\(\int \frac {1}{\sqrt {x} (1+x^2)^2} \, dx\) [324]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [B] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 90 \[ \int \frac {1}{\sqrt {x} \left (1+x^2\right )^2} \, dx=\frac {\sqrt {x}}{2 \left (1+x^2\right )}-\frac {3 \arctan \left (1-\sqrt {2} \sqrt {x}\right )}{4 \sqrt {2}}+\frac {3 \arctan \left (1+\sqrt {2} \sqrt {x}\right )}{4 \sqrt {2}}+\frac {3 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {x}}{1+x}\right )}{4 \sqrt {2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.72 \[ \int \frac {1}{\sqrt {x} \left (1+x^2\right )^2} \, dx=\frac {1}{8} \left (\frac {4 \sqrt {x}}{1+x^2}+3 \sqrt {2} \arctan \left (\frac {-1+x}{\sqrt {2} \sqrt {x}}\right )+3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {x}}{1+x}\right )\right ) \] Input:

Integrate[1/(Sqrt[x]*(1 + x^2)^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.37, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {253, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\sqrt {x} \left (x^2+1\right )^2} \, dx\)

\(\Big \downarrow \) 253

\(\displaystyle \frac {3}{4} \int \frac {1}{\sqrt {x} \left (x^2+1\right )}dx+\frac {\sqrt {x}}{2 \left (x^2+1\right )}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {3}{2} \int \frac {1}{x^2+1}d\sqrt {x}+\frac {\sqrt {x}}{2 \left (x^2+1\right )}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {3}{2} \left (\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}+\frac {1}{2} \int \frac {x+1}{x^2+1}d\sqrt {x}\right )+\frac {\sqrt {x}}{2 \left (x^2+1\right )}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {3}{2} \left (\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}+\frac {1}{2} \int \frac {1}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}\right )\right )+\frac {\sqrt {x}}{2 \left (x^2+1\right )}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {3}{2} \left (\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}+\frac {1}{2} \left (\frac {\int \frac {1}{-x-1}d\left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-x-1}d\left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}\right )\right )+\frac {\sqrt {x}}{2 \left (x^2+1\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {3}{2} \left (\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )\right )+\frac {\sqrt {x}}{2 \left (x^2+1\right )}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {3}{2} \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {x}}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {x}+1\right )}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )\right )+\frac {\sqrt {x}}{2 \left (x^2+1\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {3}{2} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {x}}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {x}+1\right )}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )\right )+\frac {\sqrt {x}}{2 \left (x^2+1\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{2} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {x}}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {x}+1}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )\right )+\frac {\sqrt {x}}{2 \left (x^2+1\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {3}{2} \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (x+\sqrt {2} \sqrt {x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (x-\sqrt {2} \sqrt {x}+1\right )}{2 \sqrt {2}}\right )\right )+\frac {\sqrt {x}}{2 \left (x^2+1\right )}\)

Input:

Int[1/(Sqrt[x]*(1 + x^2)^2),x]
 

Output:

Sqrt[x]/(2*(1 + x^2)) + (3*((-(ArcTan[1 - Sqrt[2]*Sqrt[x]]/Sqrt[2]) + ArcT 
an[1 + Sqrt[2]*Sqrt[x]]/Sqrt[2])/2 + (-1/2*Log[1 - Sqrt[2]*Sqrt[x] + x]/Sq 
rt[2] + Log[1 + Sqrt[2]*Sqrt[x] + x]/(2*Sqrt[2]))/2))/2
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217
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 253
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x 
)^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 
2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m 
}, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 266
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 755
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[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 1082
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[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])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1476
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
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 1479
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 
Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.77

method result size
derivativedivides \(\frac {\sqrt {x}}{2 x^{2}+2}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {x +\sqrt {2}\, \sqrt {x}+1}{x -\sqrt {2}\, \sqrt {x}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{16}\) \(69\)
default \(\frac {\sqrt {x}}{2 x^{2}+2}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {x +\sqrt {2}\, \sqrt {x}+1}{x -\sqrt {2}\, \sqrt {x}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{16}\) \(69\)
risch \(\frac {\sqrt {x}}{2 x^{2}+2}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {x +\sqrt {2}\, \sqrt {x}+1}{x -\sqrt {2}\, \sqrt {x}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{16}\) \(69\)
meijerg \(\frac {2 \sqrt {x}}{4 x^{2}+4}-\frac {3 \sqrt {x}\, \sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}+\sqrt {x^{2}}\right )}{16 \left (x^{2}\right )^{\frac {1}{4}}}+\frac {3 \sqrt {x}\, \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}\right )}{8 \left (x^{2}\right )^{\frac {1}{4}}}+\frac {3 \sqrt {x}\, \sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}+\sqrt {x^{2}}\right )}{16 \left (x^{2}\right )^{\frac {1}{4}}}+\frac {3 \sqrt {x}\, \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}\right )}{8 \left (x^{2}\right )^{\frac {1}{4}}}\) \(153\)
trager \(\frac {\sqrt {x}}{2 x^{2}+2}-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )-4 \sqrt {x}}{\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}-x -1}\right )}{8}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )+4 \sqrt {x}}{\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}+x +1}\right )}{8}\) \(193\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\sqrt {x} \left (1+x^2\right )^2} \, dx=\frac {6 \, \sqrt {2} {\left (x^{2} + 1\right )} \arctan \left (\sqrt {2} \sqrt {x} + 1\right ) + 6 \, \sqrt {2} {\left (x^{2} + 1\right )} \arctan \left (\sqrt {2} \sqrt {x} - 1\right ) + 3 \, \sqrt {2} {\left (x^{2} + 1\right )} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) - 3 \, \sqrt {2} {\left (x^{2} + 1\right )} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) + 8 \, \sqrt {x}}{16 \, {\left (x^{2} + 1\right )}} \] Input:

