\(\int e^{\coth ^{-1}(a x)} x^2 \, dx\) [2]

Optimal result

Integrand size = 10, antiderivative size = 90 \[ \int e^{\coth ^{-1}(a x)} x^2 \, dx=\frac {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 a}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3+\frac {\text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6304, 849, 821, 272, 65, 214} \[ \int e^{\coth ^{-1}(a x)} x^2 \, dx=\frac {x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{2 a}+\frac {2 x \sqrt {1-\frac {1}{a^2 x^2}}}{3 a^2}+\frac {1}{3} x^3 \sqrt {1-\frac {1}{a^2 x^2}}+\frac {\text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^3} \]

[In]

Int[E^ArcCoth[a*x]*x^2,x]

[Out]

(2*Sqrt[1 - 1/(a^2*x^2)]*x)/(3*a^2) + (Sqrt[1 - 1/(a^2*x^2)]*x^2)/(2*a) + (Sqrt[1 - 1/(a^2*x^2)]*x^3)/3 + ArcT
anh[Sqrt[1 - 1/(a^2*x^2)]]/(2*a^3)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g
))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e
^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0
] && EqQ[Simplify[m + 2*p + 3], 0]

Rule 849

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*
(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/((m + 1)*(c*d^2 + a*e^2))), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 6304

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> -Subst[Int[(1 + x/a)^((n + 1)/2)/(x^(m + 2)*(1 - x/
a)^((n - 1)/2)*Sqrt[1 - x^2/a^2]), x], x, 1/x] /; FreeQ[a, x] && IntegerQ[(n - 1)/2] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {1+\frac {x}{a}}{x^4 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3+\frac {1}{3} \text {Subst}\left (\int \frac {-\frac {3}{a}-\frac {2 x}{a^2}}{x^3 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {\sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 a}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {1}{6} \text {Subst}\left (\int \frac {\frac {4}{a^2}+\frac {3 x}{a^3}}{x^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 a}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{2 a^3} \\ & = \frac {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 a}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{4 a^3} \\ & = \frac {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 a}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3+\frac {\text {Subst}\left (\int \frac {1}{a^2-a^2 x^2} \, dx,x,\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a} \\ & = \frac {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{3 a^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 a}+\frac {1}{3} \sqrt {1-\frac {1}{a^2 x^2}} x^3+\frac {\text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.67 \[ \int e^{\coth ^{-1}(a x)} x^2 \, dx=\frac {a \sqrt {1-\frac {1}{a^2 x^2}} x \left (4+3 a x+2 a^2 x^2\right )+3 \log \left (\left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )}{6 a^3} \]

[In]

Integrate[E^ArcCoth[a*x]*x^2,x]

[Out]

(a*Sqrt[1 - 1/(a^2*x^2)]*x*(4 + 3*a*x + 2*a^2*x^2) + 3*Log[(1 + Sqrt[1 - 1/(a^2*x^2)])*x])/(6*a^3)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.21

method	result	size
risch	\(\frac {\left (2 a^{2} x^{2}+3 a x +4\right ) \left (a x -1\right )}{6 a^{3} \sqrt {\frac {a x -1}{a x +1}}}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right ) \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{2 a^{2} \sqrt {a^{2}}\, \sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right )}\)	\(109\)
default	\(\frac {\left (a x -1\right ) \left (3 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a x +2 \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-3 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a +6 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}+6 a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right )\right )}{6 \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{3} \sqrt {a^{2}}}\)	\(173\)

[In]

int(1/((a*x-1)/(a*x+1))^(1/2)*x^2,x,method=_RETURNVERBOSE)

[Out]

1/6*(2*a^2*x^2+3*a*x+4)*(a*x-1)/a^3/((a*x-1)/(a*x+1))^(1/2)+1/2/a^2*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2-1)^(1/2))/(a
^2)^(1/2)/((a*x-1)/(a*x+1))^(1/2)*((a*x-1)*(a*x+1))^(1/2)/(a*x+1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.93 \[ \int e^{\coth ^{-1}(a x)} x^2 \, dx=\frac {{\left (2 \, a^{3} x^{3} + 5 \, a^{2} x^{2} + 7 \, a x + 4\right )} \sqrt {\frac {a x - 1}{a x + 1}} + 3 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 3 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{6 \, a^{3}} \]

[In]

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

[Out]

1/6*((2*a^3*x^3 + 5*a^2*x^2 + 7*a*x + 4)*sqrt((a*x - 1)/(a*x + 1)) + 3*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 3*
log(sqrt((a*x - 1)/(a*x + 1)) - 1))/a^3

Sympy [F]

\[ \int e^{\coth ^{-1}(a x)} x^2 \, dx=\int \frac {x^{2}}{\sqrt {\frac {a x - 1}{a x + 1}}}\, dx \]

[In]

integrate(1/((a*x-1)/(a*x+1))**(1/2)*x**2,x)

[Out]

Integral(x**2/sqrt((a*x - 1)/(a*x + 1)), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (74) = 148\).

Time = 0.21 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.84 \[ \int e^{\coth ^{-1}(a x)} x^2 \, dx=-\frac {1}{6} \, a {\left (\frac {2 \, {\left (3 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} - 4 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + 9 \, \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {3 \, {\left (a x - 1\right )} a^{4}}{a x + 1} - \frac {3 \, {\left (a x - 1\right )}^{2} a^{4}}{{\left (a x + 1\right )}^{2}} + \frac {{\left (a x - 1\right )}^{3} a^{4}}{{\left (a x + 1\right )}^{3}} - a^{4}} - \frac {3 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{4}} + \frac {3 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{4}}\right )} \]

[In]

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

[Out]

-1/6*a*(2*(3*((a*x - 1)/(a*x + 1))^(5/2) - 4*((a*x - 1)/(a*x + 1))^(3/2) + 9*sqrt((a*x - 1)/(a*x + 1)))/(3*(a*
x - 1)*a^4/(a*x + 1) - 3*(a*x - 1)^2*a^4/(a*x + 1)^2 + (a*x - 1)^3*a^4/(a*x + 1)^3 - a^4) - 3*log(sqrt((a*x -
1)/(a*x + 1)) + 1)/a^4 + 3*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^4)

Giac [F(-2)]

Exception generated. \[ \int e^{\coth ^{-1}(a x)} x^2 \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.48 \[ \int e^{\coth ^{-1}(a x)} x^2 \, dx=\frac {3\,\sqrt {\frac {a\,x-1}{a\,x+1}}-\frac {4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{3}+{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{a^3+\frac {3\,a^3\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {a^3\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {3\,a^3\,\left (a\,x-1\right )}{a\,x+1}}+\frac {\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a^3} \]

[In]

int(x^2/((a*x - 1)/(a*x + 1))^(1/2),x)

[Out]

(3*((a*x - 1)/(a*x + 1))^(1/2) - (4*((a*x - 1)/(a*x + 1))^(3/2))/3 + ((a*x - 1)/(a*x + 1))^(5/2))/(a^3 + (3*a^
3*(a*x - 1)^2)/(a*x + 1)^2 - (a^3*(a*x - 1)^3)/(a*x + 1)^3 - (3*a^3*(a*x - 1))/(a*x + 1)) + atanh(((a*x - 1)/(
a*x + 1))^(1/2))/a^3