Integrand size = 16, antiderivative size = 55 \[ \int \frac {1}{x^{14} \left (1-x^2+x^4\right )} \, dx=-\frac {1}{13 x^{13}}-\frac {1}{11 x^{11}}+\frac {1}{7 x^7}+\frac {1}{5 x^5}-\frac {1}{x}+\frac {\text {arctanh}\left (\frac {\sqrt {3} x}{1+x^2}\right )}{\sqrt {3}} \] Output:
-1/13/x^13-1/11/x^11+1/7/x^7+1/5/x^5-1/x+1/3*arctanh(3^(1/2)*x/(x^2+1))*3^ (1/2)
Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.45 \[ \int \frac {1}{x^{14} \left (1-x^2+x^4\right )} \, dx=-\frac {1}{13 x^{13}}-\frac {1}{11 x^{11}}+\frac {1}{7 x^7}+\frac {1}{5 x^5}-\frac {1}{x}-\frac {\log \left (-1+\sqrt {3} x-x^2\right )}{2 \sqrt {3}}+\frac {\log \left (1+\sqrt {3} x+x^2\right )}{2 \sqrt {3}} \] Input:
Integrate[1/(x^14*(1 - x^2 + x^4)),x]
Output:
-1/13*1/x^13 - 1/(11*x^11) + 1/(7*x^7) + 1/(5*x^5) - x^(-1) - Log[-1 + Sqr t[3]*x - x^2]/(2*Sqrt[3]) + Log[1 + Sqrt[3]*x + x^2]/(2*Sqrt[3])
Time = 0.49 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.44, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {1443, 27, 1604, 27, 1443, 27, 1604, 27, 1443, 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.
\(\displaystyle \int \frac {1}{x^{14} \left (x^4-x^2+1\right )} \, dx\) |
\(\Big \downarrow \) 1443 |
\(\displaystyle \frac {1}{13} \int \frac {13 \left (1-x^2\right )}{x^{12} \left (x^4-x^2+1\right )}dx-\frac {1}{13 x^{13}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1-x^2}{x^{12} \left (x^4-x^2+1\right )}dx-\frac {1}{13 x^{13}}\) |
\(\Big \downarrow \) 1604 |
\(\displaystyle -\frac {1}{11} \int \frac {11}{x^8 \left (x^4-x^2+1\right )}dx-\frac {1}{13 x^{13}}-\frac {1}{11 x^{11}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {1}{x^8 \left (x^4-x^2+1\right )}dx-\frac {1}{13 x^{13}}-\frac {1}{11 x^{11}}\) |
\(\Big \downarrow \) 1443 |
\(\displaystyle -\frac {1}{7} \int \frac {7 \left (1-x^2\right )}{x^6 \left (x^4-x^2+1\right )}dx-\frac {1}{13 x^{13}}-\frac {1}{11 x^{11}}+\frac {1}{7 x^7}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {1-x^2}{x^6 \left (x^4-x^2+1\right )}dx-\frac {1}{13 x^{13}}-\frac {1}{11 x^{11}}+\frac {1}{7 x^7}\) |
\(\Big \downarrow \) 1604 |
\(\displaystyle \frac {1}{5} \int \frac {5}{x^2 \left (x^4-x^2+1\right )}dx-\frac {1}{13 x^{13}}-\frac {1}{11 x^{11}}+\frac {1}{7 x^7}+\frac {1}{5 x^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1}{x^2 \left (x^4-x^2+1\right )}dx-\frac {1}{13 x^{13}}-\frac {1}{11 x^{11}}+\frac {1}{7 x^7}+\frac {1}{5 x^5}\) |
\(\Big \downarrow \) 1443 |
\(\displaystyle \int \frac {1-x^2}{x^4-x^2+1}dx-\frac {1}{13 x^{13}}-\frac {1}{11 x^{11}}+\frac {1}{7 x^7}+\frac {1}{5 x^5}-\frac {1}{x}\) |
\(\Big \downarrow \) 1478 |
\(\displaystyle -\frac {\int -\frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\int -\frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {1}{13 x^{13}}-\frac {1}{11 x^{11}}+\frac {1}{7 x^7}+\frac {1}{5 x^5}-\frac {1}{x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {\int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {1}{13 x^{13}}-\frac {1}{11 x^{11}}+\frac {1}{7 x^7}+\frac {1}{5 x^5}-\frac {1}{x}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle -\frac {1}{13 x^{13}}-\frac {1}{11 x^{11}}+\frac {1}{7 x^7}+\frac {1}{5 x^5}-\frac {\log \left (x^2-\sqrt {3} x+1\right )}{2 \sqrt {3}}+\frac {\log \left (x^2+\sqrt {3} x+1\right )}{2 \sqrt {3}}-\frac {1}{x}\) |
Input:
Int[1/(x^14*(1 - x^2 + x^4)),x]
Output:
-1/13*1/x^13 - 1/(11*x^11) + 1/(7*x^7) + 1/(5*x^5) - x^(-1) - Log[1 - Sqrt [3]*x + x^2]/(2*Sqrt[3]) + Log[1 + Sqrt[3]*x + x^2]/(2*Sqrt[3])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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[((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])
