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

Optimal. Leaf size=75 \[ -\frac{\sqrt{\frac{1-a x}{a x+1}} (a x+1)^3}{3 a^3}+\frac{\left (4 \sqrt{\frac{1-a x}{a x+1}}+3\right ) (a x+1)^2}{6 a^3}-\frac{x}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.513254, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {6337, 1804, 1814, 12, 261} \[ -\frac{\sqrt{\frac{1-a x}{a x+1}} (a x+1)^3}{3 a^3}+\frac{\left (4 \sqrt{\frac{1-a x}{a x+1}}+3\right ) (a x+1)^2}{6 a^3}-\frac{x}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6337

Int[E^(ArcSech[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(1/u + Sqrt[(1 - u)/(1 + u)] + (1*Sqrt[(1 - u)/(1 +
u)])/u)^n, x] /; FreeQ[m, x] && IntegerQ[n]

Rule 1804

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x
^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x
], x, 1]}, Simp[((c*x)^m*(a + b*x^2)^(p + 1)*(a*g - b*f*x))/(2*a*b*(p + 1)), x] + Dist[c/(2*a*b*(p + 1)), Int[
(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 261

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

Rubi steps

\begin{align*} \int e^{-\text{sech}^{-1}(a x)} x^2 \, dx &=\int \frac{x^2}{\frac{1}{a x}+\sqrt{\frac{1-a x}{1+a x}}+\frac{\sqrt{\frac{1-a x}{1+a x}}}{a x}} \, dx\\ &=\frac{4 \operatorname{Subst}\left (\int \frac{(-1+x)^3 x (1+x)}{\left (1+x^2\right )^4} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{a^3}\\ &=-\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^3}{3 a^3}-\frac{2 \operatorname{Subst}\left (\int \frac{-4+6 x+12 x^2-6 x^3}{\left (1+x^2\right )^3} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{3 a^3}\\ &=-\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^3}{3 a^3}+\frac{(1+a x)^2 \left (3+4 \sqrt{\frac{1-a x}{1+a x}}\right )}{6 a^3}+\frac{\operatorname{Subst}\left (\int \frac{24 x}{\left (1+x^2\right )^2} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{6 a^3}\\ &=-\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^3}{3 a^3}+\frac{(1+a x)^2 \left (3+4 \sqrt{\frac{1-a x}{1+a x}}\right )}{6 a^3}+\frac{4 \operatorname{Subst}\left (\int \frac{x}{\left (1+x^2\right )^2} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{a^3}\\ &=-\frac{x}{a^2}-\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^3}{3 a^3}+\frac{(1+a x)^2 \left (3+4 \sqrt{\frac{1-a x}{1+a x}}\right )}{6 a^3}\\ \end{align*}

Mathematica [A]  time = 0.055459, size = 48, normalized size = 0.64 \[ \frac{3 a^2 x^2-2 (a x-1) \sqrt{\frac{1-a x}{a x+1}} (a x+1)^2}{6 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/E^ArcSech[a*x],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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Maple [C]  time = 0.496, size = 269, normalized size = 3.6 \begin{align*} -{\frac{ax-1}{6\,{x}^{5}{a}^{8}} \left ( 3\,{a}^{6}{x}^{6} \left ({\frac{ax+1}{ax}} \right ) ^{5/2} \left ( -{\frac{ax-1}{ax}} \right ) ^{3/2}+3\,{x}^{4}\ln \left ({a}^{2}{x}^{2} \right ) \left ({\frac{ax+1}{ax}} \right ) ^{5/2} \left ( -{\frac{ax-1}{ax}} \right ) ^{3/2}{a}^{4}+3\,\sqrt{-{\frac{ax-1}{ax}}} \left ({\frac{ax+1}{ax}} \right ) ^{5/2}\ln \left ({a}^{2}{x}^{2} \right ){x}^{4}{a}^{4}-2\,{a}^{7}{x}^{7}-3\,{x}^{3}\ln \left ({a}^{2}{x}^{2} \right ) \left ({\frac{ax+1}{ax}} \right ) ^{5/2}\sqrt{-{\frac{ax-1}{ax}}}{a}^{3}-2\,{x}^{6}{a}^{6}+6\,{x}^{5}{a}^{5}+6\,{x}^{4}{a}^{4}-6\,{x}^{3}{a}^{3}-6\,{a}^{2}{x}^{2}+2\,ax+2 \right ) \left ({\frac{ax+1}{ax}} \right ) ^{-{\frac{5}{2}}} \left ( -{\frac{ax-1}{ax}} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(a*x-1)/x^5*(3*a^6*x^6*((a*x+1)/a/x)^(5/2)*(-(a*x-1)/a/x)^(3/2)+3*x^4*ln(a^2*x^2)*((a*x+1)/a/x)^(5/2)*(-(
a*x-1)/a/x)^(3/2)*a^4+3*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(5/2)*ln(a^2*x^2)*x^4*a^4-2*a^7*x^7-3*x^3*ln(a^2*x^
2)*((a*x+1)/a/x)^(5/2)*(-(a*x-1)/a/x)^(1/2)*a^3-2*x^6*a^6+6*x^5*a^5+6*x^4*a^4-6*x^3*a^3-6*a^2*x^2+2*a*x+2)/a^8
/((a*x+1)/a/x)^(5/2)/(-(a*x-1)/a/x)^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{\frac{1}{a x} + 1} \sqrt{\frac{1}{a x} - 1} + \frac{1}{a x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.91037, size = 111, normalized size = 1.48 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)),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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \int \frac{x^{3}}{a x \sqrt{-1 + \frac{1}{a x}} \sqrt{1 + \frac{1}{a x}} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Integral(x**3/(a*x*sqrt(-1 + 1/(a*x))*sqrt(1 + 1/(a*x)) + 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{\frac{1}{a x} + 1} \sqrt{\frac{1}{a x} - 1} + \frac{1}{a x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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