\(\int \frac {1}{1+\sqrt {x}} \, dx\) [171]

Optimal result

Integrand size = 9, antiderivative size = 18 \[ \int \frac {1}{1+\sqrt {x}} \, dx=2 \sqrt {x}-2 \log \left (1+\sqrt {x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {196, 45} \[ \int \frac {1}{1+\sqrt {x}} \, dx=2 \sqrt {x}-2 \log \left (\sqrt {x}+1\right ) \]

[In]

Int[(1 + Sqrt[x])^(-1),x]

[Out]

2*Sqrt[x] - 2*Log[1 + Sqrt[x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 196

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{1+\sqrt {x}} \, dx=2 \sqrt {x}-2 \log \left (1+\sqrt {x}\right ) \]

[In]

Integrate[(1 + Sqrt[x])^(-1),x]

[Out]

2*Sqrt[x] - 2*Log[1 + Sqrt[x]]

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(2 \sqrt {x}-2 \ln \left (1+\sqrt {x}\right )\)	\(15\)
meijerg	\(2 \sqrt {x}-2 \ln \left (1+\sqrt {x}\right )\)	\(15\)
trager	\(2 \sqrt {x}-\ln \left (2 \sqrt {x}+1+x \right )\)	\(18\)
default	\(2 \sqrt {x}+\ln \left (-1+\sqrt {x}\right )-\ln \left (1+\sqrt {x}\right )-\ln \left (-1+x \right )\)	\(27\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1}{1+\sqrt {x}} \, dx=2 \, \sqrt {x} - 2 \, \log \left (\sqrt {x} + 1\right ) \]

[In]

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

[Out]

2*sqrt(x) - 2*log(sqrt(x) + 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {1}{1+\sqrt {x}} \, dx=2 \sqrt {x} - 2 \log {\left (\sqrt {x} + 1 \right )} \]

[In]

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

[Out]

2*sqrt(x) - 2*log(sqrt(x) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {1}{1+\sqrt {x}} \, dx=2 \, \sqrt {x} - 2 \, \log \left (\sqrt {x} + 1\right ) + 2 \]

[In]

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

[Out]

2*sqrt(x) - 2*log(sqrt(x) + 1) + 2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1}{1+\sqrt {x}} \, dx=2 \, \sqrt {x} - 2 \, \log \left (\sqrt {x} + 1\right ) \]

[In]

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

[Out]

2*sqrt(x) - 2*log(sqrt(x) + 1)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1}{1+\sqrt {x}} \, dx=2\,\sqrt {x}-2\,\ln \left (\sqrt {x}+1\right ) \]

[In]

int(1/(x^(1/2) + 1),x)

[Out]

2*x^(1/2) - 2*log(x^(1/2) + 1)