\(\int e^{2 \text {sech}^{-1}(a x)} x \, dx\) [20]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.05 \[ \int e^{2 \text {sech}^{-1}(a x)} x \, dx=\frac {-a^2 x^2+4 \sqrt {\frac {1-a x}{1+a x}} (1+a x)+8 \log (x)-4 \log \left (1+\sqrt {\frac {1-a x}{1+a x}}+a x \sqrt {\frac {1-a x}{1+a x}}\right )}{2 a^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.24 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.36, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6891, 7268, 25, 2178, 27, 2027, 2178, 27, 657, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 6891

\(\displaystyle \int x \left (\frac {\sqrt {\frac {1-a x}{a x+1}}}{a x}+\sqrt {\frac {1-a x}{a x+1}}+\frac {1}{a x}\right )^2dx\)

\(\Big \downarrow \) 7268

\(\displaystyle \frac {4 \int -\frac {\sqrt {\frac {1-a x}{a x+1}} \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right ) \left (\frac {1-a x}{a x+1}+1\right )^3}d\sqrt {\frac {1-a x}{a x+1}}}{a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {4 \int \frac {\sqrt {\frac {1-a x}{a x+1}} \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )^3}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right ) \left (\frac {1-a x}{a x+1}+1\right )^3}d\sqrt {\frac {1-a x}{a x+1}}}{a^2}\)

\(\Big \downarrow \) 2178

\(\displaystyle \frac {4 \left (\frac {1}{4} \int -\frac {4 \left (\frac {1-a x}{a x+1}+3 \sqrt {\frac {1-a x}{a x+1}}\right )}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right ) \left (\frac {1-a x}{a x+1}+1\right )^2}d\sqrt {\frac {1-a x}{a x+1}}-\frac {1}{2 \left (\frac {1-a x}{a x+1}+1\right )^2}\right )}{a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4 \left (-\int \frac {\frac {1-a x}{a x+1}+3 \sqrt {\frac {1-a x}{a x+1}}}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right ) \left (\frac {1-a x}{a x+1}+1\right )^2}d\sqrt {\frac {1-a x}{a x+1}}-\frac {1}{2 \left (\frac {1-a x}{a x+1}+1\right )^2}\right )}{a^2}\)

\(\Big \downarrow \) 2027

\(\displaystyle \frac {4 \left (-\int \frac {\sqrt {\frac {1-a x}{a x+1}} \left (\sqrt {\frac {1-a x}{a x+1}}+3\right )}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right ) \left (\frac {1-a x}{a x+1}+1\right )^2}d\sqrt {\frac {1-a x}{a x+1}}-\frac {1}{2 \left (\frac {1-a x}{a x+1}+1\right )^2}\right )}{a^2}\)

\(\Big \downarrow \) 2178

\(\displaystyle \frac {4 \left (\frac {1}{2} \int -\frac {2 \left (\sqrt {\frac {1-a x}{a x+1}}+1\right )}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right ) \left (\frac {1-a x}{a x+1}+1\right )}d\sqrt {\frac {1-a x}{a x+1}}+\frac {2 \sqrt {\frac {1-a x}{a x+1}}+1}{2 \left (\frac {1-a x}{a x+1}+1\right )}-\frac {1}{2 \left (\frac {1-a x}{a x+1}+1\right )^2}\right )}{a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4 \left (-\int \frac {\sqrt {\frac {1-a x}{a x+1}}+1}{\left (1-\sqrt {\frac {1-a x}{a x+1}}\right ) \left (\frac {1-a x}{a x+1}+1\right )}d\sqrt {\frac {1-a x}{a x+1}}+\frac {2 \sqrt {\frac {1-a x}{a x+1}}+1}{2 \left (\frac {1-a x}{a x+1}+1\right )}-\frac {1}{2 \left (\frac {1-a x}{a x+1}+1\right )^2}\right )}{a^2}\)