Time = 0.14 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.09
method | result | size |
default | \(-\frac {1}{13 x^{13}}-\frac {1}{11 x^{11}}-\frac {1}{x}+\frac {1}{7 x^{7}}+\frac {1}{5 x^{5}}-\frac {\sqrt {3}\, \ln \left (x^{2}-\sqrt {3}\, x +1\right )}{6}+\frac {\sqrt {3}\, \ln \left (x^{2}+\sqrt {3}\, x +1\right )}{6}\) | \(60\) |
risch | \(\frac {-x^{12}+\frac {1}{5} x^{8}+\frac {1}{7} x^{6}-\frac {1}{11} x^{2}-\frac {1}{13}}{x^{13}}-\frac {\sqrt {3}\, \ln \left (x^{2}-\sqrt {3}\, x +1\right )}{6}+\frac {\sqrt {3}\, \ln \left (x^{2}+\sqrt {3}\, x +1\right )}{6}\) | \(61\) |
Input:
int(1/x^14/(x^4-x^2+1),x,method=_RETURNVERBOSE)
Output:
-1/13/x^13-1/11/x^11-1/x+1/7/x^7+1/5/x^5-1/6*3^(1/2)*ln(x^2-3^(1/2)*x+1)+1 /6*3^(1/2)*ln(x^2+3^(1/2)*x+1)
Time = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.25 \[ \int \frac {1}{x^{14} \left (1-x^2+x^4\right )} \, dx=\frac {5005 \, \sqrt {3} x^{13} \log \left (\frac {x^{4} + 5 \, x^{2} + 2 \, \sqrt {3} {\left (x^{3} + x\right )} + 1}{x^{4} - x^{2} + 1}\right ) - 30030 \, x^{12} + 6006 \, x^{8} + 4290 \, x^{6} - 2730 \, x^{2} - 2310}{30030 \, x^{13}} \] Input:
integrate(1/x^14/(x^4-x^2+1),x, algorithm="fricas")
Output:
1/30030*(5005*sqrt(3)*x^13*log((x^4 + 5*x^2 + 2*sqrt(3)*(x^3 + x) + 1)/(x^ 4 - x^2 + 1)) - 30030*x^12 + 6006*x^8 + 4290*x^6 - 2730*x^2 - 2310)/x^13
Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.20 \[ \int \frac {1}{x^{14} \left (1-x^2+x^4\right )} \, dx=- \frac {\sqrt {3} \log {\left (x^{2} - \sqrt {3} x + 1 \right )}}{6} + \frac {\sqrt {3} \log {\left (x^{2} + \sqrt {3} x + 1 \right )}}{6} + \frac {- 5005 x^{12} + 1001 x^{8} + 715 x^{6} - 455 x^{2} - 385}{5005 x^{13}} \] Input:
integrate(1/x**14/(x**4-x**2+1),x)
Output:
-sqrt(3)*log(x**2 - sqrt(3)*x + 1)/6 + sqrt(3)*log(x**2 + sqrt(3)*x + 1)/6 + (-5005*x**12 + 1001*x**8 + 715*x**6 - 455*x**2 - 385)/(5005*x**13)
\[ \int \frac {1}{x^{14} \left (1-x^2+x^4\right )} \, dx=\int { \frac {1}{{\left (x^{4} - x^{2} + 1\right )} x^{14}} \,d x } \] Input:
integrate(1/x^14/(x^4-x^2+1),x, algorithm="maxima")
Output:
-1/5005*(5005*x^12 - 1001*x^8 - 715*x^6 + 455*x^2 + 385)/x^13 - integrate( (x^2 - 1)/(x^4 - x^2 + 1), x)
Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x^{14} \left (1-x^2+x^4\right )} \, dx=\frac {1}{6} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) - \frac {1}{6} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) - \frac {5005 \, x^{12} - 1001 \, x^{8} - 715 \, x^{6} + 455 \, x^{2} + 385}{5005 \, x^{13}} \] Input:
integrate(1/x^14/(x^4-x^2+1),x, algorithm="giac")
Output:
1/6*sqrt(3)*log(x^2 + sqrt(3)*x + 1) - 1/6*sqrt(3)*log(x^2 - sqrt(3)*x + 1 ) - 1/5005*(5005*x^12 - 1001*x^8 - 715*x^6 + 455*x^2 + 385)/x^13
Time = 18.73 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^{14} \left (1-x^2+x^4\right )} \, dx=\frac {\sqrt {3}\,\mathrm {atanh}\left (\frac {2\,\sqrt {3}\,x}{3\,\left (\frac {2\,x^2}{3}+\frac {2}{3}\right )}\right )}{3}-\frac {x^{12}-\frac {x^8}{5}-\frac {x^6}{7}+\frac {x^2}{11}+\frac {1}{13}}{x^{13}} \] Input:
int(1/(x^14*(x^4 - x^2 + 1)),x)
Output:
(3^(1/2)*atanh((2*3^(1/2)*x)/(3*((2*x^2)/3 + 2/3))))/3 - (x^2/11 - x^6/7 - x^8/5 + x^12 + 1/13)/x^13
Time = 0.16 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^{14} \left (1-x^2+x^4\right )} \, dx=\frac {-5005 \sqrt {3}\, \mathrm {log}\left (-\sqrt {3}\, x +x^{2}+1\right ) x^{13}+5005 \sqrt {3}\, \mathrm {log}\left (\sqrt {3}\, x +x^{2}+1\right ) x^{13}-30030 x^{12}+6006 x^{8}+4290 x^{6}-2730 x^{2}-2310}{30030 x^{13}} \] Input:
int(1/x^14/(x^4-x^2+1),x)
Output:
( - 5005*sqrt(3)*log( - sqrt(3)*x + x**2 + 1)*x**13 + 5005*sqrt(3)*log(sqr t(3)*x + x**2 + 1)*x**13 - 30030*x**12 + 6006*x**8 + 4290*x**6 - 2730*x**2 - 2310)/(30030*x**13)