\(\int \frac {(2+x) (d+e x)}{(4-5 x^2+x^4)^2} \, dx\) [98]

Optimal result

Integrand size = 21, antiderivative size = 105 \[ \int \frac {(2+x) (d+e x)}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {d+e}{12 (1-x)}+\frac {d+2 e}{36 (2-x)}-\frac {d-e}{36 (1+x)}+\frac {1}{36} (2 d+5 e) \log (1-x)-\frac {1}{432} (35 d+58 e) \log (2-x)+\frac {1}{108} (2 d+e) \log (1+x)+\frac {1}{144} (d-2 e) \log (2+x) \]

[Out]

1/12*(d+e)/(1-x)+1/36*(d+2*e)/(2-x)+1/36*(-d+e)/(1+x)+1/36*(2*d+5*e)*ln(1-x)-1/432*(35*d+58*e)*ln(2-x)+1/108*(
2*d+e)*ln(1+x)+1/144*(d-2*e)*ln(2+x)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1600, 6874} \[ \int \frac {(2+x) (d+e x)}{\left (4-5 x^2+x^4\right )^2} \, dx=-\frac {d-e}{36 (x+1)}+\frac {d+e}{12 (1-x)}+\frac {d+2 e}{36 (2-x)}+\frac {1}{36} (2 d+5 e) \log (1-x)-\frac {1}{432} (35 d+58 e) \log (2-x)+\frac {1}{108} (2 d+e) \log (x+1)+\frac {1}{144} (d-2 e) \log (x+2) \]

[In]

Int[((2 + x)*(d + e*x))/(4 - 5*x^2 + x^4)^2,x]

[Out]

(d + e)/(12*(1 - x)) + (d + 2*e)/(36*(2 - x)) - (d - e)/(36*(1 + x)) + ((2*d + 5*e)*Log[1 - x])/36 - ((35*d +
58*e)*Log[2 - x])/432 + ((2*d + e)*Log[1 + x])/108 + ((d - 2*e)*Log[2 + x])/144

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {d+e x}{(2+x) \left (2-x-2 x^2+x^3\right )^2} \, dx \\ & = \int \left (\frac {d+2 e}{36 (-2+x)^2}+\frac {-35 d-58 e}{432 (-2+x)}+\frac {d+e}{12 (-1+x)^2}+\frac {2 d+5 e}{36 (-1+x)}+\frac {d-e}{36 (1+x)^2}+\frac {2 d+e}{108 (1+x)}+\frac {d-2 e}{144 (2+x)}\right ) \, dx \\ & = \frac {d+e}{12 (1-x)}+\frac {d+2 e}{36 (2-x)}-\frac {d-e}{36 (1+x)}+\frac {1}{36} (2 d+5 e) \log (1-x)-\frac {1}{432} (35 d+58 e) \log (2-x)+\frac {1}{108} (2 d+e) \log (1+x)+\frac {1}{144} (d-2 e) \log (2+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92 \[ \int \frac {(2+x) (d+e x)}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{432} \left (\frac {12 \left (d \left (5+6 x-5 x^2\right )+2 e \left (5-2 x^2\right )\right )}{2-x-2 x^2+x^3}+12 (2 d+5 e) \log (1-x)-(35 d+58 e) \log (2-x)+4 (2 d+e) \log (1+x)+3 (d-2 e) \log (2+x)\right ) \]

[In]

Integrate[((2 + x)*(d + e*x))/(4 - 5*x^2 + x^4)^2,x]

[Out]

((12*(d*(5 + 6*x - 5*x^2) + 2*e*(5 - 2*x^2)))/(2 - x - 2*x^2 + x^3) + 12*(2*d + 5*e)*Log[1 - x] - (35*d + 58*e
)*Log[2 - x] + 4*(2*d + e)*Log[1 + x] + 3*(d - 2*e)*Log[2 + x])/432

