Integrand size = 16, antiderivative size = 99 \[ \int \frac {1}{x^7 \left (1+3 x^4+x^8\right )} \, dx=-\frac {1}{6 x^6}+\frac {3}{2 x^2}-\frac {1}{2} \sqrt {\frac {1}{10} \left (123-55 \sqrt {5}\right )} \arctan \left (\sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )} x^2\right )+\frac {1}{2} \sqrt {\frac {1}{10} \left (123+55 \sqrt {5}\right )} \arctan \left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right ) \] Output:
-1/6/x^6+3/2/x^2-1/2*(5/2-11/10*5^(1/2))*arctan((1/2*5^(1/2)-1/2)*x^2)+1/2 *(5/2+11/10*5^(1/2))*arctan((1/2+1/2*5^(1/2))*x^2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^7 \left (1+3 x^4+x^8\right )} \, dx=-\frac {1}{6 x^6}+\frac {3}{2 x^2}+\frac {1}{4} \text {RootSum}\left [1+3 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {8 \log (x-\text {$\#$1})+3 \log (x-\text {$\#$1}) \text {$\#$1}^4}{3 \text {$\#$1}^2+2 \text {$\#$1}^6}\&\right ] \] Input:
Integrate[1/(x^7*(1 + 3*x^4 + x^8)),x]
Output:
-1/6*1/x^6 + 3/(2*x^2) + RootSum[1 + 3*#1^4 + #1^8 & , (8*Log[x - #1] + 3* Log[x - #1]*#1^4)/(3*#1^2 + 2*#1^6) & ]/4
Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1695, 1443, 27, 1604, 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.
\(\displaystyle \int \frac {1}{x^7 \left (x^8+3 x^4+1\right )} \, dx\) |
\(\Big \downarrow \) 1695 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^8 \left (x^8+3 x^4+1\right )}dx^2\) |
\(\Big \downarrow \) 1443 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int -\frac {3 \left (x^4+3\right )}{x^4 \left (x^8+3 x^4+1\right )}dx^2-\frac {1}{3 x^6}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (-\int \frac {x^4+3}{x^4 \left (x^8+3 x^4+1\right )}dx^2-\frac {1}{3 x^6}\right )\) |
\(\Big \downarrow \) 1604 |
\(\displaystyle \frac {1}{2} \left (\int \frac {3 x^4+8}{x^8+3 x^4+1}dx^2-\frac {1}{3 x^6}+\frac {3}{x^2}\right )\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{10} \left (15+7 \sqrt {5}\right ) \int \frac {1}{x^4+\frac {1}{2} \left (3-\sqrt {5}\right )}dx^2+\frac {1}{10} \left (15-7 \sqrt {5}\right ) \int \frac {1}{x^4+\frac {1}{2} \left (3+\sqrt {5}\right )}dx^2-\frac {1}{3 x^6}+\frac {3}{x^2}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (15-7 \sqrt {5}\right ) \arctan \left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )}{5 \sqrt {2 \left (3+\sqrt {5}\right )}}+\frac {1}{10} \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \left (15+7 \sqrt {5}\right ) \arctan \left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )-\frac {1}{3 x^6}+\frac {3}{x^2}\right )\) |
Input:
Int[1/(x^7*(1 + 3*x^4 + x^8)),x]
Output:
(-1/3*1/x^6 + 3/x^2 + ((15 - 7*Sqrt[5])*ArcTan[Sqrt[2/(3 + Sqrt[5])]*x^2]) /(5*Sqrt[2*(3 + Sqrt[5])]) + (Sqrt[(3 + Sqrt[5])/2]*(15 + 7*Sqrt[5])*ArcTa n[Sqrt[(3 + Sqrt[5])/2]*x^2])/10)/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[((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[((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])
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 = 48, normalized size of antiderivative = 0.48
method | result | size |
risch | \(\frac {\frac {3 x^{4}}{2}-\frac {1}{6}}{x^{6}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}+615 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-90 \textit {\_R}^{3}+55 x^{2}-2207 \textit {\_R} \right )\right )}{4}\) | \(48\) |
default | \(-\frac {1}{6 x^{6}}+\frac {3}{2 x^{2}}+\frac {\sqrt {5}\, \left (7+3 \sqrt {5}\right ) \arctan \left (\frac {4 x^{2}}{-2+2 \sqrt {5}}\right )}{-10+10 \sqrt {5}}+\frac {\left (-7+3 \sqrt {5}\right ) \sqrt {5}\, \arctan \left (\frac {4 x^{2}}{2+2 \sqrt {5}}\right )}{10+10 \sqrt {5}}\) | \(84\) |
Input:
int(1/x^7/(x^8+3*x^4+1),x,method=_RETURNVERBOSE)
Output:
(3/2*x^4-1/6)/x^6+1/4*sum(_R*ln(-90*_R^3+55*x^2-2207*_R),_R=RootOf(25*_Z^4 +615*_Z^2+1))
