\(\int \frac {1}{\sqrt {x}+x^{3/2}} \, dx\) [433]

Optimal result

Integrand size = 13, antiderivative size = 8 \[ \int \frac {1}{\sqrt {x}+x^{3/2}} \, dx=2 \arctan \left (\sqrt {x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1607, 65, 209} \[ \int \frac {1}{\sqrt {x}+x^{3/2}} \, dx=2 \arctan \left (\sqrt {x}\right ) \]

[In]

Int[(Sqrt[x] + x^(3/2))^(-1),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 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {x}+x^{3/2}} \, dx=2 \arctan \left (\sqrt {x}\right ) \]

[In]

Integrate[(Sqrt[x] + x^(3/2))^(-1),x]

[Out]

2*ArcTan[Sqrt[x]]

Maple [A] (verified)

Time = 1.79 (sec) , antiderivative size = 7, normalized size of antiderivative = 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	\(\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x}+x -1}{1+x}\right )\)	\(29\)

[In]

int(1/(x^(3/2)+x^(1/2)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sqrt {x}+x^{3/2}} \, dx=2 \, \arctan \left (\sqrt {x}\right ) \]

[In]

integrate(1/(x^(3/2)+x^(1/2)),x, algorithm="fricas")

[Out]

2*arctan(sqrt(x))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\sqrt {x}+x^{3/2}} \, dx=2 \operatorname {atan}{\left (\sqrt {x} \right )} \]

[In]

integrate(1/(x**(3/2)+x**(1/2)),x)

[Out]

2*atan(sqrt(x))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sqrt {x}+x^{3/2}} \, dx=2 \, \arctan \left (\sqrt {x}\right ) \]

[In]

integrate(1/(x^(3/2)+x^(1/2)),x, algorithm="maxima")

[Out]

2*arctan(sqrt(x))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sqrt {x}+x^{3/2}} \, dx=2 \, \arctan \left (\sqrt {x}\right ) \]

[In]

integrate(1/(x^(3/2)+x^(1/2)),x, algorithm="giac")

[Out]

2*arctan(sqrt(x))

Mupad [B] (verification not implemented)

Time = 9.12 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sqrt {x}+x^{3/2}} \, dx=2\,\mathrm {atan}\left (\sqrt {x}\right ) \]

[In]

int(1/(x^(1/2) + x^(3/2)),x)

[Out]

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