\(\int \sqrt {x} \arctan (x) \, dx\) [53]

Optimal result

Integrand size = 8, antiderivative size = 117 \[ \int \sqrt {x} \arctan (x) \, dx=-\frac {4 \sqrt {x}}{3}-\frac {1}{3} \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {x}\right )+\frac {1}{3} \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {x}\right )+\frac {2}{3} x^{3/2} \arctan (x)-\frac {\log \left (1-\sqrt {2} \sqrt {x}+x\right )}{3 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} \sqrt {x}+x\right )}{3 \sqrt {2}} \]

[Out]

2/3*x^(3/2)*arctan(x)-1/6*ln(1+x-2^(1/2)*x^(1/2))*2^(1/2)+1/6*ln(1+x+2^(1/2)*x^(1/2))*2^(1/2)+1/3*arctan(-1+2^
(1/2)*x^(1/2))*2^(1/2)+1/3*arctan(1+2^(1/2)*x^(1/2))*2^(1/2)-4/3*x^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {4946, 327, 335, 217, 1179, 642, 1176, 631, 210} \[ \int \sqrt {x} \arctan (x) \, dx=\frac {2}{3} x^{3/2} \arctan (x)-\frac {1}{3} \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {x}\right )+\frac {1}{3} \sqrt {2} \arctan \left (\sqrt {2} \sqrt {x}+1\right )-\frac {4 \sqrt {x}}{3}-\frac {\log \left (x-\sqrt {2} \sqrt {x}+1\right )}{3 \sqrt {2}}+\frac {\log \left (x+\sqrt {2} \sqrt {x}+1\right )}{3 \sqrt {2}} \]

[In]

Int[Sqrt[x]*ArcTan[x],x]

[Out]

(-4*Sqrt[x])/3 - (Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[x]])/3 + (Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[x]])/3 + (2*x^(3/2
)*ArcTan[x])/3 - Log[1 - Sqrt[2]*Sqrt[x] + x]/(3*Sqrt[2]) + Log[1 + Sqrt[2]*Sqrt[x] + x]/(3*Sqrt[2])

Rule 210

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

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2}{3} x^{3/2} \arctan (x)-\frac {2}{3} \int \frac {x^{3/2}}{1+x^2} \, dx \\ & = -\frac {4 \sqrt {x}}{3}+\frac {2}{3} x^{3/2} \arctan (x)+\frac {2}{3} \int \frac {1}{\sqrt {x} \left (1+x^2\right )} \, dx \\ & = -\frac {4 \sqrt {x}}{3}+\frac {2}{3} x^{3/2} \arctan (x)+\frac {4}{3} \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {4 \sqrt {x}}{3}+\frac {2}{3} x^{3/2} \arctan (x)+\frac {2}{3} \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {x}\right )+\frac {2}{3} \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {4 \sqrt {x}}{3}+\frac {2}{3} x^{3/2} \arctan (x)+\frac {1}{3} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {x}\right )+\frac {1}{3} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {x}\right )-\frac {\text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {x}\right )}{3 \sqrt {2}}-\frac {\text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {x}\right )}{3 \sqrt {2}} \\ & = -\frac {4 \sqrt {x}}{3}+\frac {2}{3} x^{3/2} \arctan (x)-\frac {\log \left (1-\sqrt {2} \sqrt {x}+x\right )}{3 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} \sqrt {x}+x\right )}{3 \sqrt {2}}+\frac {1}{3} \sqrt {2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {x}\right )-\frac {1}{3} \sqrt {2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {x}\right ) \\ & = -\frac {4 \sqrt {x}}{3}-\frac {1}{3} \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {x}\right )+\frac {1}{3} \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {x}\right )+\frac {2}{3} x^{3/2} \arctan (x)-\frac {\log \left (1-\sqrt {2} \sqrt {x}+x\right )}{3 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} \sqrt {x}+x\right )}{3 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.92 \[ \int \sqrt {x} \arctan (x) \, dx=\frac {1}{6} \left (-8 \sqrt {x}-2 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {x}\right )+2 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {x}\right )+4 x^{3/2} \arctan (x)-\sqrt {2} \log \left (1-\sqrt {2} \sqrt {x}+x\right )+\sqrt {2} \log \left (1+\sqrt {2} \sqrt {x}+x\right )\right ) \]

[In]

Integrate[Sqrt[x]*ArcTan[x],x]

[Out]

(-8*Sqrt[x] - 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[x]] + 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[x]] + 4*x^(3/2)*ArcTan
[x] - Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[x] + x] + Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[x] + x])/6

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.59

method	result	size
derivativedivides	\(\frac {2 x^{\frac {3}{2}} \arctan \left (x \right )}{3}-\frac {4 \sqrt {x}}{3}+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x +\sqrt {2}\, \sqrt {x}}{1+x -\sqrt {2}\, \sqrt {x}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{6}\)	\(69\)
default	\(\frac {2 x^{\frac {3}{2}} \arctan \left (x \right )}{3}-\frac {4 \sqrt {x}}{3}+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x +\sqrt {2}\, \sqrt {x}}{1+x -\sqrt {2}\, \sqrt {x}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{6}\)	\(69\)
meijerg	\(-\frac {4 \sqrt {x}}{3}+\frac {\sqrt {x}\, \left (-\frac {\sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}+\sqrt {x^{2}}\right )}{2 \left (x^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}\right )}{\left (x^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}+\sqrt {x^{2}}\right )}{2 \left (x^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}\right )}{\left (x^{2}\right )^{\frac {1}{4}}}\right )}{3}+\frac {2 x^{\frac {5}{2}} \arctan \left (\sqrt {x^{2}}\right )}{3 \sqrt {x^{2}}}\)	\(152\)

