### 3.66 $$\int \frac{d+e x^2+f x^4}{x^3 (a+b x^2+c x^4)^2} \, dx$$

Optimal. Leaf size=234 $-\frac{2 a^2 c e+c x^2 \left (-a b e-2 a (c d-a f)+b^2 d\right )-a b^2 e-a b (3 c d-a f)+b^3 d}{2 a^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right ) \left (6 a^2 b c e+4 a^2 c (3 c d-a f)-12 a b^2 c d-a b^3 e+2 b^4 d\right )}{2 a^3 \left (b^2-4 a c\right )^{3/2}}+\frac{(2 b d-a e) \log \left (a+b x^2+c x^4\right )}{4 a^3}-\frac{\log (x) (2 b d-a e)}{a^3}-\frac{d}{2 a^2 x^2}$

[Out]

-d/(2*a^2*x^2) - (b^3*d - a*b^2*e + 2*a^2*c*e - a*b*(3*c*d - a*f) + c*(b^2*d - a*b*e - 2*a*(c*d - a*f))*x^2)/(
2*a^2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - ((2*b^4*d - 12*a*b^2*c*d - a*b^3*e + 6*a^2*b*c*e + 4*a^2*c*(3*c*d -
a*f))*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*a^3*(b^2 - 4*a*c)^(3/2)) - ((2*b*d - a*e)*Log[x])/a^3 + ((
2*b*d - a*e)*Log[a + b*x^2 + c*x^4])/(4*a^3)

________________________________________________________________________________________

Rubi [A]  time = 0.725096, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 30, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.233, Rules used = {1663, 1646, 1628, 634, 618, 206, 628} $-\frac{2 a^2 c e+c x^2 \left (-a b e-2 a (c d-a f)+b^2 d\right )-a b^2 e-a b (3 c d-a f)+b^3 d}{2 a^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right ) \left (6 a^2 b c e+4 a^2 c (3 c d-a f)-12 a b^2 c d-a b^3 e+2 b^4 d\right )}{2 a^3 \left (b^2-4 a c\right )^{3/2}}+\frac{(2 b d-a e) \log \left (a+b x^2+c x^4\right )}{4 a^3}-\frac{\log (x) (2 b d-a e)}{a^3}-\frac{d}{2 a^2 x^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x^2 + f*x^4)/(x^3*(a + b*x^2 + c*x^4)^2),x]

[Out]

-d/(2*a^2*x^2) - (b^3*d - a*b^2*e + 2*a^2*c*e - a*b*(3*c*d - a*f) + c*(b^2*d - a*b*e - 2*a*(c*d - a*f))*x^2)/(
2*a^2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - ((2*b^4*d - 12*a*b^2*c*d - a*b^3*e + 6*a^2*b*c*e + 4*a^2*c*(3*c*d -
a*f))*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*a^3*(b^2 - 4*a*c)^(3/2)) - ((2*b*d - a*e)*Log[x])/a^3 + ((
2*b*d - a*e)*Log[a + b*x^2 + c*x^4])/(4*a^3)

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 1646

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

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(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]

