Integrand size = 13, antiderivative size = 122 \[ \int \frac {x^{7/2}}{\left (1+x^2\right )^2} \, dx=\frac {5 \sqrt {x}}{2}-\frac {x^{5/2}}{2 \left (1+x^2\right )}+\frac {5 \arctan \left (1-\sqrt {2} \sqrt {x}\right )}{4 \sqrt {2}}-\frac {5 \arctan \left (1+\sqrt {2} \sqrt {x}\right )}{4 \sqrt {2}}+\frac {5 \log \left (1-\sqrt {2} \sqrt {x}+x\right )}{8 \sqrt {2}}-\frac {5 \log \left (1+\sqrt {2} \sqrt {x}+x\right )}{8 \sqrt {2}} \]
-1/2*x^(5/2)/(x^2+1)-5/8*arctan(-1+2^(1/2)*x^(1/2))*2^(1/2)-5/8*arctan(1+2 ^(1/2)*x^(1/2))*2^(1/2)+5/16*ln(1+x-2^(1/2)*x^(1/2))*2^(1/2)-5/16*ln(1+x+2 ^(1/2)*x^(1/2))*2^(1/2)+5/2*x^(1/2)
Time = 0.14 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.59 \[ \int \frac {x^{7/2}}{\left (1+x^2\right )^2} \, dx=\frac {1}{8} \left (\frac {4 \sqrt {x} \left (5+4 x^2\right )}{1+x^2}-5 \sqrt {2} \arctan \left (\frac {-1+x}{\sqrt {2} \sqrt {x}}\right )-5 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {x}}{1+x}\right )\right ) \]
((4*Sqrt[x]*(5 + 4*x^2))/(1 + x^2) - 5*Sqrt[2]*ArcTan[(-1 + x)/(Sqrt[2]*Sq rt[x])] - 5*Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[x])/(1 + x)])/8
Time = 0.30 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {252, 262, 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.
| \(\displaystyle \int \frac {x^{7/2}}{\left (x^2+1\right )^2} \, dx\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {5}{4} \int \frac {x^{3/2}}{x^2+1}dx-\frac {x^{5/2}}{2 \left (x^2+1\right )}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {5}{4} \left (2 \sqrt {x}-\int \frac {1}{\sqrt {x} \left (x^2+1\right )}dx\right )-\frac {x^{5/2}}{2 \left (x^2+1\right )}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {5}{4} \left (2 \sqrt {x}-2 \int \frac {1}{x^2+1}d\sqrt {x}\right )-\frac {x^{5/2}}{2 \left (x^2+1\right )}\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle \frac {5}{4} \left (2 \sqrt {x}-2 \left (\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}+\frac {1}{2} \int \frac {x+1}{x^2+1}d\sqrt {x}\right )\right )-\frac {x^{5/2}}{2 \left (x^2+1\right )}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {5}{4} \left (2 \sqrt {x}-2 \left (\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}+\frac {1}{2} \int \frac {1}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}\right )\right )\right )-\frac {x^{5/2}}{2 \left (x^2+1\right )}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {5}{4} \left (2 \sqrt {x}-2 \left (\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}+\frac {1}{2} \left (\frac {\int \frac {1}{-x-1}d\left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-x-1}d\left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}\right )\right )\right )-\frac {x^{5/2}}{2 \left (x^2+1\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {5}{4} \left (2 \sqrt {x}-2 \left (\frac {1}{2} \int \frac {1-x}{x^2+1}d\sqrt {x}+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )\right )\right )-\frac {x^{5/2}}{2 \left (x^2+1\right )}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {5}{4} \left (2 \sqrt {x}-2 \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {x}}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {x}+1\right )}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )\right )\right )-\frac {x^{5/2}}{2 \left (x^2+1\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {5}{4} \left (2 \sqrt {x}-2 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {x}}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {x}+1\right )}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )\right )\right )-\frac {x^{5/2}}{2 \left (x^2+1\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5}{4} \left (2 \sqrt {x}-2 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {x}}{x-\sqrt {2} \sqrt {x}+1}d\sqrt {x}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {x}+1}{x+\sqrt {2} \sqrt {x}+1}d\sqrt {x}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )\right )\right )-\frac {x^{5/2}}{2 \left (x^2+1\right )}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {5}{4} \left (2 \sqrt {x}-2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {x}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (x+\sqrt {2} \sqrt {x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (x-\sqrt {2} \sqrt {x}+1\right )}{2 \sqrt {2}}\right )\right )\right )-\frac {x^{5/2}}{2 \left (x^2+1\right )}\) |
