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

Optimal. Leaf size=46 \[ -\frac {2}{\sqrt {\frac {1-a x}{a x+1}}+1}-2 \tan ^{-1}\left (\sqrt {\frac {1-a x}{a x+1}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6337, 801, 203} \[ -\frac {2}{\sqrt {\frac {1-a x}{a x+1}}+1}-2 \tan ^{-1}\left (\sqrt {\frac {1-a x}{a x+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcSech[a*x]*x),x]

[Out]

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

Rule 203

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

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

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Mathematica [C]  time = 0.04, size = 74, normalized size = 1.61 \[ \sqrt {\frac {1-a x}{a x+1}} \left (\frac {1}{a x}+1\right )-\frac {1}{a x}+i \log \left (2 \sqrt {\frac {1-a x}{a x+1}} (a x+1)-2 i a x\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(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
)]

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fricas [A]  time = 1.70, size = 76, normalized size = 1.65 \[ \frac {a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - a x \arctan \left (\sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}}\right ) - 1}{a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right ) x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.85, size = 184, normalized size = 4.00 \[ \ln \left (\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}+1\right )\,1{}\mathrm {i}-\ln \left (\frac {\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x}+1}-1}\right )\,1{}\mathrm {i}-\frac {1}{a\,x}-\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,8{}\mathrm {i}}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2\,\left (1+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}-\frac {2\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \int \frac {1}{a x \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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