3.143 \(\int \frac {\log ^2(x) \sqrt {1+\log ^2(x)}}{x} \, dx\)

Optimal. Leaf size=42 \[ \frac {1}{8} \sqrt {\log ^2(x)+1} \log (x)+\frac {1}{4} \sqrt {\log ^2(x)+1} \log ^3(x)-\frac {1}{8} \sinh ^{-1}(\log (x)) \]

[Out]

-1/8*arcsinh(ln(x))+1/8*ln(x)*(1+ln(x)^2)^(1/2)+1/4*ln(x)^3*(1+ln(x)^2)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {279, 321, 215} \[ \frac {1}{4} \sqrt {\log ^2(x)+1} \log ^3(x)+\frac {1}{8} \sqrt {\log ^2(x)+1} \log (x)-\frac {1}{8} \sinh ^{-1}(\log (x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcSinh[Log[x]]/8 + (Log[x]*Sqrt[1 + Log[x]^2])/8 + (Log[x]^3*Sqrt[1 + Log[x]^2])/4

Rule 215

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

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {\log ^2(x) \sqrt {1+\log ^2(x)}}{x} \, dx &=\operatorname {Subst}\left (\int x^2 \sqrt {1+x^2} \, dx,x,\log (x)\right )\\ &=\frac {1}{4} \log ^3(x) \sqrt {1+\log ^2(x)}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2}} \, dx,x,\log (x)\right )\\ &=\frac {1}{8} \log (x) \sqrt {1+\log ^2(x)}+\frac {1}{4} \log ^3(x) \sqrt {1+\log ^2(x)}-\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,\log (x)\right )\\ &=-\frac {1}{8} \sinh ^{-1}(\log (x))+\frac {1}{8} \log (x) \sqrt {1+\log ^2(x)}+\frac {1}{4} \log ^3(x) \sqrt {1+\log ^2(x)}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 31, normalized size = 0.74 \[ \frac {1}{8} \left (\log (x) \sqrt {\log ^2(x)+1} \left (2 \log ^2(x)+1\right )-\sinh ^{-1}(\log (x))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-ArcSinh[Log[x]] + Log[x]*Sqrt[1 + Log[x]^2]*(1 + 2*Log[x]^2))/8

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fricas [A]  time = 0.63, size = 36, normalized size = 0.86 \[ \frac {1}{8} \, {\left (2 \, \log \relax (x)^{3} + \log \relax (x)\right )} \sqrt {\log \relax (x)^{2} + 1} + \frac {1}{8} \, \log \left (\sqrt {\log \relax (x)^{2} + 1} - \log \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*log(x)^3 + log(x))*sqrt(log(x)^2 + 1) + 1/8*log(sqrt(log(x)^2 + 1) - log(x))

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giac [A]  time = 0.17, size = 37, normalized size = 0.88 \[ \frac {1}{8} \, {\left (2 \, \log \relax (x)^{2} + 1\right )} \sqrt {\log \relax (x)^{2} + 1} \log \relax (x) + \frac {1}{8} \, \log \left (\sqrt {\log \relax (x)^{2} + 1} - \log \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*log(x)^2 + 1)*sqrt(log(x)^2 + 1)*log(x) + 1/8*log(sqrt(log(x)^2 + 1) - log(x))

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maple [A]  time = 0.07, size = 31, normalized size = 0.74 \[ -\frac {\arcsinh \left (\ln \relax (x )\right )}{8}+\frac {\left (\ln \relax (x )^{2}+1\right )^{\frac {3}{2}} \ln \relax (x )}{4}-\frac {\sqrt {\ln \relax (x )^{2}+1}\, \ln \relax (x )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(x)^2*(1+ln(x)^2)^(1/2)/x,x)

[Out]

1/4*ln(x)*(1+ln(x)^2)^(3/2)-1/8*ln(x)*(1+ln(x)^2)^(1/2)-1/8*arcsinh(ln(x))

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maxima [A]  time = 1.49, size = 30, normalized size = 0.71 \[ \frac {1}{4} \, {\left (\log \relax (x)^{2} + 1\right )}^{\frac {3}{2}} \log \relax (x) - \frac {1}{8} \, \sqrt {\log \relax (x)^{2} + 1} \log \relax (x) - \frac {1}{8} \, \operatorname {arsinh}\left (\log \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(log(x)^2 + 1)^(3/2)*log(x) - 1/8*sqrt(log(x)^2 + 1)*log(x) - 1/8*arcsinh(log(x))

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mupad [B]  time = 0.40, size = 26, normalized size = 0.62 \[ \left (\frac {{\ln \relax (x)}^3}{4}+\frac {\ln \relax (x)}{8}\right )\,\sqrt {{\ln \relax (x)}^2+1}-\frac {\mathrm {asinh}\left (\ln \relax (x)\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)^2*(log(x)^2 + 1)^(1/2))/x,x)

[Out]

(log(x)/8 + log(x)^3/4)*(log(x)^2 + 1)^(1/2) - asinh(log(x))/8

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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