Integrand size = 16, antiderivative size = 52 \[ \int \frac {x^{16}}{1-x^2+x^4} \, dx=x-\frac {x^5}{5}-\frac {x^7}{7}+\frac {x^{11}}{11}+\frac {x^{13}}{13}-\frac {\text {arctanh}\left (\frac {\sqrt {3} x}{1+x^2}\right )}{\sqrt {3}} \] Output:
x-1/5*x^5-1/7*x^7+1/11*x^11+1/13*x^13-1/3*arctanh(3^(1/2)*x/(x^2+1))*3^(1/ 2)
Time = 0.01 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.46 \[ \int \frac {x^{16}}{1-x^2+x^4} \, dx=x-\frac {x^5}{5}-\frac {x^7}{7}+\frac {x^{11}}{11}+\frac {x^{13}}{13}+\frac {\log \left (-1+\sqrt {3} x-x^2\right )}{2 \sqrt {3}}-\frac {\log \left (1+\sqrt {3} x+x^2\right )}{2 \sqrt {3}} \] Input:
Integrate[x^16/(1 - x^2 + x^4),x]
Output:
x - x^5/5 - x^7/7 + x^11/11 + x^13/13 + Log[-1 + Sqrt[3]*x - x^2]/(2*Sqrt[ 3]) - Log[1 + Sqrt[3]*x + x^2]/(2*Sqrt[3])
Time = 0.50 (sec) , antiderivative size = 75, 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 = {1442, 27, 1602, 27, 1442, 27, 1602, 27, 1442, 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.
\(\displaystyle \int \frac {x^{16}}{x^4-x^2+1} \, dx\) |
\(\Big \downarrow \) 1442 |
\(\displaystyle \frac {x^{13}}{13}-\frac {1}{13} \int \frac {13 x^{12} \left (1-x^2\right )}{x^4-x^2+1}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^{13}}{13}-\int \frac {x^{12} \left (1-x^2\right )}{x^4-x^2+1}dx\) |
\(\Big \downarrow \) 1602 |
\(\displaystyle \frac {1}{11} \int -\frac {11 x^{10}}{x^4-x^2+1}dx+\frac {x^{13}}{13}+\frac {x^{11}}{11}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {x^{10}}{x^4-x^2+1}dx+\frac {x^{13}}{13}+\frac {x^{11}}{11}\) |
\(\Big \downarrow \) 1442 |
\(\displaystyle \frac {1}{7} \int \frac {7 x^6 \left (1-x^2\right )}{x^4-x^2+1}dx+\frac {x^{13}}{13}+\frac {x^{11}}{11}-\frac {x^7}{7}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^6 \left (1-x^2\right )}{x^4-x^2+1}dx+\frac {x^{13}}{13}+\frac {x^{11}}{11}-\frac {x^7}{7}\) |
\(\Big \downarrow \) 1602 |
\(\displaystyle -\frac {1}{5} \int -\frac {5 x^4}{x^4-x^2+1}dx+\frac {x^{13}}{13}+\frac {x^{11}}{11}-\frac {x^7}{7}-\frac {x^5}{5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^4}{x^4-x^2+1}dx+\frac {x^{13}}{13}+\frac {x^{11}}{11}-\frac {x^7}{7}-\frac {x^5}{5}\) |
\(\Big \downarrow \) 1442 |
\(\displaystyle -\int \frac {1-x^2}{x^4-x^2+1}dx+\frac {x^{13}}{13}+\frac {x^{11}}{11}-\frac {x^7}{7}-\frac {x^5}{5}+x\) |
\(\Big \downarrow \) 1478 |
\(\displaystyle \frac {\int -\frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {\int -\frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {x^{13}}{13}+\frac {x^{11}}{11}-\frac {x^7}{7}-\frac {x^5}{5}+x\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {x^{13}}{13}+\frac {x^{11}}{11}-\frac {x^7}{7}-\frac {x^5}{5}+x\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {x^{13}}{13}+\frac {x^{11}}{11}-\frac {x^7}{7}-\frac {x^5}{5}+\frac {\log \left (x^2-\sqrt {3} x+1\right )}{2 \sqrt {3}}-\frac {\log \left (x^2+\sqrt {3} x+1\right )}{2 \sqrt {3}}+x\) |
Input:
Int[x^16/(1 - x^2 + x^4),x]
Output:
x - x^5/5 - x^7/7 + x^11/11 + x^13/13 + 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^3*(d*x)^(m - 3)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[d^4/(c*(m + 4*p + 1)) Int[(d*x)^(m - 4)*Simp[a*(m - 3) + b*(m + 2*p - 1)*x^2, x]*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x ] && NeQ[b^2 - 4*a*c, 0] && GtQ[m, 3] && NeQ[m + 4*p + 1, 0] && 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[e*f*(f*x)^(m - 1)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 3))), x] - Simp[f^2/(c*(m + 4*p + 3)) Int[(f*x)^(m - 2)* (a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c , 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (IntegerQ[p] | | IntegerQ[m])
