3.766 \(\int \frac{1}{x^4 (a+b x^2)^2 \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=206 \[ \frac{\sqrt{c+d x^2} \left (-4 a^2 d^2-8 a b c d+15 b^2 c^2\right )}{6 a^3 c^2 x (b c-a d)}+\frac{b^2 (5 b c-6 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{7/2} (b c-a d)^{3/2}}-\frac{\sqrt{c+d x^2} (5 b c-2 a d)}{6 a^2 c x^3 (b c-a d)}+\frac{b \sqrt{c+d x^2}}{2 a x^3 \left (a+b x^2\right ) (b c-a d)} \]

[Out]

-((5*b*c - 2*a*d)*Sqrt[c + d*x^2])/(6*a^2*c*(b*c - a*d)*x^3) + ((15*b^2*c^2 - 8*a*b*c*d - 4*a^2*d^2)*Sqrt[c +
d*x^2])/(6*a^3*c^2*(b*c - a*d)*x) + (b*Sqrt[c + d*x^2])/(2*a*(b*c - a*d)*x^3*(a + b*x^2)) + (b^2*(5*b*c - 6*a*
d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(7/2)*(b*c - a*d)^(3/2))

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Rubi [A]  time = 0.252386, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {472, 583, 12, 377, 205} \[ \frac{\sqrt{c+d x^2} \left (-4 a^2 d^2-8 a b c d+15 b^2 c^2\right )}{6 a^3 c^2 x (b c-a d)}+\frac{b^2 (5 b c-6 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{7/2} (b c-a d)^{3/2}}-\frac{\sqrt{c+d x^2} (5 b c-2 a d)}{6 a^2 c x^3 (b c-a d)}+\frac{b \sqrt{c+d x^2}}{2 a x^3 \left (a+b x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^2)^2*Sqrt[c + d*x^2]),x]

[Out]

-((5*b*c - 2*a*d)*Sqrt[c + d*x^2])/(6*a^2*c*(b*c - a*d)*x^3) + ((15*b^2*c^2 - 8*a*b*c*d - 4*a^2*d^2)*Sqrt[c +
d*x^2])/(6*a^3*c^2*(b*c - a*d)*x) + (b*Sqrt[c + d*x^2])/(2*a*(b*c - a*d)*x^3*(a + b*x^2)) + (b^2*(5*b*c - 6*a*
d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(7/2)*(b*c - a*d)^(3/2))

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 12

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

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

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^4 \left (a+b x^2\right )^2 \sqrt{c+d x^2}} \, dx &=\frac{b \sqrt{c+d x^2}}{2 a (b c-a d) x^3 \left (a+b x^2\right )}-\frac{\int \frac{-5 b c+2 a d-4 b d x^2}{x^4 \left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{2 a (b c-a d)}\\ &=-\frac{(5 b c-2 a d) \sqrt{c+d x^2}}{6 a^2 c (b c-a d) x^3}+\frac{b \sqrt{c+d x^2}}{2 a (b c-a d) x^3 \left (a+b x^2\right )}+\frac{\int \frac{-15 b^2 c^2+8 a b c d+4 a^2 d^2-2 b d (5 b c-2 a d) x^2}{x^2 \left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{6 a^2 c (b c-a d)}\\ &=-\frac{(5 b c-2 a d) \sqrt{c+d x^2}}{6 a^2 c (b c-a d) x^3}+\frac{\left (15 b^2 c^2-8 a b c d-4 a^2 d^2\right ) \sqrt{c+d x^2}}{6 a^3 c^2 (b c-a d) x}+\frac{b \sqrt{c+d x^2}}{2 a (b c-a d) x^3 \left (a+b x^2\right )}-\frac{\int -\frac{3 b^2 c^2 (5 b c-6 a d)}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{6 a^3 c^2 (b c-a d)}\\ &=-\frac{(5 b c-2 a d) \sqrt{c+d x^2}}{6 a^2 c (b c-a d) x^3}+\frac{\left (15 b^2 c^2-8 a b c d-4 a^2 d^2\right ) \sqrt{c+d x^2}}{6 a^3 c^2 (b c-a d) x}+\frac{b \sqrt{c+d x^2}}{2 a (b c-a d) x^3 \left (a+b x^2\right )}+\frac{\left (b^2 (5 b c-6 a d)\right ) \int \frac{1}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{2 a^3 (b c-a d)}\\ &=-\frac{(5 b c-2 a d) \sqrt{c+d x^2}}{6 a^2 c (b c-a d) x^3}+\frac{\left (15 b^2 c^2-8 a b c d-4 a^2 d^2\right ) \sqrt{c+d x^2}}{6 a^3 c^2 (b c-a d) x}+\frac{b \sqrt{c+d x^2}}{2 a (b c-a d) x^3 \left (a+b x^2\right )}+\frac{\left (b^2 (5 b c-6 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{a-(-b c+a d) x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{2 a^3 (b c-a d)}\\ &=-\frac{(5 b c-2 a d) \sqrt{c+d x^2}}{6 a^2 c (b c-a d) x^3}+\frac{\left (15 b^2 c^2-8 a b c d-4 a^2 d^2\right ) \sqrt{c+d x^2}}{6 a^3 c^2 (b c-a d) x}+\frac{b \sqrt{c+d x^2}}{2 a (b c-a d) x^3 \left (a+b x^2\right )}+\frac{b^2 (5 b c-6 a d) \tan ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{7/2} (b c-a d)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 5.23383, size = 136, normalized size = 0.66 \[ \frac{\sqrt{c+d x^2} \left (\frac{3 b^3 x^4}{\left (a+b x^2\right ) (b c-a d)}+\frac{4 x^2 (a d+3 b c)}{c^2}-\frac{2 a}{c}\right )}{6 a^3 x^3}+\frac{b^2 (5 b c-6 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{7/2} (b c-a d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^2)^2*Sqrt[c + d*x^2]),x]