\(\Big \downarrow \) 657

\(\displaystyle \frac {4 \left (-\int \left (\frac {\sqrt {\frac {1-a x}{a x+1}}}{\frac {1-a x}{a x+1}+1}+\frac {1}{1-\sqrt {\frac {1-a x}{a x+1}}}\right )d\sqrt {\frac {1-a x}{a x+1}}+\frac {2 \sqrt {\frac {1-a x}{a x+1}}+1}{2 \left (\frac {1-a x}{a x+1}+1\right )}-\frac {1}{2 \left (\frac {1-a x}{a x+1}+1\right )^2}\right )}{a^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {4 \left (\frac {2 \sqrt {\frac {1-a x}{a x+1}}+1}{2 \left (\frac {1-a x}{a x+1}+1\right )}-\frac {1}{2 \left (\frac {1-a x}{a x+1}+1\right )^2}+\log \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )-\frac {1}{2} \log \left (\frac {1-a x}{a x+1}+1\right )\right )}{a^2}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 657

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2027

Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ 
(p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & 
& PosQ[s - r] &&  !(EqQ[p, 1] && EqQ[u, 1])
 

rule 2178

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x^2, x], R = Coeff[Po 
lynomialRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 0], S = Coeff[Polynomia 
lRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*S - b*R*x)*((a + 
b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(d + e*x 
)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*b*(p + 1)*Qx)/(d + e*x)^m + (b*R*( 
2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x 
] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
 

rule 6891

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] && Integer 
Q[n]
 

rule 7268

Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfQuotientOfLinears 
[u, x]}, Simp[lst[[2]]*lst[[4]]   Subst[Int[lst[[1]], x], x, lst[[3]]^(1/ls 
t[[2]])], x] /;  !FalseQ[lst]]
 

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.15

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.46 \[ \int e^{2 \text {sech}^{-1}(a x)} x \, dx=-\frac {a^{2} x^{2} - 4 \, a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 2 \, \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} + 1\right ) - 2 \, \log \left (a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 1\right ) - 4 \, \log \left (x\right )}{2 \, a^{2}} \] Input:

integrate((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2*x,x, algorithm="frica 
s")
 

Output:

-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]

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

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

Output:

(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]

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

integrate((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2*x,x, algorithm="maxim 
a")
 

Output:

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

Giac [F]

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

integrate((1/a/x+(-1+1/a/x)^(1/2)*(1+1/a/x)^(1/2))^2*x,x, algorithm="giac" 
)
 

Output:

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

Mupad [B] (verification not implemented)

Time = 25.78 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.66 \[ \int e^{2 \text {sech}^{-1}(a x)} x \, dx=\frac {2\,x\,\sqrt {\frac {1}{a\,x}-1}\,\sqrt {\frac {1}{a\,x}+1}}{a}-\frac {2\,\mathrm {acosh}\left (\frac {1}{a\,x}\right )}{a^2}-\frac {x^2}{2}-\frac {2\,\ln \left (\frac {1}{x}\right )}{a^2} \] Input:

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

Output:

(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
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.52 \[ \int e^{2 \text {sech}^{-1}(a x)} x \, dx=\frac {4 \sqrt {a x +1}\, \sqrt {-a x +1}-4 \,\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )-1\right )+4 \,\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )+1\right )-4 \,\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )-1\right )+4 \,\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )+1\right )+4 \,\mathrm {log}\left (x \right )-a^{2} x^{2}}{2 a^{2}} \] Input:

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

Output:

(4*sqrt(a*x + 1)*sqrt( - a*x + 1) - 4*log( - sqrt(2) + tan(asin(sqrt( - a* 
x + 1)/sqrt(2))/2) - 1) + 4*log( - sqrt(2) + tan(asin(sqrt( - a*x + 1)/sqr 
t(2))/2) + 1) - 4*log(sqrt(2) + tan(asin(sqrt( - a*x + 1)/sqrt(2))/2) - 1) 
 + 4*log(sqrt(2) + tan(asin(sqrt( - a*x + 1)/sqrt(2))/2) + 1) + 4*log(x) - 
 a**2*x**2)/(2*a**2)