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

Optimal. Leaf size=57 \[ -\frac {4}{a \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )}+\frac {4 \tan ^{-1}\left (\sqrt {\frac {1-a x}{a x+1}}\right )}{a}-x \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6332, 1647, 12, 801, 203} \[ -\frac {4}{a \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )}+\frac {4 \tan ^{-1}\left (\sqrt {\frac {1-a x}{a x+1}}\right )}{a}-x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 6332

Int[E^(ArcSech[u_]*(n_.)), x_Symbol] :> Int[(1/u + Sqrt[(1 - u)/(1 + u)] + (1*Sqrt[(1 - u)/(1 + u)])/u)^n, x]
/; IntegerQ[n]

Rubi steps

\begin {align*} \int e^{2 \text {sech}^{-1}(a x)} \, dx &=\int \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 )^2 \, dx\\ &=-\frac {4 \operatorname {Subst}\left (\int \frac {x (1+x)^2}{(-1+x)^2 \left (1+x^2\right )^2} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{a}\\ &=-x+\frac {2 \operatorname {Subst}\left (\int -\frac {4 x}{(-1+x)^2 \left (1+x^2\right )} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{a}\\ &=-x-\frac {8 \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 )}{a}\\ &=-x-\frac {8 \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 )}{a}\\ &=-x-\frac {4}{a \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )}+\frac {4 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {\frac {1-a x}{1+a x}}\right )}{a}\\ &=-x-\frac {4}{a \left (1-\sqrt {\frac {1-a x}{1+a x}}\right )}+\frac {4 \tan ^{-1}\left (\sqrt {\frac {1-a x}{1+a x}}\right )}{a}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 75, normalized size = 1.32 \[ -\frac {a^2 x^2+2 \sqrt {\frac {1-a x}{a x+1}} (a x+1)+2 a x \tan ^{-1}\left (\frac {a x}{\sqrt {\frac {1-a x}{a x+1}} (a x+1)}\right )+2}{a^2 x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 1.02, size = 85, normalized size = 1.49 \[ -\frac {a^{2} x^{2} + 2 \, a x \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 2 \, a x \arctan \left (\sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}}\right ) + 2}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*x^2 + 2*a*x*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))) + 2)/(a^2*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.06, size = 98, normalized size = 1.72 \[ -x -\frac {2}{a^{2} x}-\frac {2 \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, \left (\arctan \left (\frac {\mathrm {csgn}\relax (a ) a x}{\sqrt {-a^{2} x^{2}+1}}\right ) x a +\sqrt {-a^{2} x^{2}+1}\, \mathrm {csgn}\relax (a )\right ) \mathrm {csgn}\relax (a )}{a \sqrt {-a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -x + \frac {2 \, {\left (-a \arcsin \left (a x\right ) - \frac {\sqrt {-a^{2} x^{2} + 1}}{x}\right )}}{a^{2}} + \frac {-\frac {1}{x}}{a^{2}} - \frac {1}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.64, size = 162, normalized size = 2.84 \[ -x-\frac {\left (\ln \left (\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}+1\right )-\ln \left (\frac {\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x}+1}-1}\right )\right )\,2{}\mathrm {i}}{a}-\frac {2}{a^2\,x}+\frac {{\left (1+\sqrt {-\frac {a-\frac {1}{x}}{a}}\,1{}\mathrm {i}\right )}^2\,{\left (\sqrt {\frac {a+\frac {1}{x}}{a}}-1\right )}^2\,4{}\mathrm {i}}{a\,{\left (\sqrt {\frac {a+\frac {1}{x}}{a}}\,1{}\mathrm {i}+\sqrt {-\frac {a-\frac {1}{x}}{a}}-2{}\mathrm {i}\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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