\(\int \frac {d+e x+f x^2}{4-5 x^2+x^4} \, dx\) [11]
Optimal result
Integrand size = 23, antiderivative size = 51 \[
\int \frac {d+e x+f x^2}{4-5 x^2+x^4} \, dx=-\frac {1}{6} (d+4 f) \text {arctanh}\left (\frac {x}{2}\right )+\frac {1}{3} (d+f) \text {arctanh}(x)-\frac {1}{6} e \log \left (1-x^2\right )+\frac {1}{6} e \log \left (4-x^2\right )
\]
[Out]
-1/6*(d+4*f)*arctanh(1/2*x)+1/3*(d+f)*arctanh(x)-1/6*e*ln(-x^2+1)+1/6*e*ln(-x^2+4)
Rubi [A] (verified)
Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of
steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {1687, 1180, 213, 12, 1121, 630,
31} \[
\int \frac {d+e x+f x^2}{4-5 x^2+x^4} \, dx=-\frac {1}{6} \text {arctanh}\left (\frac {x}{2}\right ) (d+4 f)+\frac {1}{3} \text {arctanh}(x) (d+f)-\frac {1}{6} e \log \left (1-x^2\right )+\frac {1}{6} e \log \left (4-x^2\right )
\]
[In]
Int[(d + e*x + f*x^2)/(4 - 5*x^2 + x^4),x]
[Out]
-1/6*((d + 4*f)*ArcTanh[x/2]) + ((d + f)*ArcTanh[x])/3 - (e*Log[1 - x^2])/6 + (e*Log[4 - x^2])/6
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rule 213
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])
Rule 630
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]
Rule 1121
Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]
Rule 1180
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[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]
Rule 1687
Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && !PolyQ[Pq, x^2]
Rubi steps \begin{align*}
\text {integral}& = \int \frac {e x}{4-5 x^2+x^4} \, dx+\int \frac {d+f x^2}{4-5 x^2+x^4} \, dx \\ & = e \int \frac {x}{4-5 x^2+x^4} \, dx-\frac {1}{3} (d+f) \int \frac {1}{-1+x^2} \, dx+\frac {1}{3} (d+4 f) \int \frac {1}{-4+x^2} \, dx \\ & = -\frac {1}{6} (d+4 f) \tanh ^{-1}\left (\frac {x}{2}\right )+\frac {1}{3} (d+f) \tanh ^{-1}(x)+\frac {1}{2} e \text {Subst}\left (\int \frac {1}{4-5 x+x^2} \, dx,x,x^2\right ) \\ & = -\frac {1}{6} (d+4 f) \tanh ^{-1}\left (\frac {x}{2}\right )+\frac {1}{3} (d+f) \tanh ^{-1}(x)+\frac {1}{6} e \text {Subst}\left (\int \frac {1}{-4+x} \, dx,x,x^2\right )-\frac {1}{6} e \text {Subst}\left (\int \frac {1}{-1+x} \, dx,x,x^2\right ) \\ & = -\frac {1}{6} (d+4 f) \tanh ^{-1}\left (\frac {x}{2}\right )+\frac {1}{3} (d+f) \tanh ^{-1}(x)-\frac {1}{6} e \log \left (1-x^2\right )+\frac {1}{6} e \log \left (4-x^2\right ) \\
\end{align*}
Mathematica [A] (verified)
Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.14
\[
\int \frac {d+e x+f x^2}{4-5 x^2+x^4} \, dx=\frac {1}{12} (-2 (d+e+f) \log (1-x)+(d+2 e+4 f) \log (2-x)+2 (d-e+f) \log (1+x)-(d-2 e+4 f) \log (2+x))
\]
[In]
Integrate[(d + e*x + f*x^2)/(4 - 5*x^2 + x^4),x]
[Out]
(-2*(d + e + f)*Log[1 - x] + (d + 2*e + 4*f)*Log[2 - x] + 2*(d - e + f)*Log[1 + x] - (d - 2*e + 4*f)*Log[2 + x
])/12
