3.324 \(\int \frac{\sqrt{x}}{(1+x^2)^2} \, dx\)

Optimal. Leaf size=113 \[ \frac{x^{3/2}}{2 \left (x^2+1\right )}+\frac{\log \left (x-\sqrt{2} \sqrt{x}+1\right )}{8 \sqrt{2}}-\frac{\log \left (x+\sqrt{2} \sqrt{x}+1\right )}{8 \sqrt{2}}-\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{x}+1\right )}{4 \sqrt{2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0600187, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {290, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{x^{3/2}}{2 \left (x^2+1\right )}+\frac{\log \left (x-\sqrt{2} \sqrt{x}+1\right )}{8 \sqrt{2}}-\frac{\log \left (x+\sqrt{2} \sqrt{x}+1\right )}{8 \sqrt{2}}-\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{x}+1\right )}{4 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(1 + x^2)^2,x]

[Out]

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

Rule 290

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

Rule 329

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 297

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

Rule 1162

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 617

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 204

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

Rule 1165

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 628

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]

Rubi steps

\begin{align*} \int \frac{\sqrt{x}}{\left (1+x^2\right )^2} \, dx &=\frac{x^{3/2}}{2 \left (1+x^2\right )}+\frac{1}{4} \int \frac{\sqrt{x}}{1+x^2} \, dx\\ &=\frac{x^{3/2}}{2 \left (1+x^2\right )}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,\sqrt{x}\right )\\ &=\frac{x^{3/2}}{2 \left (1+x^2\right )}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{x}\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{x}\right )\\ &=\frac{x^{3/2}}{2 \left (1+x^2\right )}+\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{x}\right )+\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{x}\right )+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{x}\right )}{8 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{x}\right )}{8 \sqrt{2}}\\ &=\frac{x^{3/2}}{2 \left (1+x^2\right )}+\frac{\log \left (1-\sqrt{2} \sqrt{x}+x\right )}{8 \sqrt{2}}-\frac{\log \left (1+\sqrt{2} \sqrt{x}+x\right )}{8 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}\\ &=\frac{x^{3/2}}{2 \left (1+x^2\right )}-\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}+\frac{\tan ^{-1}\left (1+\sqrt{2} \sqrt{x}\right )}{4 \sqrt{2}}+\frac{\log \left (1-\sqrt{2} \sqrt{x}+x\right )}{8 \sqrt{2}}-\frac{\log \left (1+\sqrt{2} \sqrt{x}+x\right )}{8 \sqrt{2}}\\ \end{align*}

Mathematica [C]  time = 0.0036141, size = 22, normalized size = 0.19 \[ \frac{2}{3} x^{3/2} \, _2F_1\left (\frac{3}{4},2;\frac{7}{4};-x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(1 + x^2)^2,x]

[Out]

(2*x^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, -x^2])/3

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 74, normalized size = 0.7 \begin{align*}{\frac{1}{2\,{x}^{2}+2}{x}^{{\frac{3}{2}}}}+{\frac{\sqrt{2}}{8}\arctan \left ( 1+\sqrt{2}\sqrt{x} \right ) }+{\frac{\sqrt{2}}{8}\arctan \left ( -1+\sqrt{2}\sqrt{x} \right ) }+{\frac{\sqrt{2}}{16}\ln \left ({ \left ( 1+x-\sqrt{2}\sqrt{x} \right ) \left ( 1+x+\sqrt{2}\sqrt{x} \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^(3/2)/(x^2+1)+1/8*arctan(1+2^(1/2)*x^(1/2))*2^(1/2)+1/8*arctan(-1+2^(1/2)*x^(1/2))*2^(1/2)+1/16*2^(1/2)*
ln((1+x-2^(1/2)*x^(1/2))/(1+x+2^(1/2)*x^(1/2)))

________________________________________________________________________________________

Maxima [A]  time = 3.21816, size = 116, normalized size = 1.03 \begin{align*} \frac{1}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{x}\right )}\right ) + \frac{1}{8} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{x}\right )}\right ) - \frac{1}{16} \, \sqrt{2} \log \left (\sqrt{2} \sqrt{x} + x + 1\right ) + \frac{1}{16} \, \sqrt{2} \log \left (-\sqrt{2} \sqrt{x} + x + 1\right ) + \frac{x^{\frac{3}{2}}}{2 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) + 1/8*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(x)))
- 1/16*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) + 1/16*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1) + 1/2*x^(3/2)/(x^2 +
1)

