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

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

[Out]

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

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Rubi [A]  time = 0.198374, 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} \[ -\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{3/2} (b c-a d)^2}+\frac{d^{3/2} (5 b c-3 a d) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 c^{5/2} (b c-a d)^2}-\frac{2 b c-3 a d}{2 a c^2 x (b c-a d)}-\frac{d}{2 c x \left (c+d x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.249047, size = 123, normalized size = 0.85 \[ -\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{3/2} (a d-b c)^2}+\frac{d^2 x}{2 c^2 \left (c+d x^2\right ) (b c-a d)}+\frac{d^{3/2} (5 b c-3 a d) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 c^{5/2} (b c-a d)^2}-\frac{1}{a c^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 169, normalized size = 1.2 \begin{align*} -{\frac{{d}^{3}xa}{2\,{c}^{2} \left ( ad-bc \right ) ^{2} \left ( d{x}^{2}+c \right ) }}+{\frac{b{d}^{2}x}{2\,c \left ( ad-bc \right ) ^{2} \left ( d{x}^{2}+c \right ) }}-{\frac{3\,a{d}^{3}}{2\,{c}^{2} \left ( ad-bc \right ) ^{2}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}+{\frac{5\,b{d}^{2}}{2\,c \left ( ad-bc \right ) ^{2}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{1}{a{c}^{2}x}}-{\frac{{b}^{3}}{a \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)/(d*x^2+c)^2,x)

[Out]

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

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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)/(d*x^2+c)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.6902, 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)/(d*x^2+c)^2,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 71.0031, 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)/(d*x**2+c)**2,x)

[Out]

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

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Giac [A]  time = 1.13505, size = 221, normalized size = 1.53 \begin{align*} -\frac{b^{3} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt{a b}} + \frac{{\left (5 \, b c d^{2} - 3 \, a d^{3}\right )} \arctan \left (\frac{d x}{\sqrt{c d}}\right )}{2 \,{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )} \sqrt{c d}} - \frac{2 \, b c d x^{2} - 3 \, a d^{2} x^{2} + 2 \, b c^{2} - 2 \, a c d}{2 \,{\left (a b c^{3} - a^{2} c^{2} d\right )}{\left (d x^{3} + c x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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