3.295 \(\int \frac{1}{x^2 (a+b x^2)^2 (c+d x^2)} \, dx\)

Optimal. Leaf size=144 \[ -\frac{b^{3/2} (3 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2} (b c-a d)^2}-\frac{3 b c-2 a d}{2 a^2 c x (b c-a d)}-\frac{d^{5/2} \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{c^{3/2} (b c-a d)^2}+\frac{b}{2 a x \left (a+b x^2\right ) (b c-a d)} \]

[Out]

-(3*b*c - 2*a*d)/(2*a^2*c*(b*c - a*d)*x) + b/(2*a*(b*c - a*d)*x*(a + b*x^2)) - (b^(3/2)*(3*b*c - 5*a*d)*ArcTan
[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(5/2)*(b*c - a*d)^2) - (d^(5/2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(c^(3/2)*(b*c - a*d)^
2)

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Rubi [A]  time = 0.206279, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {472, 583, 522, 205} \[ -\frac{b^{3/2} (3 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2} (b c-a d)^2}-\frac{3 b c-2 a d}{2 a^2 c x (b c-a d)}-\frac{d^{5/2} \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{c^{3/2} (b c-a d)^2}+\frac{b}{2 a x \left (a+b x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^2)^2*(c + d*x^2)),x]

[Out]

-(3*b*c - 2*a*d)/(2*a^2*c*(b*c - a*d)*x) + b/(2*a*(b*c - a*d)*x*(a + b*x^2)) - (b^(3/2)*(3*b*c - 5*a*d)*ArcTan
[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(5/2)*(b*c - a*d)^2) - (d^(5/2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(c^(3/2)*(b*c - a*d)^
2)

Rule 472

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p
, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{x^2 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx &=\frac{b}{2 a (b c-a d) x \left (a+b x^2\right )}-\frac{\int \frac{-3 b c+2 a d-3 b d x^2}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{2 a (b c-a d)}\\ &=-\frac{3 b c-2 a d}{2 a^2 c (b c-a d) x}+\frac{b}{2 a (b c-a d) x \left (a+b x^2\right )}+\frac{\int \frac{-3 b^2 c^2+2 a b c d+2 a^2 d^2-b d (3 b c-2 a d) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{2 a^2 c (b c-a d)}\\ &=-\frac{3 b c-2 a d}{2 a^2 c (b c-a d) x}+\frac{b}{2 a (b c-a d) x \left (a+b x^2\right )}-\frac{d^3 \int \frac{1}{c+d x^2} \, dx}{c (b c-a d)^2}-\frac{\left (b^2 (3 b c-5 a d)\right ) \int \frac{1}{a+b x^2} \, dx}{2 a^2 (b c-a d)^2}\\ &=-\frac{3 b c-2 a d}{2 a^2 c (b c-a d) x}+\frac{b}{2 a (b c-a d) x \left (a+b x^2\right )}-\frac{b^{3/2} (3 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2} (b c-a d)^2}-\frac{d^{5/2} \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{c^{3/2} (b c-a d)^2}\\ \end{align*}

Mathematica [A]  time = 0.188553, size = 123, normalized size = 0.85 \[ \frac{b^2 x}{2 a^2 \left (a+b x^2\right ) (a d-b c)}+\frac{b^{3/2} (5 a d-3 b c) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2} (a d-b c)^2}-\frac{1}{a^2 c x}-\frac{d^{5/2} \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{c^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x^2)^2*(c + d*x^2)),x]

[Out]

-(1/(a^2*c*x)) + (b^2*x)/(2*a^2*(-(b*c) + a*d)*(a + b*x^2)) + (b^(3/2)*(-3*b*c + 5*a*d)*ArcTan[(Sqrt[b]*x)/Sqr
t[a]])/(2*a^(5/2)*(-(b*c) + a*d)^2) - (d^(5/2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(c^(3/2)*(b*c - a*d)^2)

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Maple [A]  time = 0.015, size = 169, normalized size = 1.2 \begin{align*} -{\frac{{d}^{3}}{c \left ( ad-bc \right ) ^{2}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{1}{{a}^{2}cx}}+{\frac{{b}^{2}xd}{2\,a \left ( ad-bc \right ) ^{2} \left ( b{x}^{2}+a \right ) }}-{\frac{{b}^{3}xc}{2\,{a}^{2} \left ( ad-bc \right ) ^{2} \left ( b{x}^{2}+a \right ) }}+{\frac{5\,{b}^{2}d}{2\,a \left ( ad-bc \right ) ^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{3\,{b}^{3}c}{2\,{a}^{2} \left ( ad-bc \right ) ^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b*x^2+a)^2/(d*x^2+c),x)