Rubi steps

\begin{align*} \int \frac{d+e x^2+f x^4}{x^3 \left (a+b x^2+c x^4\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{d+e x+f x^2}{x^2 \left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )\\ &=-\frac{b^3 d-a b^2 e+2 a^2 c e-a b (3 c d-a f)+c \left (b^2 d-a b e-2 a (c d-a f)\right ) x^2}{2 a^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \frac{-\left (\frac{b^2}{a}-4 c\right ) d+\frac{\left (b^2-4 a c\right ) (b d-a e) x}{a^2}+\frac{c \left (b^2 d-a b e-2 a (c d-a f)\right ) x^2}{a^2}}{x^2 \left (a+b x+c x^2\right )} \, dx,x,x^2\right )}{2 \left (b^2-4 a c\right )}\\ &=-\frac{b^3 d-a b^2 e+2 a^2 c e-a b (3 c d-a f)+c \left (b^2 d-a b e-2 a (c d-a f)\right ) x^2}{2 a^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \left (\frac{\left (-b^2+4 a c\right ) d}{a^2 x^2}+\frac{\left (-b^2+4 a c\right ) (-2 b d+a e)}{a^3 x}+\frac{-2 b^4 d+10 a b^2 c d+a b^3 e-5 a^2 b c e-2 a^2 c (3 c d-a f)-c \left (b^2-4 a c\right ) (2 b d-a e) x}{a^3 \left (a+b x+c x^2\right )}\right ) \, dx,x,x^2\right )}{2 \left (b^2-4 a c\right )}\\ &=-\frac{d}{2 a^2 x^2}-\frac{b^3 d-a b^2 e+2 a^2 c e-a b (3 c d-a f)+c \left (b^2 d-a b e-2 a (c d-a f)\right ) x^2}{2 a^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{(2 b d-a e) \log (x)}{a^3}-\frac{\operatorname{Subst}\left (\int \frac{-2 b^4 d+10 a b^2 c d+a b^3 e-5 a^2 b c e-2 a^2 c (3 c d-a f)-c \left (b^2-4 a c\right ) (2 b d-a e) x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 a^3 \left (b^2-4 a c\right )}\\ &=-\frac{d}{2 a^2 x^2}-\frac{b^3 d-a b^2 e+2 a^2 c e-a b (3 c d-a f)+c \left (b^2 d-a b e-2 a (c d-a f)\right ) x^2}{2 a^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{(2 b d-a e) \log (x)}{a^3}+\frac{(2 b d-a e) \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 a^3}+\frac{\left (2 b^4 d-12 a b^2 c d-a b^3 e+6 a^2 b c e+4 a^2 c (3 c d-a f)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 a^3 \left (b^2-4 a c\right )}\\ &=-\frac{d}{2 a^2 x^2}-\frac{b^3 d-a b^2 e+2 a^2 c e-a b (3 c d-a f)+c \left (b^2 d-a b e-2 a (c d-a f)\right ) x^2}{2 a^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{(2 b d-a e) \log (x)}{a^3}+\frac{(2 b d-a e) \log \left (a+b x^2+c x^4\right )}{4 a^3}-\frac{\left (2 b^4 d-12 a b^2 c d-a b^3 e+6 a^2 b c e+4 a^2 c (3 c d-a f)\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 a^3 \left (b^2-4 a c\right )}\\ &=-\frac{d}{2 a^2 x^2}-\frac{b^3 d-a b^2 e+2 a^2 c e-a b (3 c d-a f)+c \left (b^2 d-a b e-2 a (c d-a f)\right ) x^2}{2 a^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\left (2 b^4 d-12 a b^2 c d-a b^3 e+6 a^2 b c e+4 a^2 c (3 c d-a f)\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 a^3 \left (b^2-4 a c\right )^{3/2}}-\frac{(2 b d-a e) \log (x)}{a^3}+\frac{(2 b d-a e) \log \left (a+b x^2+c x^4\right )}{4 a^3}\\ \end{align*}

Mathematica [A]  time = 0.691677, size = 403, normalized size = 1.72 $\frac{\frac{\log \left (-\sqrt{b^2-4 a c}+b+2 c x^2\right ) \left (4 a^2 c \left (e \sqrt{b^2-4 a c}-a f+3 c d\right )+b^3 \left (2 d \sqrt{b^2-4 a c}-a e\right )-a b^2 \left (e \sqrt{b^2-4 a c}+12 c d\right )+2 a b c \left (3 a e-4 d \sqrt{b^2-4 a c}\right )+2 b^4 d\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac{\log \left (\sqrt{b^2-4 a c}+b+2 c x^2\right ) \left (4 a^2 c \left (e \sqrt{b^2-4 a c}+a f-3 c d\right )+b^3 \left (2 d \sqrt{b^2-4 a c}+a e\right )+a b^2 \left (12 c d-e \sqrt{b^2-4 a c}\right )-2 a b c \left (4 d \sqrt{b^2-4 a c}+3 a e\right )-2 b^4 d\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac{2 a \left (b^2 \left (c d x^2-a e\right )+a b \left (a f-c \left (3 d+e x^2\right )\right )+2 a c \left (a \left (e+f x^2\right )-c d x^2\right )+b^3 d\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+4 \log (x) (a e-2 b d)-\frac{2 a d}{x^2}}{4 a^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x^2 + f*x^4)/(x^3*(a + b*x^2 + c*x^4)^2),x]

