\(\int e^{2 \text {sech}^{-1}(a x)} x \, dx\) [68]

Optimal result

Integrand size = 10, antiderivative size = 85 \[ \int e^{2 \text {sech}^{-1}(a x)} x \, dx=-\frac {(1+a x)^2}{2 a^2}+\frac {(1+a x) \left (1+2 \sqrt {\frac {1-a x}{1+a x}}\right )}{a^2}+\frac {2 \log (1+a x)}{a^2}+\frac {4 \log \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )}{a^2} \]

[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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6472, 1661, 1607, 815, 266} \[ \int e^{2 \text {sech}^{-1}(a x)} x \, dx=-\frac {(a x+1)^2}{2 a^2}+\frac {\left (2 \sqrt {\frac {1-a x}{a x+1}}+1\right ) (a x+1)}{a^2}+\frac {2 \log (a x+1)}{a^2}+\frac {4 \log \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )}{a^2} \]

[In]

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

[Out]

-1/2*(1 + a*x)^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 266

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 815

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 1607

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 1661

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 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*} \text {integral}& = \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 \text {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 {\text {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 {\text {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 {\text {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 {\text {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 \text {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*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.05 \[ \int e^{2 \text {sech}^{-1}(a x)} x \, dx=\frac {-a^2 x^2+4 \sqrt {\frac {1-a x}{1+a x}} (1+a x)+8 \log (x)-4 \log \left (1+\sqrt {\frac {1-a x}{1+a x}}+a x \sqrt {\frac {1-a x}{1+a x}}\right )}{2 a^2} \]

[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)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.15

method	result	size
default	\(\frac {-\frac {a^{2} x^{2}}{2}+\ln \left (x \right )}{a^{2}}-\frac {2 \sqrt {\frac {a x +1}{a x}}\, x \sqrt {-\frac {a x -1}{a x}}\, \left (-\sqrt {-a^{2} x^{2}+1}+\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )}{a \sqrt {-a^{2} x^{2}+1}}+\frac {\ln \left (x \right )}{a^{2}}\)	\(98\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.46 \[ \int e^{2 \text {sech}^{-1}(a x)} x \, dx=-\frac {a^{2} x^{2} - 4 \, a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 2 \, \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 1\right ) - 2 \, \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 1\right ) - 4 \, \log \left (x\right )}{2 \, a^{2}} \]

[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

Sympy [F]

\[ \int e^{2 \text {sech}^{-1}(a x)} x \, dx=\frac {\int \frac {2}{x}\, dx + \int \left (- a^{2} x\right )\, dx + \int 2 a \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}}\, dx}{a^{2}} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int e^{2 \text {sech}^{-1}(a x)} x \, dx=\int { x {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}^{2} \,d x } \]

[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)

Giac [F]

\[ \int e^{2 \text {sech}^{-1}(a x)} x \, dx=\int { x {\left (\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}\right )}^{2} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 7.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.66 \[ \int e^{2 \text {sech}^{-1}(a x)} x \, dx=\frac {2\,x\,\sqrt {\frac {1}{a\,x}-1}\,\sqrt {\frac {1}{a\,x}+1}}{a}-\frac {2\,\mathrm {acosh}\left (\frac {1}{a\,x}\right )}{a^2}-\frac {x^2}{2}-\frac {2\,\ln \left (\frac {1}{x}\right )}{a^2} \]

[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