\(\int x \sqrt {4+x^2} \log (x) \, dx\) [275]

Optimal result

Integrand size = 13, antiderivative size = 60 \[ \int x \sqrt {4+x^2} \log (x) \, dx=-\frac {4}{3} \sqrt {4+x^2}-\frac {1}{9} \left (4+x^2\right )^{3/2}+\frac {8}{3} \text {arctanh}\left (\frac {\sqrt {4+x^2}}{2}\right )+\frac {1}{3} \left (4+x^2\right )^{3/2} \log (x) \]

[Out]

-1/9*(x^2+4)^(3/2)+8/3*arctanh(1/2*(x^2+4)^(1/2))+1/3*(x^2+4)^(3/2)*ln(x)-4/3*(x^2+4)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2376, 272, 52, 65, 213} \[ \int x \sqrt {4+x^2} \log (x) \, dx=\frac {8}{3} \text {arctanh}\left (\frac {\sqrt {x^2+4}}{2}\right )-\frac {1}{9} \left (x^2+4\right )^{3/2}-\frac {4 \sqrt {x^2+4}}{3}+\frac {1}{3} \left (x^2+4\right )^{3/2} \log (x) \]

[In]

Int[x*Sqrt[4 + x^2]*Log[x],x]

[Out]

(-4*Sqrt[4 + x^2])/3 - (4 + x^2)^(3/2)/9 + (8*ArcTanh[Sqrt[4 + x^2]/2])/3 + ((4 + x^2)^(3/2)*Log[x])/3

Rule 52

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 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 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 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 2376

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \left (4+x^2\right )^{3/2} \log (x)-\frac {1}{3} \int \frac {\left (4+x^2\right )^{3/2}}{x} \, dx \\ & = \frac {1}{3} \left (4+x^2\right )^{3/2} \log (x)-\frac {1}{6} \text {Subst}\left (\int \frac {(4+x)^{3/2}}{x} \, dx,x,x^2\right ) \\ & = -\frac {1}{9} \left (4+x^2\right )^{3/2}+\frac {1}{3} \left (4+x^2\right )^{3/2} \log (x)-\frac {2}{3} \text {Subst}\left (\int \frac {\sqrt {4+x}}{x} \, dx,x,x^2\right ) \\ & = -\frac {4}{3} \sqrt {4+x^2}-\frac {1}{9} \left (4+x^2\right )^{3/2}+\frac {1}{3} \left (4+x^2\right )^{3/2} \log (x)-\frac {8}{3} \text {Subst}\left (\int \frac {1}{x \sqrt {4+x}} \, dx,x,x^2\right ) \\ & = -\frac {4}{3} \sqrt {4+x^2}-\frac {1}{9} \left (4+x^2\right )^{3/2}+\frac {1}{3} \left (4+x^2\right )^{3/2} \log (x)-\frac {16}{3} \text {Subst}\left (\int \frac {1}{-4+x^2} \, dx,x,\sqrt {4+x^2}\right ) \\ & = -\frac {4}{3} \sqrt {4+x^2}-\frac {1}{9} \left (4+x^2\right )^{3/2}+\frac {8}{3} \tanh ^{-1}\left (\frac {\sqrt {4+x^2}}{2}\right )+\frac {1}{3} \left (4+x^2\right )^{3/2} \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.88 \[ \int x \sqrt {4+x^2} \log (x) \, dx=\frac {1}{3} \left (-\frac {1}{3} \sqrt {4+x^2} \left (16+x^2\right )-8 \log (x)+\left (4+x^2\right )^{3/2} \log (x)+8 \log \left (2+\sqrt {4+x^2}\right )\right ) \]

[In]

Integrate[x*Sqrt[4 + x^2]*Log[x],x]

[Out]

(-1/3*(Sqrt[4 + x^2]*(16 + x^2)) - 8*Log[x] + (4 + x^2)^(3/2)*Log[x] + 8*Log[2 + Sqrt[4 + x^2]])/3

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.25

method	result	size
meijerg	\(\left (-\frac {2 \sqrt {1+\frac {x^{2}}{4}}}{9}+\frac {2 \ln \left (x \right ) \sqrt {1+\frac {x^{2}}{4}}}{3}\right ) x^{2}+\frac {32}{9}-\frac {32 \sqrt {1+\frac {x^{2}}{4}}}{9}+\ln \left (x \right ) \left (-\frac {8}{3}+\frac {8 \sqrt {1+\frac {x^{2}}{4}}}{3}\right )+\frac {8 \ln \left (\frac {1}{2}+\frac {\sqrt {1+\frac {x^{2}}{4}}}{2}\right )}{3}\)	\(75\)

