Integrand size = 23, antiderivative size = 89 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{2 x^2}+\frac {1}{4} \arctan \left (\sqrt {3}-2 x^2\right )-\frac {1}{4} \arctan \left (\sqrt {3}+2 x^2\right )-\frac {\log \left (1-\sqrt {3} x^2+x^4\right )}{8 \sqrt {3}}+\frac {\log \left (1+\sqrt {3} x^2+x^4\right )}{8 \sqrt {3}} \]
-1/2/x^2-1/4*arctan(2*x^2-3^(1/2))-1/4*arctan(2*x^2+3^(1/2))-1/24*ln(1+x^4 -x^2*3^(1/2))*3^(1/2)+1/24*ln(1+x^4+x^2*3^(1/2))*3^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.55 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{2 x^2}-\frac {1}{4} \text {RootSum}\left [1-\text {$\#$1}^4+\text {$\#$1}^8\&,\frac {\log (x-\text {$\#$1}) \text {$\#$1}^2}{-1+2 \text {$\#$1}^4}\&\right ] \]
Time = 0.32 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {1814, 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.
\(\displaystyle \int \frac {1-x^4}{x^3 \left (x^8-x^4+1\right )} \, dx\) |
\(\Big \downarrow \) 1814 |
\(\displaystyle \frac {1}{2} \int \frac {1-x^4}{x^4 \left (x^8-x^4+1\right )}dx^2\) |
\(\Big \downarrow \) 1604 |
\(\displaystyle \frac {1}{2} \left (-\int \frac {x^4}{x^8-x^4+1}dx^2-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 1447 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^4}{x^8-x^4+1}dx^2-\frac {1}{2} \int \frac {x^4+1}{x^8-x^4+1}dx^2-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 1475 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^4-\sqrt {3} x^2+1}dx^2-\frac {1}{2} \int \frac {1}{x^4+\sqrt {3} x^2+1}dx^2\right )+\frac {1}{2} \int \frac {1-x^4}{x^8-x^4+1}dx^2-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\int \frac {1}{-x^4-1}d\left (2 x^2-\sqrt {3}\right )+\int \frac {1}{-x^4-1}d\left (2 x^2+\sqrt {3}\right )\right )+\frac {1}{2} \int \frac {1-x^4}{x^8-x^4+1}dx^2-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^4}{x^8-x^4+1}dx^2+\frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x^2\right )-\arctan \left (2 x^2+\sqrt {3}\right )\right )-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 1478 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {3}-2 x^2}{x^4-\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}-\frac {\int -\frac {2 x^2+\sqrt {3}}{x^4+\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x^2\right )-\arctan \left (2 x^2+\sqrt {3}\right )\right )-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {3}-2 x^2}{x^4-\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}+\frac {\int \frac {2 x^2+\sqrt {3}}{x^4+\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x^2\right )-\arctan \left (2 x^2+\sqrt {3}\right )\right )-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x^2\right )-\arctan \left (2 x^2+\sqrt {3}\right )\right )-\frac {1}{x^2}+\frac {1}{2} \left (\frac {\log \left (x^4+\sqrt {3} x^2+1\right )}{2 \sqrt {3}}-\frac {\log \left (x^4-\sqrt {3} x^2+1\right )}{2 \sqrt {3}}\right )\right )\) |
(-x^(-2) + (ArcTan[Sqrt[3] - 2*x^2] - ArcTan[Sqrt[3] + 2*x^2])/2 + (-1/2*L og[1 - Sqrt[3]*x^2 + x^4]/Sqrt[3] + Log[1 + Sqrt[3]*x^2 + x^4]/(2*Sqrt[3]) )/2)/2
3.1.59.3.1 Defintions of rubi rules used
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[(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])