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88

method	result	size
default	\(\left (\frac {d}{144}-\frac {e}{72}\right ) \ln \left (x +2\right )-\frac {\frac {d}{36}-\frac {e}{36}}{x +1}+\left (\frac {d}{54}+\frac {e}{108}\right ) \ln \left (x +1\right )-\frac {\frac {d}{12}+\frac {e}{12}}{x -1}+\left (\frac {d}{18}+\frac {5 e}{36}\right ) \ln \left (x -1\right )+\left (-\frac {35 d}{432}-\frac {29 e}{216}\right ) \ln \left (x -2\right )-\frac {\frac {d}{36}+\frac {e}{18}}{x -2}\)	\(92\)
norman	\(\frac {\left (-\frac {5 d}{36}-\frac {e}{9}\right ) x^{3}+\left (\frac {17 d}{36}+\frac {5 e}{18}\right ) x +\left (-\frac {d}{9}-\frac {2 e}{9}\right ) x^{2}+\frac {5 d}{18}+\frac {5 e}{9}}{x^{4}-5 x^{2}+4}+\left (-\frac {35 d}{432}-\frac {29 e}{216}\right ) \ln \left (x -2\right )+\left (\frac {d}{18}+\frac {5 e}{36}\right ) \ln \left (x -1\right )+\left (\frac {d}{54}+\frac {e}{108}\right ) \ln \left (x +1\right )+\left (\frac {d}{144}-\frac {e}{72}\right ) \ln \left (x +2\right )\)	\(101\)
risch	\(\frac {\left (-\frac {5 d}{36}-\frac {e}{9}\right ) x^{2}+\frac {d x}{6}+\frac {5 d}{36}+\frac {5 e}{18}}{x^{3}-2 x^{2}-x +2}+\frac {\ln \left (x +2\right ) d}{144}-\frac {\ln \left (x +2\right ) e}{72}+\frac {\ln \left (x -1\right ) d}{18}+\frac {5 \ln \left (x -1\right ) e}{36}-\frac {35 \ln \left (2-x \right ) d}{432}-\frac {29 \ln \left (2-x \right ) e}{216}+\frac {\ln \left (-x -1\right ) d}{54}+\frac {\ln \left (-x -1\right ) e}{108}\)	\(104\)
parallelrisch	\(-\frac {60 d \,x^{2}-60 d -120 e -72 d x -60 \ln \left (x -1\right ) x^{3} e -8 \ln \left (x +1\right ) x^{3} d -4 \ln \left (x +1\right ) x^{3} e -3 \ln \left (x +2\right ) x^{3} d +6 \ln \left (x +2\right ) x^{3} e +70 \ln \left (x -2\right ) d +116 \ln \left (x -2\right ) e -48 \ln \left (x -1\right ) d -120 \ln \left (x -1\right ) e +48 e \,x^{2}+58 \ln \left (x -2\right ) x^{3} e -24 \ln \left (x -1\right ) x^{3} d -6 \ln \left (x +2\right ) x e -58 \ln \left (x -2\right ) x e +24 \ln \left (x -1\right ) x d +60 \ln \left (x -1\right ) x e +8 \ln \left (x +1\right ) x d +4 \ln \left (x +1\right ) x e +3 \ln \left (x +2\right ) x d -116 \ln \left (x -2\right ) x^{2} e +48 \ln \left (x -1\right ) x^{2} d +120 \ln \left (x -1\right ) x^{2} e +16 \ln \left (x +1\right ) x^{2} d +8 \ln \left (x +1\right ) x^{2} e +6 \ln \left (x +2\right ) x^{2} d -12 \ln \left (x +2\right ) x^{2} e -6 \ln \left (x +2\right ) d +12 \ln \left (x +2\right ) e -16 \ln \left (x +1\right ) d -8 \ln \left (x +1\right ) e -35 \ln \left (x -2\right ) x d +35 \ln \left (x -2\right ) x^{3} d -70 \ln \left (x -2\right ) x^{2} d}{432 \left (x^{3}-2 x^{2}-x +2\right )}\)	\(321\)

