3.30 \(\int \frac{d+e x+f x^2+g x^3+h x^4+i x^5}{(4-5 x^2+x^4)^2} \, dx\)

Optimal. Leaf size=162 \[ \frac{x \left (x^2 (-(5 d+8 f+20 h))+17 d+20 f+32 h\right )}{72 \left (x^4-5 x^2+4\right )}+\frac{1}{432} \tanh ^{-1}\left (\frac{x}{2}\right ) (19 d+52 f+112 h)-\frac{1}{54} \tanh ^{-1}(x) (d+7 f+13 h)+\frac{x^2 (-(2 e+5 g+17 i))+5 e+8 g+20 i}{18 \left (x^4-5 x^2+4\right )}+\frac{1}{54} \log \left (1-x^2\right ) (2 e+5 g+8 i)-\frac{1}{54} \log \left (4-x^2\right ) (2 e+5 g+8 i) \]

[Out]

(x*(17*d + 20*f + 32*h - (5*d + 8*f + 20*h)*x^2))/(72*(4 - 5*x^2 + x^4)) + (5*e + 8*g + 20*i - (2*e + 5*g + 17
*i)*x^2)/(18*(4 - 5*x^2 + x^4)) + ((19*d + 52*f + 112*h)*ArcTanh[x/2])/432 - ((d + 7*f + 13*h)*ArcTanh[x])/54
+ ((2*e + 5*g + 8*i)*Log[1 - x^2])/54 - ((2*e + 5*g + 8*i)*Log[4 - x^2])/54

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Rubi [A]  time = 0.231837, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.237, Rules used = {1673, 1678, 1166, 207, 1663, 1660, 12, 616, 31} \[ \frac{x \left (x^2 (-(5 d+8 f+20 h))+17 d+20 f+32 h\right )}{72 \left (x^4-5 x^2+4\right )}+\frac{1}{432} \tanh ^{-1}\left (\frac{x}{2}\right ) (19 d+52 f+112 h)-\frac{1}{54} \tanh ^{-1}(x) (d+7 f+13 h)+\frac{x^2 (-(2 e+5 g+17 i))+5 e+8 g+20 i}{18 \left (x^4-5 x^2+4\right )}+\frac{1}{54} \log \left (1-x^2\right ) (2 e+5 g+8 i)-\frac{1}{54} \log \left (4-x^2\right ) (2 e+5 g+8 i) \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*x^2 + g*x^3 + h*x^4 + i*x^5)/(4 - 5*x^2 + x^4)^2,x]

[Out]

(x*(17*d + 20*f + 32*h - (5*d + 8*f + 20*h)*x^2))/(72*(4 - 5*x^2 + x^4)) + (5*e + 8*g + 20*i - (2*e + 5*g + 17
*i)*x^2)/(18*(4 - 5*x^2 + x^4)) + ((19*d + 52*f + 112*h)*ArcTanh[x/2])/432 - ((d + 7*f + 13*h)*ArcTanh[x])/54
+ ((2*e + 5*g + 8*i)*Log[1 - x^2])/54 - ((2*e + 5*g + 8*i)*Log[4 - x^2])/54

Rule 1673

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]

Rule 1678

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainder[Pq, a +
b*x^2 + c*x^4, x], x, 0], e = Coeff[PolynomialRemainder[Pq, a + b*x^2 + c*x^4, x], x, 2]}, Simp[(x*(a + b*x^2
+ c*x^4)^(p + 1)*(a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*
a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuot
ient[Pq, a + b*x^2 + c*x^4, x] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4*p + 7)*(b*d - 2*a*e)*x^2,
x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 1166

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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 1663

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)
*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && Inte
gerQ[(m - 1)/2]