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e _.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Sub st[Int[x^((m + 1)/k - 1)*(d + e*x^(n/k))^q*(a + b*x^(n/k) + c*x^(2*(n/k)))^ p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, c, d, e, p, q}, 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 = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.45
method | result | size |
risch | \(-\frac {1}{2 x^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+3 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-6 \textit {\_R}^{3}+x^{2}-\textit {\_R} \right )\right )}{4}\) | \(40\) |
default | \(-\frac {1}{2 x^{2}}+\frac {\sqrt {3}\, \left (-\frac {\ln \left (1+x^{4}-x^{2} \sqrt {3}\right )}{2}-\sqrt {3}\, \arctan \left (2 x^{2}-\sqrt {3}\right )\right )}{12}+\frac {\sqrt {3}\, \left (\frac {\ln \left (1+x^{4}+x^{2} \sqrt {3}\right )}{2}-\sqrt {3}\, \arctan \left (2 x^{2}+\sqrt {3}\right )\right )}{12}\) | \(82\) |
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.90 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=-\frac {\sqrt {6} x^{2} \sqrt {i \, \sqrt {3} - 1} \log \left (6 \, x^{2} + i \, \sqrt {6} \sqrt {3} \sqrt {i \, \sqrt {3} - 1}\right ) - \sqrt {6} x^{2} \sqrt {i \, \sqrt {3} - 1} \log \left (6 \, x^{2} - i \, \sqrt {6} \sqrt {3} \sqrt {i \, \sqrt {3} - 1}\right ) - \sqrt {6} x^{2} \sqrt {-i \, \sqrt {3} - 1} \log \left (6 \, x^{2} + i \, \sqrt {6} \sqrt {3} \sqrt {-i \, \sqrt {3} - 1}\right ) + \sqrt {6} x^{2} \sqrt {-i \, \sqrt {3} - 1} \log \left (6 \, x^{2} - i \, \sqrt {6} \sqrt {3} \sqrt {-i \, \sqrt {3} - 1}\right ) + 12}{24 \, x^{2}} \]
-1/24*(sqrt(6)*x^2*sqrt(I*sqrt(3) - 1)*log(6*x^2 + I*sqrt(6)*sqrt(3)*sqrt( I*sqrt(3) - 1)) - sqrt(6)*x^2*sqrt(I*sqrt(3) - 1)*log(6*x^2 - I*sqrt(6)*sq rt(3)*sqrt(I*sqrt(3) - 1)) - sqrt(6)*x^2*sqrt(-I*sqrt(3) - 1)*log(6*x^2 + I*sqrt(6)*sqrt(3)*sqrt(-I*sqrt(3) - 1)) + sqrt(6)*x^2*sqrt(-I*sqrt(3) - 1) *log(6*x^2 - I*sqrt(6)*sqrt(3)*sqrt(-I*sqrt(3) - 1)) + 12)/x^2
Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=- \frac {\sqrt {3} \log {\left (x^{4} - \sqrt {3} x^{2} + 1 \right )}}{24} + \frac {\sqrt {3} \log {\left (x^{4} + \sqrt {3} x^{2} + 1 \right )}}{24} - \frac {\operatorname {atan}{\left (2 x^{2} - \sqrt {3} \right )}}{4} - \frac {\operatorname {atan}{\left (2 x^{2} + \sqrt {3} \right )}}{4} - \frac {1}{2 x^{2}} \]
-sqrt(3)*log(x**4 - sqrt(3)*x**2 + 1)/24 + sqrt(3)*log(x**4 + sqrt(3)*x**2 + 1)/24 - atan(2*x**2 - sqrt(3))/4 - atan(2*x**2 + sqrt(3))/4 - 1/(2*x**2 )
\[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=\int { -\frac {x^{4} - 1}{{\left (x^{8} - x^{4} + 1\right )} x^{3}} \,d x } \]
Time = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.91 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{24} \, \sqrt {3} x^{4} \log \left (x^{4} + \sqrt {3} x^{2} + 1\right ) + \frac {1}{24} \, \sqrt {3} x^{4} \log \left (x^{4} - \sqrt {3} x^{2} + 1\right ) - \frac {1}{4} \, x^{4} \arctan \left (2 \, x^{2} + \sqrt {3}\right ) - \frac {1}{4} \, x^{4} \arctan \left (2 \, x^{2} - \sqrt {3}\right ) - \frac {1}{2 \, x^{2}} \]
-1/24*sqrt(3)*x^4*log(x^4 + sqrt(3)*x^2 + 1) + 1/24*sqrt(3)*x^4*log(x^4 - sqrt(3)*x^2 + 1) - 1/4*x^4*arctan(2*x^2 + sqrt(3)) - 1/4*x^4*arctan(2*x^2 - sqrt(3)) - 1/2/x^2
Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.63 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=\mathrm {atan}\left (\frac {2\,x^2}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\mathrm {atan}\left (\frac {2\,x^2}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\frac {1}{2\,x^2} \]