[Out]

-1/c*d^3/(a*d-b*c)^2/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))-1/a^2/c/x+1/2*b^2/a/(a*d-b*c)^2*x/(b*x^2+a)*d-1/2*b^3
/a^2/(a*d-b*c)^2*x/(b*x^2+a)*c+5/2*b^2/a/(a*d-b*c)^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*d-3/2*b^3/a^2/(a*d-b*
c)^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*c

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.67996, size = 2049, normalized size = 14.23 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x^2+a)^2/(d*x^2+c),x, algorithm="fricas")

[Out]

[-1/4*(4*a*b^2*c^2 - 8*a^2*b*c*d + 4*a^3*d^2 + 2*(3*b^3*c^2 - 5*a*b^2*c*d + 2*a^2*b*d^2)*x^2 + ((3*b^3*c^2 - 5
*a*b^2*c*d)*x^3 + (3*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) -
2*(a^2*b*d^2*x^3 + a^3*d^2*x)*sqrt(-d/c)*log((d*x^2 - 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)))/((a^2*b^3*c^3 - 2*a^
3*b^2*c^2*d + a^4*b*c*d^2)*x^3 + (a^3*b^2*c^3 - 2*a^4*b*c^2*d + a^5*c*d^2)*x), -1/4*(4*a*b^2*c^2 - 8*a^2*b*c*d
+ 4*a^3*d^2 + 2*(3*b^3*c^2 - 5*a*b^2*c*d + 2*a^2*b*d^2)*x^2 + 4*(a^2*b*d^2*x^3 + a^3*d^2*x)*sqrt(d/c)*arctan(
x*sqrt(d/c)) + ((3*b^3*c^2 - 5*a*b^2*c*d)*x^3 + (3*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(-b/a)*log((b*x^2 + 2*a*x*s
qrt(-b/a) - a)/(b*x^2 + a)))/((a^2*b^3*c^3 - 2*a^3*b^2*c^2*d + a^4*b*c*d^2)*x^3 + (a^3*b^2*c^3 - 2*a^4*b*c^2*d
+ a^5*c*d^2)*x), -1/2*(2*a*b^2*c^2 - 4*a^2*b*c*d + 2*a^3*d^2 + (3*b^3*c^2 - 5*a*b^2*c*d + 2*a^2*b*d^2)*x^2 +
((3*b^3*c^2 - 5*a*b^2*c*d)*x^3 + (3*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(b/a)*arctan(x*sqrt(b/a)) - (a^2*b*d^2*x^3
+ a^3*d^2*x)*sqrt(-d/c)*log((d*x^2 - 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)))/((a^2*b^3*c^3 - 2*a^3*b^2*c^2*d + a^
4*b*c*d^2)*x^3 + (a^3*b^2*c^3 - 2*a^4*b*c^2*d + a^5*c*d^2)*x), -1/2*(2*a*b^2*c^2 - 4*a^2*b*c*d + 2*a^3*d^2 + (
3*b^3*c^2 - 5*a*b^2*c*d + 2*a^2*b*d^2)*x^2 + ((3*b^3*c^2 - 5*a*b^2*c*d)*x^3 + (3*a*b^2*c^2 - 5*a^2*b*c*d)*x)*s
qrt(b/a)*arctan(x*sqrt(b/a)) + 2*(a^2*b*d^2*x^3 + a^3*d^2*x)*sqrt(d/c)*arctan(x*sqrt(d/c)))/((a^2*b^3*c^3 - 2*
a^3*b^2*c^2*d + a^4*b*c*d^2)*x^3 + (a^3*b^2*c^3 - 2*a^4*b*c^2*d + a^5*c*d^2)*x)]

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Sympy [B]  time = 127.24, size = 2526, normalized size = 17.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(b*x**2+a)**2/(d*x**2+c),x)

[Out]