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 616

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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{d+e x+f x^2+g x^3+h x^4+30 x^5}{\left (4-5 x^2+x^4\right )^2} \, dx &=\int \frac{x \left (e+g x^2+30 x^4\right )}{\left (4-5 x^2+x^4\right )^2} \, dx+\int \frac{d+f x^2+h x^4}{\left (4-5 x^2+x^4\right )^2} \, dx\\ &=\frac{x \left (17 d+20 f+32 h-(5 d+8 f+20 h) x^2\right )}{72 \left (4-5 x^2+x^4\right )}-\frac{1}{72} \int \frac{-d+20 f+32 h+(5 d+8 f+20 h) x^2}{4-5 x^2+x^4} \, dx+\frac{1}{2} \operatorname{Subst}\left (\int \frac{e+g x+30 x^2}{\left (4-5 x+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{600+5 e+8 g-(510+2 e+5 g) x^2}{18 \left (4-5 x^2+x^4\right )}+\frac{x \left (17 d+20 f+32 h-(5 d+8 f+20 h) x^2\right )}{72 \left (4-5 x^2+x^4\right )}-\frac{1}{18} \operatorname{Subst}\left (\int \frac{240+2 e+5 g}{4-5 x+x^2} \, dx,x,x^2\right )-\frac{1}{54} (-d-7 f-13 h) \int \frac{1}{-1+x^2} \, dx-\frac{1}{216} (19 d+52 f+112 h) \int \frac{1}{-4+x^2} \, dx\\ &=\frac{600+5 e+8 g-(510+2 e+5 g) x^2}{18 \left (4-5 x^2+x^4\right )}+\frac{x \left (17 d+20 f+32 h-(5 d+8 f+20 h) x^2\right )}{72 \left (4-5 x^2+x^4\right )}+\frac{1}{432} (19 d+52 f+112 h) \tanh ^{-1}\left (\frac{x}{2}\right )-\frac{1}{54} (d+7 f+13 h) \tanh ^{-1}(x)-\frac{1}{18} (240+2 e+5 g) \operatorname{Subst}\left (\int \frac{1}{4-5 x+x^2} \, dx,x,x^2\right )\\ &=\frac{600+5 e+8 g-(510+2 e+5 g) x^2}{18 \left (4-5 x^2+x^4\right )}+\frac{x \left (17 d+20 f+32 h-(5 d+8 f+20 h) x^2\right )}{72 \left (4-5 x^2+x^4\right )}+\frac{1}{432} (19 d+52 f+112 h) \tanh ^{-1}\left (\frac{x}{2}\right )-\frac{1}{54} (d+7 f+13 h) \tanh ^{-1}(x)-\frac{1}{54} (-240-2 e-5 g) \operatorname{Subst}\left (\int \frac{1}{-1+x} \, dx,x,x^2\right )-\frac{1}{54} (240+2 e+5 g) \operatorname{Subst}\left (\int \frac{1}{-4+x} \, dx,x,x^2\right )\\ &=\frac{600+5 e+8 g-(510+2 e+5 g) x^2}{18 \left (4-5 x^2+x^4\right )}+\frac{x \left (17 d+20 f+32 h-(5 d+8 f+20 h) x^2\right )}{72 \left (4-5 x^2+x^4\right )}+\frac{1}{432} (19 d+52 f+112 h) \tanh ^{-1}\left (\frac{x}{2}\right )-\frac{1}{54} (d+7 f+13 h) \tanh ^{-1}(x)+\frac{1}{54} (240+2 e+5 g) \log \left (1-x^2\right )-\frac{1}{54} (240+2 e+5 g) \log \left (4-x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0940909, size = 185, normalized size = 1.14 \[ \frac{-5 d x^3+17 d x-8 e x^2+20 e-8 f x^3+20 f x-20 g x^2+32 g-20 h x^3+32 h x-68 i x^2+80 i}{72 \left (x^4-5 x^2+4\right )}+\frac{1}{108} \log (1-x) (d+4 e+7 f+10 g+13 h+16 i)+\frac{1}{864} \log (2-x) (-19 d-32 e-52 f-80 g-112 h-128 i)+\frac{1}{108} \log (x+1) (-d+4 e-7 f+10 g-13 h+16 i)+\frac{1}{864} \log (x+2) (19 d-32 e+52 f-80 g+112 h-128 i) \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*x^2 + g*x^3 + h*x^4 + i*x^5)/(4 - 5*x^2 + x^4)^2,x]

[Out]

