Integrand size = 18, antiderivative size = 45 \[ \int \frac {d+e x}{4-5 x^2+x^4} \, dx=-\frac {1}{6} d \text {arctanh}\left (\frac {x}{2}\right )+\frac {1}{3} d \text {arctanh}(x)-\frac {1}{6} e \log \left (1-x^2\right )+\frac {1}{6} e \log \left (4-x^2\right ) \]
Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.11 \[ \int \frac {d+e x}{4-5 x^2+x^4} \, dx=\frac {1}{12} (-2 (d+e) \log (1-x)+(d+2 e) \log (2-x)+2 (d-e) \log (1+x)-(d-2 e) \log (2+x)) \]
(-2*(d + e)*Log[1 - x] + (d + 2*e)*Log[2 - x] + 2*(d - e)*Log[1 + x] - (d - 2*e)*Log[2 + x])/12
Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {2202, 27, 1406, 220, 1432, 1081, 2009}
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.
\(\displaystyle \int \frac {d+e x}{x^4-5 x^2+4} \, dx\) |
\(\Big \downarrow \) 2202 |
\(\displaystyle \int \frac {d}{x^4-5 x^2+4}dx+\int \frac {e x}{x^4-5 x^2+4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle d \int \frac {1}{x^4-5 x^2+4}dx+e \int \frac {x}{x^4-5 x^2+4}dx\) |
\(\Big \downarrow \) 1406 |
\(\displaystyle d \left (\frac {1}{3} \int \frac {1}{x^2-4}dx-\frac {1}{3} \int \frac {1}{x^2-1}dx\right )+e \int \frac {x}{x^4-5 x^2+4}dx\) |
\(\Big \downarrow \) 220 |
\(\displaystyle e \int \frac {x}{x^4-5 x^2+4}dx+d \left (\frac {\text {arctanh}(x)}{3}-\frac {1}{6} \text {arctanh}\left (\frac {x}{2}\right )\right )\) |
\(\Big \downarrow \) 1432 |
\(\displaystyle \frac {1}{2} e \int \frac {1}{x^4-5 x^2+4}dx^2+d \left (\frac {\text {arctanh}(x)}{3}-\frac {1}{6} \text {arctanh}\left (\frac {x}{2}\right )\right )\) |
\(\Big \downarrow \) 1081 |
\(\displaystyle \frac {1}{2} e \int \left (\frac {1}{3 \left (1-x^2\right )}-\frac {1}{3 \left (4-x^2\right )}\right )dx^2+d \left (\frac {\text {arctanh}(x)}{3}-\frac {1}{6} \text {arctanh}\left (\frac {x}{2}\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle d \left (\frac {\text {arctanh}(x)}{3}-\frac {1}{6} \text {arctanh}\left (\frac {x}{2}\right )\right )+\frac {1}{2} e \left (\frac {1}{3} \log \left (4-x^2\right )-\frac {1}{3} \log \left (1-x^2\right )\right )\) |
3.1.10.3.1 Defintions of rubi rules used
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[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[c Int[ExpandIntegrand[1/((b/2 - q/2 + c*x)*(b/2 + q/2 + c*x)), x], x], x]] /; FreeQ[{a, b, c}, x] && NiceSqrtQ[b^2 - 4*a*c]
Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^ 2 - 4*a*c, 2]}, Simp[c/q Int[1/(b/2 - q/2 + c*x^2), x], x] - Simp[c/q I nt[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c , 0] && PosQ[b^2 - 4*a*c]
Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 Subst[Int[(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x]
Int[(Pn_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{n = Expon[Pn, x], k}, Int[Sum[Coeff[Pn, x, 2*k]*x^(2*k), {k, 0, n/2}]*(a + b *x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pn, x, 2*k + 1]*x^(2*k), {k, 0, (n - 1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pn, x] && !PolyQ[Pn, x^2]
Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.11
method | result | size |
