3.66 \(\int e^{2 \text{sech}^{-1}(a x)} x^3 \, dx\)

Optimal. Leaf size=117 \[ \frac{(1-a x) (a x+1)^3}{4 a^4}+\frac{\sqrt{\frac{1-a x}{a x+1}} \left (4-3 \sqrt{\frac{1-a x}{a x+1}}\right ) (a x+1)^3}{6 a^4}+\frac{\left (3-8 \sqrt{\frac{1-a x}{a x+1}}\right ) (a x+1)^2}{6 a^4}-\frac{x}{a^3} \]

[Out]

-(x/a^3) + ((1 - a*x)*(1 + a*x)^3)/(4*a^4) + ((1 + a*x)^2*(3 - 8*Sqrt[(1 - a*x)/(1 + a*x)]))/(6*a^4) + (Sqrt[(
1 - a*x)/(1 + a*x)]*(1 + a*x)^3*(4 - 3*Sqrt[(1 - a*x)/(1 + a*x)]))/(6*a^4)

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Rubi [A]  time = 0.546433, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6337, 1804, 1811, 1814, 12, 261} \[ \frac{(1-a x) (a x+1)^3}{4 a^4}+\frac{\sqrt{\frac{1-a x}{a x+1}} \left (4-3 \sqrt{\frac{1-a x}{a x+1}}\right ) (a x+1)^3}{6 a^4}+\frac{\left (3-8 \sqrt{\frac{1-a x}{a x+1}}\right ) (a x+1)^2}{6 a^4}-\frac{x}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcSech[a*x])*x^3,x]

[Out]

-(x/a^3) + ((1 - a*x)*(1 + a*x)^3)/(4*a^4) + ((1 + a*x)^2*(3 - 8*Sqrt[(1 - a*x)/(1 + a*x)]))/(6*a^4) + (Sqrt[(
1 - a*x)/(1 + a*x)]*(1 + a*x)^3*(4 - 3*Sqrt[(1 - a*x)/(1 + a*x)]))/(6*a^4)

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 1811

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[x*PolynomialQuotient[Pq, x, x]*(a + b*x^2)^p, x] /; Fre
eQ[{a, b, p}, x] && PolyQ[Pq, x] && EqQ[Coeff[Pq, x, 0], 0] &&  !MatchQ[Pq, x^(m_.)*(u_.) /; IntegerQ[m]]

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^{2 \text{sech}^{-1}(a x)} x^3 \, dx &=\int x^3 \left (\frac{1}{a x}+\sqrt{\frac{1-a x}{1+a x}}+\frac{\sqrt{\frac{1-a x}{1+a x}}}{a x}\right )^2 \, dx\\ &=\frac{4 \operatorname{Subst}\left (\int \frac{(-1+x) x (1+x)^5}{\left (1+x^2\right )^5} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{a^4}\\ &=\frac{(1-a x) (1+a x)^3}{4 a^4}-\frac{\operatorname{Subst}\left (\int \frac{24 x+32 x^2-32 x^3-32 x^4-8 x^5}{\left (1+x^2\right )^4} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{2 a^4}\\ &=\frac{(1-a x) (1+a x)^3}{4 a^4}-\frac{\operatorname{Subst}\left (\int \frac{x \left (24+32 x-32 x^2-32 x^3-8 x^4\right )}{\left (1+x^2\right )^4} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{2 a^4}\\ &=\frac{(1-a x) (1+a x)^3}{4 a^4}+\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^3 \left (4-3 \sqrt{\frac{1-a x}{1+a x}}\right )}{6 a^4}+\frac{\operatorname{Subst}\left (\int \frac{-64-48 x+192 x^2+48 x^3}{\left (1+x^2\right )^3} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{12 a^4}\\ &=\frac{(1-a x) (1+a x)^3}{4 a^4}+\frac{(1+a x)^2 \left (3-8 \sqrt{\frac{1-a x}{1+a x}}\right )}{6 a^4}+\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^3 \left (4-3 \sqrt{\frac{1-a x}{1+a x}}\right )}{6 a^4}-\frac{\operatorname{Subst}\left (\int -\frac{192 x}{\left (1+x^2\right )^2} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{48 a^4}\\ &=\frac{(1-a x) (1+a x)^3}{4 a^4}+\frac{(1+a x)^2 \left (3-8 \sqrt{\frac{1-a x}{1+a x}}\right )}{6 a^4}+\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^3 \left (4-3 \sqrt{\frac{1-a x}{1+a x}}\right )}{6 a^4}+\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^4}\\ &=-\frac{x}{a^3}+\frac{(1-a x) (1+a x)^3}{4 a^4}+\frac{(1+a x)^2 \left (3-8 \sqrt{\frac{1-a x}{1+a x}}\right )}{6 a^4}+\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^3 \left (4-3 \sqrt{\frac{1-a x}{1+a x}}\right )}{6 a^4}\\ \end{align*}

Mathematica [A]  time = 0.0679191, size = 52, normalized size = 0.44 \[ \frac{x^2}{a^2}+\frac{2 (a x-1) \sqrt{\frac{1-a x}{a x+1}} (a x+1)^2}{3 a^4}-\frac{x^4}{4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcSech[a*x])*x^3,x]

[Out]

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

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Maple [A]  time = 0.181, size = 72, normalized size = 0.6 \begin{align*}{\frac{1}{{a}^{2}} \left ( -{\frac{{a}^{2}{x}^{4}}{4}}+{\frac{{x}^{2}}{2}} \right ) }+{\frac{2\,x \left ({a}^{2}{x}^{2}-1 \right ) }{3\,{a}^{3}}\sqrt{-{\frac{ax-1}{ax}}}\sqrt{{\frac{ax+1}{ax}}}}+{\frac{{x}^{2}}{2\,{a}^{2}}} \end{align*}

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))^2*x^3,x)

[Out]

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

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Maxima [A]  time = 1.04463, size = 57, normalized size = 0.49 \begin{align*} -\frac{1}{4} \, x^{4} + \frac{x^{2}}{a^{2}} + \frac{2 \,{\left (a^{2} x^{2} - 1\right )} \sqrt{a x + 1} \sqrt{-a x + 1}}{3 \, a^{4}} \end{align*}

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))^2*x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.9465, size = 131, normalized size = 1.12 \begin{align*} -\frac{3 \, a^{3} x^{4} - 12 \, a x^{2} - 8 \,{\left (a^{2} x^{3} - x\right )} \sqrt{\frac{a x + 1}{a x}} \sqrt{-\frac{a x - 1}{a x}}}{12 \, a^{3}} \end{align*}

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))^2*x^3,x, algorithm="fricas")

[Out]

-1/12*(3*a^3*x^4 - 12*a*x^2 - 8*(a^2*x^3 - x)*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)))/a^3

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

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))**2*x**3,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3}{\left (\sqrt{\frac{1}{a x} + 1} \sqrt{\frac{1}{a x} - 1} + \frac{1}{a x}\right )}^{2}\,{d x} \end{align*}

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))^2*x^3,x, algorithm="giac")

[Out]

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