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^7 \left (1+3 x^4+x^8\right )} \, dx=\frac {90 \, x^{4} + 3 \, {\left (11 \, \sqrt {5} x^{6} + 25 \, x^{6}\right )} \arctan \left (\frac {1}{2} \, \sqrt {5} x^{2} + \frac {1}{2} \, x^{2}\right ) + 3 \, {\left (11 \, \sqrt {5} x^{6} - 25 \, x^{6}\right )} \arctan \left (\frac {1}{2} \, \sqrt {5} x^{2} - \frac {1}{2} \, x^{2}\right ) - 10}{60 \, x^{6}} \] Input:
integrate(1/x^7/(x^8+3*x^4+1),x, algorithm="fricas")
Output:
1/60*(90*x^4 + 3*(11*sqrt(5)*x^6 + 25*x^6)*arctan(1/2*sqrt(5)*x^2 + 1/2*x^ 2) + 3*(11*sqrt(5)*x^6 - 25*x^6)*arctan(1/2*sqrt(5)*x^2 - 1/2*x^2) - 10)/x ^6
Time = 0.15 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.66 \[ \int \frac {1}{x^7 \left (1+3 x^4+x^8\right )} \, dx=2 \cdot \left (\frac {11 \sqrt {5}}{40} + \frac {5}{8}\right ) \operatorname {atan}{\left (\frac {2 x^{2}}{-1 + \sqrt {5}} \right )} - 2 \cdot \left (\frac {5}{8} - \frac {11 \sqrt {5}}{40}\right ) \operatorname {atan}{\left (\frac {2 x^{2}}{1 + \sqrt {5}} \right )} + \frac {9 x^{4} - 1}{6 x^{6}} \] Input:
integrate(1/x**7/(x**8+3*x**4+1),x)
Output:
2*(11*sqrt(5)/40 + 5/8)*atan(2*x**2/(-1 + sqrt(5))) - 2*(5/8 - 11*sqrt(5)/ 40)*atan(2*x**2/(1 + sqrt(5))) + (9*x**4 - 1)/(6*x**6)
\[ \int \frac {1}{x^7 \left (1+3 x^4+x^8\right )} \, dx=\int { \frac {1}{{\left (x^{8} + 3 \, x^{4} + 1\right )} x^{7}} \,d x } \] Input:
integrate(1/x^7/(x^8+3*x^4+1),x, algorithm="maxima")
Output:
1/6*(9*x^4 - 1)/x^6 + integrate((3*x^4 + 8)*x/(x^8 + 3*x^4 + 1), x)
Time = 0.16 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^7 \left (1+3 x^4+x^8\right )} \, dx=\frac {1}{20} \, {\left (3 \, x^{4} {\left (\sqrt {5} - 5\right )} + 8 \, \sqrt {5} - 40\right )} \arctan \left (\frac {2 \, x^{2}}{\sqrt {5} + 1}\right ) + \frac {1}{20} \, {\left (3 \, x^{4} {\left (\sqrt {5} + 5\right )} + 8 \, \sqrt {5} + 40\right )} \arctan \left (\frac {2 \, x^{2}}{\sqrt {5} - 1}\right ) + \frac {9 \, x^{4} - 1}{6 \, x^{6}} \] Input:
integrate(1/x^7/(x^8+3*x^4+1),x, algorithm="giac")
Output:
1/20*(3*x^4*(sqrt(5) - 5) + 8*sqrt(5) - 40)*arctan(2*x^2/(sqrt(5) + 1)) + 1/20*(3*x^4*(sqrt(5) + 5) + 8*sqrt(5) + 40)*arctan(2*x^2/(sqrt(5) - 1)) + 1/6*(9*x^4 - 1)/x^6
Time = 18.86 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.37 \[ \int \frac {1}{x^7 \left (1+3 x^4+x^8\right )} \, dx=2\,\mathrm {atanh}\left (\frac {3327500\,x^2\,\sqrt {\frac {11\,\sqrt {5}}{32}-\frac {123}{160}}}{1140425\,\sqrt {5}-2550075}-\frac {1488300\,\sqrt {5}\,x^2\,\sqrt {\frac {11\,\sqrt {5}}{32}-\frac {123}{160}}}{1140425\,\sqrt {5}-2550075}\right )\,\sqrt {\frac {11\,\sqrt {5}}{32}-\frac {123}{160}}-2\,\mathrm {atanh}\left (\frac {3327500\,x^2\,\sqrt {-\frac {11\,\sqrt {5}}{32}-\frac {123}{160}}}{1140425\,\sqrt {5}+2550075}+\frac {1488300\,\sqrt {5}\,x^2\,\sqrt {-\frac {11\,\sqrt {5}}{32}-\frac {123}{160}}}{1140425\,\sqrt {5}+2550075}\right )\,\sqrt {-\frac {11\,\sqrt {5}}{32}-\frac {123}{160}}+\frac {\frac {3\,x^4}{2}-\frac {1}{6}}{x^6} \] Input:
int(1/(x^7*(3*x^4 + x^8 + 1)),x)
Output:
2*atanh((3327500*x^2*((11*5^(1/2))/32 - 123/160)^(1/2))/(1140425*5^(1/2) - 2550075) - (1488300*5^(1/2)*x^2*((11*5^(1/2))/32 - 123/160)^(1/2))/(11404 25*5^(1/2) - 2550075))*((11*5^(1/2))/32 - 123/160)^(1/2) - 2*atanh((332750 0*x^2*(- (11*5^(1/2))/32 - 123/160)^(1/2))/(1140425*5^(1/2) + 2550075) + ( 1488300*5^(1/2)*x^2*(- (11*5^(1/2))/32 - 123/160)^(1/2))/(1140425*5^(1/2) + 2550075))*(- (11*5^(1/2))/32 - 123/160)^(1/2) + ((3*x^4)/2 - 1/6)/x^6
\[ \int \frac {1}{x^7 \left (1+3 x^4+x^8\right )} \, dx=\frac {18 \left (\int \frac {x^{5}}{x^{8}+3 x^{4}+1}d x \right ) x^{6}+48 \left (\int \frac {x}{x^{8}+3 x^{4}+1}d x \right ) x^{6}+9 x^{4}-1}{6 x^{6}} \] Input:
int(1/x^7/(x^8+3*x^4+1),x)
Output:
(18*int(x**5/(x**8 + 3*x**4 + 1),x)*x**6 + 48*int(x/(x**8 + 3*x**4 + 1),x) *x**6 + 9*x**4 - 1)/(6*x**6)