\(\int \frac {1}{x^{7/2} (1+x^2)} \, dx\) [319]

Optimal result

Integrand size = 13, antiderivative size = 83 \[ \int \frac {1}{x^{7/2} \left (1+x^2\right )} \, dx=-\frac {2}{5 x^{5/2}}+\frac {2}{\sqrt {x}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {x}}{1+x}\right )}{\sqrt {2}} \] Output:

-2/5/x^(5/2)+2/x^(1/2)+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 = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^{7/2} \left (1+x^2\right )} \, dx=\frac {2 \left (-1+5 x^2\right )}{5 x^{5/2}}+\frac {\arctan \left (\frac {-1+x}{\sqrt {2} \sqrt {x}}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {x}}{1+x}\right )}{\sqrt {2}} \] Input:

Integrate[1/(x^(7/2)*(1 + x^2)),x]
 

Output:

(2*(-1 + 5*x^2))/(5*x^(5/2)) + ArcTan[(-1 + x)/(Sqrt[2]*Sqrt[x])]/Sqrt[2] 
- ArcTanh[(Sqrt[2]*Sqrt[x])/(1 + x)]/Sqrt[2]
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.46, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {264, 264, 266, 826, 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.

\(\displaystyle \int \frac {1}{x^{7/2} \left (x^2+1\right )} \, dx\)

\(\Big \downarrow \) 264

\(\displaystyle -\int \frac {1}{x^{3/2} \left (x^2+1\right )}dx-\frac {2}{5 x^{5/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle \int \frac {\sqrt {x}}{x^2+1}dx-\frac {2}{5 x^{5/2}}+\frac {2}{\sqrt {x}}\)

\(\Big \downarrow \) 266

\(\displaystyle 2 \int \frac {x}{x^2+1}d\sqrt {x}-\frac {2}{5 x^{5/2}}+\frac {2}{\sqrt {x}}\)

\(\Big \downarrow \) 826

\(\displaystyle 2 \left (\frac {1}{2} \int \frac {x+1}{x^2+1}d\sqrt {x}-\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}\right )-\frac {2}{5 x^{5/2}}+\frac {2}{\sqrt {x}}\)

\(\Big \downarrow \) 1476

\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}+\frac {1}{2} \int \frac {1}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}\right )-\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}\right )-\frac {2}{5 x^{5/2}}+\frac {2}{\sqrt {x}}\)

\(\Big \downarrow \) 1082

\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-x-1}d\left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-x-1}d\left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}\right )-\frac {2}{5 x^{5/2}}+\frac {2}{\sqrt {x}}\)

\(\Big \downarrow \) 217

\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}\right )-\frac {2}{5 x^{5/2}}+\frac {2}{\sqrt {x}}\)

\(\Big \downarrow \) 1479

\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-2 \sqrt {x}}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {x}+1\right )}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )\right )-\frac {2}{5 x^{5/2}}+\frac {2}{\sqrt {x}}\)

\(\Big \downarrow \) 25

\(\displaystyle 2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 \sqrt {x}}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {x}+1\right )}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )\right )-\frac {2}{5 x^{5/2}}+\frac {2}{\sqrt {x}}\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 \sqrt {x}}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}-\frac {1}{2} \int \frac {\sqrt {2} \sqrt {x}+1}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )\right )-\frac {2}{5 x^{5/2}}+\frac {2}{\sqrt {x}}\)

\(\Big \downarrow \) 1103

\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (x-\sqrt {2} \sqrt {x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (x+\sqrt {2} \sqrt {x}+1\right )}{2 \sqrt {2}}\right )\right )-\frac {2}{5 x^{5/2}}+\frac {2}{\sqrt {x}}\)

Input:

Int[1/(x^(7/2)*(1 + x^2)),x]
 

Output:

-2/(5*x^(5/2)) + 2/Sqrt[x] + 2*((-(ArcTan[1 - Sqrt[2]*Sqrt[x]]/Sqrt[2]) + 
ArcTan[1 + Sqrt[2]*Sqrt[x]]/Sqrt[2])/2 + (Log[1 - Sqrt[2]*Sqrt[x] + x]/(2* 
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 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   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 = 0.77 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {x -\sqrt {2}\, \sqrt {x}+1}{x +\sqrt {2}\, \sqrt {x}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{4}-\frac {2}{5 x^{\frac {5}{2}}}+\frac {2}{\sqrt {x}}\)	\(67\)
default	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {x -\sqrt {2}\, \sqrt {x}+1}{x +\sqrt {2}\, \sqrt {x}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{4}-\frac {2}{5 x^{\frac {5}{2}}}+\frac {2}{\sqrt {x}}\)	\(67\)
risch	\(\frac {2 x^{2}-\frac {2}{5}}{x^{\frac {5}{2}}}+\frac {\sqrt {2}\, \left (\ln \left (\frac {x -\sqrt {2}\, \sqrt {x}+1}{x +\sqrt {2}\, \sqrt {x}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{4}\)	\(69\)
meijerg	\(\frac {2}{\sqrt {x}}-\frac {2}{5 x^{\frac {5}{2}}}+\frac {x^{\frac {3}{2}} \left (\frac {\sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}+\sqrt {x^{2}}\right )}{2 \left (x^{2}\right )^{\frac {3}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}\right )}{\left (x^{2}\right )^{\frac {3}{4}}}-\frac {\sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}+\sqrt {x^{2}}\right )}{2 \left (x^{2}\right )^{\frac {3}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}\right )}{\left (x^{2}\right )^{\frac {3}{4}}}\right )}{2}\)	\(141\)
trager	\(\frac {2 x^{2}-\frac {2}{5}}{x^{\frac {5}{2}}}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )-4 \sqrt {x}}{\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}+x +1}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )+4 \sqrt {x}}{\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}-x -1}\right )}{2}\)	\(193\)