[In]

int((x+2)*(e*x+d)/(x^4-5*x^2+4)^2,x,method=_RETURNVERBOSE)

[Out]

(1/144*d-1/72*e)*ln(x+2)-(1/36*d-1/36*e)/(x+1)+(1/54*d+1/108*e)*ln(x+1)-(1/12*d+1/12*e)/(x-1)+(1/18*d+5/36*e)*
ln(x-1)+(-35/432*d-29/216*e)*ln(x-2)-(1/36*d+1/18*e)/(x-2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (87) = 174\).

Time = 0.28 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.01 \[ \int \frac {(2+x) (d+e x)}{\left (4-5 x^2+x^4\right )^2} \, dx=-\frac {12 \, {\left (5 \, d + 4 \, e\right )} x^{2} - 72 \, d x - 3 \, {\left ({\left (d - 2 \, e\right )} x^{3} - 2 \, {\left (d - 2 \, e\right )} x^{2} - {\left (d - 2 \, e\right )} x + 2 \, d - 4 \, e\right )} \log \left (x + 2\right ) - 4 \, {\left ({\left (2 \, d + e\right )} x^{3} - 2 \, {\left (2 \, d + e\right )} x^{2} - {\left (2 \, d + e\right )} x + 4 \, d + 2 \, e\right )} \log \left (x + 1\right ) - 12 \, {\left ({\left (2 \, d + 5 \, e\right )} x^{3} - 2 \, {\left (2 \, d + 5 \, e\right )} x^{2} - {\left (2 \, d + 5 \, e\right )} x + 4 \, d + 10 \, e\right )} \log \left (x - 1\right ) + {\left ({\left (35 \, d + 58 \, e\right )} x^{3} - 2 \, {\left (35 \, d + 58 \, e\right )} x^{2} - {\left (35 \, d + 58 \, e\right )} x + 70 \, d + 116 \, e\right )} \log \left (x - 2\right ) - 60 \, d - 120 \, e}{432 \, {\left (x^{3} - 2 \, x^{2} - x + 2\right )}} \]

[In]

integrate((2+x)*(e*x+d)/(x^4-5*x^2+4)^2,x, algorithm="fricas")

[Out]

-1/432*(12*(5*d + 4*e)*x^2 - 72*d*x - 3*((d - 2*e)*x^3 - 2*(d - 2*e)*x^2 - (d - 2*e)*x + 2*d - 4*e)*log(x + 2)
- 4*((2*d + e)*x^3 - 2*(2*d + e)*x^2 - (2*d + e)*x + 4*d + 2*e)*log(x + 1) - 12*((2*d + 5*e)*x^3 - 2*(2*d + 5
*e)*x^2 - (2*d + 5*e)*x + 4*d + 10*e)*log(x - 1) + ((35*d + 58*e)*x^3 - 2*(35*d + 58*e)*x^2 - (35*d + 58*e)*x
+ 70*d + 116*e)*log(x - 2) - 60*d - 120*e)/(x^3 - 2*x^2 - x + 2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1034 vs. \(2 (82) = 164\).

Time = 5.46 (sec) , antiderivative size = 1034, normalized size of antiderivative = 9.85 \[ \int \frac {(2+x) (d+e x)}{\left (4-5 x^2+x^4\right )^2} \, dx=\text {Too large to display} \]

[In]

integrate((2+x)*(e*x+d)/(x**4-5*x**2+4)**2,x)

[Out]

