Integrand size = 16, antiderivative size = 88 \[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=-\frac {1}{2 x^2}+\frac {\arctan \left (\sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )} x^2\right )}{2 \sqrt {5 \left (9+4 \sqrt {5}\right )}}-\frac {\arctan \left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )}{2 \sqrt {5 \left (9-4 \sqrt {5}\right )}} \] Output:
-1/2/x^2+1/2*arctan((1/2*5^(1/2)-1/2)*x^2)/(5+2*5^(1/2))-1/2*arctan((1/2+1 /2*5^(1/2))*x^2)/(5-2*5^(1/2))
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=-\frac {1}{2 x^2}-\frac {1}{4} \text {RootSum}\left [1+3 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {3 \log (x-\text {$\#$1})+\log (x-\text {$\#$1}) \text {$\#$1}^4}{3 \text {$\#$1}^2+2 \text {$\#$1}^6}\&\right ] \] Input:
Integrate[1/(x^3*(1 + 3*x^4 + x^8)),x]
Output:
-1/2*1/x^2 - RootSum[1 + 3*#1^4 + #1^8 & , (3*Log[x - #1] + Log[x - #1]*#1 ^4)/(3*#1^2 + 2*#1^6) & ]/4
Time = 0.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.18, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1695, 1443, 25, 1480, 216}
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.
\(\displaystyle \int \frac {1}{x^3 \left (x^8+3 x^4+1\right )} \, dx\) |
\(\Big \downarrow \) 1695 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (x^8+3 x^4+1\right )}dx^2\) |
\(\Big \downarrow \) 1443 |
\(\displaystyle \frac {1}{2} \left (\int -\frac {x^4+3}{x^8+3 x^4+1}dx^2-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (-\int \frac {x^4+3}{x^8+3 x^4+1}dx^2-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{10} \left (5+3 \sqrt {5}\right ) \int \frac {1}{x^4+\frac {1}{2} \left (3-\sqrt {5}\right )}dx^2-\frac {1}{10} \left (5-3 \sqrt {5}\right ) \int \frac {1}{x^4+\frac {1}{2} \left (3+\sqrt {5}\right )}dx^2-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (-\frac {\left (5-3 \sqrt {5}\right ) \arctan \left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )}{5 \sqrt {2 \left (3+\sqrt {5}\right )}}-\frac {1}{10} \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \left (5+3 \sqrt {5}\right ) \arctan \left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )-\frac {1}{x^2}\right )\) |
Input:
Int[1/(x^3*(1 + 3*x^4 + x^8)),x]
Output:
(-x^(-2) - ((5 - 3*Sqrt[5])*ArcTan[Sqrt[2/(3 + Sqrt[5])]*x^2])/(5*Sqrt[2*( 3 + Sqrt[5])]) - (Sqrt[(3 + Sqrt[5])/2]*(5 + 3*Sqrt[5])*ArcTan[Sqrt[(3 + S qrt[5])/2]*x^2])/10)/2
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[((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[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b *x^(n/k) + c*x^(2*(n/k)))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IntegerQ[m]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.48
method | result | size |
risch | \(-\frac {1}{2 x^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}+90 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (35 \textit {\_R}^{3}+8 x^{2}+123 \textit {\_R} \right )\right )}{4}\) | \(42\) |
default | \(-\frac {1}{2 x^{2}}-\frac {\left (3+\sqrt {5}\right ) \sqrt {5}\, \arctan \left (\frac {4 x^{2}}{-2+2 \sqrt {5}}\right )}{5 \left (-2+2 \sqrt {5}\right )}-\frac {\left (\sqrt {5}-3\right ) \sqrt {5}\, \arctan \left (\frac {4 x^{2}}{2+2 \sqrt {5}}\right )}{5 \left (2+2 \sqrt {5}\right )}\) | \(75\) |
Input:
int(1/x^3/(x^8+3*x^4+1),x,method=_RETURNVERBOSE)
Output:
-1/2/x^2+1/4*sum(_R*ln(35*_R^3+8*x^2+123*_R),_R=RootOf(25*_Z^4+90*_Z^2+1))
