\(\int \frac {\log ^2(x) \sqrt {1+\log ^2(x)}}{x} \, dx\) [143]

Optimal result

Integrand size = 18, antiderivative size = 42 \[ \int \frac {\log ^2(x) \sqrt {1+\log ^2(x)}}{x} \, dx=-\frac {1}{8} \text {arcsinh}(\log (x))+\frac {1}{8} \log (x) \sqrt {1+\log ^2(x)}+\frac {1}{4} \log ^3(x) \sqrt {1+\log ^2(x)} \]

[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)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {285, 327, 221} \[ \int \frac {\log ^2(x) \sqrt {1+\log ^2(x)}}{x} \, dx=-\frac {1}{8} \text {arcsinh}(\log (x))+\frac {1}{8} \sqrt {\log ^2(x)+1} \log (x)+\frac {1}{4} \sqrt {\log ^2(x)+1} \log ^3(x) \]

[In]

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

[Out]

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

Rule 221

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 285

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 327

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*} \text {integral}& = \text {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} \text {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} \text {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*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.74 \[ \int \frac {\log ^2(x) \sqrt {1+\log ^2(x)}}{x} \, dx=\frac {1}{8} \left (-\text {arcsinh}(\log (x))+\log (x) \sqrt {1+\log ^2(x)} \left (1+2 \log ^2(x)\right )\right ) \]

[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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {\ln \left (x \right ) \left (1+\ln \left (x \right )^{2}\right )^{\frac {3}{2}}}{4}-\frac {\ln \left (x \right ) \sqrt {1+\ln \left (x \right )^{2}}}{8}-\frac {\operatorname {arcsinh}\left (\ln \left (x \right )\right )}{8}\)	\(31\)
default	\(\frac {\ln \left (x \right ) \left (1+\ln \left (x \right )^{2}\right )^{\frac {3}{2}}}{4}-\frac {\ln \left (x \right ) \sqrt {1+\ln \left (x \right )^{2}}}{8}-\frac {\operatorname {arcsinh}\left (\ln \left (x \right )\right )}{8}\)	\(31\)

[In]

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

[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))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {\log ^2(x) \sqrt {1+\log ^2(x)}}{x} \, dx=\frac {1}{8} \, {\left (2 \, \log \left (x\right )^{3} + \log \left (x\right )\right )} \sqrt {\log \left (x\right )^{2} + 1} + \frac {1}{8} \, \log \left (\sqrt {\log \left (x\right )^{2} + 1} - \log \left (x\right )\right ) \]

[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))

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.64 \[ \int \frac {\log ^2(x) \sqrt {1+\log ^2(x)}}{x} \, dx=\sqrt {\log {\left (x \right )}^{2} + 1} \left (\frac {\log {\left (x \right )}^{3}}{4} + \frac {\log {\left (x \right )}}{8}\right ) - \frac {\operatorname {asinh}{\left (\log {\left (x \right )} \right )}}{8} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.71 \[ \int \frac {\log ^2(x) \sqrt {1+\log ^2(x)}}{x} \, dx=\frac {1}{4} \, {\left (\log \left (x\right )^{2} + 1\right )}^{\frac {3}{2}} \log \left (x\right ) - \frac {1}{8} \, \sqrt {\log \left (x\right )^{2} + 1} \log \left (x\right ) - \frac {1}{8} \, \operatorname {arsinh}\left (\log \left (x\right )\right ) \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int \frac {\log ^2(x) \sqrt {1+\log ^2(x)}}{x} \, dx=\frac {1}{8} \, {\left (2 \, \log \left (x\right )^{2} + 1\right )} \sqrt {\log \left (x\right )^{2} + 1} \log \left (x\right ) + \frac {1}{8} \, \log \left (\sqrt {\log \left (x\right )^{2} + 1} - \log \left (x\right )\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 1.47 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.62 \[ \int \frac {\log ^2(x) \sqrt {1+\log ^2(x)}}{x} \, dx=\left (\frac {{\ln \left (x\right )}^3}{4}+\frac {\ln \left (x\right )}{8}\right )\,\sqrt {{\ln \left (x\right )}^2+1}-\frac {\mathrm {asinh}\left (\ln \left (x\right )\right )}{8} \]

[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