### 3.65 $$\int \frac{d+e x^2+f x^4}{x (a+b x^2+c x^4)^2} \, dx$$

Optimal. Leaf size=166 $\frac{\tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right ) \left (4 a^2 c e-2 a b (a f+3 c d)+b^3 d\right )}{2 a^2 \left (b^2-4 a c\right )^{3/2}}-\frac{d \log \left (a+b x^2+c x^4\right )}{4 a^2}+\frac{d \log (x)}{a^2}+\frac{x^2 (a b f-2 a c e+b c d)-a b e-2 a (c d-a f)+b^2 d}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}$

[Out]

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

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Rubi [A]  time = 0.393699, antiderivative size = 166, 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, 800, 634, 618, 206, 628} $\frac{\tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right ) \left (4 a^2 c e-2 a b (a f+3 c d)+b^3 d\right )}{2 a^2 \left (b^2-4 a c\right )^{3/2}}-\frac{d \log \left (a+b x^2+c x^4\right )}{4 a^2}+\frac{d \log (x)}{a^2}+\frac{x^2 (a b f-2 a c e+b c d)-a b e-2 a (c d-a f)+b^2 d}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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

Mathematica [A]  time = 0.469313, size = 268, normalized size = 1.61 $-\frac{-\frac{2 a \left (b \left (-a e+a f x^2+c d x^2\right )+2 a \left (a f-c \left (d+e x^2\right )\right )+b^2 d\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{\log \left (-\sqrt{b^2-4 a c}+b+2 c x^2\right ) \left (4 a c \left (a e-d \sqrt{b^2-4 a c}\right )+b^2 d \sqrt{b^2-4 a c}-2 a b (a f+3 c d)+b^3 d\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac{\log \left (\sqrt{b^2-4 a c}+b+2 c x^2\right ) \left (-4 a c \left (d \sqrt{b^2-4 a c}+a e\right )+b^2 d \sqrt{b^2-4 a c}+2 a b (a f+3 c d)+b^3 (-d)\right )}{\left (b^2-4 a c\right )^{3/2}}-4 d \log (x)}{4 a^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.018, size = 462, normalized size = 2.8 \begin{align*}{\frac{d\ln \left ( x \right ) }{{a}^{2}}}-{\frac{b{x}^{2}f}{ \left ( 2\,c{x}^{4}+2\,b{x}^{2}+2\,a \right ) \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{c{x}^{2}e}{ \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{b{x}^{2}cd}{2\,a \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{af}{ \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{be}{ \left ( 2\,c{x}^{4}+2\,b{x}^{2}+2\,a \right ) \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{cd}{ \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{{b}^{2}d}{2\,a \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{c\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) d}{a \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ){b}^{2}d}{4\,{a}^{2} \left ( 4\,ac-{b}^{2} \right ) }}-{bf\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ) \left ( 4\,ac-{b}^{2} \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{ce}{ \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-3\,{\frac{bcd}{a \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{{b}^{3}d}{2\,{a}^{2}}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ) \left ( 4\,ac-{b}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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/(c*x^4+b*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 7.60329, size = 2333, normalized size = 14.05 \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/(c*x^4+b*x^2+a)^2,x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 19.5116, size = 306, normalized size = 1.84 \begin{align*} -\frac{{\left (b^{3} d - 6 \, a b c d - 2 \, a^{2} b f + 4 \, a^{2} c e\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \,{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{d \log \left (c x^{4} + b x^{2} + a\right )}{4 \, a^{2}} + \frac{d \log \left (x^{2}\right )}{2 \, a^{2}} + \frac{b^{2} c d x^{4} - 4 \, a c^{2} d x^{4} + b^{3} d x^{2} - 2 \, a b c d x^{2} + 2 \, a^{2} b f x^{2} - 4 \, a^{2} c x^{2} e + 3 \, a b^{2} d - 8 \, a^{2} c d + 4 \, a^{3} f - 2 \, a^{2} b e}{4 \,{\left (c x^{4} + b x^{2} + a\right )}{\left (a^{2} b^{2} - 4 \, a^{3} c\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^3*d - 6*a*b*c*d - 2*a^2*b*f + 4*a^2*c*e)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/((a^2*b^2 - 4*a^3*c)
*sqrt(-b^2 + 4*a*c)) - 1/4*d*log(c*x^4 + b*x^2 + a)/a^2 + 1/2*d*log(x^2)/a^2 + 1/4*(b^2*c*d*x^4 - 4*a*c^2*d*x^
4 + b^3*d*x^2 - 2*a*b*c*d*x^2 + 2*a^2*b*f*x^2 - 4*a^2*c*x^2*e + 3*a*b^2*d - 8*a^2*c*d + 4*a^3*f - 2*a^2*b*e)/(
(c*x^4 + b*x^2 + a)*(a^2*b^2 - 4*a^3*c))