Maple [A] (verified)
Time = 0.06 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.22
| | |
| method | result | size |
| | |
| default |
\(\left (-\frac {d}{12}+\frac {e}{6}-\frac {f}{3}\right ) \ln \left (x +2\right )+\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}\right ) \ln \left (x +1\right )+\left (-\frac {d}{6}-\frac {e}{6}-\frac {f}{6}\right ) \ln \left (x -1\right )+\left (\frac {d}{12}+\frac {e}{6}+\frac {f}{3}\right ) \ln \left (x -2\right )\) |
\(62\) |
| norman |
\(\left (-\frac {d}{12}+\frac {e}{6}-\frac {f}{3}\right ) \ln \left (x +2\right )+\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}\right ) \ln \left (x +1\right )+\left (-\frac {d}{6}-\frac {e}{6}-\frac {f}{6}\right ) \ln \left (x -1\right )+\left (\frac {d}{12}+\frac {e}{6}+\frac {f}{3}\right ) \ln \left (x -2\right )\) |
\(62\) |
| parallelrisch |
\(\frac {\ln \left (x -2\right ) d}{12}+\frac {\ln \left (x -2\right ) e}{6}+\frac {\ln \left (x -2\right ) f}{3}-\frac {\ln \left (x -1\right ) d}{6}-\frac {\ln \left (x -1\right ) e}{6}-\frac {\ln \left (x -1\right ) f}{6}+\frac {\ln \left (x +1\right ) d}{6}-\frac {\ln \left (x +1\right ) e}{6}+\frac {\ln \left (x +1\right ) f}{6}-\frac {\ln \left (x +2\right ) d}{12}+\frac {\ln \left (x +2\right ) e}{6}-\frac {\ln \left (x +2\right ) f}{3}\) |
\(86\) |
| risch |
\(-\frac {\ln \left (1-x \right ) d}{6}-\frac {\ln \left (1-x \right ) e}{6}-\frac {\ln \left (1-x \right ) f}{6}+\frac {\ln \left (x +1\right ) d}{6}-\frac {\ln \left (x +1\right ) e}{6}+\frac {\ln \left (x +1\right ) f}{6}+\frac {\ln \left (2-x \right ) d}{12}+\frac {\ln \left (2-x \right ) e}{6}+\frac {\ln \left (2-x \right ) f}{3}-\frac {\ln \left (x +2\right ) d}{12}+\frac {\ln \left (x +2\right ) e}{6}-\frac {\ln \left (x +2\right ) f}{3}\) |
\(98\) |
| | |
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[In]
int((f*x^2+e*x+d)/(x^4-5*x^2+4),x,method=_RETURNVERBOSE)
[Out]
(-1/12*d+1/6*e-1/3*f)*ln(x+2)+(1/6*d-1/6*e+1/6*f)*ln(x+1)+(-1/6*d-1/6*e-1/6*f)*ln(x-1)+(1/12*d+1/6*e+1/3*f)*ln
(x-2)
Fricas [A] (verification not implemented)
none
Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00
\[
\int \frac {d+e x+f x^2}{4-5 x^2+x^4} \, dx=-\frac {1}{12} \, {\left (d - 2 \, e + 4 \, f\right )} \log \left (x + 2\right ) + \frac {1}{6} \, {\left (d - e + f\right )} \log \left (x + 1\right ) - \frac {1}{6} \, {\left (d + e + f\right )} \log \left (x - 1\right ) + \frac {1}{12} \, {\left (d + 2 \, e + 4 \, f\right )} \log \left (x - 2\right )
\]
[In]
integrate((f*x^2+e*x+d)/(x^4-5*x^2+4),x, algorithm="fricas")
[Out]
-1/12*(d - 2*e + 4*f)*log(x + 2) + 1/6*(d - e + f)*log(x + 1) - 1/6*(d + e + f)*log(x - 1) + 1/12*(d + 2*e + 4
*f)*log(x - 2)
Sympy [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 2195 vs. \(2 (44) = 88\).
