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

Optimal. Leaf size=38 \[ -\frac {x e^{\text {sech}^{-1}(a x)}}{3 a^2}+\frac {1}{3} x^3 e^{\text {sech}^{-1}(a x)}+\frac {x^2}{6 a} \]

[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

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Rubi [A]  time = 0.02, antiderivative size = 52, normalized size of antiderivative = 1.37, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6335, 30, 74} \[ -\frac {\sqrt {1-a x}}{3 a^3 \sqrt {\frac {1}{a x+1}}}+\frac {x^2}{6 a}+\frac {1}{3} x^3 e^{\text {sech}^{-1}(a x)} \]

Warning: Unable to verify antiderivative.

[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 74

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 6335

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]*Sqrt[1/(1 + a*x^p)])/(a*(m + 1)), 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*} \int e^{\text {sech}^{-1}(a x)} x^2 \, dx &=\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*}

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Mathematica [A]  time = 0.06, size = 48, normalized size = 1.26 \[ \frac {3 a^2 x^2+2 (a x-1) \sqrt {\frac {1-a x}{a x+1}} (a x+1)^2}{6 a^3} \]

Warning: Unable to verify antiderivative.

[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)

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fricas [A]  time = 0.76, size = 54, normalized size = 1.42 \[ \frac {3 \, a x^{2} + 2 \, {\left (a^{2} x^{3} - x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}}}{6 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[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,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]Warning, choosing root of [1,0,%%%{-4,[1,1]%%%},0,%%%{4,[4,4]%%%}] at parameters values [86,-97]Warning,
choosing root of [1,0,%%%{4,[1,1]%%%},0,%%%{4,[4,4]%%%}] at parameters values [-82,7]Warning, choosing root of
[1,0,%%%{-4,[1,1]%%%},0,%%%{4,[4,4]%%%}] at parameters values [-27,26]Warning, choosing root of [1,0,%%%{4,[1
,1]%%%},0,%%%{4,[4,4]%%%}] at parameters values [-89,63]Unable to divide, perhaps due to rounding error%%%{1,[
1,2,2,0,0]%%%}+%%%{1,[0,0,1,1,1]%%%} / %%%{1,[0,2,3,0,0]%%%} Error: Bad Argument Value

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maple [A]  time = 0.05, size = 54, normalized size = 1.42 \[ \frac {\sqrt {-\frac {a x -1}{a x}}\, x \sqrt {\frac {a x +1}{a x}}\, \left (a^{2} x^{2}-1\right )}{3 a^{2}}+\frac {x^{2}}{2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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maxima [A]  time = 0.40, size = 38, normalized size = 1.00 \[ \frac {x^{2}}{2 \, a} + \frac {{\left (a^{2} x^{2} - 1\right )} \sqrt {a x + 1} \sqrt {-a x + 1}}{3 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.43, size = 55, normalized size = 1.45 \[ \sqrt {\frac {1}{a\,x}-1}\,\left (\frac {x^3\,\sqrt {\frac {1}{a\,x}+1}}{3}-\frac {x\,\sqrt {\frac {1}{a\,x}+1}}{3\,a^2}\right )+\frac {x^2}{2\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int x\, dx + \int a x^{2} \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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