(20*e + 32*g + 80*i + 17*d*x + 20*f*x + 32*h*x - 8*e*x^2 - 20*g*x^2 - 68*i*x^2 - 5*d*x^3 - 8*f*x^3 - 20*h*x^3)
/(72*(4 - 5*x^2 + x^4)) + ((d + 4*e + 7*f + 10*g + 13*h + 16*i)*Log[1 - x])/108 + ((-19*d - 32*e - 52*f - 80*g
- 112*h - 128*i)*Log[2 - x])/864 + ((-d + 4*e - 7*f + 10*g - 13*h + 16*i)*Log[1 + x])/108 + ((19*d - 32*e + 5
2*f - 80*g + 112*h - 128*i)*Log[2 + x])/864

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Maple [B]  time = 0.019, size = 362, normalized size = 2.2 \begin{align*}{\frac{19\,\ln \left ( 2+x \right ) d}{864}}-{\frac{\ln \left ( 2+x \right ) e}{27}}-{\frac{\ln \left ( 1+x \right ) d}{108}}+{\frac{\ln \left ( 1+x \right ) e}{27}}-{\frac{19\,\ln \left ( x-2 \right ) d}{864}}-{\frac{\ln \left ( x-2 \right ) e}{27}}+{\frac{\ln \left ( x-1 \right ) d}{108}}+{\frac{\ln \left ( x-1 \right ) e}{27}}-{\frac{2\,i}{9\,x-18}}-{\frac{i}{36\,x-36}}+{\frac{i}{36+36\,x}}+{\frac{2\,i}{18+9\,x}}-{\frac{h}{9\,x-18}}-{\frac{h}{36\,x-36}}-{\frac{h}{36+36\,x}}-{\frac{h}{18+9\,x}}+{\frac{g}{36+18\,x}}-{\frac{d}{36+36\,x}}+{\frac{e}{36+36\,x}}-{\frac{g}{18\,x-36}}-{\frac{d}{144\,x-288}}-{\frac{e}{72\,x-144}}-{\frac{g}{36\,x-36}}-{\frac{d}{36\,x-36}}-{\frac{e}{36\,x-36}}-{\frac{d}{288+144\,x}}+{\frac{e}{144+72\,x}}+{\frac{g}{36+36\,x}}-{\frac{f}{36+36\,x}}-{\frac{f}{36\,x-72}}-{\frac{f}{36\,x-36}}-{\frac{f}{72+36\,x}}-{\frac{4\,\ln \left ( x-2 \right ) i}{27}}+{\frac{4\,\ln \left ( x-1 \right ) i}{27}}-{\frac{4\,\ln \left ( 2+x \right ) i}{27}}+{\frac{4\,\ln \left ( 1+x \right ) i}{27}}-{\frac{5\,\ln \left ( 2+x \right ) g}{54}}+{\frac{5\,\ln \left ( 1+x \right ) g}{54}}-{\frac{5\,\ln \left ( x-2 \right ) g}{54}}+{\frac{5\,\ln \left ( x-1 \right ) g}{54}}+{\frac{7\,\ln \left ( 2+x \right ) h}{54}}-{\frac{13\,\ln \left ( 1+x \right ) h}{108}}-{\frac{7\,\ln \left ( x-2 \right ) h}{54}}+{\frac{13\,\ln \left ( x-1 \right ) h}{108}}-{\frac{13\,\ln \left ( x-2 \right ) f}{216}}+{\frac{7\,\ln \left ( x-1 \right ) f}{108}}+{\frac{13\,\ln \left ( 2+x \right ) f}{216}}-{\frac{7\,\ln \left ( 1+x \right ) f}{108}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4)^2,x)

[Out]

19/864*ln(2+x)*d-1/27*ln(2+x)*e-1/108*ln(1+x)*d+1/27*ln(1+x)*e-19/864*ln(x-2)*d-1/27*ln(x-2)*e+1/108*ln(x-1)*d
+1/27*ln(x-1)*e-2/9/(x-2)*i-1/36/(x-1)*i+1/36/(1+x)*i+2/9/(2+x)*i-1/9/(x-2)*h-1/36/(x-1)*h-1/36/(1+x)*h-1/9/(2
+x)*h+1/18/(2+x)*g-1/36/(1+x)*d+1/36/(1+x)*e-1/18/(x-2)*g-1/144/(x-2)*d-1/72/(x-2)*e-1/36/(x-1)*g-1/36/(x-1)*d
-1/36/(x-1)*e-1/144/(2+x)*d+1/72/(2+x)*e+1/36/(1+x)*g-1/36/(1+x)*f-1/36/(x-2)*f-1/36/(x-1)*f-1/36/(2+x)*f-4/27
*ln(x-2)*i+4/27*ln(x-1)*i-4/27*ln(2+x)*i+4/27*ln(1+x)*i-5/54*ln(2+x)*g+5/54*ln(1+x)*g-5/54*ln(x-2)*g+5/54*ln(x
-1)*g+7/54*ln(2+x)*h-13/108*ln(1+x)*h-7/54*ln(x-2)*h+13/108*ln(x-1)*h-13/216*ln(x-2)*f+7/108*ln(x-1)*f+13/216*
ln(2+x)*f-7/108*ln(1+x)*f

