Integrand size = 16, antiderivative size = 65 \[ \int \frac {1}{x^4 \left (1-x^2+x^4\right )} \, dx=-\frac {1}{3 x^3}-\frac {1}{x}+\frac {1}{2} \arctan \left (\sqrt {3}-2 x\right )-\frac {1}{2} \arctan \left (\sqrt {3}+2 x\right )+\frac {\text {arctanh}\left (\frac {\sqrt {3} x}{1+x^2}\right )}{2 \sqrt {3}} \] Output:
-1/3/x^3-1/x-1/2*arctan(-3^(1/2)+2*x)-1/2*arctan(3^(1/2)+2*x)+1/6*arctanh( 3^(1/2)*x/(x^2+1))*3^(1/2)
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.52 \[ \int \frac {1}{x^4 \left (1-x^2+x^4\right )} \, dx=-\frac {1}{3 x^3}-\frac {1}{x}-\frac {\left (i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x\right )}{\sqrt {-6+6 i \sqrt {3}}}-\frac {\left (-i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x\right )}{\sqrt {-6-6 i \sqrt {3}}} \] Input:
Integrate[1/(x^4*(1 - x^2 + x^4)),x]
Output:
-1/3*1/x^3 - x^(-1) - ((I + Sqrt[3])*ArcTan[((1 - I*Sqrt[3])*x)/2])/Sqrt[- 6 + (6*I)*Sqrt[3]] - ((-I + Sqrt[3])*ArcTan[((1 + I*Sqrt[3])*x)/2])/Sqrt[- 6 - (6*I)*Sqrt[3]]
Time = 0.50 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.38, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {1443, 27, 1604, 1447, 1475, 1083, 217, 1478, 25, 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.
\(\displaystyle \int \frac {1}{x^4 \left (x^4-x^2+1\right )} \, dx\) |
\(\Big \downarrow \) 1443 |
\(\displaystyle \frac {1}{3} \int \frac {3 \left (1-x^2\right )}{x^2 \left (x^4-x^2+1\right )}dx-\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1-x^2}{x^2 \left (x^4-x^2+1\right )}dx-\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 1604 |
\(\displaystyle -\int \frac {x^2}{x^4-x^2+1}dx-\frac {1}{3 x^3}-\frac {1}{x}\) |
\(\Big \downarrow \) 1447 |
\(\displaystyle \frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx-\frac {1}{2} \int \frac {x^2+1}{x^4-x^2+1}dx-\frac {1}{3 x^3}-\frac {1}{x}\) |
\(\Big \downarrow \) 1475 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^2-\sqrt {3} x+1}dx-\frac {1}{2} \int \frac {1}{x^2+\sqrt {3} x+1}dx\right )+\frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx-\frac {1}{3 x^3}-\frac {1}{x}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx+\frac {1}{2} \left (\int \frac {1}{-\left (2 x-\sqrt {3}\right )^2-1}d\left (2 x-\sqrt {3}\right )+\int \frac {1}{-\left (2 x+\sqrt {3}\right )^2-1}d\left (2 x+\sqrt {3}\right )\right )-\frac {1}{3 x^3}-\frac {1}{x}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx+\frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x\right )-\arctan \left (2 x+\sqrt {3}\right )\right )-\frac {1}{3 x^3}-\frac {1}{x}\) |
\(\Big \downarrow \) 1478 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int -\frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\int -\frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x\right )-\arctan \left (2 x+\sqrt {3}\right )\right )-\frac {1}{3 x^3}-\frac {1}{x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {\int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x\right )-\arctan \left (2 x+\sqrt {3}\right )\right )-\frac {1}{3 x^3}-\frac {1}{x}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x\right )-\arctan \left (2 x+\sqrt {3}\right )\right )-\frac {1}{3 x^3}+\frac {1}{2} \left (\frac {\log \left (x^2+\sqrt {3} x+1\right )}{2 \sqrt {3}}-\frac {\log \left (x^2-\sqrt {3} x+1\right )}{2 \sqrt {3}}\right )-\frac {1}{x}\) |
Input:
Int[1/(x^4*(1 - x^2 + x^4)),x]
Output:
-1/3*1/x^3 - x^(-1) + (ArcTan[Sqrt[3] - 2*x] - ArcTan[Sqrt[3] + 2*x])/2 + (-1/2*Log[1 - Sqrt[3]*x + x^2]/Sqrt[3] + Log[1 + Sqrt[3]*x + x^2]/(2*Sqrt[ 3]))/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_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{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_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1)/(a*d*(m + 1))), x] - Sim p[1/(a*d^2*(m + 1)) Int[(d*x)^(m + 2)*(b*(m + 2*p + 3) + c*(m + 4*p + 5)* x^2)*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Int[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ a/c, 2]}, Simp[1/2 Int[(q + x^2)/(a + b*x^2 + c*x^4), x], x] - Simp[1/2 Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && LtQ[b ^2 - 4*a*c, 0] && PosQ[a*c]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[2*(d/e) - b/c, 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]] /; F reeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !LtQ[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2] , 0]))