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

Output:

1/16*(6*sqrt(2)*(x^2 + 1)*arctan(sqrt(2)*sqrt(x) + 1) + 6*sqrt(2)*(x^2 + 1 
)*arctan(sqrt(2)*sqrt(x) - 1) + 3*sqrt(2)*(x^2 + 1)*log(sqrt(2)*sqrt(x) + 
x + 1) - 3*sqrt(2)*(x^2 + 1)*log(-sqrt(2)*sqrt(x) + x + 1) + 8*sqrt(x))/(x 
^2 + 1)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (78) = 156\).

Time = 0.52 (sec) , antiderivative size = 264, normalized size of antiderivative = 2.93 \[ \int \frac {1}{\sqrt {x} \left (1+x^2\right )^2} \, dx=\frac {8 \sqrt {x}}{16 x^{2} + 16} - \frac {3 \sqrt {2} x^{2} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} + \frac {3 \sqrt {2} x^{2} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} + \frac {6 \sqrt {2} x^{2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{16 x^{2} + 16} + \frac {6 \sqrt {2} x^{2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{16 x^{2} + 16} - \frac {3 \sqrt {2} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} + \frac {3 \sqrt {2} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} + \frac {6 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{16 x^{2} + 16} + \frac {6 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{16 x^{2} + 16} \] Input:

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

Output:

8*sqrt(x)/(16*x**2 + 16) - 3*sqrt(2)*x**2*log(-4*sqrt(2)*sqrt(x) + 4*x + 4 
)/(16*x**2 + 16) + 3*sqrt(2)*x**2*log(4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x** 
2 + 16) + 6*sqrt(2)*x**2*atan(sqrt(2)*sqrt(x) - 1)/(16*x**2 + 16) + 6*sqrt 
(2)*x**2*atan(sqrt(2)*sqrt(x) + 1)/(16*x**2 + 16) - 3*sqrt(2)*log(-4*sqrt( 
2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) + 3*sqrt(2)*log(4*sqrt(2)*sqrt(x) + 4 
*x + 4)/(16*x**2 + 16) + 6*sqrt(2)*atan(sqrt(2)*sqrt(x) - 1)/(16*x**2 + 16 
) + 6*sqrt(2)*atan(sqrt(2)*sqrt(x) + 1)/(16*x**2 + 16)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\sqrt {x} \left (1+x^2\right )^2} \, dx=\frac {3}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) + \frac {3}{8} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) + \frac {3}{16} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {3}{16} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) + \frac {\sqrt {x}}{2 \, {\left (x^{2} + 1\right )}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\sqrt {x} \left (1+x^2\right )^2} \, dx=\frac {3}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) + \frac {3}{8} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) + \frac {3}{16} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {3}{16} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) + \frac {\sqrt {x}}{2 \, {\left (x^{2} + 1\right )}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.56 \[ \int \frac {1}{\sqrt {x} \left (1+x^2\right )^2} \, dx=\frac {\sqrt {x}}{2\,\left (x^2+1\right )}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {3}{8}+\frac {3}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {3}{8}-\frac {3}{8}{}\mathrm {i}\right ) \] Input:

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

Output:

2^(1/2)*atan(2^(1/2)*x^(1/2)*(1/2 - 1i/2))*(3/8 + 3i/8) + 2^(1/2)*atan(2^( 
1/2)*x^(1/2)*(1/2 + 1i/2))*(3/8 - 3i/8) + x^(1/2)/(2*(x^2 + 1))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.70 \[ \int \frac {1}{\sqrt {x} \left (1+x^2\right )^2} \, dx=\frac {6 \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {x}-\sqrt {2}}{\sqrt {2}}\right ) x^{2}+6 \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {x}-\sqrt {2}}{\sqrt {2}}\right )+6 \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {x}+\sqrt {2}}{\sqrt {2}}\right ) x^{2}+6 \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {x}+\sqrt {2}}{\sqrt {2}}\right )+8 \sqrt {x}-3 \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, \sqrt {2}+x +1\right ) x^{2}-3 \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, \sqrt {2}+x +1\right )+3 \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, \sqrt {2}+x +1\right ) x^{2}+3 \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, \sqrt {2}+x +1\right )}{16 x^{2}+16} \] Input:

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

Output:

(6*sqrt(2)*atan((2*sqrt(x) - sqrt(2))/sqrt(2))*x**2 + 6*sqrt(2)*atan((2*sq 
rt(x) - sqrt(2))/sqrt(2)) + 6*sqrt(2)*atan((2*sqrt(x) + sqrt(2))/sqrt(2))* 
x**2 + 6*sqrt(2)*atan((2*sqrt(x) + sqrt(2))/sqrt(2)) + 8*sqrt(x) - 3*sqrt( 
2)*log( - sqrt(x)*sqrt(2) + x + 1)*x**2 - 3*sqrt(2)*log( - sqrt(x)*sqrt(2) 
 + x + 1) + 3*sqrt(2)*log(sqrt(x)*sqrt(2) + x + 1)*x**2 + 3*sqrt(2)*log(sq 
rt(x)*sqrt(2) + x + 1))/(16*(x**2 + 1))