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]

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

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Rubi [A]  time = 0.31, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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 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 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 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 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*}

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Mathematica [C]  time = 0.07, size = 75, normalized size = 0.80 \[ -\frac {-2 a x+a x \sqrt {\frac {1-a x}{a x+1}} (a x+1)+i \log \left (2 \sqrt {\frac {1-a x}{a x+1}} (a x+1)-2 i a x\right )}{2 a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*(-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)])/a^2

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fricas [A]  time = 0.85, size = 79, normalized size = 0.84 \[ -\frac {a^{2} x^{2} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 2 \, a x - \arctan \left (\sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}}\right )}{2 \, a^{2}} \]

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

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)

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

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {\frac {1}{a x} + 1} \sqrt {\frac {1}{a x} - 1} + \frac {1}{a x}}\,{d x} \]

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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mupad [B]  time = 9.22, size = 407, normalized size = 4.33 \[ \frac {x}{a}-\frac {\ln \left (\frac {\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x}+1}-1}\right )\,1{}\mathrm {i}}{2\,a^2}-\frac {\frac {1{}\mathrm {i}}{32\,a^2}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{16\,a^2\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}-\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4\,15{}\mathrm {i}}{32\,a^2\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}}{\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}+\frac {2\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^6}}-\frac {\left (\ln \left (\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}+1\right )-\ln \left (\frac {\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x}+1}-1}\right )\right )\,1{}\mathrm {i}}{a^2}+\frac {\ln \left (\frac {2\,a\,\sqrt {\frac {a+\frac {1}{x}}{a}}-\frac {2}{x}+a\,\sqrt {-\frac {a-\frac {1}{x}}{a}}\,2{}\mathrm {i}}{2\,a+\frac {1}{x}-2\,a\,\sqrt {\frac {a+\frac {1}{x}}{a}}}\right )\,1{}\mathrm {i}}{2\,a^2}-\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{32\,a^2\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/a - (log(((1/(a*x) - 1)^(1/2) - 1i)/((1/(a*x) + 1)^(1/2) - 1))*1i)/(2*a^2) - (1i/(32*a^2) + (((1/(a*x) - 1)^
(1/2) - 1i)^2*1i)/(16*a^2*((1/(a*x) + 1)^(1/2) - 1)^2) - (((1/(a*x) - 1)^(1/2) - 1i)^4*15i)/(32*a^2*((1/(a*x)
+ 1)^(1/2) - 1)^4))/(((1/(a*x) - 1)^(1/2) - 1i)^2/((1/(a*x) + 1)^(1/2) - 1)^2 + (2*((1/(a*x) - 1)^(1/2) - 1i)^
4)/((1/(a*x) + 1)^(1/2) - 1)^4 + ((1/(a*x) - 1)^(1/2) - 1i)^6/((1/(a*x) + 1)^(1/2) - 1)^6) - ((log(((1/(a*x) -
1)^(1/2) - 1i)^2/((1/(a*x) + 1)^(1/2) - 1)^2 + 1) - log(((1/(a*x) - 1)^(1/2) - 1i)/((1/(a*x) + 1)^(1/2) - 1))
)*1i)/a^2 + (log((a*(-(a - 1/x)/a)^(1/2)*2i - 2/x + 2*a*((a + 1/x)/a)^(1/2))/(2*a + 1/x - 2*a*((a + 1/x)/a)^(1
/2)))*1i)/(2*a^2) - (((1/(a*x) - 1)^(1/2) - 1i)^2*1i)/(32*a^2*((1/(a*x) + 1)^(1/2) - 1)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \int \frac {x^{2}}{a x \sqrt {-1 + \frac {1}{a x}} \sqrt {1 + \frac {1}{a x}} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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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