\(\int \frac {\sqrt {x}}{1+x} \, dx\) [203]

Optimal result

Integrand size = 11, antiderivative size = 16 \[ \int \frac {\sqrt {x}}{1+x} \, dx=2 \sqrt {x}-2 \arctan \left (\sqrt {x}\right ) \]

[Out]

2*x^(1/2)-2*arctan(x^(1/2))

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {52, 65, 209} \[ \int \frac {\sqrt {x}}{1+x} \, dx=2 \sqrt {x}-2 \arctan \left (\sqrt {x}\right ) \]

[In]

Int[Sqrt[x]/(1 + x),x]

[Out]

2*Sqrt[x] - 2*ArcTan[Sqrt[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 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 209

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {x}}{1+x} \, dx=2 \sqrt {x}-2 \arctan \left (\sqrt {x}\right ) \]

[In]

Integrate[Sqrt[x]/(1 + x),x]

[Out]

2*Sqrt[x] - 2*ArcTan[Sqrt[x]]

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(2 \sqrt {x}-2 \arctan \left (\sqrt {x}\right )\)	\(13\)
default	\(2 \sqrt {x}-2 \arctan \left (\sqrt {x}\right )\)	\(13\)
meijerg	\(2 \sqrt {x}-2 \arctan \left (\sqrt {x}\right )\)	\(13\)
risch	\(2 \sqrt {x}-2 \arctan \left (\sqrt {x}\right )\)	\(13\)
trager	\(2 \sqrt {x}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x}-x +1}{1+x}\right )\)	\(38\)

[In]

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

[Out]

2*x^(1/2)-2*arctan(x^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt {x}}{1+x} \, dx=2 \, \sqrt {x} - 2 \, \arctan \left (\sqrt {x}\right ) \]

[In]

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

[Out]

2*sqrt(x) - 2*arctan(sqrt(x))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {x}}{1+x} \, dx=2 \sqrt {x} - 2 \operatorname {atan}{\left (\sqrt {x} \right )} \]

[In]

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

[Out]

2*sqrt(x) - 2*atan(sqrt(x))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt {x}}{1+x} \, dx=2 \, \sqrt {x} - 2 \, \arctan \left (\sqrt {x}\right ) \]

[In]

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

[Out]

2*sqrt(x) - 2*arctan(sqrt(x))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt {x}}{1+x} \, dx=2 \, \sqrt {x} - 2 \, \arctan \left (\sqrt {x}\right ) \]

[In]

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

[Out]

2*sqrt(x) - 2*arctan(sqrt(x))

Mupad [B] (verification not implemented)

Time = 15.91 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt {x}}{1+x} \, dx=2\,\sqrt {x}-2\,\mathrm {atan}\left (\sqrt {x}\right ) \]

[In]

int(x^(1/2)/(x + 1),x)

[Out]

2*x^(1/2) - 2*atan(x^(1/2))