3.6.75 \(\int \frac {\sqrt {x}}{x+x^2} \, dx\) [575]

Optimal. Leaf size=8 \[ 2 \tan ^{-1}\left (\sqrt {x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {661, 65, 209} \begin {gather*} 2 \text {ArcTan}\left (\sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(x + x^2),x]

[Out]

2*ArcTan[Sqrt[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])

Rule 661

Int[((e_.)*(x_))^(m_.)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e^p, Int[(e*x)^(m + p)*(b + c*x)
^p, x], x] /; FreeQ[{b, c, e, m}, x] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 8, normalized size = 1.00 \begin {gather*} 2 \tan ^{-1}\left (\sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(x + x^2),x]

[Out]

2*ArcTan[Sqrt[x]]

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Maple [A]
time = 0.20, size = 7, normalized size = 0.88

method	result	size
derivativedivides	\(2 \arctan \left (\sqrt {x}\right )\)	\(7\)
default	\(2 \arctan \left (\sqrt {x}\right )\)	\(7\)
meijerg	\(2 \arctan \left (\sqrt {x}\right )\)	\(7\)
trager	\(\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x}+x -1}{1+x}\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 6, normalized size = 0.75 \begin {gather*} 2 \, \arctan \left (\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(sqrt(x))

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Fricas [A]
time = 0.34, size = 6, normalized size = 0.75 \begin {gather*} 2 \, \arctan \left (\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(sqrt(x))

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Sympy [A]
time = 0.17, size = 7, normalized size = 0.88 \begin {gather*} 2 \operatorname {atan}{\left (\sqrt {x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*atan(sqrt(x))

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Giac [A]
time = 3.61, size = 6, normalized size = 0.75 \begin {gather*} 2 \, \arctan \left (\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(sqrt(x))

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Mupad [B]
time = 0.14, size = 6, normalized size = 0.75 \begin {gather*} 2\,\mathrm {atan}\left (\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(x + x^2),x)

[Out]

2*atan(x^(1/2))

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