\(\int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))} \, dx\) [160]

Optimal result

Integrand size = 13, antiderivative size = 56 \[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))} \, dx=\frac {x^2}{2 b}+\frac {x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^2}+\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2190, 2189, 2188, 29} \[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))} \, dx=\frac {\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac {x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^2}+\frac {x^2}{2 b} \]

[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 29

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

Rule 2188

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 2189

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 2190

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]

Rubi steps \begin{align*} \text {integral}& = \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 \text {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] (verified)

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.98 \[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))} \, dx=\frac {x^2}{2 b}-\frac {x \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}{b^2}+\frac {\left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^3} \]

[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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.37 (sec) , antiderivative size = 28786, normalized size of antiderivative = 514.04

method	result	size
risch	\(\text {Expression too large to display}\)	\(28786\)

[In]

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

[Out]

result too large to display

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.91 \[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))} \, dx=\frac {2 \, b^{2} x^{2} - 2 i \, \pi b x - 4 \, a b x - {\left (\pi ^{2} - 4 i \, \pi a - 4 \, a^{2}\right )} \log \left (i \, \pi + 2 \, b x + 2 \, a\right )}{4 \, b^{3}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))} \, dx=\int \frac {x^{2}}{\operatorname {acoth}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.37 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.91 \[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))} \, dx=\frac {b x^{2} + {\left (i \, \pi - 2 \, a\right )} x}{2 \, b^{2}} - \frac {{\left (\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}\right )} \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{4 \, b^{3}} \]

[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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.89 \[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))} \, dx=\frac {x^{2}}{2 \, b} - \frac {{\left (i \, \pi + 2 \, a\right )} x}{2 \, b^{2}} - \frac {{\left (\pi ^{2} - 4 i \, \pi a - 4 \, a^{2}\right )} \log \left (\pi - 2 i \, b x - 2 i \, a\right )}{4 \, b^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 234, normalized size of antiderivative = 4.18 \[ \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))} \, dx=\frac {x^2}{2\,b}+\frac {x\,\left (\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}{2\,b^2}+\frac {\ln \left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\right )\,\left ({\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^2-4\,a\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )+4\,a^2\right )}{4\,b^3} \]

[In]

int(x^2/acoth(tanh(a + b*x)),x)

[Out]

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