Time = 0.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.08
method | result | size |
default | \(\frac {x^{13}}{13}+\frac {x^{11}}{11}-\frac {x^{7}}{7}-\frac {x^{5}}{5}+x +\frac {\sqrt {3}\, \ln \left (x^{2}-\sqrt {3}\, x +1\right )}{6}-\frac {\sqrt {3}\, \ln \left (x^{2}+\sqrt {3}\, x +1\right )}{6}\) | \(56\) |
risch | \(\frac {x^{13}}{13}+\frac {x^{11}}{11}-\frac {x^{7}}{7}-\frac {x^{5}}{5}+x +\frac {\sqrt {3}\, \ln \left (x^{2}-\sqrt {3}\, x +1\right )}{6}-\frac {\sqrt {3}\, \ln \left (x^{2}+\sqrt {3}\, x +1\right )}{6}\) | \(56\) |
Input:
int(x^16/(x^4-x^2+1),x,method=_RETURNVERBOSE)
Output:
1/13*x^13+1/11*x^11-1/7*x^7-1/5*x^5+x+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 = 61, normalized size of antiderivative = 1.17 \[ \int \frac {x^{16}}{1-x^2+x^4} \, dx=\frac {1}{13} \, x^{13} + \frac {1}{11} \, x^{11} - \frac {1}{7} \, x^{7} - \frac {1}{5} \, x^{5} + \frac {1}{6} \, \sqrt {3} \log \left (\frac {x^{4} + 5 \, x^{2} - 2 \, \sqrt {3} {\left (x^{3} + x\right )} + 1}{x^{4} - x^{2} + 1}\right ) + x \] Input:
integrate(x^16/(x^4-x^2+1),x, algorithm="fricas")
Output:
1/13*x^13 + 1/11*x^11 - 1/7*x^7 - 1/5*x^5 + 1/6*sqrt(3)*log((x^4 + 5*x^2 - 2*sqrt(3)*(x^3 + x) + 1)/(x^4 - x^2 + 1)) + x
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.17 \[ \int \frac {x^{16}}{1-x^2+x^4} \, dx=\frac {x^{13}}{13} + \frac {x^{11}}{11} - \frac {x^{7}}{7} - \frac {x^{5}}{5} + x + \frac {\sqrt {3} \log {\left (x^{2} - \sqrt {3} x + 1 \right )}}{6} - \frac {\sqrt {3} \log {\left (x^{2} + \sqrt {3} x + 1 \right )}}{6} \] Input:
integrate(x**16/(x**4-x**2+1),x)
Output:
x**13/13 + x**11/11 - x**7/7 - x**5/5 + x + sqrt(3)*log(x**2 - sqrt(3)*x + 1)/6 - sqrt(3)*log(x**2 + sqrt(3)*x + 1)/6
\[ \int \frac {x^{16}}{1-x^2+x^4} \, dx=\int { \frac {x^{16}}{x^{4} - x^{2} + 1} \,d x } \] Input:
integrate(x^16/(x^4-x^2+1),x, algorithm="maxima")
Output:
1/13*x^13 + 1/11*x^11 - 1/7*x^7 - 1/5*x^5 + x + integrate((x^2 - 1)/(x^4 - x^2 + 1), x)
Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.06 \[ \int \frac {x^{16}}{1-x^2+x^4} \, dx=\frac {1}{13} \, x^{13} + \frac {1}{11} \, x^{11} - \frac {1}{7} \, x^{7} - \frac {1}{5} \, x^{5} - \frac {1}{6} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{6} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + x \] Input:
integrate(x^16/(x^4-x^2+1),x, algorithm="giac")
Output:
1/13*x^13 + 1/11*x^11 - 1/7*x^7 - 1/5*x^5 - 1/6*sqrt(3)*log(x^2 + sqrt(3)* x + 1) + 1/6*sqrt(3)*log(x^2 - sqrt(3)*x + 1) + x
Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87 \[ \int \frac {x^{16}}{1-x^2+x^4} \, dx=x-\frac {x^5}{5}-\frac {x^7}{7}+\frac {x^{11}}{11}+\frac {x^{13}}{13}+\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}\,x\,2{}\mathrm {i}}{3\,\left (\frac {2\,x^2}{3}+\frac {2}{3}\right )}\right )\,1{}\mathrm {i}}{3} \] Input:
int(x^16/(x^4 - x^2 + 1),x)
Output:
x + (3^(1/2)*atan((3^(1/2)*x*2i)/(3*((2*x^2)/3 + 2/3)))*1i)/3 - x^5/5 - x^ 7/7 + x^11/11 + x^13/13
Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98 \[ \int \frac {x^{16}}{1-x^2+x^4} \, dx=\frac {\sqrt {3}\, \mathrm {log}\left (-\sqrt {3}\, x +x^{2}+1\right )}{6}-\frac {\sqrt {3}\, \mathrm {log}\left (\sqrt {3}\, x +x^{2}+1\right )}{6}+\frac {x^{13}}{13}+\frac {x^{11}}{11}-\frac {x^{7}}{7}-\frac {x^{5}}{5}+x \] Input:
int(x^16/(x^4-x^2+1),x)
Output:
(5005*sqrt(3)*log( - sqrt(3)*x + x**2 + 1) - 5005*sqrt(3)*log(sqrt(3)*x + x**2 + 1) + 2310*x**13 + 2730*x**11 - 4290*x**7 - 6006*x**5 + 30030*x)/300 30