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

Optimal. Leaf size=34 \[ -\sqrt{x^2+1}+\sqrt{x^2+1} \log (x)+\tanh ^{-1}\left (\sqrt{x^2+1}\right ) \]

[Out]

-Sqrt[1 + x^2] + ArcTanh[Sqrt[1 + x^2]] + Sqrt[1 + x^2]*Log[x]

________________________________________________________________________________________

Rubi [A]  time = 0.0346594, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2338, 266, 50, 63, 207} \[ -\sqrt{x^2+1}+\sqrt{x^2+1} \log (x)+\tanh ^{-1}\left (\sqrt{x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x*Log[x])/Sqrt[1 + x^2],x]

[Out]

-Sqrt[1 + x^2] + ArcTanh[Sqrt[1 + x^2]] + Sqrt[1 + x^2]*Log[x]

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_))^(q_.), x_Symbol] :
> Simp[(f^m*(d + e*x^r)^(q + 1)*(a + b*Log[c*x^n])^p)/(e*r*(q + 1)), x] - Dist[(b*f^m*n*p)/(e*r*(q + 1)), Int[
((d + e*x^r)^(q + 1)*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[
m, r - 1] && IGtQ[p, 0] && (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n] && NeQ[q, -1]

Rule 266

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

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 207

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

Rubi steps

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

Mathematica [A]  time = 0.0171028, size = 40, normalized size = 1.18 \[ -\sqrt{x^2+1}+\sqrt{x^2+1} \log (x)+\log \left (\sqrt{x^2+1}+1\right )-\log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Log[x])/Sqrt[1 + x^2],x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.025, size = 39, normalized size = 1.2 \begin{align*} 1-\sqrt{{x}^{2}+1}+{\frac{\ln \left ( x \right ) }{2} \left ( -2+2\,\sqrt{{x}^{2}+1} \right ) }+\ln \left ({\frac{1}{2}}+{\frac{1}{2}\sqrt{{x}^{2}+1}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1-(x^2+1)^(1/2)+1/2*ln(x)*(-2+2*(x^2+1)^(1/2))+ln(1/2+1/2*(x^2+1)^(1/2))

________________________________________________________________________________________

Maxima [A]  time = 1.427, size = 34, normalized size = 1. \begin{align*} \sqrt{x^{2} + 1} \log \left (x\right ) - \sqrt{x^{2} + 1} + \operatorname{arsinh}\left (\frac{1}{{\left | x \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.06166, size = 119, normalized size = 3.5 \begin{align*} \sqrt{x^{2} + 1}{\left (\log \left (x\right ) - 1\right )} + \log \left (-x + \sqrt{x^{2} + 1} + 1\right ) - \log \left (-x + \sqrt{x^{2} + 1} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 4.54774, size = 41, normalized size = 1.21 \begin{align*} - \frac{x}{\sqrt{1 + \frac{1}{x^{2}}}} + \sqrt{x^{2} + 1} \log{\left (x \right )} + \operatorname{asinh}{\left (\frac{1}{x} \right )} - \frac{1}{x \sqrt{1 + \frac{1}{x^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.08961, size = 59, normalized size = 1.74 \begin{align*} \sqrt{x^{2} + 1} \log \left (x\right ) - \sqrt{x^{2} + 1} + \frac{1}{2} \, \log \left (\sqrt{x^{2} + 1} + 1\right ) - \frac{1}{2} \, \log \left (\sqrt{x^{2} + 1} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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