[Out]

(Sqrt[c + d*x^2]*((-2*a)/c + (4*(3*b*c + a*d)*x^2)/c^2 + (3*b^3*x^4)/((b*c - a*d)*(a + b*x^2))))/(6*a^3*x^3) +
(b^2*(5*b*c - 6*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(7/2)*(b*c - a*d)^(3/2))

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Maple [B]  time = 0.015, size = 893, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(b*x^2+a)^2/(d*x^2+c)^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{2} \sqrt{d x^{2} + c} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^2*sqrt(d*x^2 + c)*x^4), x)

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Fricas [A]  time = 5.08281, size = 1536, normalized size = 7.46 \begin{align*} \left [-\frac{3 \,{\left ({\left (5 \, b^{4} c^{3} - 6 \, a b^{3} c^{2} d\right )} x^{5} +{\left (5 \, a b^{3} c^{3} - 6 \, a^{2} b^{2} c^{2} d\right )} x^{3}\right )} \sqrt{-a b c + a^{2} d} \log \left (\frac{{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \,{\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \,{\left ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt{-a b c + a^{2} d} \sqrt{d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \,{\left (2 \, a^{3} b^{2} c^{3} - 4 \, a^{4} b c^{2} d + 2 \, a^{5} c d^{2} -{\left (15 \, a b^{4} c^{3} - 23 \, a^{2} b^{3} c^{2} d + 4 \, a^{3} b^{2} c d^{2} + 4 \, a^{4} b d^{3}\right )} x^{4} - 2 \,{\left (5 \, a^{2} b^{3} c^{3} - 8 \, a^{3} b^{2} c^{2} d + a^{4} b c d^{2} + 2 \, a^{5} d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{24 \,{\left ({\left (a^{4} b^{3} c^{4} - 2 \, a^{5} b^{2} c^{3} d + a^{6} b c^{2} d^{2}\right )} x^{5} +{\left (a^{5} b^{2} c^{4} - 2 \, a^{6} b c^{3} d + a^{7} c^{2} d^{2}\right )} x^{3}\right )}}, \frac{3 \,{\left ({\left (5 \, b^{4} c^{3} - 6 \, a b^{3} c^{2} d\right )} x^{5} +{\left (5 \, a b^{3} c^{3} - 6 \, a^{2} b^{2} c^{2} d\right )} x^{3}\right )} \sqrt{a b c - a^{2} d} \arctan \left (\frac{\sqrt{a b c - a^{2} d}{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c}}{2 \,{\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} +{\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) - 2 \,{\left (2 \, a^{3} b^{2} c^{3} - 4 \, a^{4} b c^{2} d + 2 \, a^{5} c d^{2} -{\left (15 \, a b^{4} c^{3} - 23 \, a^{2} b^{3} c^{2} d + 4 \, a^{3} b^{2} c d^{2} + 4 \, a^{4} b d^{3}\right )} x^{4} - 2 \,{\left (5 \, a^{2} b^{3} c^{3} - 8 \, a^{3} b^{2} c^{2} d + a^{4} b c d^{2} + 2 \, a^{5} d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{12 \,{\left ({\left (a^{4} b^{3} c^{4} - 2 \, a^{5} b^{2} c^{3} d + a^{6} b c^{2} d^{2}\right )} x^{5} +{\left (a^{5} b^{2} c^{4} - 2 \, a^{6} b c^{3} d + a^{7} c^{2} d^{2}\right )} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/24*(3*((5*b^4*c^3 - 6*a*b^3*c^2*d)*x^5 + (5*a*b^3*c^3 - 6*a^2*b^2*c^2*d)*x^3)*sqrt(-a*b*c + a^2*d)*log(((b
^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 - 4*((b*c - 2*a*d)*x^3 - a*c*x)*