(d - 2*e)*log(x + (8710660*d**5 + 91884504*d**4*e - 7579779*d**4*(d - 2*e)/4 + 364910432*d**3*e**2 - 18128055*
d**3*e*(d - 2*e) - 83772*d**3*(d - 2*e)**2 + 686697536*d**2*e**3 - 60296868*d**2*e**2*(d - 2*e) - 597816*d**2*
e*(d - 2*e)**2 + 65907*d**2*(d - 2*e)**3/4 + 614357568*d*e**4 - 85949220*d*e**3*(d - 2*e) - 1500048*d*e**2*(d
- 2*e)**2 + 105840*d*e*(d - 2*e)**3 + 208470400*e**5 - 45136356*e**4*(d - 2*e) - 1196064*e**3*(d - 2*e)**2 + 1
28277*e**2*(d - 2*e)**3)/(3374210*d**5 + 38645295*d**4*e + 170558380*d**3*e**2 + 362061760*d**2*e**3 + 3702981
60*d*e**4 + 146466320*e**5))/144 + (2*d + e)*log(x + (8710660*d**5 + 91884504*d**4*e - 2526593*d**4*(2*d + e)
+ 364910432*d**3*e**2 - 24170740*d**3*e*(2*d + e) - 148928*d**3*(2*d + e)**2 + 686697536*d**2*e**3 - 80395824*
d**2*e**2*(2*d + e) - 1062784*d**2*e*(2*d + e)**2 + 39056*d**2*(2*d + e)**3 + 614357568*d*e**4 - 114598960*d*e
**3*(2*d + e) - 2666752*d*e**2*(2*d + e)**2 + 250880*d*e*(2*d + e)**3 + 208470400*e**5 - 60181808*e**4*(2*d +
e) - 2126336*e**3*(2*d + e)**2 + 304064*e**2*(2*d + e)**3)/(3374210*d**5 + 38645295*d**4*e + 170558380*d**3*e*
*2 + 362061760*d**2*e**3 + 370298160*d*e**4 + 146466320*e**5))/108 + (2*d + 5*e)*log(x + (8710660*d**5 + 91884
504*d**4*e - 7579779*d**4*(2*d + 5*e) + 364910432*d**3*e**2 - 72512220*d**3*e*(2*d + 5*e) - 1340352*d**3*(2*d
+ 5*e)**2 + 686697536*d**2*e**3 - 241187472*d**2*e**2*(2*d + 5*e) - 9565056*d**2*e*(2*d + 5*e)**2 + 1054512*d*
*2*(2*d + 5*e)**3 + 614357568*d*e**4 - 343796880*d*e**3*(2*d + 5*e) - 24000768*d*e**2*(2*d + 5*e)**2 + 6773760
*d*e*(2*d + 5*e)**3 + 208470400*e**5 - 180545424*e**4*(2*d + 5*e) - 19137024*e**3*(2*d + 5*e)**2 + 8209728*e**
2*(2*d + 5*e)**3)/(3374210*d**5 + 38645295*d**4*e + 170558380*d**3*e**2 + 362061760*d**2*e**3 + 370298160*d*e*
*4 + 146466320*e**5))/36 - (35*d + 58*e)*log(x + (8710660*d**5 + 91884504*d**4*e + 2526593*d**4*(35*d + 58*e)/
4 + 364910432*d**3*e**2 + 6042685*d**3*e*(35*d + 58*e) - 9308*d**3*(35*d + 58*e)**2 + 686697536*d**2*e**3 + 20
098956*d**2*e**2*(35*d + 58*e) - 66424*d**2*e*(35*d + 58*e)**2 - 2441*d**2*(35*d + 58*e)**3/4 + 614357568*d*e*
*4 + 28649740*d*e**3*(35*d + 58*e) - 166672*d*e**2*(35*d + 58*e)**2 - 3920*d*e*(35*d + 58*e)**3 + 208470400*e*
*5 + 15045452*e**4*(35*d + 58*e) - 132896*e**3*(35*d + 58*e)**2 - 4751*e**2*(35*d + 58*e)**3)/(3374210*d**5 +
38645295*d**4*e + 170558380*d**3*e**2 + 362061760*d**2*e**3 + 370298160*d*e**4 + 146466320*e**5))/432 + (6*d*x
+ 5*d + 10*e + x**2*(-5*d - 4*e))/(36*x**3 - 72*x**2 - 36*x + 72)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.84 \[ \int \frac {(2+x) (d+e x)}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{144} \, {\left (d - 2 \, e\right )} \log \left (x + 2\right ) + \frac {1}{108} \, {\left (2 \, d + e\right )} \log \left (x + 1\right ) + \frac {1}{36} \, {\left (2 \, d + 5 \, e\right )} \log \left (x - 1\right ) - \frac {1}{432} \, {\left (35 \, d + 58 \, e\right )} \log \left (x - 2\right ) - \frac {{\left (5 \, d + 4 \, e\right )} x^{2} - 6 \, d x - 5 \, d - 10 \, e}{36 \, {\left (x^{3} - 2 \, x^{2} - x + 2\right )}} \]