-1/2*x^(5/2)/(1 + x^2) + (5*(2*Sqrt[x] - 2*((-(ArcTan[1 - Sqrt[2]*Sqrt[x]] /Sqrt[2]) + ArcTan[1 + Sqrt[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)))/4
3.4.21.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
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]
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]]))
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]
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]
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]
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]
Time = 1.84 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.61
| method | result | size |
| derivativedivides | \(2 \sqrt {x}+\frac {\sqrt {x}}{2 x^{2}+2}-\frac {5 \sqrt {2}\, \left (\ln \left (\frac {1+x +\sqrt {2}\, \sqrt {x}}{1+x -\sqrt {2}\, \sqrt {x}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{16}\) | \(74\) |
| default | \(2 \sqrt {x}+\frac {\sqrt {x}}{2 x^{2}+2}-\frac {5 \sqrt {2}\, \left (\ln \left (\frac {1+x +\sqrt {2}\, \sqrt {x}}{1+x -\sqrt {2}\, \sqrt {x}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{16}\) | \(74\) |
| risch | \(\frac {\left (4 x^{2}+5\right ) \sqrt {x}}{2 x^{2}+2}-\frac {5 \sqrt {2}\, \left (\ln \left (\frac {1+x +\sqrt {2}\, \sqrt {x}}{1+x -\sqrt {2}\, \sqrt {x}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {x}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {x}\right )\right )}{16}\) | \(76\) |
| meijerg | \(\frac {\sqrt {x}\, \left (36 x^{2}+45\right )}{18 x^{2}+18}-\frac {5 \sqrt {x}\, \left (-\frac {\sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}+\sqrt {x^{2}}\right )}{2 \left (x^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}\right )}{\left (x^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}+\sqrt {x^{2}}\right )}{2 \left (x^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{2}\right )^{\frac {1}{4}}}\right )}{\left (x^{2}\right )^{\frac {1}{4}}}\right )}{8}\) | \(150\) |
| trager | \(\frac {\left (4 x^{2}+5\right ) \sqrt {x}}{2 x^{2}+2}-\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x +4 \sqrt {x}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}+x +1}\right )}{8}-\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}-\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x +4 \sqrt {x}-\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2}-x -1}\right )}{8}\) | \(201\) |
2*x^(1/2)+1/2*x^(1/2)/(x^2+1)-5/16*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)) )
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.97 \[ \int \frac {x^{7/2}}{\left (1+x^2\right )^2} \, dx=-\frac {5 \, \sqrt {2} {\left (\left (i + 1\right ) \, x^{2} + i + 1\right )} \log \left (\left (i + 1\right ) \, \sqrt {2} + 2 \, \sqrt {x}\right ) + 5 \, \sqrt {2} {\left (-\left (i - 1\right ) \, x^{2} - i + 1\right )} \log \left (-\left (i - 1\right ) \, \sqrt {2} + 2 \, \sqrt {x}\right ) + 5 \, \sqrt {2} {\left (\left (i - 1\right ) \, x^{2} + i - 1\right )} \log \left (\left (i - 1\right ) \, \sqrt {2} + 2 \, \sqrt {x}\right ) + 5 \, \sqrt {2} {\left (-\left (i + 1\right ) \, x^{2} - i - 1\right )} \log \left (-\left (i + 1\right ) \, \sqrt {2} + 2 \, \sqrt {x}\right ) - 8 \, {\left (4 \, x^{2} + 5\right )} \sqrt {x}}{16 \, {\left (x^{2} + 1\right )}} \]
-1/16*(5*sqrt(2)*((I + 1)*x^2 + I + 1)*log((I + 1)*sqrt(2) + 2*sqrt(x)) + 5*sqrt(2)*(-(I - 1)*x^2 - I + 1)*log(-(I - 1)*sqrt(2) + 2*sqrt(x)) + 5*sqr t(2)*((I - 1)*x^2 + I - 1)*log((I - 1)*sqrt(2) + 2*sqrt(x)) + 5*sqrt(2)*(- (I + 1)*x^2 - I - 1)*log(-(I + 1)*sqrt(2) + 2*sqrt(x)) - 8*(4*x^2 + 5)*sqr t(x))/(x^2 + 1)
Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (110) = 220\).