________________________________________________________________________________________

Fricas [A]  time = 1.46522, size = 439, normalized size = 3.88 \begin{align*} -\frac{4 \, \sqrt{2}{\left (x^{2} + 1\right )} \arctan \left (\sqrt{2} \sqrt{\sqrt{2} \sqrt{x} + x + 1} - \sqrt{2} \sqrt{x} - 1\right ) + 4 \, \sqrt{2}{\left (x^{2} + 1\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} \sqrt{-4 \, \sqrt{2} \sqrt{x} + 4 \, x + 4} - \sqrt{2} \sqrt{x} + 1\right ) + \sqrt{2}{\left (x^{2} + 1\right )} \log \left (4 \, \sqrt{2} \sqrt{x} + 4 \, x + 4\right ) - \sqrt{2}{\left (x^{2} + 1\right )} \log \left (-4 \, \sqrt{2} \sqrt{x} + 4 \, x + 4\right ) - 8 \, x^{\frac{3}{2}}}{16 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(4*sqrt(2)*(x^2 + 1)*arctan(sqrt(2)*sqrt(sqrt(2)*sqrt(x) + x + 1) - sqrt(2)*sqrt(x) - 1) + 4*sqrt(2)*(x^
2 + 1)*arctan(1/2*sqrt(2)*sqrt(-4*sqrt(2)*sqrt(x) + 4*x + 4) - sqrt(2)*sqrt(x) + 1) + sqrt(2)*(x^2 + 1)*log(4*
sqrt(2)*sqrt(x) + 4*x + 4) - sqrt(2)*(x^2 + 1)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4) - 8*x^(3/2))/(x^2 + 1)

________________________________________________________________________________________

Sympy [B]  time = 2.85631, size = 257, normalized size = 2.27 \begin{align*} \frac{8 x^{\frac{3}{2}}}{16 x^{2} + 16} + \frac{\sqrt{2} x^{2} \log{\left (- 4 \sqrt{2} \sqrt{x} + 4 x + 4 \right )}}{16 x^{2} + 16} - \frac{\sqrt{2} x^{2} \log{\left (4 \sqrt{2} \sqrt{x} + 4 x + 4 \right )}}{16 x^{2} + 16} + \frac{2 \sqrt{2} x^{2} \operatorname{atan}{\left (\sqrt{2} \sqrt{x} - 1 \right )}}{16 x^{2} + 16} + \frac{2 \sqrt{2} x^{2} \operatorname{atan}{\left (\sqrt{2} \sqrt{x} + 1 \right )}}{16 x^{2} + 16} + \frac{\sqrt{2} \log{\left (- 4 \sqrt{2} \sqrt{x} + 4 x + 4 \right )}}{16 x^{2} + 16} - \frac{\sqrt{2} \log{\left (4 \sqrt{2} \sqrt{x} + 4 x + 4 \right )}}{16 x^{2} + 16} + \frac{2 \sqrt{2} \operatorname{atan}{\left (\sqrt{2} \sqrt{x} - 1 \right )}}{16 x^{2} + 16} + \frac{2 \sqrt{2} \operatorname{atan}{\left (\sqrt{2} \sqrt{x} + 1 \right )}}{16 x^{2} + 16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*x**(3/2)/(16*x**2 + 16) + sqrt(2)*x**2*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) - sqrt(2)*x**2*log(4
*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) + 2*sqrt(2)*x**2*atan(sqrt(2)*sqrt(x) - 1)/(16*x**2 + 16) + 2*sqrt(
2)*x**2*atan(sqrt(2)*sqrt(x) + 1)/(16*x**2 + 16) + sqrt(2)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) -
sqrt(2)*log(4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) + 2*sqrt(2)*atan(sqrt(2)*sqrt(x) - 1)/(16*x**2 + 16) +
2*sqrt(2)*atan(sqrt(2)*sqrt(x) + 1)/(16*x**2 + 16)

________________________________________________________________________________________

Giac [A]  time = 1.88885, size = 116, normalized size = 1.03 \begin{align*} \frac{1}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{x}\right )}\right ) + \frac{1}{8} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{x}\right )}\right ) - \frac{1}{16} \, \sqrt{2} \log \left (\sqrt{2} \sqrt{x} + x + 1\right ) + \frac{1}{16} \, \sqrt{2} \log \left (-\sqrt{2} \sqrt{x} + x + 1\right ) + \frac{x^{\frac{3}{2}}}{2 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) + 1/8*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(x)))
- 1/16*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) + 1/16*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1) + 1/2*x^(3/2)/(x^2 +
1)