sqrt(-a*b*c + a^2*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*(2*a^3*b^2*c^3 - 4*a^4*b*c^2*d + 2*a^5*
c*d^2 - (15*a*b^4*c^3 - 23*a^2*b^3*c^2*d + 4*a^3*b^2*c*d^2 + 4*a^4*b*d^3)*x^4 - 2*(5*a^2*b^3*c^3 - 8*a^3*b^2*c
^2*d + a^4*b*c*d^2 + 2*a^5*d^3)*x^2)*sqrt(d*x^2 + c))/((a^4*b^3*c^4 - 2*a^5*b^2*c^3*d + a^6*b*c^2*d^2)*x^5 + (
a^5*b^2*c^4 - 2*a^6*b*c^3*d + a^7*c^2*d^2)*x^3), 1/12*(3*((5*b^4*c^3 - 6*a*b^3*c^2*d)*x^5 + (5*a*b^3*c^3 - 6*a
^2*b^2*c^2*d)*x^3)*sqrt(a*b*c - a^2*d)*arctan(1/2*sqrt(a*b*c - a^2*d)*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c
)/((a*b*c*d - a^2*d^2)*x^3 + (a*b*c^2 - a^2*c*d)*x)) - 2*(2*a^3*b^2*c^3 - 4*a^4*b*c^2*d + 2*a^5*c*d^2 - (15*a*
b^4*c^3 - 23*a^2*b^3*c^2*d + 4*a^3*b^2*c*d^2 + 4*a^4*b*d^3)*x^4 - 2*(5*a^2*b^3*c^3 - 8*a^3*b^2*c^2*d + a^4*b*c
*d^2 + 2*a^5*d^3)*x^2)*sqrt(d*x^2 + c))/((a^4*b^3*c^4 - 2*a^5*b^2*c^3*d + a^6*b*c^2*d^2)*x^5 + (a^5*b^2*c^4 -
2*a^6*b*c^3*d + a^7*c^2*d^2)*x^3)]

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Sympy [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**4/(b*x**2+a)**2/(d*x**2+c)**(1/2),x)

[Out]

Exception raised: ValueError

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Giac [B]  time = 6.43584, size = 506, normalized size = 2.46 \begin{align*} \frac{1}{6} \, d^{\frac{7}{2}}{\left (\frac{3 \,{\left (5 \, b^{3} c - 6 \, a b^{2} d\right )} \arctan \left (-\frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt{a b c d - a^{2} d^{2}}}\right )}{{\left (a^{3} b c d^{3} - a^{4} d^{4}\right )} \sqrt{a b c d - a^{2} d^{2}}} - \frac{6 \,{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b^{3} c - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b^{2} d - b^{3} c^{2}\right )}}{{\left (a^{3} b c d^{3} - a^{4} d^{4}\right )}{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c + 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a d + b c^{2}\right )}} - \frac{8 \,{\left (3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b - 6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c - 3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a d + 3 \, b c^{2} + a c d\right )}}{{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c\right )}^{3} a^{3} d^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

1/6*d^(7/2)*(3*(5*b^3*c - 6*a*b^2*d)*arctan(-1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*c*
d - a^2*d^2))/((a^3*b*c*d^3 - a^4*d^4)*sqrt(a*b*c*d - a^2*d^2)) - 6*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b^3*c - 2
*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b^2*d - b^3*c^2)/((a^3*b*c*d^3 - a^4*d^4)*((sqrt(d)*x - sqrt(d*x^2 + c))^4*
b - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c + 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d + b*c^2)) - 8*(3*(sqrt(d)*x
- sqrt(d*x^2 + c))^4*b - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c - 3*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d + 3*b*c
^2 + a*c*d)/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^3*a^3*d^3))