[In]

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

[Out]

2/3*x^(3/2)*arctan(x)-4/3*x^(1/2)+1/6*2^(1/2)*(ln((1+x+2^(1/2)*x^(1/2))/(1+x-2^(1/2)*x^(1/2)))+2*arctan(1+2^(1
/2)*x^(1/2))+2*arctan(-1+2^(1/2)*x^(1/2)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.68 \[ \int \sqrt {x} \arctan (x) \, dx=\frac {2}{3} \, {\left (x \arctan \left (x\right ) - 2\right )} \sqrt {x} + \left (\frac {1}{6} i + \frac {1}{6}\right ) \, \sqrt {2} \log \left (\left (i + 1\right ) \, \sqrt {2} + 2 \, \sqrt {x}\right ) - \left (\frac {1}{6} i - \frac {1}{6}\right ) \, \sqrt {2} \log \left (-\left (i - 1\right ) \, \sqrt {2} + 2 \, \sqrt {x}\right ) + \left (\frac {1}{6} i - \frac {1}{6}\right ) \, \sqrt {2} \log \left (\left (i - 1\right ) \, \sqrt {2} + 2 \, \sqrt {x}\right ) - \left (\frac {1}{6} i + \frac {1}{6}\right ) \, \sqrt {2} \log \left (-\left (i + 1\right ) \, \sqrt {2} + 2 \, \sqrt {x}\right ) \]

[In]

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

[Out]

2/3*(x*arctan(x) - 2)*sqrt(x) + (1/6*I + 1/6)*sqrt(2)*log((I + 1)*sqrt(2) + 2*sqrt(x)) - (1/6*I - 1/6)*sqrt(2)
*log(-(I - 1)*sqrt(2) + 2*sqrt(x)) + (1/6*I - 1/6)*sqrt(2)*log((I - 1)*sqrt(2) + 2*sqrt(x)) - (1/6*I + 1/6)*sq
rt(2)*log(-(I + 1)*sqrt(2) + 2*sqrt(x))

Sympy [A] (verification not implemented)

Time = 1.95 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.94 \[ \int \sqrt {x} \arctan (x) \, dx=\frac {2 x^{\frac {3}{2}} \operatorname {atan}{\left (x \right )}}{3} - \frac {4 \sqrt {x}}{3} - \frac {\sqrt {2} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{6} + \frac {\sqrt {2} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{6} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{3} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{3} \]

[In]

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

[Out]

2*x**(3/2)*atan(x)/3 - 4*sqrt(x)/3 - sqrt(2)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)/6 + sqrt(2)*log(4*sqrt(2)*sqrt(
x) + 4*x + 4)/6 + sqrt(2)*atan(sqrt(2)*sqrt(x) - 1)/3 + sqrt(2)*atan(sqrt(2)*sqrt(x) + 1)/3

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.74 \[ \int \sqrt {x} \arctan (x) \, dx=\frac {2}{3} \, x^{\frac {3}{2}} \arctan \left (x\right ) + \frac {1}{3} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) + \frac {1}{3} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) + \frac {1}{6} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {1}{6} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {4}{3} \, \sqrt {x} \]

[In]

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

[Out]

2/3*x^(3/2)*arctan(x) + 1/3*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) + 1/3*sqrt(2)*arctan(-1/2*sqrt(2
)*(sqrt(2) - 2*sqrt(x))) + 1/6*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) - 1/6*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1
) - 4/3*sqrt(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.74 \[ \int \sqrt {x} \arctan (x) \, dx=\frac {2}{3} \, x^{\frac {3}{2}} \arctan \left (x\right ) + \frac {1}{3} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) + \frac {1}{3} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) + \frac {1}{6} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {1}{6} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {4}{3} \, \sqrt {x} \]

[In]

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

[Out]

2/3*x^(3/2)*arctan(x) + 1/3*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) + 1/3*sqrt(2)*arctan(-1/2*sqrt(2
)*(sqrt(2) - 2*sqrt(x))) + 1/6*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) - 1/6*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1
) - 4/3*sqrt(x)

Mupad [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.42 \[ \int \sqrt {x} \arctan (x) \, dx=\frac {2\,x^{3/2}\,\mathrm {atan}\left (x\right )}{3}-\frac {4\,\sqrt {x}}{3}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{3}+\frac {1}{3}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{3}-\frac {1}{3}{}\mathrm {i}\right ) \]

[In]

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

[Out]

(2*x^(3/2)*atan(x))/3 + 2^(1/2)*atan(2^(1/2)*x^(1/2)*(1/2 - 1i/2))*(1/3 + 1i/3) + 2^(1/2)*atan(2^(1/2)*x^(1/2)
*(1/2 + 1i/2))*(1/3 - 1i/3) - (4*x^(1/2))/3