default | \(\left (-\frac {d}{12}+\frac {e}{6}\right ) \ln \left (x +2\right )+\left (\frac {d}{6}-\frac {e}{6}\right ) \ln \left (x +1\right )+\left (-\frac {d}{6}-\frac {e}{6}\right ) \ln \left (x -1\right )+\left (\frac {d}{12}+\frac {e}{6}\right ) \ln \left (x -2\right )\) | \(50\) |
norman | \(\left (-\frac {d}{12}+\frac {e}{6}\right ) \ln \left (x +2\right )+\left (\frac {d}{6}-\frac {e}{6}\right ) \ln \left (x +1\right )+\left (-\frac {d}{6}-\frac {e}{6}\right ) \ln \left (x -1\right )+\left (\frac {d}{12}+\frac {e}{6}\right ) \ln \left (x -2\right )\) | \(50\) |
parallelrisch | \(\frac {\ln \left (x -2\right ) d}{12}+\frac {\ln \left (x -2\right ) e}{6}-\frac {\ln \left (x -1\right ) d}{6}-\frac {\ln \left (x -1\right ) e}{6}+\frac {\ln \left (x +1\right ) d}{6}-\frac {\ln \left (x +1\right ) e}{6}-\frac {\ln \left (x +2\right ) d}{12}+\frac {\ln \left (x +2\right ) e}{6}\) | \(58\) |
risch | \(\frac {\ln \left (2-x \right ) d}{12}+\frac {\ln \left (2-x \right ) e}{6}-\frac {\ln \left (x +2\right ) d}{12}+\frac {\ln \left (x +2\right ) e}{6}-\frac {\ln \left (1-x \right ) d}{6}-\frac {\ln \left (1-x \right ) e}{6}+\frac {\ln \left (x +1\right ) d}{6}-\frac {\ln \left (x +1\right ) e}{6}\) | \(66\) |
Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96 \[ \int \frac {d+e x}{4-5 x^2+x^4} \, dx=-\frac {1}{12} \, {\left (d - 2 \, e\right )} \log \left (x + 2\right ) + \frac {1}{6} \, {\left (d - e\right )} \log \left (x + 1\right ) - \frac {1}{6} \, {\left (d + e\right )} \log \left (x - 1\right ) + \frac {1}{12} \, {\left (d + 2 \, e\right )} \log \left (x - 2\right ) \]
-1/12*(d - 2*e)*log(x + 2) + 1/6*(d - e)*log(x + 1) - 1/6*(d + e)*log(x - 1) + 1/12*(d + 2*e)*log(x - 2)
Leaf count of result is larger than twice the leaf count of optimal. 515 vs. \(2 (34) = 68\).
Time = 1.81 (sec) , antiderivative size = 515, normalized size of antiderivative = 11.44 \[ \int \frac {d+e x}{4-5 x^2+x^4} \, dx=- \frac {\left (d - 2 e\right ) \log {\left (x + \frac {- 35 d^{4} e + \frac {51 d^{4} \left (d - 2 e\right )}{2} - 180 d^{2} e^{3} - 90 d^{2} e^{2} \left (d - 2 e\right ) + 41 d^{2} e \left (d - 2 e\right )^{2} - \frac {15 d^{2} \left (d - 2 e\right )^{3}}{2} + 320 e^{5} - 96 e^{4} \left (d - 2 e\right ) - 80 e^{3} \left (d - 2 e\right )^{2} + 24 e^{2} \left (d - 2 e\right )^{3}}{9 d^{5} - 160 d^{3} e^{2} + 256 d e^{4}} \right )}}{12} + \frac {\left (d - e\right ) \log {\left (x + \frac {- 35 d^{4} e - 51 d^{4} \left (d - e\right ) - 180 d^{2} e^{3} + 180 d^{2} e^{2} \left (d - e\right ) + 164 d^{2} e \left (d - e\right )^{2} + 60 d^{2} \left (d - e\right )^{3} + 320 e^{5} + 192 e^{4} \left (d - e\right ) - 320 e^{3} \left (d - e\right )^{2} - 192 e^{2} \left (d - e\right )^{3}}{9 d^{5} - 160 d^{3} e^{2} + 256 d e^{4}} \right )}}{6} - \frac {\left (d + e\right ) \log {\left (x + \frac {- 35 d^{4} e + 51 d^{4} \left (d + e\right ) - 180 d^{2} e^{3} - 180 d^{2} e^{2} \left (d + e\right ) + 164 d^{2} e \left (d + e\right )^{2} - 60 d^{2} \left (d + e\right )^{3} + 320 e^{5} - 192 e^{4} \left (d + e\right ) - 