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[-2*(d/e) - b/c, 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]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ [c*d^2 - a*e^2, 0] && !GtQ[b^2 - 4*a*c, 0]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( x_)^4)^(p_), x_Symbol] :> Simp[d*(f*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1) /(a*f*(m + 1))), x] + Simp[1/(a*f^2*(m + 1)) Int[(f*x)^(m + 2)*(a + b*x^2 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x ], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[ m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.68
method | result | size |
risch | \(\frac {-x^{2}-\frac {1}{3}}{x^{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+3 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-6 \textit {\_R}^{3}-\textit {\_R} +x \right )\right )}{2}\) | \(44\) |
default | \(-\frac {1}{3 x^{3}}-\frac {1}{x}+\frac {\sqrt {3}\, \left (-\frac {\ln \left (x^{2}-\sqrt {3}\, x +1\right )}{2}-\sqrt {3}\, \arctan \left (2 x -\sqrt {3}\right )\right )}{6}+\frac {\sqrt {3}\, \left (\frac {\ln \left (x^{2}+\sqrt {3}\, x +1\right )}{2}-\sqrt {3}\, \arctan \left (2 x +\sqrt {3}\right )\right )}{6}\) | \(79\) |
Input:
int(1/x^4/(x^4-x^2+1),x,method=_RETURNVERBOSE)
Output:
(-x^2-1/3)/x^3+1/2*sum(_R*ln(-6*_R^3-_R+x),_R=RootOf(9*_Z^4+3*_Z^2+1))
Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x^4 \left (1-x^2+x^4\right )} \, dx=\frac {\sqrt {3} x^{3} \log \left (x^{2} + \sqrt {3} x + 1\right ) - \sqrt {3} x^{3} \log \left (x^{2} - \sqrt {3} x + 1\right ) - 6 \, x^{3} \arctan \left (2 \, x + \sqrt {3}\right ) + 6 \, x^{3} \arctan \left (-2 \, x + \sqrt {3}\right ) - 12 \, x^{2} - 4}{12 \, x^{3}} \] Input:
integrate(1/x^4/(x^4-x^2+1),x, algorithm="fricas")
Output:
1/12*(sqrt(3)*x^3*log(x^2 + sqrt(3)*x + 1) - sqrt(3)*x^3*log(x^2 - sqrt(3) *x + 1) - 6*x^3*arctan(2*x + sqrt(3)) + 6*x^3*arctan(-2*x + sqrt(3)) - 12* x^2 - 4)/x^3
Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x^4 \left (1-x^2+x^4\right )} \, dx=- \frac {\sqrt {3} \log {\left (x^{2} - \sqrt {3} x + 1 \right )}}{12} + \frac {\sqrt {3} \log {\left (x^{2} + \sqrt {3} x + 1 \right )}}{12} - \frac {\operatorname {atan}{\left (2 x - \sqrt {3} \right )}}{2} - \frac {\operatorname {atan}{\left (2 x + \sqrt {3} \right )}}{2} + \frac {- 3 x^{2} - 1}{3 x^{3}} \] Input:
integrate(1/x**4/(x**4-x**2+1),x)
Output:
-sqrt(3)*log(x**2 - sqrt(3)*x + 1)/12 + sqrt(3)*log(x**2 + sqrt(3)*x + 1)/ 12 - atan(2*x - sqrt(3))/2 - atan(2*x + sqrt(3))/2 + (-3*x**2 - 1)/(3*x**3 )
\[ \int \frac {1}{x^4 \left (1-x^2+x^4\right )} \, dx=\int { \frac {1}{{\left (x^{4} - x^{2} + 1\right )} x^{4}} \,d x } \] Input:
integrate(1/x^4/(x^4-x^2+1),x, algorithm="maxima")
Output:
-1/3*(3*x^2 + 1)/x^3 - integrate(x^2/(x^4 - x^2 + 1), x)
Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^4 \left (1-x^2+x^4\right )} \, dx=-\frac {1}{12} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{12} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) - \frac {3 \, x^{2} + 1}{3 \, x^{3}} - \frac {1}{2} \, \arctan \left (2 \, x + \sqrt {3}\right ) - \frac {1}{2} \, \arctan \left (2 \, x - \sqrt {3}\right ) \] Input:
integrate(1/x^4/(x^4-x^2+1),x, algorithm="giac")
Output:
-1/12*sqrt(3)*log(x^2 + sqrt(3)*x + 1) + 1/12*sqrt(3)*log(x^2 - sqrt(3)*x + 1) - 1/3*(3*x^2 + 1)/x^3 - 1/2*arctan(2*x + sqrt(3)) - 1/2*arctan(2*x - sqrt(3))
Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^4 \left (1-x^2+x^4\right )} \, dx=\mathrm {atan}\left (\frac {x}{2}-\frac {\sqrt {3}\,x\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\mathrm {atan}\left (\frac {x}{2}+\frac {\sqrt {3}\,x\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\frac {x^2+\frac {1}{3}}{x^3} \] Input:
int(1/(x^4*(x^4 - x^2 + 1)),x)
Output:
atan(x/2 - (3^(1/2)*x*1i)/2)*((3^(1/2)*1i)/6 - 1/2) - atan(x/2 + (3^(1/2)* x*1i)/2)*((3^(1/2)*1i)/6 + 1/2) - (x^2 + 1/3)/x^3
Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08 \[ \int \frac {1}{x^4 \left (1-x^2+x^4\right )} \, dx=\frac {6 \mathit {atan} \left (\sqrt {3}-2 x \right ) x^{3}-6 \mathit {atan} \left (\sqrt {3}+2 x \right ) x^{3}-\sqrt {3}\, \mathrm {log}\left (-\sqrt {3}\, x +x^{2}+1\right ) x^{3}+\sqrt {3}\, \mathrm {log}\left (\sqrt {3}\, x +x^{2}+1\right ) x^{3}-12 x^{2}-4}{12 x^{3}} \] Input:
int(1/x^4/(x^4-x^2+1),x)
Output:
(6*atan(sqrt(3) - 2*x)*x**3 - 6*atan(sqrt(3) + 2*x)*x**3 - sqrt(3)*log( - sqrt(3)*x + x**2 + 1)*x**3 + sqrt(3)*log(sqrt(3)*x + x**2 + 1)*x**3 - 12*x **2 - 4)/(12*x**3)