3.1.69 \(\int e^{2 \text {sech}^{-1}(a x)} \, dx\) [69]

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

[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 = 0.11, 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 = {6467, 1661, 12, 815, 209} \begin {gather*} \frac {4 \text {ArcTan}\left (\sqrt {\frac {1-a x}{a x+1}}\right )}{a}-\frac {4}{a \left (1-\sqrt {\frac {1-a x}{a x+1}}\right )}-x \end {gather*}

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 209

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

Rule 815

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 1661

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 6467

Int[E^(ArcSech[u_]*(n_.)), x_Symbol] :> Int[(1/u + Sqrt[(1 - u)/(1 + u)] + (1/u)*Sqrt[(1 - u)/(1 + 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 \text {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 \text {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 \text {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 \text {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 \text {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 = 0.07, size = 75, normalized size = 1.32 \begin {gather*} -\frac {2+a^2 x^2+2 \sqrt {\frac {1-a x}{1+a x}} (1+a x)+2 a x \text {ArcTan}\left (\frac {a x}{\sqrt {\frac {1-a x}{1+a x}} (1+a x)}\right )}{a^2 x} \end {gather*}

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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.04, size = 111, normalized size = 1.95

method	result	size
default	\(\frac {-a^{2} x -\frac {1}{x}}{a^{2}}-\frac {2 \sqrt {\frac {a x +1}{a x}}\, \sqrt {-\frac {a x -1}{a x}}\, \left (\arctan \left (\frac {\mathrm {csgn}\left (a \right ) a x}{\sqrt {-a^{2} x^{2}+1}}\right ) a x +\mathrm {csgn}\left (a \right ) \sqrt {-a^{2} x^{2}+1}\right ) \mathrm {csgn}\left (a \right )}{a \sqrt {-a^{2} x^{2}+1}}-\frac {1}{a^{2} x}\)	\(111\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [A]
time = 0.48, size = 85, normalized size = 1.49 \begin {gather*} -\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} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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}} \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [B]
time = 4.64, size = 162, normalized size = 2.84 \begin {gather*} -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} \end {gather*}

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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