Time = 95.81 (sec) , antiderivative size = 2195, normalized size of antiderivative = 43.04
\[
\int \frac {d+e x+f x^2}{4-5 x^2+x^4} \, dx=\text {Too large to display}
\]
[In]
integrate((f*x**2+e*x+d)/(x**4-5*x**2+4),x)
[Out]
-(d - 2*e + 4*f)*log(x + (-35*d**5*e + 51*d**5*(d - 2*e + 4*f)/2 - 820*d**4*e*f + 90*d**4*f*(d - 2*e + 4*f) -
180*d**3*e**3 - 90*d**3*e**2*(d - 2*e + 4*f) - 4100*d**3*e*f**2 + 41*d**3*e*(d - 2*e + 4*f)**2 + 42*d**3*f**2*
(d - 2*e + 4*f) - 15*d**3*(d - 2*e + 4*f)**3/2 - 432*d**2*e**2*f*(d - 2*e + 4*f) - 8000*d**2*e*f**3 + 240*d**2
*e*f*(d - 2*e + 4*f)**2 - 240*d**2*f**3*(d - 2*e + 4*f) - 12*d**2*f*(d - 2*e + 4*f)**3 + 320*d*e**5 - 96*d*e**
4*(d - 2*e + 4*f) + 720*d*e**3*f**2 - 80*d*e**3*(d - 2*e + 4*f)**2 - 1080*d*e**2*f**2*(d - 2*e + 4*f) + 24*d*e
**2*(d - 2*e + 4*f)**3 - 6400*d*e*f**4 + 492*d*e*f**2*(d - 2*e + 4*f)**2 - 576*d*f**4*(d - 2*e + 4*f) + 30*d*f
**2*(d - 2*e + 4*f)**3 + 512*e**5*f - 128*e**3*f*(d - 2*e + 4*f)**2 - 576*e**2*f**3*(d - 2*e + 4*f) - 1472*e*f
**5 + 320*e*f**3*(d - 2*e + 4*f)**2 - 480*f**5*(d - 2*e + 4*f) + 48*f**3*(d - 2*e + 4*f)**3)/(9*d**6 + 45*d**5
*f - 160*d**4*e**2 - 36*d**4*f**2 - 1312*d**3*e**2*f - 360*d**3*f**3 + 256*d**2*e**4 - 3840*d**2*e**2*f**2 - 1
44*d**2*f**4 + 1280*d*e**4*f - 5248*d*e**2*f**3 + 720*d*f**5 + 1024*e**4*f**2 - 2560*e**2*f**4 + 576*f**6))/12
+ (d - e + f)*log(x + (-35*d**5*e - 51*d**5*(d - e + f) - 820*d**4*e*f - 180*d**4*f*(d - e + f) - 180*d**3*e*
*3 + 180*d**3*e**2*(d - e + f) - 4100*d**3*e*f**2 + 164*d**3*e*(d - e + f)**2 - 84*d**3*f**2*(d - e + f) + 60*
d**3*(d - e + f)**3 + 864*d**2*e**2*f*(d - e + f) - 8000*d**2*e*f**3 + 960*d**2*e*f*(d - e + f)**2 + 480*d**2*
f**3*(d - e + f) + 96*d**2*f*(d - e + f)**3 + 320*d*e**5 + 192*d*e**4*(d - e + f) + 720*d*e**3*f**2 - 320*d*e*
*3*(d - e + f)**2 + 2160*d*e**2*f**2*(d - e + f) - 192*d*e**2*(d - e + f)**3 - 6400*d*e*f**4 + 1968*d*e*f**2*(
d - e + f)**2 + 1152*d*f**4*(d - e + f) - 240*d*f**2*(d - e + f)**3 + 512*e**5*f - 512*e**3*f*(d - e + f)**2 +
1152*e**2*f**3*(d - e + f) - 1472*e*f**5 + 1280*e*f**3*(d - e + f)**2 + 960*f**5*(d - e + f) - 384*f**3*(d -
e + f)**3)/(9*d**6 + 45*d**5*f - 160*d**4*e**2 - 36*d**4*f**2 - 1312*d**3*e**2*f - 360*d**3*f**3 + 256*d**2*e*
