\(\int \log (x+\sqrt {1+x^2}) \, dx\) [231]

Optimal result

Integrand size = 12, antiderivative size = 26 \[ \int \log \left (x+\sqrt {1+x^2}\right ) \, dx=-\sqrt {1+x^2}+x \log \left (x+\sqrt {1+x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2614, 267} \[ \int \log \left (x+\sqrt {1+x^2}\right ) \, dx=x \log \left (\sqrt {x^2+1}+x\right )-\sqrt {x^2+1} \]

[In]

Int[Log[x + Sqrt[1 + x^2]],x]

[Out]

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

Rule 267

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

Rule 2614

Int[Log[(d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (c_.)*(x_)^2]], x_Symbol] :> Simp[x*Log[d + e*x + f*Sqrt[a + c
*x^2]], x] - Dist[a*c*f^2, Int[x/(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}, x] && EqQ[e^2 - c*f^2, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \log \left (x+\sqrt {1+x^2}\right ) \, dx=-\sqrt {1+x^2}+x \log \left (x+\sqrt {1+x^2}\right ) \]

[In]

Integrate[Log[x + Sqrt[1 + x^2]],x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88

method	result	size
default	\(x \ln \left (x +\sqrt {x^{2}+1}\right )-\sqrt {x^{2}+1}\)	\(23\)
parts	\(x \ln \left (x +\sqrt {x^{2}+1}\right )+\frac {x^{2} \sqrt {x^{2}+1}}{3}-\frac {2 \sqrt {x^{2}+1}}{3}-\frac {\left (x^{2}+1\right )^{\frac {3}{2}}}{3}\)	\(44\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \log \left (x+\sqrt {1+x^2}\right ) \, dx=x \log \left (x + \sqrt {x^{2} + 1}\right ) - \sqrt {x^{2} + 1} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 3.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \log \left (x+\sqrt {1+x^2}\right ) \, dx=x \log {\left (x + \sqrt {x^{2} + 1} \right )} - \sqrt {x^{2} + 1} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \log \left (x+\sqrt {1+x^2}\right ) \, dx=\int { \log \left (x + \sqrt {x^{2} + 1}\right ) \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \log \left (x+\sqrt {1+x^2}\right ) \, dx=x \log \left (x + \sqrt {x^{2} + 1}\right ) - \sqrt {x^{2} + 1} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \log \left (x+\sqrt {1+x^2}\right ) \, dx=x\,\ln \left (x+\sqrt {x^2+1}\right )-\sqrt {x^2+1} \]

[In]

int(log(x + (x^2 + 1)^(1/2)),x)

[Out]

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