\(\int \frac {1}{x^{5/2} (1+x^2)} \, dx\) [318]

Optimal result

Integrand size = 13, antiderivative size = 76 \[ \int \frac {1}{x^{5/2} \left (1+x^2\right )} \, dx=-\frac {2}{3 x^{3/2}}+\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {x}}{1+x}\right )}{\sqrt {2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^{5/2} \left (1+x^2\right )} \, dx=-\frac {2}{3 x^{3/2}}-\frac {\arctan \left (\frac {-1+x}{\sqrt {2} \sqrt {x}}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {x}}{1+x}\right )}{\sqrt {2}} \] Input:

Integrate[1/(x^(5/2)*(1 + x^2)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.50, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {264, 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.

\(\displaystyle \int \frac {1}{x^{5/2} \left (x^2+1\right )} \, dx\)

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[1/(x^(5/2)*(1 + x^2)),x]
 

Output:

-2/(3*x^(3/2)) - 2*((-(ArcTan[1 - Sqrt[2]*Sqrt[x]]/Sqrt[2]) + ArcTan[1 + S 
qrt[2]*Sqrt[x]]/Sqrt[2])/2 + (-1/2*Log[1 - Sqrt[2]*Sqrt[x] + x]/Sqrt[2] + 
Log[1 + Sqrt[2]*Sqrt[x] + x]/(2*Sqrt[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 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -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 = 0.50 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(-\frac {\sqrt {2}\, \left (\ln \left (\frac {x +\sqrt {2}\, \sqrt {x}+1}{x -\sqrt {2}\, \sqrt {x}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{4}-\frac {2}{3 x^{\frac {3}{2}}}\)	\(62\)
default	\(-\frac {\sqrt {2}\, \left (\ln \left (\frac {x +\sqrt {2}\, \sqrt {x}+1}{x -\sqrt {2}\, \sqrt {x}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{4}-\frac {2}{3 x^{\frac {3}{2}}}\)	\(62\)
risch	\(-\frac {\sqrt {2}\, \left (\ln \left (\frac {x +\sqrt {2}\, \sqrt {x}+1}{x -\sqrt {2}\, \sqrt {x}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{4}-\frac {2}{3 x^{\frac {3}{2}}}\)	\(62\)
meijerg	\(-\frac {2}{3 x^{\frac {3}{2}}}-\frac {\sqrt {x}\, \left (-\frac {\sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}+\sqrt {x^{2}}\right )}{2 \left (x^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}\right )}{\left (x^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}+\sqrt {x^{2}}\right )}{2 \left (x^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}\right )}{\left (x^{2}\right )^{\frac {1}{4}}}\right )}{2}\)	\(136\)
trager	\(-\frac {2}{3 x^{\frac {3}{2}}}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )+4 \sqrt {x}}{\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}+x +1}\right )}{2}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )+4 \sqrt {x}}{\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}-x -1}\right )}{2}\)	\(187\)

Input:

int(1/x^(5/2)/(x^2+1),x,method=_RETURNVERBOSE)
 

Output:

-1/4*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)))-2/3/x^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^{5/2} \left (1+x^2\right )} \, dx=-\frac {6 \, \sqrt {2} x^{2} \arctan \left (\sqrt {2} \sqrt {x} + 1\right ) + 6 \, \sqrt {2} x^{2} \arctan \left (\sqrt {2} \sqrt {x} - 1\right ) + 3 \, \sqrt {2} x^{2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) - 3 \, \sqrt {2} x^{2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) + 8 \, \sqrt {x}}{12 \, x^{2}} \] Input:

integrate(1/x^(5/2)/(x^2+1),x, algorithm="fricas")
 

Output:

-1/12*(6*sqrt(2)*x^2*arctan(sqrt(2)*sqrt(x) + 1) + 6*sqrt(2)*x^2*arctan(sq 
rt(2)*sqrt(x) - 1) + 3*sqrt(2)*x^2*log(sqrt(2)*sqrt(x) + x + 1) - 3*sqrt(2 
)*x^2*log(-sqrt(2)*sqrt(x) + x + 1) + 8*sqrt(x))/x^2
 

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.30 \[ \int \frac {1}{x^{5/2} \left (1+x^2\right )} \, dx=\frac {\sqrt {2} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{4} - \frac {\sqrt {2} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{4} - \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{2} - \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{2} - \frac {2}{3 x^{\frac {3}{2}}} \] Input:

integrate(1/x**(5/2)/(x**2+1),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^{5/2} \left (1+x^2\right )} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {2}{3 \, x^{\frac {3}{2}}} \] Input:

integrate(1/x^(5/2)/(x^2+1),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^{5/2} \left (1+x^2\right )} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) - \frac {2}{3 \, x^{\frac {3}{2}}} \] Input:

integrate(1/x^(5/2)/(x^2+1),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.55 \[ \int \frac {1}{x^{5/2} \left (1+x^2\right )} \, dx=-\frac {2}{3\,x^{3/2}}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right ) \] Input:

int(1/(x^(5/2)*(x^2 + 1)),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^{5/2} \left (1+x^2\right )} \, dx=\frac {-6 \sqrt {x}\, \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {x}-\sqrt {2}}{\sqrt {2}}\right ) x -6 \sqrt {x}\, \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {x}+\sqrt {2}}{\sqrt {2}}\right ) x +3 \sqrt {x}\, \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, \sqrt {2}+x +1\right ) x -3 \sqrt {x}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, \sqrt {2}+x +1\right ) x -8}{12 \sqrt {x}\, x} \] Input:

int(1/x^(5/2)/(x^2+1),x)
 

Output:

( - 6*sqrt(x)*sqrt(2)*atan((2*sqrt(x) - sqrt(2))/sqrt(2))*x - 6*sqrt(x)*sq 
rt(2)*atan((2*sqrt(x) + sqrt(2))/sqrt(2))*x + 3*sqrt(x)*sqrt(2)*log( - sqr 
t(x)*sqrt(2) + x + 1)*x - 3*sqrt(x)*sqrt(2)*log(sqrt(x)*sqrt(2) + x + 1)*x 
 - 8)/(12*sqrt(x)*x)