*4 - 3840*d**2*e**2*f**2 - 144*d**2*f**4 + 1280*d*e**4*f - 5248*d*e**2*f**3 + 720*d*f**5 + 1024*e**4*f**2 - 25
60*e**2*f**4 + 576*f**6))/6 - (d + e + f)*log(x + (-35*d**5*e + 51*d**5*(d + e + f) - 820*d**4*e*f + 180*d**4*
f*(d + e + f) - 180*d**3*e**3 - 180*d**3*e**2*(d + e + f) - 4100*d**3*e*f**2 + 164*d**3*e*(d + e + f)**2 + 84*
d**3*f**2*(d + e + f) - 60*d**3*(d + e + f)**3 - 864*d**2*e**2*f*(d + e + f) - 8000*d**2*e*f**3 + 960*d**2*e*f
*(d + e + f)**2 - 480*d**2*f**3*(d + e + f) - 96*d**2*f*(d + e + f)**3 + 320*d*e**5 - 192*d*e**4*(d + e + f) +
720*d*e**3*f**2 - 320*d*e**3*(d + e + f)**2 - 2160*d*e**2*f**2*(d + e + f) + 192*d*e**2*(d + e + f)**3 - 6400
*d*e*f**4 + 1968*d*e*f**2*(d + e + f)**2 - 1152*d*f**4*(d + e + f) + 240*d*f**2*(d + e + f)**3 + 512*e**5*f -
512*e**3*f*(d + e + f)**2 - 1152*e**2*f**3*(d + e + f) - 1472*e*f**5 + 1280*e*f**3*(d + e + f)**2 - 960*f**5*(
d + e + f) + 384*f**3*(d + e + f)**3)/(9*d**6 + 45*d**5*f - 160*d**4*e**2 - 36*d**4*f**2 - 1312*d**3*e**2*f -
360*d**3*f**3 + 256*d**2*e**4 - 3840*d**2*e**2*f**2 - 144*d**2*f**4 + 1280*d*e**4*f - 5248*d*e**2*f**3 + 720*d
*f**5 + 1024*e**4*f**2 - 2560*e**2*f**4 + 576*f**6))/6 + (d + 2*e + 4*f)*log(x + (-35*d**5*e - 51*d**5*(d + 2*
e + 4*f)/2 - 820*d**4*e*f - 90*d**4*f*(d + 2*e + 4*f) - 180*d**3*e**3 + 90*d**3*e**2*(d + 2*e + 4*f) - 4100*d*
*3*e*f**2 + 41*d**3*e*(d + 2*e + 4*f)**2 - 42*d**3*f**2*(d + 2*e + 4*f) + 15*d**3*(d + 2*e + 4*f)**3/2 + 432*d
**2*e**2*f*(d + 2*e + 4*f) - 8000*d**2*e*f**3 + 240*d**2*e*f*(d + 2*e + 4*f)**2 + 240*d**2*f**3*(d + 2*e + 4*f
) + 12*d**2*f*(d + 2*e + 4*f)**3 + 320*d*e**5 + 96*d*e**4*(d + 2*e + 4*f) + 720*d*e**3*f**2 - 80*d*e**3*(d + 2
*e + 4*f)**2 + 1080*d*e**2*f**2*(d + 2*e + 4*f) - 24*d*e**2*(d + 2*e + 4*f)**3 - 6400*d*e*f**4 + 492*d*e*f**2*
(d + 2*e + 4*f)**2 + 576*d*f**4*(d + 2*e + 4*f) - 30*d*f**2*(d + 2*e + 4*f)**3 + 512*e**5*f - 128*e**3*f*(d +
2*e + 4*f)**2 + 576*e**2*f**3*(d + 2*e + 4*f) - 1472*e*f**5 + 320*e*f**3*(d + 2*e + 4*f)**2 + 480*f**5*(d + 2*
e + 4*f) - 48*f**3*(d + 2*e + 4*f)**3)/(9*d**6 + 45*d**5*f - 160*d**4*e**2 - 36*d**4*f**2 - 1312*d**3*e**2*f -