Input:

int(1/x^(7/2)/(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)))-2/5/x^(5/2)+2/x^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x^{7/2} \left (1+x^2\right )} \, dx=\frac {10 \, \sqrt {2} x^{3} \arctan \left (\sqrt {2} \sqrt {x} + 1\right ) + 10 \, \sqrt {2} x^{3} \arctan \left (\sqrt {2} \sqrt {x} - 1\right ) - 5 \, \sqrt {2} x^{3} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) + 5 \, \sqrt {2} x^{3} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) + 8 \, {\left (5 \, x^{2} - 1\right )} \sqrt {x}}{20 \, x^{3}} \] Input:

integrate(1/x^(7/2)/(x^2+1),x, algorithm="fricas")
 

Output:

1/20*(10*sqrt(2)*x^3*arctan(sqrt(2)*sqrt(x) + 1) + 10*sqrt(2)*x^3*arctan(s 
qrt(2)*sqrt(x) - 1) - 5*sqrt(2)*x^3*log(sqrt(2)*sqrt(x) + x + 1) + 5*sqrt( 
2)*x^3*log(-sqrt(2)*sqrt(x) + x + 1) + 8*(5*x^2 - 1)*sqrt(x))/x^3
 

Sympy [A] (verification not implemented)

Time = 0.64 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.27 \[ \int \frac {1}{x^{7/2} \left (1+x^2\right )} \, dx=\frac {\sqrt {2} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{4} - \frac {\sqrt {2} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{4} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{2} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{2} + \frac {2}{\sqrt {x}} - \frac {2}{5 x^{\frac {5}{2}}} \] Input:

integrate(1/x**(7/2)/(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/sqrt(x) - 2/(5*x**(5/2))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^{7/2} \left (1+x^2\right )} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) + \frac {2 \, {\left (5 \, x^{2} - 1\right )}}{5 \, x^{\frac {5}{2}}} \] Input:

integrate(1/x^(7/2)/(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)*arctan 
(-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) + 2/5*(5*x^2 - 1)/x^(5/2 
)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^{7/2} \left (1+x^2\right )} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) + \frac {2 \, {\left (5 \, x^{2} - 1\right )}}{5 \, x^{\frac {5}{2}}} \] Input:

integrate(1/x^(7/2)/(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)*arctan 
(-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) + 2/5*(5*x^2 - 1)/x^(5/2 
)
 

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.58 \[ \int \frac {1}{x^{7/2} \left (1+x^2\right )} \, dx=\frac {2\,x^2-\frac {2}{5}}{x^{5/2}}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right ) \] Input:

int(1/(x^(7/2)*(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*x^2 - 2/5)/x^(5/2)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.19 \[ \int \frac {1}{x^{7/2} \left (1+x^2\right )} \, dx=\frac {10 \sqrt {x}\, \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {x}-\sqrt {2}}{\sqrt {2}}\right ) x^{2}+10 \sqrt {x}\, \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {x}+\sqrt {2}}{\sqrt {2}}\right ) x^{2}+5 \sqrt {x}\, \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, \sqrt {2}+x +1\right ) x^{2}-5 \sqrt {x}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, \sqrt {2}+x +1\right ) x^{2}+40 x^{2}-8}{20 \sqrt {x}\, x^{2}} \] Input:

int(1/x^(7/2)/(x^2+1),x)
 

Output:

(10*sqrt(x)*sqrt(2)*atan((2*sqrt(x) - sqrt(2))/sqrt(2))*x**2 + 10*sqrt(x)* 
sqrt(2)*atan((2*sqrt(x) + sqrt(2))/sqrt(2))*x**2 + 5*sqrt(x)*sqrt(2)*log( 
- sqrt(x)*sqrt(2) + x + 1)*x**2 - 5*sqrt(x)*sqrt(2)*log(sqrt(x)*sqrt(2) + 
x + 1)*x**2 + 40*x**2 - 8)/(20*sqrt(x)*x**2)