-sqrt(-b**3/a**5)*(5*a*d - 3*b*c)*log(x + (-a**12*c**3*d**7*(-b**3/a**5)**(3/2)*(5*a*d - 3*b*c)**3/(a*d - b*c)
**6 + 4*a**11*b*c**4*d**6*(-b**3/a**5)**(3/2)*(5*a*d - 3*b*c)**3/(a*d - b*c)**6 - 17*a**10*b**2*c**5*d**5*(-b*
*3/a**5)**(3/2)*(5*a*d - 3*b*c)**3/(2*(a*d - b*c)**6) + 31*a**9*b**3*c**6*d**4*(-b**3/a**5)**(3/2)*(5*a*d - 3*
b*c)**3/(2*(a*d - b*c)**6) - 22*a**8*b**4*c**7*d**3*(-b**3/a**5)**(3/2)*(5*a*d - 3*b*c)**3/(a*d - b*c)**6 - 4*
a**8*d**8*sqrt(-b**3/a**5)*(5*a*d - 3*b*c)/(a*d - b*c)**2 + 19*a**7*b**5*c**8*d**2*(-b**3/a**5)**(3/2)*(5*a*d
- 3*b*c)**3/(a*d - b*c)**6 - 17*a**6*b**6*c**9*d*(-b**3/a**5)**(3/2)*(5*a*d - 3*b*c)**3/(2*(a*d - b*c)**6) + 3
*a**5*b**7*c**10*(-b**3/a**5)**(3/2)*(5*a*d - 3*b*c)**3/(2*(a*d - b*c)**6) - 125*a**3*b**5*c**5*d**3*sqrt(-b**
3/a**5)*(5*a*d - 3*b*c)/(2*(a*d - b*c)**2) + 225*a**2*b**6*c**6*d**2*sqrt(-b**3/a**5)*(5*a*d - 3*b*c)/(2*(a*d
- b*c)**2) - 135*a*b**7*c**7*d*sqrt(-b**3/a**5)*(5*a*d - 3*b*c)/(2*(a*d - b*c)**2) + 27*b**8*c**8*sqrt(-b**3/a
**5)*(5*a*d - 3*b*c)/(2*(a*d - b*c)**2))/(20*a**4*b**2*d**7 + 28*a**3*b**3*c*d**6 + 36*a**2*b**4*c**2*d**5 - 8
1*a*b**5*c**3*d**4 + 27*b**6*c**4*d**3))/(4*(a*d - b*c)**2) + sqrt(-b**3/a**5)*(5*a*d - 3*b*c)*log(x + (a**12*
c**3*d**7*(-b**3/a**5)**(3/2)*(5*a*d - 3*b*c)**3/(a*d - b*c)**6 - 4*a**11*b*c**4*d**6*(-b**3/a**5)**(3/2)*(5*a
*d - 3*b*c)**3/(a*d - b*c)**6 + 17*a**10*b**2*c**5*d**5*(-b**3/a**5)**(3/2)*(5*a*d - 3*b*c)**3/(2*(a*d - b*c)*
*6) - 31*a**9*b**3*c**6*d**4*(-b**3/a**5)**(3/2)*(5*a*d - 3*b*c)**3/(2*(a*d - b*c)**6) + 22*a**8*b**4*c**7*d**
3*(-b**3/a**5)**(3/2)*(5*a*d - 3*b*c)**3/(a*d - b*c)**6 + 4*a**8*d**8*sqrt(-b**3/a**5)*(5*a*d - 3*b*c)/(a*d -
b*c)**2 - 19*a**7*b**5*c**8*d**2*(-b**3/a**5)**(3/2)*(5*a*d - 3*b*c)**3/(a*d - b*c)**6 + 17*a**6*b**6*c**9*d*(
-b**3/a**5)**(3/2)*(5*a*d - 3*b*c)**3/(2*(a*d - b*c)**6) - 3*a**5*b**7*c**10*(-b**3/a**5)**(3/2)*(5*a*d - 3*b*
c)**3/(2*(a*d - b*c)**6) + 125*a**3*b**5*c**5*d**3*sqrt(-b**3/a**5)*(5*a*d - 3*b*c)/(2*(a*d - b*c)**2) - 225*a
**2*b**6*c**6*d**2*sqrt(-b**3/a**5)*(5*a*d - 3*b*c)/(2*(a*d - b*c)**2) + 135*a*b**7*c**7*d*sqrt(-b**3/a**5)*(5
*a*d - 3*b*c)/(2*(a*d - b*c)**2) - 27*b**8*c**8*sqrt(-b**3/a**5)*(5*a*d - 3*b*c)/(2*(a*d - b*c)**2))/(20*a**4*
b**2*d**7 + 28*a**3*b**3*c*d**6 + 36*a**2*b**4*c**2*d**5 - 81*a*b**5*c**3*d**4 + 27*b**6*c**4*d**3))/(4*(a*d -
b*c)**2) - sqrt(-d**5/c**3)*log(x + (-8*a**12*c**3*d**7*(-d**5/c**3)**(3/2)/(a*d - b*c)**6 + 32*a**11*b*c**4*
d**6*(-d**5/c**3)**(3/2)/(a*d - b*c)**6 - 68*a**10*b**2*c**5*d**5*(-d**5/c**3)**(3/2)/(a*d - b*c)**6 + 124*a**
9*b**3*c**6*d**4*(-d**5/c**3)**(3/2)/(a*d - b*c)**6 - 176*a**8*b**4*c**7*d**3*(-d**5/c**3)**(3/2)/(a*d - b*c)*
