\(\int \frac {e^{\text {sech}^{-1}(a x)}}{x} \, dx\) [37]

Optimal result

Integrand size = 10, antiderivative size = 48 \[ \int \frac {e^{\text {sech}^{-1}(a x)}}{x} \, dx=-\frac {2}{1-\sqrt {\frac {1-a x}{1+a x}}}+2 \arctan \left (\sqrt {\frac {1-a x}{1+a x}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.33, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6469, 99, 12, 41, 222} \[ \int \frac {e^{\text {sech}^{-1}(a x)}}{x} \, dx=-\sqrt {\frac {1}{a x+1}} \sqrt {a x+1} \arcsin (a x)-\frac {\sqrt {1-a x}}{a x \sqrt {\frac {1}{a x+1}}}-\frac {1}{a x} \]

[In]

Int[E^ArcSech[a*x]/x,x]

[Out]

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

Rule 12

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

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 6469

Int[E^ArcSech[(a_.)*(x_)^(p_.)]/(x_), x_Symbol] :> -Simp[(a*p*x^p)^(-1), x] + Dist[(Sqrt[1 + a*x^p]/a)*Sqrt[1/
(1 + a*x^p)], Int[Sqrt[1 + a*x^p]*(Sqrt[1 - a*x^p]/x^(p + 1)), x], x] /; FreeQ[{a, p}, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{a x}+\frac {\left (\sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \int \frac {\sqrt {1-a x} \sqrt {1+a x}}{x^2} \, dx}{a} \\ & = -\frac {1}{a x}-\frac {\sqrt {1-a x}}{a x \sqrt {\frac {1}{1+a x}}}-\frac {\left (\sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \int \frac {a^2}{\sqrt {1-a x} \sqrt {1+a x}} \, dx}{a} \\ & = -\frac {1}{a x}-\frac {\sqrt {1-a x}}{a x \sqrt {\frac {1}{1+a x}}}-\left (a \sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \int \frac {1}{\sqrt {1-a x} \sqrt {1+a x}} \, dx \\ & = -\frac {1}{a x}-\frac {\sqrt {1-a x}}{a x \sqrt {\frac {1}{1+a x}}}-\left (a \sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx \\ & = -\frac {1}{a x}-\frac {\sqrt {1-a x}}{a x \sqrt {\frac {1}{1+a x}}}-\sqrt {\frac {1}{1+a x}} \sqrt {1+a x} \arcsin (a x) \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.06 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.56 \[ \int \frac {e^{\text {sech}^{-1}(a x)}}{x} \, dx=-\frac {1}{a x}+\left (-1-\frac {1}{a x}\right ) \sqrt {\frac {1-a x}{1+a x}}-i \log \left (-2 i a x+2 \sqrt {\frac {1-a x}{1+a x}} (1+a x)\right ) \]

[In]

Integrate[E^ArcSech[a*x]/x,x]

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.05 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.92

method	result	size
default	\(-\frac {\sqrt {\frac {a x +1}{a x}}\, \sqrt {-\frac {a x -1}{a x}}\, \left (\arctan \left (\frac {\operatorname {csgn}\left (a \right ) a x}{\sqrt {-a^{2} x^{2}+1}}\right ) a x +\operatorname {csgn}\left (a \right ) \sqrt {-a^{2} x^{2}+1}\right ) \operatorname {csgn}\left (a \right )}{\sqrt {-a^{2} x^{2}+1}}-\frac {1}{a x}\)	\(92\)

[In]

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

[Out]

-((a*x+1)/a/x)^(1/2)*(-(a*x-1)/a/x)^(1/2)*(arctan(csgn(a)*a*x/(-a^2*x^2+1)^(1/2))*a*x+csgn(a)*(-a^2*x^2+1)^(1/
2))*csgn(a)/(-a^2*x^2+1)^(1/2)-1/a/x

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.60 \[ \int \frac {e^{\text {sech}^{-1}(a x)}}{x} \, dx=-\frac {a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - a x \arctan \left (\sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}}\right ) + 1}{a x} \]

[In]

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

[Out]

-(a*x*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) - a*x*arctan(sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)))
+ 1)/(a*x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.37 (sec) , antiderivative size = 184, normalized size of antiderivative = 3.83 \[ \int \frac {e^{\text {sech}^{-1}(a x)}}{x} \, dx=-\ln \left (\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}+1\right )\,1{}\mathrm {i}+\ln \left (\frac {\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x}+1}-1}\right )\,1{}\mathrm {i}-\frac {1}{a\,x}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,8{}\mathrm {i}}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2\,\left (1+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}-\frac {2\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}\right )} \]

[In]

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

[Out]

log(((1/(a*x) - 1)^(1/2) - 1i)/((1/(a*x) + 1)^(1/2) - 1))*1i - log(((1/(a*x) - 1)^(1/2) - 1i)^2/((1/(a*x) + 1)
^(1/2) - 1)^2 + 1)*1i - 1/(a*x) + (((1/(a*x) - 1)^(1/2) - 1i)^2*8i)/(((1/(a*x) + 1)^(1/2) - 1)^2*(((1/(a*x) -
1)^(1/2) - 1i)^4/((1/(a*x) + 1)^(1/2) - 1)^4 - (2*((1/(a*x) - 1)^(1/2) - 1i)^2)/((1/(a*x) + 1)^(1/2) - 1)^2 +
1))