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Maxima [A]  time = 1.00125, size = 220, normalized size = 1.36 \begin{align*} \frac{1}{864} \,{\left (19 \, d - 32 \, e + 52 \, f - 80 \, g + 112 \, h - 128 \, i\right )} \log \left (x + 2\right ) - \frac{1}{108} \,{\left (d - 4 \, e + 7 \, f - 10 \, g + 13 \, h - 16 \, i\right )} \log \left (x + 1\right ) + \frac{1}{108} \,{\left (d + 4 \, e + 7 \, f + 10 \, g + 13 \, h + 16 \, i\right )} \log \left (x - 1\right ) - \frac{1}{864} \,{\left (19 \, d + 32 \, e + 52 \, f + 80 \, g + 112 \, h + 128 \, i\right )} \log \left (x - 2\right ) - \frac{{\left (5 \, d + 8 \, f + 20 \, h\right )} x^{3} + 4 \,{\left (2 \, e + 5 \, g + 17 \, i\right )} x^{2} -{\left (17 \, d + 20 \, f + 32 \, h\right )} x - 20 \, e - 32 \, g - 80 \, i}{72 \,{\left (x^{4} - 5 \, x^{2} + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4)^2,x, algorithm="maxima")

[Out]

1/864*(19*d - 32*e + 52*f - 80*g + 112*h - 128*i)*log(x + 2) - 1/108*(d - 4*e + 7*f - 10*g + 13*h - 16*i)*log(
x + 1) + 1/108*(d + 4*e + 7*f + 10*g + 13*h + 16*i)*log(x - 1) - 1/864*(19*d + 32*e + 52*f + 80*g + 112*h + 12
8*i)*log(x - 2) - 1/72*((5*d + 8*f + 20*h)*x^3 + 4*(2*e + 5*g + 17*i)*x^2 - (17*d + 20*f + 32*h)*x - 20*e - 32
*g - 80*i)/(x^4 - 5*x^2 + 4)

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Fricas [B]  time = 65.6791, size = 1007, normalized size = 6.22 \begin{align*} -\frac{12 \,{\left (5 \, d + 8 \, f + 20 \, h\right )} x^{3} + 48 \,{\left (2 \, e + 5 \, g + 17 \, i\right )} x^{2} - 12 \,{\left (17 \, d + 20 \, f + 32 \, h\right )} x -{\left ({\left (19 \, d - 32 \, e + 52 \, f - 80 \, g + 112 \, h - 128 \, i\right )} x^{4} - 5 \,{\left (19 \, d - 32 \, e + 52 \, f - 80 \, g + 112 \, h - 128 \, i\right )} x^{2} + 76 \, d - 128 \, e + 208 \, f - 320 \, g + 448 \, h - 512 \, i\right )} \log \left (x + 2\right ) + 8 \,{\left ({\left (d - 4 \, e + 7 \, f - 10 \, g + 13 \, h - 16 \, i\right )} x^{4} - 5 \,{\left (d - 4 \, e + 7 \, f - 10 \, g + 13 \, h - 16 \, i\right )} x^{2} + 4 \, d - 16 \, e + 28 \, f - 40 \, g + 52 \, h - 64 \, i\right )} \log \left (x + 1\right ) - 8 \,{\left ({\left (d + 4 \, e + 7 \, f + 10 \, g + 13 \, h + 16 \, i\right )} x^{4} - 5 \,{\left (d + 4 \, e + 7 \, f + 10 \, g + 13 \, h + 16 \, i\right )} x^{2} + 4 \, d + 16 \, e + 28 \, f + 40 \, g + 52 \, h + 64 \, i\right )} \log \left (x - 1\right ) +{\left ({\left (19 \, d + 32 \, e + 52 \, f + 80 \, g + 112 \, h + 128 \, i\right )} x^{4} - 5 \,{\left (19 \, d + 32 \, e + 52 \, f + 80 \, g + 112 \, h + 128 \, i\right )} x^{2} + 76 \, d + 128 \, e + 208 \, f + 320 \, g + 448 \, h + 512 \, i\right )} \log \left (x - 2\right ) - 240 \, e - 384 \, g - 960 \, i}{864 \,{\left (x^{4} - 5 \, x^{2} + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4)^2,x, algorithm="fricas")

