}\left (\tanh \left (b x + a\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/arccoth(tanh(b*x + a)), x)