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

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Rubi [A]  time = 0.51, antiderivative size = 75, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.06, 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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fricas [A]  time = 0.54, size = 54, normalized size = 0.72 \[ \frac {3 \, a x^{2} - 2 \, {\left (a^{2} x^{3} - x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}}}{6 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}}\,{d x} \]

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)

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maple [C]  time = 0.32, size = 269, normalized size = 3.59 \[ -\frac {\left (a x -1\right ) \left (3 a^{6} x^{6} \left (\frac {a x +1}{a x}\right )^{\frac {5}{2}} \left (-\frac {a x -1}{a x}\right )^{\frac {3}{2}}+3 x^{4} \ln \left (a^{2} x^{2}\right ) \left (\frac {a x +1}{a x}\right )^{\frac {5}{2}} \left (-\frac {a x -1}{a x}\right )^{\frac {3}{2}} a^{4}+3 \ln \left (a^{2} x^{2}\right ) \sqrt {-\frac {a x -1}{a x}}\, \left (\frac {a x +1}{a x}\right )^{\frac {5}{2}} x^{4} a^{4}-2 x^{7} a^{7}-3 x^{3} \ln \left (a^{2} x^{2}\right ) \left (\frac {a x +1}{a x}\right )^{\frac {5}{2}} \sqrt {-\frac {a x -1}{a x}}\, a^{3}-2 x^{6} a^{6}+6 x^{5} a^{5}+6 x^{4} a^{4}-6 x^{3} a^{3}-6 a^{2} x^{2}+2 a x +2\right )}{6 x^{5} a^{8} \left (\frac {a x +1}{a x}\right )^{\frac {5}{2}} \left (-\frac {a x -1}{a x}\right )^{\frac {5}{2}}} \]

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*ln(a^2*x^2)*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(5/2)*x^4*a^4-2*x^7*a^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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}}\,{d x} \]

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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mupad [B]  time = 2.06, size = 57, normalized size = 0.76 \[ \frac {x^2}{2\,a}+\frac {\sqrt {\frac {1}{a\,x}-1}\,\left (\frac {x}{3\,a^2}+\frac {1}{3\,a^3}-\frac {x^3}{3}-\frac {x^2}{3\,a}\right )}{\sqrt {\frac {1}{a\,x}+1}} \]

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]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \int \frac {x^{3}}{a x \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}} + 1}\, dx \]

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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