[Out]

((-2*a*d)/x^2 - (2*a*(b^3*d + b^2*(-(a*e) + c*d*x^2) + a*b*(a*f - c*(3*d + e*x^2)) + 2*a*c*(-(c*d*x^2) + a*(e
+ f*x^2))))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + 4*(-2*b*d + a*e)*Log[x] + ((2*b^4*d + b^3*(2*Sqrt[b^2 - 4*a*
c]*d - a*e) + 2*a*b*c*(-4*Sqrt[b^2 - 4*a*c]*d + 3*a*e) - a*b^2*(12*c*d + Sqrt[b^2 - 4*a*c]*e) + 4*a^2*c*(3*c*d
+ Sqrt[b^2 - 4*a*c]*e - a*f))*Log[b - Sqrt[b^2 - 4*a*c] + 2*c*x^2])/(b^2 - 4*a*c)^(3/2) + ((-2*b^4*d + b^3*(2
*Sqrt[b^2 - 4*a*c]*d + a*e) - 2*a*b*c*(4*Sqrt[b^2 - 4*a*c]*d + 3*a*e) + a*b^2*(12*c*d - Sqrt[b^2 - 4*a*c]*e) +
4*a^2*c*(-3*c*d + Sqrt[b^2 - 4*a*c]*e + a*f))*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x^2])/(b^2 - 4*a*c)^(3/2))/(4*a
^3)

________________________________________________________________________________________

Maple [B]  time = 0.023, size = 722, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^4+e*x^2+d)/x^3/(c*x^4+b*x^2+a)^2,x)

[Out]

-1/2*d/a^2/x^2+1/a^2*ln(x)*e-2/a^3*ln(x)*b*d+1/(c*x^4+b*x^2+a)*c/(4*a*c-b^2)*x^2*f-1/2/a/(c*x^4+b*x^2+a)*c/(4*
a*c-b^2)*x^2*b*e-1/a/(c*x^4+b*x^2+a)*c^2/(4*a*c-b^2)*x^2*d+1/2/a^2/(c*x^4+b*x^2+a)*c/(4*a*c-b^2)*x^2*b^2*d+1/2
/(c*x^4+b*x^2+a)/(4*a*c-b^2)*b*f+1/(c*x^4+b*x^2+a)/(4*a*c-b^2)*c*e-1/2/a/(c*x^4+b*x^2+a)/(4*a*c-b^2)*b^2*e-3/2
/a/(c*x^4+b*x^2+a)/(4*a*c-b^2)*b*c*d+1/2/a^2/(c*x^4+b*x^2+a)/(4*a*c-b^2)*b^3*d-1/a/(4*a*c-b^2)*c*ln(c*x^4+b*x^
2+a)*e+1/4/a^2/(4*a*c-b^2)*ln(c*x^4+b*x^2+a)*b^2*e+2/a^2/(4*a*c-b^2)*c*ln(c*x^4+b*x^2+a)*b*d-1/2/a^3/(4*a*c-b^
2)*ln(c*x^4+b*x^2+a)*b^3*d+2/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*c*f-3/a/(4*a*c-b^2)^(3/2)
*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b*c*e-6/a/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*c^2*d
+1/2/a^2/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b^3*e+6/a^2/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2
+b)/(4*a*c-b^2)^(1/2))*b^2*c*d-1/a^3/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b^4*d

