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

Optimal. Leaf size=94 \[ \frac{(a x+1)^2 \left (1-\sqrt{\frac{1-a x}{a x+1}}\right )^2}{4 a^2}+\frac{(a x+1) \left (\sqrt{\frac{1-a x}{a x+1}}+1\right )}{2 a^2}+\frac{\tan ^{-1}\left (\sqrt{\frac{1-a x}{a x+1}}\right )}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.308749, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6337, 819, 639, 203} \[ \frac{(a x+1)^2 \left (1-\sqrt{\frac{1-a x}{a x+1}}\right )^2}{4 a^2}+\frac{(a x+1) \left (\sqrt{\frac{1-a x}{a x+1}}+1\right )}{2 a^2}+\frac{\tan ^{-1}\left (\sqrt{\frac{1-a x}{a x+1}}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

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])

Rubi steps

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

Mathematica [C]  time = 0.0679862, size = 75, normalized size = 0.8 \[ -\frac{-2 a x+a x \sqrt{\frac{1-a x}{a x+1}} (a x+1)+i \log \left (2 \sqrt{\frac{1-a x}{a x+1}} (a x+1)-2 i a x\right )}{2 a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{x \left ({\frac{1}{ax}}+\sqrt{{\frac{1}{ax}}-1}\sqrt{1+{\frac{1}{ax}}} \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{\frac{1}{a x} + 1} \sqrt{\frac{1}{a x} - 1} + \frac{1}{a x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.02363, size = 173, normalized size = 1.84 \begin{align*} -\frac{a^{2} x^{2} \sqrt{\frac{a x + 1}{a x}} \sqrt{-\frac{a x - 1}{a x}} - 2 \, a x - \arctan \left (\sqrt{\frac{a x + 1}{a x}} \sqrt{-\frac{a x - 1}{a x}}\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \int \frac{x^{2}}{a x \sqrt{-1 + \frac{1}{a x}} \sqrt{1 + \frac{1}{a x}} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{\frac{1}{a x} + 1} \sqrt{\frac{1}{a x} - 1} + \frac{1}{a x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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