3.1.51 \(\int e^{\text {sech}^{-1}(a x^2)} x \, dx\) [51]

Optimal. Leaf size=68 \[ \frac {1}{2} e^{\text {sech}^{-1}\left (a x^2\right )} x^2-\frac {\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} \tanh ^{-1}\left (\sqrt {1-a^2 x^4}\right )}{2 a}+\frac {\log (x)}{a} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6470, 29, 265, 272, 65, 214} \begin {gather*} -\frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \tanh ^{-1}\left (\sqrt {1-a^2 x^4}\right )}{2 a}+\frac {1}{2} x^2 e^{\text {sech}^{-1}\left (a x^2\right )}+\frac {\log (x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^ArcSech[a*x^2]*x^2)/2 - (Sqrt[(1 + a*x^2)^(-1)]*Sqrt[1 + a*x^2]*ArcTanh[Sqrt[1 - a^2*x^4]])/(2*a) + Log[x]/
a

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rule 265

Int[((c_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[(c*x)
^m*(a1*a2 + b1*b2*x^(2*n))^p, x] /; FreeQ[{a1, b1, a2, b2, c, m, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && (Intege
rQ[p] || (GtQ[a1, 0] && GtQ[a2, 0]))

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6470

Int[E^ArcSech[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*(E^ArcSech[a*x^p]/(m + 1)), x] + (Dist
[p/(a*(m + 1)), Int[x^(m - p), x], x] + Dist[p*(Sqrt[1 + a*x^p]/(a*(m + 1)))*Sqrt[1/(1 + a*x^p)], Int[x^(m - p
)/(Sqrt[1 + a*x^p]*Sqrt[1 - a*x^p]), x], x]) /; FreeQ[{a, m, p}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int e^{\text {sech}^{-1}\left (a x^2\right )} x \, dx &=\frac {1}{2} e^{\text {sech}^{-1}\left (a x^2\right )} x^2+\frac {\int \frac {1}{x} \, dx}{a}+\frac {\left (\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \int \frac {1}{x \sqrt {1-a x^2} \sqrt {1+a x^2}} \, dx}{a}\\ &=\frac {1}{2} e^{\text {sech}^{-1}\left (a x^2\right )} x^2+\frac {\log (x)}{a}+\frac {\left (\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \int \frac {1}{x \sqrt {1-a^2 x^4}} \, dx}{a}\\ &=\frac {1}{2} e^{\text {sech}^{-1}\left (a x^2\right )} x^2+\frac {\log (x)}{a}+\frac {\left (\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^4\right )}{4 a}\\ &=\frac {1}{2} e^{\text {sech}^{-1}\left (a x^2\right )} x^2+\frac {\log (x)}{a}-\frac {\left (\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \text {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2}} \, dx,x,\sqrt {1-a^2 x^4}\right )}{2 a^3}\\ &=\frac {1}{2} e^{\text {sech}^{-1}\left (a x^2\right )} x^2-\frac {\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} \tanh ^{-1}\left (\sqrt {1-a^2 x^4}\right )}{2 a}+\frac {\log (x)}{a}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 100, normalized size = 1.47 \begin {gather*} \frac {\sqrt {\frac {1-a x^2}{1+a x^2}} \left (1+a x^2\right )+2 \log \left (a x^2\right )-\log \left (1+\sqrt {\frac {1-a x^2}{1+a x^2}}+a x^2 \sqrt {\frac {1-a x^2}{1+a x^2}}\right )}{2 a} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.10, size = 127, normalized size = 1.87

method	result	size
default	\(\frac {\sqrt {-\frac {a \,x^{2}-1}{a \,x^{2}}}\, x^{2} \sqrt {\frac {a \,x^{2}+1}{a \,x^{2}}}\, \left (\mathrm {csgn}\left (\frac {1}{a}\right ) a \sqrt {-\frac {a^{2} x^{4}-1}{a^{2}}}-\ln \left (\frac {2 \,\mathrm {csgn}\left (\frac {1}{a}\right ) a \sqrt {-\frac {a^{2} x^{4}-1}{a^{2}}}+2}{a^{2} x^{2}}\right )\right ) \mathrm {csgn}\left (\frac {1}{a}\right )}{2 a \sqrt {-\frac {a^{2} x^{4}-1}{a^{2}}}}+\frac {\ln \left (x \right )}{a}\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (61) = 122\).
time = 0.36, size = 133, normalized size = 1.96 \begin {gather*} \frac {2 \, a x^{2} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} - \log \left (a x^{2} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} + 1\right ) + \log \left (a x^{2} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} - 1\right ) + 4 \, \log \left (x\right )}{4 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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Mupad [B]
time = 3.20, size = 182, normalized size = 2.68 \begin {gather*} \frac {\ln \left (x\right )}{a}-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x^2}+1}-1}\right )}{a}+\frac {\frac {5\,{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^2}+1}{\frac {8\,a\,\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}{\sqrt {\frac {1}{a\,x^2}+1}-1}+\frac {8\,a\,{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^3}{{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^3}}+\frac {\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}}{8\,a\,\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)/a - (2*atanh(((1/(a*x^2) - 1)^(1/2) - 1i)/((1/(a*x^2) + 1)^(1/2) - 1)))/a + ((5*((1/(a*x^2) - 1)^(1/2)
- 1i)^2)/((1/(a*x^2) + 1)^(1/2) - 1)^2 + 1)/((8*a*((1/(a*x^2) - 1)^(1/2) - 1i))/((1/(a*x^2) + 1)^(1/2) - 1) +
(8*a*((1/(a*x^2) - 1)^(1/2) - 1i)^3)/((1/(a*x^2) + 1)^(1/2) - 1)^3) + ((1/(a*x^2) - 1)^(1/2) - 1i)/(8*a*((1/(a
*x^2) + 1)^(1/2) - 1))

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