360*d**3*f**3 + 256*d**2*e**4 - 3840*d**2*e**2*f**2 - 144*d**2*f**4 + 1280*d*e**4*f - 5248*d*e**2*f**3 + 720*
d*f**5 + 1024*e**4*f**2 - 2560*e**2*f**4 + 576*f**6))/12
Maxima [A] (verification not implemented)
none
Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00
\[
\int \frac {d+e x+f x^2}{4-5 x^2+x^4} \, dx=-\frac {1}{12} \, {\left (d - 2 \, e + 4 \, f\right )} \log \left (x + 2\right ) + \frac {1}{6} \, {\left (d - e + f\right )} \log \left (x + 1\right ) - \frac {1}{6} \, {\left (d + e + f\right )} \log \left (x - 1\right ) + \frac {1}{12} \, {\left (d + 2 \, e + 4 \, f\right )} \log \left (x - 2\right )
\]
[In]
integrate((f*x^2+e*x+d)/(x^4-5*x^2+4),x, algorithm="maxima")
[Out]
-1/12*(d - 2*e + 4*f)*log(x + 2) + 1/6*(d - e + f)*log(x + 1) - 1/6*(d + e + f)*log(x - 1) + 1/12*(d + 2*e + 4
*f)*log(x - 2)
Giac [A] (verification not implemented)
none
Time = 0.36 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08
\[
\int \frac {d+e x+f x^2}{4-5 x^2+x^4} \, dx=-\frac {1}{12} \, {\left (d - 2 \, e + 4 \, f\right )} \log \left ({\left | x + 2 \right |}\right ) + \frac {1}{6} \, {\left (d - e + f\right )} \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{6} \, {\left (d + e + f\right )} \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{12} \, {\left (d + 2 \, e + 4 \, f\right )} \log \left ({\left | x - 2 \right |}\right )
\]
[In]
integrate((f*x^2+e*x+d)/(x^4-5*x^2+4),x, algorithm="giac")
[Out]
-1/12*(d - 2*e + 4*f)*log(abs(x + 2)) + 1/6*(d - e + f)*log(abs(x + 1)) - 1/6*(d + e + f)*log(abs(x - 1)) + 1/
12*(d + 2*e + 4*f)*log(abs(x - 2))
Mupad [B] (verification not implemented)
Time = 7.76 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.24
\[
\int \frac {d+e x+f x^2}{4-5 x^2+x^4} \, dx=\ln \left (x+1\right )\,\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}\right )-\ln \left (x-1\right )\,\left (\frac {d}{6}+\frac {e}{6}+\frac {f}{6}\right )+\ln \left (x-2\right )\,\left (\frac {d}{12}+\frac {e}{6}+\frac {f}{3}\right )-\ln \left (x+2\right )\,\left (\frac {d}{12}-\frac {e}{6}+\frac {f}{3}\right )
\]
[In]
int((d + e*x + f*x^2)/(x^4 - 5*x^2 + 4),x)
[Out]
log(x + 1)*(d/6 - e/6 + f/6) - log(x - 1)*(d/6 + e/6 + f/6) + log(x - 2)*(d/12 + e/6 + f/3) - log(x + 2)*(d/12
- e/6 + f/3)