Integrand size = 31, antiderivative size = 35 \[ \int \frac {-4+8 x-4 x^2+4 x^3-x^4+x^5}{\left (2+x^2\right )^3} \, dx=-\frac {1}{\left (2+x^2\right )^2}-\frac {\arctan \left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2}}+\frac {1}{2} \log \left (2+x^2\right ) \] Output:
-1/(x^2+2)^2-1/2*arctan(1/2*x*2^(1/2))*2^(1/2)+1/2*ln(x^2+2)
Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {-4+8 x-4 x^2+4 x^3-x^4+x^5}{\left (2+x^2\right )^3} \, dx=-\frac {1}{\left (2+x^2\right )^2}-\frac {\arctan \left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2}}+\frac {1}{2} \log \left (2+x^2\right ) \] Input:
Integrate[(-4 + 8*x - 4*x^2 + 4*x^3 - x^4 + x^5)/(2 + x^2)^3,x]
Output:
-(2 + x^2)^(-2) - ArcTan[x/Sqrt[2]]/Sqrt[2] + Log[2 + x^2]/2
Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {2345, 27, 2019, 452, 216, 240}
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.
\(\displaystyle \int \frac {x^5-x^4+4 x^3-4 x^2+8 x-4}{\left (x^2+2\right )^3} \, dx\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {1}{8} \int \frac {8 \left (-x^3+x^2-2 x+2\right )}{\left (x^2+2\right )^2}dx-\frac {1}{\left (x^2+2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {-x^3+x^2-2 x+2}{\left (x^2+2\right )^2}dx-\frac {1}{\left (x^2+2\right )^2}\) |
\(\Big \downarrow \) 2019 |
\(\displaystyle -\int \frac {1-x}{x^2+2}dx-\frac {1}{\left (x^2+2\right )^2}\) |
\(\Big \downarrow \) 452 |
\(\displaystyle -\int \frac {1}{x^2+2}dx+\int \frac {x}{x^2+2}dx-\frac {1}{\left (x^2+2\right )^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \int \frac {x}{x^2+2}dx-\frac {\arctan \left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {1}{\left (x^2+2\right )^2}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle -\frac {\arctan \left (\frac {x}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {1}{\left (x^2+2\right )^2}+\frac {1}{2} \log \left (x^2+2\right )\) |
Input:
Int[(-4 + 8*x - 4*x^2 + 4*x^3 - x^4 + x^5)/(2 + x^2)^3,x]
Output:
-(2 + x^2)^(-2) - ArcTan[x/Sqrt[2]]/Sqrt[2] + Log[2 + x^2]/2
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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c Int[1/ (a + b*x^2), x], x] + Simp[d Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c^2 + a*d^2, 0]
Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px , Qx, x]^p*Qx^(p + q), x] /; FreeQ[q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {1}{\left (x^{2}+2\right )^{2}}-\frac {\arctan \left (\frac {x \sqrt {2}}{2}\right ) \sqrt {2}}{2}+\frac {\ln \left (x^{2}+2\right )}{2}\) | \(31\) |
risch | \(-\frac {1}{\left (x^{2}+2\right )^{2}}-\frac {\arctan \left (\frac {x \sqrt {2}}{2}\right ) \sqrt {2}}{2}+\frac {\ln \left (x^{2}+2\right )}{2}\) | \(31\) |
meijerg | \(-\frac {\sqrt {2}\, \left (\frac {x \sqrt {2}\, \left (\frac {3 x^{2}}{2}+5\right )}{4 \left (\frac {x^{2}}{2}+1\right )^{2}}+\frac {3 \arctan \left (\frac {x \sqrt {2}}{2}\right )}{2}\right )}{8}-\frac {x^{2} \left (\frac {9 x^{2}}{2}+6\right )}{24 \left (\frac {x^{2}}{2}+1\right )^{2}}+\frac {\ln \left (\frac {x^{2}}{2}+1\right )}{2}-\frac {\sqrt {2}\, \left (-\frac {x \sqrt {2}\, \left (\frac {25 x^{2}}{2}+15\right )}{20 \left (\frac {x^{2}}{2}+1\right )^{2}}+\frac {3 \arctan \left (\frac {x \sqrt {2}}{2}\right )}{2}\right )}{8}+\frac {x^{4}}{8 \left (\frac {x^{2}}{2}+1\right )^{2}}-\frac {\sqrt {2}\, \left (-\frac {x \sqrt {2}\, \left (-\frac {3 x^{2}}{2}+3\right )}{12 \left (\frac {x^{2}}{2}+1\right )^{2}}+\frac {\arctan \left (\frac {x \sqrt {2}}{2}\right )}{2}\right )}{4}+\frac {x^{2} \left (\frac {x^{2}}{2}+2\right )}{4 \left (\frac {x^{2}}{2}+1\right )^{2}}\) | \(179\) |
