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

Optimal. Leaf size=85 \[ -\frac {(a x+1)^2}{2 a^2}+\frac {\left (2 \sqrt {\frac {1-a x}{a x+1}}+1\right ) (a x+1)}{a^2}+\frac {2 \log (a x+1)}{a^2}+\frac {4 \log \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.43, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6337, 1647, 1593, 801, 260} \[ -\frac {(a x+1)^2}{2 a^2}+\frac {\left (2 \sqrt {\frac {1-a x}{a x+1}}+1\right ) (a x+1)}{a^2}+\frac {2 \log (a x+1)}{a^2}+\frac {4 \log \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 + a*x)^2/(2*a^2) + ((1 + a*x)*(1 + 2*Sqrt[(1 - a*x)/(1 + a*x)]))/a^2 + (2*Log[1 + a*x])/a^2 + (4*Log[1 - S
qrt[(1 - a*x)/(1 + a*x)]])/a^2

Rule 260

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

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1647

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

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

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Mathematica [A]  time = 0.06, size = 89, normalized size = 1.05 \[ \frac {-a^2 x^2+4 \sqrt {\frac {1-a x}{a x+1}} (a x+1)-4 \log \left (a x \sqrt {\frac {1-a x}{a x+1}}+\sqrt {\frac {1-a x}{a x+1}}+1\right )+8 \log (x)}{2 a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(a^2*x^2) + 4*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x) + 8*Log[x] - 4*Log[1 + Sqrt[(1 - a*x)/(1 + a*x)] + a*x*Sqr
t[(1 - a*x)/(1 + a*x)]])/(2*a^2)

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fricas [A]  time = 0.50, size = 124, normalized size = 1.46 \[ -\frac {a^{2} x^{2} - 4 \, a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 2 \, \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 1\right ) - 2 \, \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 1\right ) - 4 \, \log \relax (x)}{2 \, a^{2}} \]

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))^2*x,x, algorithm="fricas")

[Out]

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

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

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,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 89, normalized size = 1.05 \[ -\frac {x^{2}}{2}+\frac {2 \ln \relax (x )}{a^{2}}-\frac {2 \sqrt {-\frac {a x -1}{a x}}\, x \sqrt {\frac {a x +1}{a x}}\, \left (-\sqrt {-a^{2} x^{2}+1}+\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )}{a \sqrt {-a^{2} x^{2}+1}} \]

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,x)

[Out]

-1/2*x^2+2*ln(x)/a^2-2/a*(-(a*x-1)/a/x)^(1/2)*x*((a*x+1)/a/x)^(1/2)*(-(-a^2*x^2+1)^(1/2)+arctanh(1/(-a^2*x^2+1
)^(1/2)))/(-a^2*x^2+1)^(1/2)

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

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,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.69, size = 56, normalized size = 0.66 \[ \frac {2\,x\,\sqrt {\frac {1}{a\,x}-1}\,\sqrt {\frac {1}{a\,x}+1}}{a}-\frac {2\,\mathrm {acosh}\left (\frac {1}{a\,x}\right )}{a^2}-\frac {x^2}{2}-\frac {2\,\ln \left (\frac {1}{x}\right )}{a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x*(1/(a*x) - 1)^(1/2)*(1/(a*x) + 1)^(1/2))/a - (2*acosh(1/(a*x)))/a^2 - x^2/2 - (2*log(1/x))/a^2

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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,x)

[Out]

Timed out

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