*6 - 8*a**8*d**8*sqrt(-d**5/c**3)/(a*d - b*c)**2 + 152*a**7*b**5*c**8*d**2*(-d**5/c**3)**(3/2)/(a*d - b*c)**6
- 68*a**6*b**6*c**9*d*(-d**5/c**3)**(3/2)/(a*d - b*c)**6 + 12*a**5*b**7*c**10*(-d**5/c**3)**(3/2)/(a*d - b*c)*
*6 - 125*a**3*b**5*c**5*d**3*sqrt(-d**5/c**3)/(a*d - b*c)**2 + 225*a**2*b**6*c**6*d**2*sqrt(-d**5/c**3)/(a*d -
b*c)**2 - 135*a*b**7*c**7*d*sqrt(-d**5/c**3)/(a*d - b*c)**2 + 27*b**8*c**8*sqrt(-d**5/c**3)/(a*d - b*c)**2)/(
20*a**4*b**2*d**7 + 28*a**3*b**3*c*d**6 + 36*a**2*b**4*c**2*d**5 - 81*a*b**5*c**3*d**4 + 27*b**6*c**4*d**3))/(
2*(a*d - b*c)**2) + sqrt(-d**5/c**3)*log(x + (8*a**12*c**3*d**7*(-d**5/c**3)**(3/2)/(a*d - b*c)**6 - 32*a**11*
b*c**4*d**6*(-d**5/c**3)**(3/2)/(a*d - b*c)**6 + 68*a**10*b**2*c**5*d**5*(-d**5/c**3)**(3/2)/(a*d - b*c)**6 -
124*a**9*b**3*c**6*d**4*(-d**5/c**3)**(3/2)/(a*d - b*c)**6 + 176*a**8*b**4*c**7*d**3*(-d**5/c**3)**(3/2)/(a*d
- b*c)**6 + 8*a**8*d**8*sqrt(-d**5/c**3)/(a*d - b*c)**2 - 152*a**7*b**5*c**8*d**2*(-d**5/c**3)**(3/2)/(a*d - b
*c)**6 + 68*a**6*b**6*c**9*d*(-d**5/c**3)**(3/2)/(a*d - b*c)**6 - 12*a**5*b**7*c**10*(-d**5/c**3)**(3/2)/(a*d
- b*c)**6 + 125*a**3*b**5*c**5*d**3*sqrt(-d**5/c**3)/(a*d - b*c)**2 - 225*a**2*b**6*c**6*d**2*sqrt(-d**5/c**3)
/(a*d - b*c)**2 + 135*a*b**7*c**7*d*sqrt(-d**5/c**3)/(a*d - b*c)**2 - 27*b**8*c**8*sqrt(-d**5/c**3)/(a*d - b*c
)**2)/(20*a**4*b**2*d**7 + 28*a**3*b**3*c*d**6 + 36*a**2*b**4*c**2*d**5 - 81*a*b**5*c**3*d**4 + 27*b**6*c**4*d
**3))/(2*(a*d - b*c)**2) - (2*a**2*d - 2*a*b*c + x**2*(2*a*b*d - 3*b**2*c))/(x**3*(2*a**3*b*c*d - 2*a**2*b**2*
c**2) + x*(2*a**4*c*d - 2*a**3*b*c**2))

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Giac [A]  time = 1.13578, size = 221, normalized size = 1.53 \begin{align*} -\frac{d^{3} \arctan \left (\frac{d x}{\sqrt{c d}}\right )}{{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt{c d}} - \frac{{\left (3 \, b^{3} c - 5 \, a b^{2} d\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \,{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \sqrt{a b}} - \frac{3 \, b^{2} c x^{2} - 2 \, a b d x^{2} + 2 \, a b c - 2 \, a^{2} d}{2 \,{\left (a^{2} b c^{2} - a^{3} c d\right )}{\left (b x^{3} + a x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x^2+a)^2/(d*x^2+c),x, algorithm="giac")

[Out]

-d^3*arctan(d*x/sqrt(c*d))/((b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(c*d)) - 1/2*(3*b^3*c - 5*a*b^2*d)*arctan(
b*x/sqrt(a*b))/((a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)*sqrt(a*b)) - 1/2*(3*b^2*c*x^2 - 2*a*b*d*x^2 + 2*a*b*c -
2*a^2*d)/((a^2*b*c^2 - a^3*c*d)*(b*x^3 + a*x))