\(\int \frac {\arctan (\sqrt {x})}{\sqrt {x} (1+x)} \, dx\) [96]

Optimal result

Integrand size = 17, antiderivative size = 8 \[ \int \frac {\arctan \left (\sqrt {x}\right )}{\sqrt {x} (1+x)} \, dx=\arctan \left (\sqrt {x}\right )^2 \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {65, 209, 6818} \[ \int \frac {\arctan \left (\sqrt {x}\right )}{\sqrt {x} (1+x)} \, dx=\arctan \left (\sqrt {x}\right )^2 \]

[In]

Int[ArcTan[Sqrt[x]]/(Sqrt[x]*(1 + x)),x]

[Out]

ArcTan[Sqrt[x]]^2

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 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \arctan \left (\sqrt {x}\right )^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {\arctan \left (\sqrt {x}\right )}{\sqrt {x} (1+x)} \, dx=\arctan \left (\sqrt {x}\right )^2 \]

[In]

Integrate[ArcTan[Sqrt[x]]/(Sqrt[x]*(1 + x)),x]

[Out]

ArcTan[Sqrt[x]]^2

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\arctan \left (\sqrt {x}\right )^{2}\)	\(7\)
default	\(\arctan \left (\sqrt {x}\right )^{2}\)	\(7\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {\arctan \left (\sqrt {x}\right )}{\sqrt {x} (1+x)} \, dx=\arctan \left (\sqrt {x}\right )^{2} \]

[In]

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

[Out]

arctan(sqrt(x))^2

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88 \[ \int \frac {\arctan \left (\sqrt {x}\right )}{\sqrt {x} (1+x)} \, dx=\operatorname {atan}^{2}{\left (\sqrt {x} \right )} \]

[In]

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

[Out]

atan(sqrt(x))**2

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {\arctan \left (\sqrt {x}\right )}{\sqrt {x} (1+x)} \, dx=\arctan \left (\sqrt {x}\right )^{2} \]

[In]

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

[Out]

arctan(sqrt(x))^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {\arctan \left (\sqrt {x}\right )}{\sqrt {x} (1+x)} \, dx=\arctan \left (\sqrt {x}\right )^{2} \]

[In]

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

[Out]

arctan(sqrt(x))^2

Mupad [B] (verification not implemented)

Time = 0.67 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {\arctan \left (\sqrt {x}\right )}{\sqrt {x} (1+x)} \, dx={\mathrm {atan}\left (\sqrt {x}\right )}^2 \]

[In]

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

[Out]

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