\(\int e^{\text {sech}^{-1}(a x)} x^2 \, dx\) [34]

Optimal result

Integrand size = 10, antiderivative size = 38 \[ \int e^{\text {sech}^{-1}(a x)} x^2 \, dx=-\frac {e^{\text {sech}^{-1}(a x)} x}{3 a^2}+\frac {x^2}{6 a}+\frac {1}{3} e^{\text {sech}^{-1}(a x)} x^3 \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.37, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6470, 30, 75} \[ \int e^{\text {sech}^{-1}(a x)} x^2 \, dx=-\frac {\sqrt {1-a x}}{3 a^3 \sqrt {\frac {1}{a x+1}}}+\frac {1}{3} x^3 e^{\text {sech}^{-1}(a x)}+\frac {x^2}{6 a} \]

[In]

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

[Out]

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

Rule 30

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

Rule 75

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 6470

Int[E^ArcSech[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*(E^ArcSech[a*x^p]/(m + 1)), x] + (Dist
[p/(a*(m + 1)), Int[x^(m - p), x], x] + Dist[p*(Sqrt[1 + a*x^p]/(a*(m + 1)))*Sqrt[1/(1 + a*x^p)], Int[x^(m - p
)/(Sqrt[1 + a*x^p]*Sqrt[1 - a*x^p]), x], x]) /; FreeQ[{a, m, p}, x] && NeQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int e^{\text {sech}^{-1}(a x)} x^2 \, dx=\frac {3 a^2 x^2+2 (-1+a x) \sqrt {\frac {1-a x}{1+a x}} (1+a x)^2}{6 a^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.42

method	result	size
default	\(\frac {\sqrt {\frac {a x +1}{a x}}\, x \sqrt {-\frac {a x -1}{a x}}\, \left (a^{2} x^{2}-1\right )}{3 a^{2}}+\frac {x^{2}}{2 a}\)	\(54\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.42 \[ \int e^{\text {sech}^{-1}(a x)} x^2 \, dx=\frac {3 \, a x^{2} + 2 \, {\left (a^{2} x^{3} - x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}}}{6 \, a^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int e^{\text {sech}^{-1}(a x)} x^2 \, dx=\frac {x^{2}}{2 \, a} + \frac {{\left (a^{2} x^{2} - 1\right )} \sqrt {a x + 1} \sqrt {-a x + 1}}{3 \, a^{3}} \]

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

integrate((1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(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:Unable to divide, perhaps due to rounding error%%%{1,[0,1,2,2,0,0]%%%}+%%%{1,[0,0,0,1,1,1]%%%} / %%%{1,[0,0
,2,3,0,0]%%

Mupad [B] (verification not implemented)

Time = 5.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.45 \[ \int e^{\text {sech}^{-1}(a x)} x^2 \, dx=\sqrt {\frac {1}{a\,x}-1}\,\left (\frac {x^3\,\sqrt {\frac {1}{a\,x}+1}}{3}-\frac {x\,\sqrt {\frac {1}{a\,x}+1}}{3\,a^2}\right )+\frac {x^2}{2\,a} \]

[In]

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

[Out]

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