3.1.66 \(\int e^{2 \text {sech}^{-1}(a x)} x^3 \, dx\) [66]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.37, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6472, 1818, 1825, 1828, 12, 267} \begin {gather*} \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} \end {gather*}

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 12

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

Rule 267

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 1818

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 1825

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 1828

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 6472

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

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 \text {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 {\text {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 {\text {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 {\text {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 {\text {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 \text {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 = 0.05, size = 52, normalized size = 0.44 \begin {gather*} \frac {x^2}{a^2}-\frac {x^4}{4}+\frac {2 (-1+a x) \sqrt {\frac {1-a x}{1+a x}} (1+a x)^2}{3 a^4} \end {gather*}

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)

________________________________________________________________________________________

Maple [A]
time = 0.03, size = 72, normalized size = 0.62

method	result	size
default	\(\frac {-\frac {1}{4} a^{2} x^{4}+\frac {1}{2} x^{2}}{a^{2}}+\frac {2 \sqrt {\frac {a x +1}{a x}}\, x \sqrt {-\frac {a x -1}{a x}}\, \left (a^{2} x^{2}-1\right )}{3 a^{3}}+\frac {x^{2}}{2 a^{2}}\)	\(72\)

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,method=_RETURNVERBOSE)

[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

________________________________________________________________________________________

Maxima [A]
time = 0.27, size = 42, normalized size = 0.36 \begin {gather*} -\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 {gather*}

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

________________________________________________________________________________________

Fricas [A]
time = 0.39, size = 62, normalized size = 0.53 \begin {gather*} -\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 {gather*}

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int 2 x\, dx + \int \left (- a^{2} x^{3}\right )\, dx + \int 2 a x^{2} \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}}\, dx}{a^{2}} \end {gather*}

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [B]
time = 1.75, size = 63, normalized size = 0.54 \begin {gather*} \frac {x^2}{a^2}-\frac {x^4}{4}-\sqrt {\frac {1}{a\,x}-1}\,\left (\frac {2\,x\,\sqrt {\frac {1}{a\,x}+1}}{3\,a^3}-\frac {2\,x^3\,\sqrt {\frac {1}{a\,x}+1}}{3\,a}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________