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^4+e*x^2+d)/x^3/(c*x^4+b*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 16.6412, size = 3637, normalized size = 15.54 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^4+e*x^2+d)/x^3/(c*x^4+b*x^2+a)^2,x, algorithm="fricas")

[Out]

[-1/4*(2*(2*(a*b^4*c - 7*a^2*b^2*c^2 + 12*a^3*c^3)*d - (a^2*b^3*c - 4*a^3*b*c^2)*e + 2*(a^3*b^2*c - 4*a^4*c^2)
*f)*x^4 + 2*((2*a*b^5 - 15*a^2*b^3*c + 28*a^3*b*c^2)*d - (a^2*b^4 - 6*a^3*b^2*c + 8*a^4*c^2)*e + (a^3*b^3 - 4*
a^4*b*c)*f)*x^2 + ((4*a^3*c^2*f - 2*(b^4*c - 6*a*b^2*c^2 + 6*a^2*c^3)*d + (a*b^3*c - 6*a^2*b*c^2)*e)*x^6 + (4*
a^3*b*c*f - 2*(b^5 - 6*a*b^3*c + 6*a^2*b*c^2)*d + (a*b^4 - 6*a^2*b^2*c)*e)*x^4 + (4*a^4*c*f - 2*(a*b^4 - 6*a^2
*b^2*c + 6*a^3*c^2)*d + (a^2*b^3 - 6*a^3*b*c)*e)*x^2)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^4 + 2*b*c*x^2 + b^2 - 2*a
*c - (2*c*x^2 + b)*sqrt(b^2 - 4*a*c))/(c*x^4 + b*x^2 + a)) + 2*(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)*d - ((2*(b
^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d - (a*b^4*c - 8*a^2*b^2*c^2 + 16*a^3*c^3)*e)*x^6 + (2*(b^6 - 8*a*b^4*c + 1
6*a^2*b^2*c^2)*d - (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*e)*x^4 + (2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d - (
a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)*e)*x^2)*log(c*x^4 + b*x^2 + a) + 4*((2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3
)*d - (a*b^4*c - 8*a^2*b^2*c^2 + 16*a^3*c^3)*e)*x^6 + (2*(b^6 - 8*a*b^4*c + 16*a^2*b^2*c^2)*d - (a*b^5 - 8*a^2
*b^3*c + 16*a^3*b*c^2)*e)*x^4 + (2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d - (a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^
2)*e)*x^2)*log(x))/((a^3*b^4*c - 8*a^4*b^2*c^2 + 16*a^5*c^3)*x^6 + (a^3*b^5 - 8*a^4*b^3*c + 16*a^5*b*c^2)*x^4
+ (a^4*b^4 - 8*a^5*b^2*c + 16*a^6*c^2)*x^2), -1/4*(2*(2*(a*b^4*c - 7*a^2*b^2*c^2 + 12*a^3*c^3)*d - (a^2*b^3*c
- 4*a^3*b*c^2)*e + 2*(a^3*b^2*c - 4*a^4*c^2)*f)*x^4 + 2*((2*a*b^5 - 15*a^2*b^3*c + 28*a^3*b*c^2)*d - (a^2*b^4
- 6*a^3*b^2*c + 8*a^4*c^2)*e + (a^3*b^3 - 4*a^4*b*c)*f)*x^2 - 2*((4*a^3*c^2*f - 2*(b^4*c - 6*a*b^2*c^2 + 6*a^2
*c^3)*d + (a*b^3*c - 6*a^2*b*c^2)*e)*x^6 + (4*a^3*b*c*f - 2*(b^5 - 6*a*b^3*c + 6*a^2*b*c^2)*d + (a*b^4 - 6*a^2
*b^2*c)*e)*x^4 + (4*a^4*c*f - 2*(a*b^4 - 6*a^2*b^2*c + 6*a^3*c^2)*d + (a^2*b^3 - 6*a^3*b*c)*e)*x^2)*sqrt(-b^2
+ 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) + 2*(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)*d -
((2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d - (a*b^4*c - 8*a^2*b^2*c^2 + 16*a^3*c^3)*e)*x^6 + (2*(b^6 - 8*a*b^4
*c + 16*a^2*b^2*c^2)*d - (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*e)*x^4 + (2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)
*d - (a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)*e)*x^2)*log(c*x^4 + b*x^2 + a) + 4*((2*(b^5*c - 8*a*b^3*c^2 + 16*a^2
*b*c^3)*d - (a*b^4*c - 8*a^2*b^2*c^2 + 16*a^3*c^3)*e)*x^6 + (2*(b^6 - 8*a*b^4*c + 16*a^2*b^2*c^2)*d - (a*b^5 -
8*a^2*b^3*c + 16*a^3*b*c^2)*e)*x^4 + (2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d - (a^2*b^4 - 8*a^3*b^2*c + 16*
a^4*c^2)*e)*x^2)*log(x))/((a^3*b^4*c - 8*a^4*b^2*c^2 + 16*a^5*c^3)*x^6 + (a^3*b^5 - 8*a^4*b^3*c + 16*a^5*b*c^2
)*x^4 + (a^4*b^4 - 8*a^5*b^2*c + 16*a^6*c^2)*x^2)]

