3.224 \(\int \frac{\sqrt{x}}{1+x} \, dx\)

Optimal. Leaf size=16 \[ 2 \sqrt{x}-2 \tan ^{-1}\left (\sqrt{x}\right ) \]

[Out]

2*Sqrt[x] - 2*ArcTan[Sqrt[x]]

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Rubi [A]  time = 0.0030161, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {50, 63, 203} \[ 2 \sqrt{x}-2 \tan ^{-1}\left (\sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(1 + x),x]

[Out]

2*Sqrt[x] - 2*ArcTan[Sqrt[x]]

Rule 50

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 63

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 203

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

Rubi steps

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

Mathematica [A]  time = 0.0033632, size = 16, normalized size = 1. \[ 2 \sqrt{x}-2 \tan ^{-1}\left (\sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(1 + x),x]

[Out]

2*Sqrt[x] - 2*ArcTan[Sqrt[x]]

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Maple [A]  time = 0.003, size = 13, normalized size = 0.8 \begin{align*} -2\,\arctan \left ( \sqrt{x} \right ) +2\,\sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.40177, size = 16, normalized size = 1. \begin{align*} 2 \, \sqrt{x} - 2 \, \arctan \left (\sqrt{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x) - 2*arctan(sqrt(x))

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Fricas [A]  time = 1.91908, size = 42, normalized size = 2.62 \begin{align*} 2 \, \sqrt{x} - 2 \, \arctan \left (\sqrt{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x) - 2*arctan(sqrt(x))

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Sympy [A]  time = 0.149241, size = 14, normalized size = 0.88 \begin{align*} 2 \sqrt{x} - 2 \operatorname{atan}{\left (\sqrt{x} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x) - 2*atan(sqrt(x))

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Giac [A]  time = 1.04868, size = 16, normalized size = 1. \begin{align*} 2 \, \sqrt{x} - 2 \, \arctan \left (\sqrt{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x) - 2*arctan(sqrt(x))