[Out]

-1/864*(12*(5*d + 8*f + 20*h)*x^3 + 48*(2*e + 5*g + 17*i)*x^2 - 12*(17*d + 20*f + 32*h)*x - ((19*d - 32*e + 52
*f - 80*g + 112*h - 128*i)*x^4 - 5*(19*d - 32*e + 52*f - 80*g + 112*h - 128*i)*x^2 + 76*d - 128*e + 208*f - 32
0*g + 448*h - 512*i)*log(x + 2) + 8*((d - 4*e + 7*f - 10*g + 13*h - 16*i)*x^4 - 5*(d - 4*e + 7*f - 10*g + 13*h
- 16*i)*x^2 + 4*d - 16*e + 28*f - 40*g + 52*h - 64*i)*log(x + 1) - 8*((d + 4*e + 7*f + 10*g + 13*h + 16*i)*x^
4 - 5*(d + 4*e + 7*f + 10*g + 13*h + 16*i)*x^2 + 4*d + 16*e + 28*f + 40*g + 52*h + 64*i)*log(x - 1) + ((19*d +
32*e + 52*f + 80*g + 112*h + 128*i)*x^4 - 5*(19*d + 32*e + 52*f + 80*g + 112*h + 128*i)*x^2 + 76*d + 128*e +
208*f + 320*g + 448*h + 512*i)*log(x - 2) - 240*e - 384*g - 960*i)/(x^4 - 5*x^2 + 4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x**5+h*x**4+g*x**3+f*x**2+e*x+d)/(x**4-5*x**2+4)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.08324, size = 242, normalized size = 1.49 \begin{align*} \frac{1}{864} \,{\left (19 \, d + 52 \, f - 80 \, g + 112 \, h - 128 \, i - 32 \, e\right )} \log \left ({\left | x + 2 \right |}\right ) - \frac{1}{108} \,{\left (d + 7 \, f - 10 \, g + 13 \, h - 16 \, i - 4 \, e\right )} \log \left ({\left | x + 1 \right |}\right ) + \frac{1}{108} \,{\left (d + 7 \, f + 10 \, g + 13 \, h + 16 \, i + 4 \, e\right )} \log \left ({\left | x - 1 \right |}\right ) - \frac{1}{864} \,{\left (19 \, d + 52 \, f + 80 \, g + 112 \, h + 128 \, i + 32 \, e\right )} \log \left ({\left | x - 2 \right |}\right ) - \frac{5 \, d x^{3} + 8 \, f x^{3} + 20 \, h x^{3} + 20 \, g x^{2} + 68 \, i x^{2} + 8 \, x^{2} e - 17 \, d x - 20 \, f x - 32 \, h x - 32 \, g - 80 \, i - 20 \, e}{72 \,{\left (x^{4} - 5 \, x^{2} + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4)^2,x, algorithm="giac")

[Out]

1/864*(19*d + 52*f - 80*g + 112*h - 128*i - 32*e)*log(abs(x + 2)) - 1/108*(d + 7*f - 10*g + 13*h - 16*i - 4*e)
*log(abs(x + 1)) + 1/108*(d + 7*f + 10*g + 13*h + 16*i + 4*e)*log(abs(x - 1)) - 1/864*(19*d + 52*f + 80*g + 11
2*h + 128*i + 32*e)*log(abs(x - 2)) - 1/72*(5*d*x^3 + 8*f*x^3 + 20*h*x^3 + 20*g*x^2 + 68*i*x^2 + 8*x^2*e - 17*
d*x - 20*f*x - 32*h*x - 32*g - 80*i - 20*e)/(x^4 - 5*x^2 + 4)