320 e^{3} \left (d + e\right )^{2} + 192 e^{2} \left (d + e\right )^{3}}{9 d^{5} - 160 d^{3} e^{2} + 256 d e^{4}} \right )}}{6} + \frac {\left (d + 2 e\right ) \log {\left (x + \frac {- 35 d^{4} e - \frac {51 d^{4} \left (d + 2 e\right )}{2} - 180 d^{2} e^{3} + 90 d^{2} e^{2} \left (d + 2 e\right ) + 41 d^{2} e \left (d + 2 e\right )^{2} + \frac {15 d^{2} \left (d + 2 e\right )^{3}}{2} + 320 e^{5} + 96 e^{4} \left (d + 2 e\right ) - 80 e^{3} \left (d + 2 e\right )^{2} - 24 e^{2} \left (d + 2 e\right )^{3}}{9 d^{5} - 160 d^{3} e^{2} + 256 d e^{4}} \right )}}{12} \]
-(d - 2*e)*log(x + (-35*d**4*e + 51*d**4*(d - 2*e)/2 - 180*d**2*e**3 - 90* d**2*e**2*(d - 2*e) + 41*d**2*e*(d - 2*e)**2 - 15*d**2*(d - 2*e)**3/2 + 32 0*e**5 - 96*e**4*(d - 2*e) - 80*e**3*(d - 2*e)**2 + 24*e**2*(d - 2*e)**3)/ (9*d**5 - 160*d**3*e**2 + 256*d*e**4))/12 + (d - e)*log(x + (-35*d**4*e - 51*d**4*(d - e) - 180*d**2*e**3 + 180*d**2*e**2*(d - e) + 164*d**2*e*(d - e)**2 + 60*d**2*(d - e)**3 + 320*e**5 + 192*e**4*(d - e) - 320*e**3*(d - e )**2 - 192*e**2*(d - e)**3)/(9*d**5 - 160*d**3*e**2 + 256*d*e**4))/6 - (d + e)*log(x + (-35*d**4*e + 51*d**4*(d + e) - 180*d**2*e**3 - 180*d**2*e**2 *(d + e) + 164*d**2*e*(d + e)**2 - 60*d**2*(d + e)**3 + 320*e**5 - 192*e** 4*(d + e) - 320*e**3*(d + e)**2 + 192*e**2*(d + e)**3)/(9*d**5 - 160*d**3* e**2 + 256*d*e**4))/6 + (d + 2*e)*log(x + (-35*d**4*e - 51*d**4*(d + 2*e)/ 2 - 180*d**2*e**3 + 90*d**2*e**2*(d + 2*e) + 41*d**2*e*(d + 2*e)**2 + 15*d **2*(d + 2*e)**3/2 + 320*e**5 + 96*e**4*(d + 2*e) - 80*e**3*(d + 2*e)**2 - 24*e**2*(d + 2*e)**3)/(9*d**5 - 160*d**3*e**2 + 256*d*e**4))/12
Time = 0.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96 \[ \int \frac {d+e x}{4-5 x^2+x^4} \, dx=-\frac {1}{12} \, {\left (d - 2 \, e\right )} \log \left (x + 2\right ) + \frac {1}{6} \, {\left (d - e\right )} \log \left (x + 1\right ) - \frac {1}{6} \, {\left (d + e\right )} \log \left (x - 1\right ) + \frac {1}{12} \, {\left (d + 2 \, e\right )} \log \left (x - 2\right ) \]
-1/12*(d - 2*e)*log(x + 2) + 1/6*(d - e)*log(x + 1) - 1/6*(d + e)*log(x - 1) + 1/12*(d + 2*e)*log(x - 2)
Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.04 \[ \int \frac {d+e x}{4-5 x^2+x^4} \, dx=-\frac {1}{12} \, {\left (d - 2 \, e\right )} \log \left ({\left | x + 2 \right |}\right ) + \frac {1}{6} \, {\left (d - e\right )} \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{6} \, {\left (d + e\right )} \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{12} \, {\left (d + 2 \, e\right )} \log \left ({\left | x - 2 \right |}\right ) \]
-1/12*(d - 2*e)*log(abs(x + 2)) + 1/6*(d - e)*log(abs(x + 1)) - 1/6*(d + e )*log(abs(x - 1)) + 1/12*(d + 2*e)*log(abs(x - 2))
Time = 7.81 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.13 \[ \int \frac {d+e x}{4-5 x^2+x^4} \, dx=\ln \left (x+1\right )\,\left (\frac {d}{6}-\frac {e}{6}\right )-\ln \left (x-1\right )\,\left (\frac {d}{6}+\frac {e}{6}\right )+\ln \left (x-2\right )\,\left (\frac {d}{12}+\frac {e}{6}\right )-\ln \left (x+2\right )\,\left (\frac {d}{12}-\frac {e}{6}\right ) \]