3.7.22 \(\int \frac {1}{x \sqrt {a^2+\log ^2(x)}} \, dx\) [622]

Optimal. Leaf size=16 \[ \tanh ^{-1}\left (\frac {\log (x)}{\sqrt {a^2+\log ^2(x)}}\right ) \]

[Out]

arctanh(ln(x)/(a^2+ln(x)^2)^(1/2))

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Rubi [A]
time = 0.02, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {223, 212} \begin {gather*} \tanh ^{-1}\left (\frac {\log (x)}{\sqrt {a^2+\log ^2(x)}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[a^2 + Log[x]^2]),x]

[Out]

ArcTanh[Log[x]/Sqrt[a^2 + Log[x]^2]]

Rule 212

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(46\) vs. \(2(16)=32\).
time = 0.01, size = 46, normalized size = 2.88 \begin {gather*} -\frac {1}{2} \log \left (1-\frac {\log (x)}{\sqrt {a^2+\log ^2(x)}}\right )+\frac {1}{2} \log \left (1+\frac {\log (x)}{\sqrt {a^2+\log ^2(x)}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.02, size = 15, normalized size = 0.94

method	result	size
derivativedivides	\(\ln \left (\ln \left (x \right )+\sqrt {a^{2}+\ln \left (x \right )^{2}}\right )\)	\(15\)
default	\(\ln \left (\ln \left (x \right )+\sqrt {a^{2}+\ln \left (x \right )^{2}}\right )\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(ln(x)+(a^2+ln(x)^2)^(1/2))

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Maxima [A]
time = 1.60, size = 7, normalized size = 0.44 \begin {gather*} \operatorname {arsinh}\left (\frac {\log \left (x\right )}{a}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsinh(log(x)/a)

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Fricas [A]
time = 0.74, size = 18, normalized size = 1.12 \begin {gather*} -\log \left (\sqrt {a^{2} + \log \left (x\right )^{2}} - \log \left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(sqrt(a^2 + log(x)^2) - log(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.34, size = 18, normalized size = 1.12 \begin {gather*} -\log \left (\sqrt {a^{2} + \log \left (x\right )^{2}} - \log \left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(sqrt(a^2 + log(x)^2) - log(x))

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Mupad [B]
time = 0.43, size = 14, normalized size = 0.88 \begin {gather*} \ln \left (\ln \left (x\right )+\sqrt {a^2+{\ln \left (x\right )}^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(log(x)^2 + a^2)^(1/2)),x)

[Out]

log(log(x) + (log(x)^2 + a^2)^(1/2))

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