[In]

int(x*ln(x)*(x^2+4)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(-2/9*(1+1/4*x^2)^(1/2)+2/3*ln(x)*(1+1/4*x^2)^(1/2))*x^2+32/9-32/9*(1+1/4*x^2)^(1/2)+ln(x)*(-8/3+8/3*(1+1/4*x^
2)^(1/2))+8/3*ln(1/2+1/2*(1+1/4*x^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90 \[ \int x \sqrt {4+x^2} \log (x) \, dx=-\frac {1}{9} \, {\left (x^{2} - 3 \, {\left (x^{2} + 4\right )} \log \left (x\right ) + 16\right )} \sqrt {x^{2} + 4} + \frac {8}{3} \, \log \left (-x + \sqrt {x^{2} + 4} + 2\right ) - \frac {8}{3} \, \log \left (-x + \sqrt {x^{2} + 4} - 2\right ) \]

[In]

integrate(x*log(x)*(x^2+4)^(1/2),x, algorithm="fricas")

[Out]

-1/9*(x^2 - 3*(x^2 + 4)*log(x) + 16)*sqrt(x^2 + 4) + 8/3*log(-x + sqrt(x^2 + 4) + 2) - 8/3*log(-x + sqrt(x^2 +
4) - 2)

Sympy [A] (verification not implemented)

Time = 7.76 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.18 \[ \int x \sqrt {4+x^2} \log (x) \, dx=\left (\frac {x^{2}}{3} + \frac {4}{3}\right ) \sqrt {x^{2} + 4} \log {\left (x \right )} - \frac {\left (x^{2} + 4\right )^{\frac {3}{2}}}{9} - \frac {4 \sqrt {x^{2} + 4}}{3} - \frac {4 \log {\left (\sqrt {x^{2} + 4} - 2 \right )}}{3} + \frac {4 \log {\left (\sqrt {x^{2} + 4} + 2 \right )}}{3} \]

[In]

integrate(x*ln(x)*(x**2+4)**(1/2),x)

[Out]

(x**2/3 + 4/3)*sqrt(x**2 + 4)*log(x) - (x**2 + 4)**(3/2)/9 - 4*sqrt(x**2 + 4)/3 - 4*log(sqrt(x**2 + 4) - 2)/3
+ 4*log(sqrt(x**2 + 4) + 2)/3

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.65 \[ \int x \sqrt {4+x^2} \log (x) \, dx=\frac {1}{3} \, {\left (x^{2} + 4\right )}^{\frac {3}{2}} \log \left (x\right ) - \frac {1}{9} \, {\left (x^{2} + 4\right )}^{\frac {3}{2}} - \frac {4}{3} \, \sqrt {x^{2} + 4} + \frac {8}{3} \, \operatorname {arsinh}\left (\frac {2}{{\left | x \right |}}\right ) \]

[In]

integrate(x*log(x)*(x^2+4)^(1/2),x, algorithm="maxima")

[Out]

1/3*(x^2 + 4)^(3/2)*log(x) - 1/9*(x^2 + 4)^(3/2) - 4/3*sqrt(x^2 + 4) + 8/3*arcsinh(2/abs(x))

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90 \[ \int x \sqrt {4+x^2} \log (x) \, dx=\frac {1}{3} \, {\left (x^{2} + 4\right )}^{\frac {3}{2}} \log \left (x\right ) - \frac {1}{9} \, {\left (x^{2} + 4\right )}^{\frac {3}{2}} - \frac {4}{3} \, \sqrt {x^{2} + 4} + \frac {4}{3} \, \log \left (\sqrt {x^{2} + 4} + 2\right ) - \frac {4}{3} \, \log \left (\sqrt {x^{2} + 4} - 2\right ) \]

[In]

integrate(x*log(x)*(x^2+4)^(1/2),x, algorithm="giac")

[Out]

1/3*(x^2 + 4)^(3/2)*log(x) - 1/9*(x^2 + 4)^(3/2) - 4/3*sqrt(x^2 + 4) + 4/3*log(sqrt(x^2 + 4) + 2) - 4/3*log(sq
rt(x^2 + 4) - 2)

Mupad [F(-1)]

Timed out. \[ \int x \sqrt {4+x^2} \log (x) \, dx=\int x\,\ln \left (x\right )\,\sqrt {x^2+4} \,d x \]

[In]

int(x*log(x)*(x^2 + 4)^(1/2),x)

[Out]

int(x*log(x)*(x^2 + 4)^(1/2), x)