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

Optimal. Leaf size=40 \[ -\frac {1}{4} \sqrt {x^2+1} x+\frac {1}{2} x^2 \log \left (\sqrt {x^2+1}+x\right )+\frac {1}{4} \sinh ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2536, 321, 215} \[ -\frac {1}{4} \sqrt {x^2+1} x+\frac {1}{2} x^2 \log \left (\sqrt {x^2+1}+x\right )+\frac {1}{4} \sinh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[x*Log[x + Sqrt[1 + x^2]],x]

[Out]

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

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 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]

Rule 2536

Int[Log[(d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (c_.)*(x_)^2]]*((g_.)*(x_))^(m_.), x_Symbol] :> Simp[((g*x)^(m
+ 1)*Log[d + e*x + f*Sqrt[a + c*x^2]])/(g*(m + 1)), x] - Dist[(a*c*f^2)/(g*(m + 1)), Int[(g*x)^(m + 1)/(d*e*(
a + c*x^2) + f*(a*e - c*d*x)*Sqrt[a + c*x^2]), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && EqQ[e^2 - c*f^2, 0
] && NeQ[m, -1] && IntegerQ[2*m]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 36, normalized size = 0.90 \[ \frac {1}{4} \left (-\sqrt {x^2+1} x+2 x^2 \log \left (\sqrt {x^2+1}+x\right )+\sinh ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Log[x + Sqrt[1 + x^2]],x]

[Out]

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

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fricas [A]  time = 0.41, size = 30, normalized size = 0.75 \[ \frac {1}{4} \, {\left (2 \, x^{2} + 1\right )} \log \left (x + \sqrt {x^{2} + 1}\right ) - \frac {1}{4} \, \sqrt {x^{2} + 1} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.94, size = 40, normalized size = 1.00 \[ \frac {1}{2} \, x^{2} \log \left (x + \sqrt {x^{2} + 1}\right ) - \frac {1}{4} \, \sqrt {x^{2} + 1} x - \frac {1}{4} \, \log \left (-x + \sqrt {x^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int x \ln \left (x +\sqrt {x^{2}+1}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, x^{2} \log \left (x + \sqrt {x^{2} + 1}\right ) - \frac {1}{4} \, x^{2} - \int \frac {x^{2}}{2 \, {\left (x^{3} + {\left (x^{2} + 1\right )}^{\frac {3}{2}} + x\right )}}\,{d x} + \frac {1}{4} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*log(x + sqrt(x^2 + 1)) - 1/4*x^2 - integrate(1/2*x^2/(x^3 + (x^2 + 1)^(3/2) + x), x) + 1/4*log(x^2 + 1
)

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mupad [B]  time = 0.04, size = 32, normalized size = 0.80 \[ x\,\ln \left (x+\sqrt {x^2+1}\right )\,\left (\frac {x}{2}+\frac {1}{4\,x}\right )-\frac {x\,\sqrt {x^2+1}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*log(x + (x^2 + 1)^(1/2)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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