Time = 1.21 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.27 \[ \int \frac {x^{7/2}}{\left (1+x^2\right )^2} \, dx=\frac {32 x^{\frac {5}{2}}}{16 x^{2} + 16} + \frac {40 \sqrt {x}}{16 x^{2} + 16} + \frac {5 \sqrt {2} x^{2} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} - \frac {5 \sqrt {2} x^{2} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} - \frac {10 \sqrt {2} x^{2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{16 x^{2} + 16} - \frac {10 \sqrt {2} x^{2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{16 x^{2} + 16} + \frac {5 \sqrt {2} \log {\left (- 4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} - \frac {5 \sqrt {2} \log {\left (4 \sqrt {2} \sqrt {x} + 4 x + 4 \right )}}{16 x^{2} + 16} - \frac {10 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} - 1 \right )}}{16 x^{2} + 16} - \frac {10 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} \sqrt {x} + 1 \right )}}{16 x^{2} + 16} \]
32*x**(5/2)/(16*x**2 + 16) + 40*sqrt(x)/(16*x**2 + 16) + 5*sqrt(2)*x**2*lo g(-4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) - 5*sqrt(2)*x**2*log(4*sqrt (2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) - 10*sqrt(2)*x**2*atan(sqrt(2)*sqrt( x) - 1)/(16*x**2 + 16) - 10*sqrt(2)*x**2*atan(sqrt(2)*sqrt(x) + 1)/(16*x** 2 + 16) + 5*sqrt(2)*log(-4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) - 5*s qrt(2)*log(4*sqrt(2)*sqrt(x) + 4*x + 4)/(16*x**2 + 16) - 10*sqrt(2)*atan(s qrt(2)*sqrt(x) - 1)/(16*x**2 + 16) - 10*sqrt(2)*atan(sqrt(2)*sqrt(x) + 1)/ (16*x**2 + 16)
Time = 0.28 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.75 \[ \int \frac {x^{7/2}}{\left (1+x^2\right )^2} \, dx=-\frac {5}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) - \frac {5}{8} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) - \frac {5}{16} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) + \frac {5}{16} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) + 2 \, \sqrt {x} + \frac {\sqrt {x}}{2 \, {\left (x^{2} + 1\right )}} \]
-5/8*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) - 5/8*sqrt(2)*arcta n(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(x))) - 5/16*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) + 5/16*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1) + 2*sqrt(x) + 1/2*sqr t(x)/(x^2 + 1)
Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.75 \[ \int \frac {x^{7/2}}{\left (1+x^2\right )^2} \, dx=-\frac {5}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {x}\right )}\right ) - \frac {5}{8} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {x}\right )}\right ) - \frac {5}{16} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {x} + x + 1\right ) + \frac {5}{16} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {x} + x + 1\right ) + 2 \, \sqrt {x} + \frac {\sqrt {x}}{2 \, {\left (x^{2} + 1\right )}} \]
-5/8*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(x))) - 5/8*sqrt(2)*arcta n(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(x))) - 5/16*sqrt(2)*log(sqrt(2)*sqrt(x) + x + 1) + 5/16*sqrt(2)*log(-sqrt(2)*sqrt(x) + x + 1) + 2*sqrt(x) + 1/2*sqr t(x)/(x^2 + 1)
Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.45 \[ \int \frac {x^{7/2}}{\left (1+x^2\right )^2} \, dx=\frac {\sqrt {x}}{2\,\left (x^2+1\right )}+2\,\sqrt {x}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {5}{8}-\frac {5}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {x}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {5}{8}+\frac {5}{8}{}\mathrm {i}\right ) \]