[In]

integrate((2+x)*(e*x+d)/(x^4-5*x^2+4)^2,x, algorithm="maxima")

[Out]

1/144*(d - 2*e)*log(x + 2) + 1/108*(2*d + e)*log(x + 1) + 1/36*(2*d + 5*e)*log(x - 1) - 1/432*(35*d + 58*e)*lo
g(x - 2) - 1/36*((5*d + 4*e)*x^2 - 6*d*x - 5*d - 10*e)/(x^3 - 2*x^2 - x + 2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int \frac {(2+x) (d+e x)}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{144} \, {\left (d - 2 \, e\right )} \log \left ({\left | x + 2 \right |}\right ) + \frac {1}{108} \, {\left (2 \, d + e\right )} \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{36} \, {\left (2 \, d + 5 \, e\right )} \log \left ({\left | x - 1 \right |}\right ) - \frac {1}{432} \, {\left (35 \, d + 58 \, e\right )} \log \left ({\left | x - 2 \right |}\right ) - \frac {{\left (5 \, d + 4 \, e\right )} x^{2} - 6 \, d x - 5 \, d - 10 \, e}{36 \, {\left (x + 1\right )} {\left (x - 1\right )} {\left (x - 2\right )}} \]

[In]

integrate((2+x)*(e*x+d)/(x^4-5*x^2+4)^2,x, algorithm="giac")

[Out]

1/144*(d - 2*e)*log(abs(x + 2)) + 1/108*(2*d + e)*log(abs(x + 1)) + 1/36*(2*d + 5*e)*log(abs(x - 1)) - 1/432*(
35*d + 58*e)*log(abs(x - 2)) - 1/36*((5*d + 4*e)*x^2 - 6*d*x - 5*d - 10*e)/((x + 1)*(x - 1)*(x - 2))

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.86 \[ \int \frac {(2+x) (d+e x)}{\left (4-5 x^2+x^4\right )^2} \, dx=\ln \left (x-1\right )\,\left (\frac {d}{18}+\frac {5\,e}{36}\right )-\frac {\left (-\frac {5\,d}{36}-\frac {e}{9}\right )\,x^2+\frac {d\,x}{6}+\frac {5\,d}{36}+\frac {5\,e}{18}}{-x^3+2\,x^2+x-2}+\ln \left (x+1\right )\,\left (\frac {d}{54}+\frac {e}{108}\right )+\ln \left (x+2\right )\,\left (\frac {d}{144}-\frac {e}{72}\right )-\ln \left (x-2\right )\,\left (\frac {35\,d}{432}+\frac {29\,e}{216}\right ) \]

[In]

int(((x + 2)*(d + e*x))/(x^4 - 5*x^2 + 4)^2,x)

[Out]

log(x - 1)*(d/18 + (5*e)/36) - ((5*d)/36 + (5*e)/18 - x^2*((5*d)/36 + e/9) + (d*x)/6)/(x + 2*x^2 - x^3 - 2) +
log(x + 1)*(d/54 + e/108) + log(x + 2)*(d/144 - e/72) - log(x - 2)*((35*d)/432 + (29*e)/216)