\(\int \log (\sqrt {x}+\sqrt {1+x}) \, dx\) [234]

Optimal result

Integrand size = 14, antiderivative size = 43 \[ \int \log \left (\sqrt {x}+\sqrt {1+x}\right ) \, dx=-\frac {1}{2} \sqrt {x} \sqrt {1+x}+\frac {\text {arcsinh}\left (\sqrt {x}\right )}{2}+x \log \left (\sqrt {x}+\sqrt {1+x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2628, 12, 1978, 52, 56, 221} \[ \int \log \left (\sqrt {x}+\sqrt {1+x}\right ) \, dx=\frac {\text {arcsinh}\left (\sqrt {x}\right )}{2}-\frac {1}{2} \sqrt {x} \sqrt {x+1}+x \log \left (\sqrt {x}+\sqrt {x+1}\right ) \]

[In]

Int[Log[Sqrt[x] + Sqrt[1 + x]],x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 56

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

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 1978

Int[(u_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Int[u*((a*e + b*e*
x^n)^p/(c + d*x^n)^p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && GtQ[b*d*e, 0] && GtQ[c - a*(d/b), 0]

Rule 2628

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*(D[u, x]/u), x], x] /; InverseFunctionFr
eeQ[u, x]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \log \left (\sqrt {x}+\sqrt {1+x}\right ) \, dx=-\frac {1}{2} \sqrt {x} \sqrt {1+x}+\frac {\text {arcsinh}\left (\sqrt {x}\right )}{2}+x \log \left (\sqrt {x}+\sqrt {1+x}\right ) \]

[In]

Integrate[Log[Sqrt[x] + Sqrt[1 + x]],x]

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.21

method	result	size
default	\(x \ln \left (\sqrt {x}+\sqrt {x +1}\right )-\frac {\sqrt {x}\, \sqrt {x +1}}{2}+\frac {\sqrt {x \left (x +1\right )}\, \ln \left (\frac {1}{2}+x +\sqrt {x^{2}+x}\right )}{4 \sqrt {x}\, \sqrt {x +1}}\)	\(52\)
parts	\(x \ln \left (\sqrt {x}+\sqrt {x +1}\right )-\frac {\sqrt {x}\, \left (x +1\right )^{\frac {3}{2}}}{4}-\frac {\sqrt {x}\, \sqrt {x +1}}{4}+\frac {\sqrt {x \left (x +1\right )}\, \ln \left (\frac {1}{2}+x +\sqrt {x^{2}+x}\right )}{4 \sqrt {x}\, \sqrt {x +1}}+\frac {x^{\frac {3}{2}} \sqrt {x +1}}{4}\)	\(72\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.65 \[ \int \log \left (\sqrt {x}+\sqrt {1+x}\right ) \, dx=\frac {1}{2} \, {\left (2 \, x + 1\right )} \log \left (\sqrt {x + 1} + \sqrt {x}\right ) - \frac {1}{2} \, \sqrt {x + 1} \sqrt {x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.93 \[ \int \log \left (\sqrt {x}+\sqrt {1+x}\right ) \, dx=x \log \left (\sqrt {x + 1} + \sqrt {x}\right ) - \frac {1}{2} \, \sqrt {x^{2} + x} - \frac {1}{4} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} + x} - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 2.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \log \left (\sqrt {x}+\sqrt {1+x}\right ) \, dx=\mathrm {atanh}\left (\frac {\sqrt {x}}{\sqrt {x+1}-1}\right )-\frac {\sqrt {x}\,\sqrt {x+1}}{2}+x\,\ln \left (\sqrt {x+1}+\sqrt {x}\right ) \]

[In]

int(log((x + 1)^(1/2) + x^(1/2)),x)

[Out]

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