Integrand size = 20, antiderivative size = 115 \[ \int \frac {1}{2-4 x+6 x^2-4 x^3+x^4} \, dx=-\frac {\arctan \left (1+\sqrt {2}-\sqrt {2} x\right )}{2 \sqrt {2}}+\frac {\arctan \left (1-\sqrt {2}+\sqrt {2} x\right )}{2 \sqrt {2}}-\frac {\log \left (2+\sqrt {2}-2 x-\sqrt {2} x+x^2\right )}{4 \sqrt {2}}+\frac {\log \left (2-\sqrt {2}-2 x+\sqrt {2} x+x^2\right )}{4 \sqrt {2}} \] Output:
1/4*arctan(-1-2^(1/2)+x*2^(1/2))*2^(1/2)+1/4*arctan(1-2^(1/2)+x*2^(1/2))*2 ^(1/2)-1/8*ln(2+2^(1/2)-2*x-x*2^(1/2)+x^2)*2^(1/2)+1/8*ln(2-2^(1/2)-2*x+x* 2^(1/2)+x^2)*2^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.48 \[ \int \frac {1}{2-4 x+6 x^2-4 x^3+x^4} \, dx=\frac {1}{4} \text {RootSum}\left [2-4 \text {$\#$1}+6 \text {$\#$1}^2-4 \text {$\#$1}^3+\text {$\#$1}^4\&,\frac {\log (x-\text {$\#$1})}{-1+3 \text {$\#$1}-3 \text {$\#$1}^2+\text {$\#$1}^3}\&\right ] \] Input:
Integrate[(2 - 4*x + 6*x^2 - 4*x^3 + x^4)^(-1),x]
Output:
RootSum[2 - 4*#1 + 6*#1^2 - 4*#1^3 + #1^4 & , Log[x - #1]/(-1 + 3*#1 - 3*# 1^2 + #1^3) & ]/4
Time = 0.52 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.50, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {2458, 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.
\(\displaystyle \int \frac {1}{x^4-4 x^3+6 x^2-4 x+2} \, dx\) |
\(\Big \downarrow \) 2458 |
\(\displaystyle \int \frac {1}{(x-1)^4+1}d(x-1)\) |
\(\Big \downarrow \) 755 |
\(\displaystyle \frac {1}{2} \int \frac {1-(x-1)^2}{(x-1)^4+1}d(x-1)+\frac {1}{2} \int \frac {(x-1)^2+1}{(x-1)^4+1}d(x-1)\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{(x-1)^2-\sqrt {2} (x-1)+1}d(x-1)+\frac {1}{2} \int \frac {1}{(x-1)^2+\sqrt {2} (x-1)+1}d(x-1)\right )+\frac {1}{2} \int \frac {1-(x-1)^2}{(x-1)^4+1}d(x-1)\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {1}{-\left (1-\sqrt {2} (x-1)\right )^2-1}d\left (1-\sqrt {2} (x-1)\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\left (\sqrt {2} (x-1)+1\right )^2-1}d\left (\sqrt {2} (x-1)+1\right )}{\sqrt {2}}\right )+\frac {1}{2} \int \frac {1-(x-1)^2}{(x-1)^4+1}d(x-1)\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \int \frac {1-(x-1)^2}{(x-1)^4+1}d(x-1)\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 (x-1)}{(x-1)^2-\sqrt {2} (x-1)+1}d(x-1)}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} (x-1)+1\right )}{(x-1)^2+\sqrt {2} (x-1)+1}d(x-1)}{2 \sqrt {2}}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 (x-1)}{(x-1)^2-\sqrt {2} (x-1)+1}d(x-1)}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} (x-1)+1\right )}{(x-1)^2+\sqrt {2} (x-1)+1}d(x-1)}{2 \sqrt {2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 (x-1)}{(x-1)^2-\sqrt {2} (x-1)+1}d(x-1)}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} (x-1)+1}{(x-1)^2+\sqrt {2} (x-1)+1}d(x-1)\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {\log \left ((x-1)^2+\sqrt {2} (x-1)+1\right )}{2 \sqrt {2}}-\frac {\log \left ((x-1)^2-\sqrt {2} (x-1)+1\right )}{2 \sqrt {2}}\right )\) |
Input:
Int[(2 - 4*x + 6*x^2 - 4*x^3 + x^4)^(-1),x]
Output:
(-1/2*Log[1 - Sqrt[2]*(-1 + x) + (-1 + x)^2]/Sqrt[2] + Log[1 + Sqrt[2]*(-1 + x) + (-1 + x)^2]/(2*Sqrt[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[(-(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[((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]
Int[(Pn_)^(p_.), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1]/(Exp on[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p, x], x, x + S] /; BinomialQ[Pn /. x -> x - S, x] || (IntegerQ[Exp on[Pn, x]/2] && TrinomialQ[Pn /. x -> x - S, x])] /; FreeQ[p, x] && PolyQ[P n, x] && GtQ[Expon[Pn, x], 2] && NeQ[Coeff[Pn, x, Expon[Pn, x] - 1], 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.41
method | result | size |
default | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-4 \textit {\_Z}^{3}+6 \textit {\_Z}^{2}-4 \textit {\_Z} +2\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}-3 \textit {\_R}^{2}+3 \textit {\_R} -1}\right )}{4}\) | \(47\) |
risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-4 \textit {\_Z}^{3}+6 \textit {\_Z}^{2}-4 \textit {\_Z} +2\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}-3 \textit {\_R}^{2}+3 \textit {\_R} -1}\right )}{4}\) | \(47\) |
Input:
int(1/(x^4-4*x^3+6*x^2-4*x+2),x,method=_RETURNVERBOSE)
Output:
1/4*sum(1/(_R^3-3*_R^2+3*_R-1)*ln(x-_R),_R=RootOf(_Z^4-4*_Z^3+6*_Z^2-4*_Z+ 2))
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.64 \[ \int \frac {1}{2-4 x+6 x^2-4 x^3+x^4} \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\sqrt {2} {\left (x - 1\right )} + 1\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (\sqrt {2} {\left (x - 1\right )} - 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} {\left (x - 1\right )} - 2 \, x + 2\right ) - \frac {1}{8} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} {\left (x - 1\right )} - 2 \, x + 2\right ) \] Input:
integrate(1/(x^4-4*x^3+6*x^2-4*x+2),x, algorithm="fricas")
Output:
1/4*sqrt(2)*arctan(sqrt(2)*(x - 1) + 1) + 1/4*sqrt(2)*arctan(sqrt(2)*(x - 1) - 1) + 1/8*sqrt(2)*log(x^2 + sqrt(2)*(x - 1) - 2*x + 2) - 1/8*sqrt(2)*l og(x^2 - sqrt(2)*(x - 1) - 2*x + 2)
Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.86 \[ \int \frac {1}{2-4 x+6 x^2-4 x^3+x^4} \, dx=- \frac {\sqrt {2} \log {\left (x^{2} + x \left (-2 - \sqrt {2}\right ) + \sqrt {2} + 2 \right )}}{8} + \frac {\sqrt {2} \log {\left (x^{2} + x \left (-2 + \sqrt {2}\right ) - \sqrt {2} + 2 \right )}}{8} - \frac {\sqrt {2} \operatorname {atan}{\left (- \sqrt {2} x + 1 + \sqrt {2} \right )}}{4} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - \sqrt {2} + 1 \right )}}{4} \] Input:
integrate(1/(x**4-4*x**3+6*x**2-4*x+2),x)
Output:
-sqrt(2)*log(x**2 + x*(-2 - sqrt(2)) + sqrt(2) + 2)/8 + sqrt(2)*log(x**2 + x*(-2 + sqrt(2)) - sqrt(2) + 2)/8 - sqrt(2)*atan(-sqrt(2)*x + 1 + sqrt(2) )/4 + sqrt(2)*atan(sqrt(2)*x - sqrt(2) + 1)/4
\[ \int \frac {1}{2-4 x+6 x^2-4 x^3+x^4} \, dx=\int { \frac {1}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 2} \,d x } \] Input:
integrate(1/(x^4-4*x^3+6*x^2-4*x+2),x, algorithm="maxima")
Output:
integrate(1/(x^4 - 4*x^3 + 6*x^2 - 4*x + 2), x)
Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.75 \[ \int \frac {1}{2-4 x+6 x^2-4 x^3+x^4} \, dx=\frac {1}{16} \, \sqrt {2} {\left (\pi + 4 \, \arctan \left (\sqrt {2} x - \sqrt {2} - 1\right )\right )} - \frac {1}{16} \, \sqrt {2} {\left (\pi + 4 \, \arctan \left (-\sqrt {2} x + \sqrt {2} - 1\right )\right )} + \frac {1}{8} \, \sqrt {2} \log \left ({\left (x + \sqrt {2} - 1\right )}^{2} + {\left (x - 1\right )}^{2}\right ) - \frac {1}{8} \, \sqrt {2} \log \left ({\left (x - \sqrt {2} - 1\right )}^{2} + {\left (x - 1\right )}^{2}\right ) \] Input:
integrate(1/(x^4-4*x^3+6*x^2-4*x+2),x, algorithm="giac")
Output:
1/16*sqrt(2)*(pi + 4*arctan(sqrt(2)*x - sqrt(2) - 1)) - 1/16*sqrt(2)*(pi + 4*arctan(-sqrt(2)*x + sqrt(2) - 1)) + 1/8*sqrt(2)*log((x + sqrt(2) - 1)^2 + (x - 1)^2) - 1/8*sqrt(2)*log((x - sqrt(2) - 1)^2 + (x - 1)^2)
Time = 0.09 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.43 \[ \int \frac {1}{2-4 x+6 x^2-4 x^3+x^4} \, dx=\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\left (-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\left (-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right ) \] Input:
int(1/(6*x^2 - 4*x - 4*x^3 + x^4 + 2),x)
Output:
2^(1/2)*atan(2^(1/2)*x*(1/2 - 1i/2) - 2^(1/2)*(1/2 - 1i/2))*(1/4 + 1i/4) + 2^(1/2)*atan(2^(1/2)*x*(1/2 + 1i/2) - 2^(1/2)*(1/2 + 1i/2))*(1/4 - 1i/4)
Time = 0.17 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.61 \[ \int \frac {1}{2-4 x+6 x^2-4 x^3+x^4} \, dx=\frac {\sqrt {2}\, \left (-2 \mathit {atan} \left (\frac {\sqrt {2}-2 x +2}{\sqrt {2}}\right )+2 \mathit {atan} \left (\frac {\sqrt {2}+2 x -2}{\sqrt {2}}\right )-\mathrm {log}\left (-\sqrt {2}\, x +\sqrt {2}+x^{2}-2 x +2\right )+\mathrm {log}\left (\sqrt {2}\, x -\sqrt {2}+x^{2}-2 x +2\right )\right )}{8} \] Input:
int(1/(x^4-4*x^3+6*x^2-4*x+2),x)
Output:
(sqrt(2)*( - 2*atan((sqrt(2) - 2*x + 2)/sqrt(2)) + 2*atan((sqrt(2) + 2*x - 2)/sqrt(2)) - log( - sqrt(2)*x + sqrt(2) + x**2 - 2*x + 2) + log(sqrt(2)* x - sqrt(2) + x**2 - 2*x + 2)))/8