Input:
int((x^5-x^4+4*x^3-4*x^2+8*x-4)/(x^2+2)^3,x,method=_RETURNVERBOSE)
Output:
-1/(x^2+2)^2-1/2*arctan(1/2*x*2^(1/2))*2^(1/2)+1/2*ln(x^2+2)
Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.57 \[ \int \frac {-4+8 x-4 x^2+4 x^3-x^4+x^5}{\left (2+x^2\right )^3} \, dx=-\frac {\sqrt {2} {\left (x^{4} + 4 \, x^{2} + 4\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) - {\left (x^{4} + 4 \, x^{2} + 4\right )} \log \left (x^{2} + 2\right ) + 2}{2 \, {\left (x^{4} + 4 \, x^{2} + 4\right )}} \] Input:
integrate((x^5-x^4+4*x^3-4*x^2+8*x-4)/(x^2+2)^3,x, algorithm="fricas")
Output:
-1/2*(sqrt(2)*(x^4 + 4*x^2 + 4)*arctan(1/2*sqrt(2)*x) - (x^4 + 4*x^2 + 4)* log(x^2 + 2) + 2)/(x^4 + 4*x^2 + 4)
Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {-4+8 x-4 x^2+4 x^3-x^4+x^5}{\left (2+x^2\right )^3} \, dx=\frac {\log {\left (x^{2} + 2 \right )}}{2} - \frac {\sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} \right )}}{2} - \frac {1}{x^{4} + 4 x^{2} + 4} \] Input:
integrate((x**5-x**4+4*x**3-4*x**2+8*x-4)/(x**2+2)**3,x)
Output:
log(x**2 + 2)/2 - sqrt(2)*atan(sqrt(2)*x/2)/2 - 1/(x**4 + 4*x**2 + 4)
Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {-4+8 x-4 x^2+4 x^3-x^4+x^5}{\left (2+x^2\right )^3} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) - \frac {1}{x^{4} + 4 \, x^{2} + 4} + \frac {1}{2} \, \log \left (x^{2} + 2\right ) \] Input:
integrate((x^5-x^4+4*x^3-4*x^2+8*x-4)/(x^2+2)^3,x, algorithm="maxima")
Output:
-1/2*sqrt(2)*arctan(1/2*sqrt(2)*x) - 1/(x^4 + 4*x^2 + 4) + 1/2*log(x^2 + 2 )
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int \frac {-4+8 x-4 x^2+4 x^3-x^4+x^5}{\left (2+x^2\right )^3} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) - \frac {1}{{\left (x^{2} + 2\right )}^{2}} + \frac {1}{2} \, \log \left (x^{2} + 2\right ) \] Input:
integrate((x^5-x^4+4*x^3-4*x^2+8*x-4)/(x^2+2)^3,x, algorithm="giac")
Output:
-1/2*sqrt(2)*arctan(1/2*sqrt(2)*x) - 1/(x^2 + 2)^2 + 1/2*log(x^2 + 2)
Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {-4+8 x-4 x^2+4 x^3-x^4+x^5}{\left (2+x^2\right )^3} \, dx=\frac {\ln \left (x^2+2\right )}{2}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,x}{2}\right )}{2}-\frac {1}{x^4+4\,x^2+4} \] Input:
int((8*x - 4*x^2 + 4*x^3 - x^4 + x^5 - 4)/(x^2 + 2)^3,x)
Output:
log(x^2 + 2)/2 - (2^(1/2)*atan((2^(1/2)*x)/2))/2 - 1/(4*x^2 + x^4 + 4)
Time = 0.16 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.43 \[ \int \frac {-4+8 x-4 x^2+4 x^3-x^4+x^5}{\left (2+x^2\right )^3} \, dx=\frac {-\sqrt {2}\, \mathit {atan} \left (\frac {x}{\sqrt {2}}\right ) x^{4}-4 \sqrt {2}\, \mathit {atan} \left (\frac {x}{\sqrt {2}}\right ) x^{2}-4 \sqrt {2}\, \mathit {atan} \left (\frac {x}{\sqrt {2}}\right )+\mathrm {log}\left (x^{2}+2\right ) x^{4}+4 \,\mathrm {log}\left (x^{2}+2\right ) x^{2}+4 \,\mathrm {log}\left (x^{2}+2\right )-2}{2 x^{4}+8 x^{2}+8} \] Input:
int((x^5-x^4+4*x^3-4*x^2+8*x-4)/(x^2+2)^3,x)
Output:
( - sqrt(2)*atan(x/sqrt(2))*x**4 - 4*sqrt(2)*atan(x/sqrt(2))*x**2 - 4*sqrt (2)*atan(x/sqrt(2)) + log(x**2 + 2)*x**4 + 4*log(x**2 + 2)*x**2 + 4*log(x* *2 + 2) - 2)/(2*(x**4 + 4*x**2 + 4))