Time = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=-\frac {{\left (2 \, \sqrt {5} x^{2} + 5 \, x^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {5} x^{2} + \frac {1}{2} \, x^{2}\right ) + {\left (2 \, \sqrt {5} x^{2} - 5 \, x^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {5} x^{2} - \frac {1}{2} \, x^{2}\right ) + 5}{10 \, x^{2}} \] Input:
integrate(1/x^3/(x^8+3*x^4+1),x, algorithm="fricas")
Output:
-1/10*((2*sqrt(5)*x^2 + 5*x^2)*arctan(1/2*sqrt(5)*x^2 + 1/2*x^2) + (2*sqrt (5)*x^2 - 5*x^2)*arctan(1/2*sqrt(5)*x^2 - 1/2*x^2) + 5)/x^2
Time = 0.14 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=- 2 \left (\frac {\sqrt {5}}{10} + \frac {1}{4}\right ) \operatorname {atan}{\left (\frac {2 x^{2}}{-1 + \sqrt {5}} \right )} + 2 \cdot \left (\frac {1}{4} - \frac {\sqrt {5}}{10}\right ) \operatorname {atan}{\left (\frac {2 x^{2}}{1 + \sqrt {5}} \right )} - \frac {1}{2 x^{2}} \] Input:
integrate(1/x**3/(x**8+3*x**4+1),x)
Output:
-2*(sqrt(5)/10 + 1/4)*atan(2*x**2/(-1 + sqrt(5))) + 2*(1/4 - sqrt(5)/10)*a tan(2*x**2/(1 + sqrt(5))) - 1/(2*x**2)
\[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=\int { \frac {1}{{\left (x^{8} + 3 \, x^{4} + 1\right )} x^{3}} \,d x } \] Input:
integrate(1/x^3/(x^8+3*x^4+1),x, algorithm="maxima")
Output:
-1/2/x^2 - integrate((x^4 + 3)*x/(x^8 + 3*x^4 + 1), x)
Time = 0.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=-\frac {1}{20} \, {\left (x^{4} {\left (\sqrt {5} - 5\right )} + 3 \, \sqrt {5} - 15\right )} \arctan \left (\frac {2 \, x^{2}}{\sqrt {5} + 1}\right ) - \frac {1}{20} \, {\left (x^{4} {\left (\sqrt {5} + 5\right )} + 3 \, \sqrt {5} + 15\right )} \arctan \left (\frac {2 \, x^{2}}{\sqrt {5} - 1}\right ) - \frac {1}{2 \, x^{2}} \] Input:
integrate(1/x^3/(x^8+3*x^4+1),x, algorithm="giac")
Output:
-1/20*(x^4*(sqrt(5) - 5) + 3*sqrt(5) - 15)*arctan(2*x^2/(sqrt(5) + 1)) - 1 /20*(x^4*(sqrt(5) + 5) + 3*sqrt(5) + 15)*arctan(2*x^2/(sqrt(5) - 1)) - 1/2 /x^2
Time = 18.86 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.48 \[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=2\,\mathrm {atanh}\left (\frac {26880\,x^2\,\sqrt {-\frac {\sqrt {5}}{20}-\frac {9}{80}}}{3520\,\sqrt {5}+7872}+\frac {12032\,\sqrt {5}\,x^2\,\sqrt {-\frac {\sqrt {5}}{20}-\frac {9}{80}}}{3520\,\sqrt {5}+7872}\right )\,\sqrt {-\frac {\sqrt {5}}{20}-\frac {9}{80}}-2\,\mathrm {atanh}\left (\frac {26880\,x^2\,\sqrt {\frac {\sqrt {5}}{20}-\frac {9}{80}}}{3520\,\sqrt {5}-7872}-\frac {12032\,\sqrt {5}\,x^2\,\sqrt {\frac {\sqrt {5}}{20}-\frac {9}{80}}}{3520\,\sqrt {5}-7872}\right )\,\sqrt {\frac {\sqrt {5}}{20}-\frac {9}{80}}-\frac {1}{2\,x^2} \] Input:
int(1/(x^3*(3*x^4 + x^8 + 1)),x)
Output:
2*atanh((26880*x^2*(- 5^(1/2)/20 - 9/80)^(1/2))/(3520*5^(1/2) + 7872) + (1 2032*5^(1/2)*x^2*(- 5^(1/2)/20 - 9/80)^(1/2))/(3520*5^(1/2) + 7872))*(- 5^ (1/2)/20 - 9/80)^(1/2) - 2*atanh((26880*x^2*(5^(1/2)/20 - 9/80)^(1/2))/(35 20*5^(1/2) - 7872) - (12032*5^(1/2)*x^2*(5^(1/2)/20 - 9/80)^(1/2))/(3520*5 ^(1/2) - 7872))*(5^(1/2)/20 - 9/80)^(1/2) - 1/(2*x^2)
\[ \int \frac {1}{x^3 \left (1+3 x^4+x^8\right )} \, dx=\frac {-2 \left (\int \frac {x^{5}}{x^{8}+3 x^{4}+1}d x \right ) x^{2}-6 \left (\int \frac {x}{x^{8}+3 x^{4}+1}d x \right ) x^{2}-1}{2 x^{2}} \] Input:
int(1/x^3/(x^8+3*x^4+1),x)
Output:
( - 2*int(x**5/(x**8 + 3*x**4 + 1),x)*x**2 - 6*int(x/(x**8 + 3*x**4 + 1),x )*x**2 - 1)/(2*x**2)