________________________________________________________________________________________

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((f*x**4+e*x**2+d)/x**3/(c*x**4+b*x**2+a)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 21.9743, size = 387, normalized size = 1.65 \begin{align*} \frac{{\left (2 \, b^{4} d - 12 \, a b^{2} c d + 12 \, a^{2} c^{2} d - 4 \, a^{3} c f - a b^{3} e + 6 \, a^{2} b c e\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \,{\left (a^{3} b^{2} - 4 \, a^{4} c\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{2 \, b^{2} c d x^{4} - 6 \, a c^{2} d x^{4} + 2 \, a^{2} c f x^{4} - a b c x^{4} e + 2 \, b^{3} d x^{2} - 7 \, a b c d x^{2} + a^{2} b f x^{2} - a b^{2} x^{2} e + 2 \, a^{2} c x^{2} e + a b^{2} d - 4 \, a^{2} c d}{2 \,{\left (c x^{6} + b x^{4} + a x^{2}\right )}{\left (a^{2} b^{2} - 4 \, a^{3} c\right )}} + \frac{{\left (2 \, b d - a e\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, a^{3}} - \frac{{\left (2 \, b d - a e\right )} \log \left (x^{2}\right )}{2 \, a^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^4+e*x^2+d)/x^3/(c*x^4+b*x^2+a)^2,x, algorithm="giac")

[Out]

1/2*(2*b^4*d - 12*a*b^2*c*d + 12*a^2*c^2*d - 4*a^3*c*f - a*b^3*e + 6*a^2*b*c*e)*arctan((2*c*x^2 + b)/sqrt(-b^2
+ 4*a*c))/((a^3*b^2 - 4*a^4*c)*sqrt(-b^2 + 4*a*c)) - 1/2*(2*b^2*c*d*x^4 - 6*a*c^2*d*x^4 + 2*a^2*c*f*x^4 - a*b
*c*x^4*e + 2*b^3*d*x^2 - 7*a*b*c*d*x^2 + a^2*b*f*x^2 - a*b^2*x^2*e + 2*a^2*c*x^2*e + a*b^2*d - 4*a^2*c*d)/((c*
x^6 + b*x^4 + a*x^2)*(a^2*b^2 - 4*a^3*c)) + 1/4*(2*b*d - a*e)*log(c*x^4 + b*x^2 + a)/a^3 - 1/2*(2*b*d - a*e)*l
og(x^2)/a^3