3.777 \(\int \frac{x^3}{(a+b x^2)^2 (c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=170 \[ \frac{a}{2 b \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}+\frac{3 a d+2 b c}{2 \sqrt{c+d x^2} (b c-a d)^3}+\frac{3 a d+2 b c}{6 b \left (c+d x^2\right )^{3/2} (b c-a d)^2}-\frac{\sqrt{b} (3 a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 (b c-a d)^{7/2}} \]

[Out]

(2*b*c + 3*a*d)/(6*b*(b*c - a*d)^2*(c + d*x^2)^(3/2)) + a/(2*b*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)^(3/2)) + (2
*b*c + 3*a*d)/(2*(b*c - a*d)^3*Sqrt[c + d*x^2]) - (Sqrt[b]*(2*b*c + 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/S
qrt[b*c - a*d]])/(2*(b*c - a*d)^(7/2))

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Rubi [A]  time = 0.164626, antiderivative size = 170, 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 = {446, 78, 51, 63, 208} \[ \frac{a}{2 b \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}+\frac{3 a d+2 b c}{2 \sqrt{c+d x^2} (b c-a d)^3}+\frac{3 a d+2 b c}{6 b \left (c+d x^2\right )^{3/2} (b c-a d)^2}-\frac{\sqrt{b} (3 a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 (b c-a d)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/((a + b*x^2)^2*(c + d*x^2)^(5/2)),x]

[Out]

(2*b*c + 3*a*d)/(6*b*(b*c - a*d)^2*(c + d*x^2)^(3/2)) + a/(2*b*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)^(3/2)) + (2
*b*c + 3*a*d)/(2*(b*c - a*d)^3*Sqrt[c + d*x^2]) - (Sqrt[b]*(2*b*c + 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/S
qrt[b*c - a*d]])/(2*(b*c - a*d)^(7/2))

Rule 446

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

Rule 78

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [C]  time = 0.0380696, size = 93, normalized size = 0.55 \[ \frac{\left (a+b x^2\right ) (3 a d+2 b c) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b \left (d x^2+c\right )}{b c-a d}\right )+3 a (b c-a d)}{6 b \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/((a + b*x^2)^2*(c + d*x^2)^(5/2)),x]

[Out]

(3*a*(b*c - a*d) + (2*b*c + 3*a*d)*(a + b*x^2)*Hypergeometric2F1[-3/2, 1, -1/2, (b*(c + d*x^2))/(b*c - a*d)])/
(6*b*(b*c - a*d)^2*(a + b*x^2)*(c + d*x^2)^(3/2))

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Maple [B]  time = 0.015, size = 2400, normalized size = 14.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x^2+a)^2/(d*x^2+c)^(5/2),x)

[Out]

-5/12/b^2*(-a*b)^(1/2)*d^2*a/(a*d-b*c)^2/c/((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)^(3/2)*x-5/6/b^2*(-a*b)^(1/2)*d^2*a/(a*d-b*c)^2/c^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+5/4/b*(-a*b)^(1/2)*d^2*a/(a*d-b*c)^3/c/((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+5/6/b^2*(-a*b)^(1/2)*d^2*a/(a*d-b*c)^2/c^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+5/12/b^2*(-a*b)^(1/2)*d^2*a/(a*d-b
*c)^2/c/((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)^(3/2)*x+1/2/(a*d-b*c)^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)+1/2/(a*d-b*c)^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/b^2*(-a*b)^(1/2)*d/(a*d-b*c)/c^
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-1/2/b^2*(-a*b)^(1/2)*
d/(a*d-b*c)/c/((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)^(3/2)*x+1/2/b^2*d
*(-a*b)^(1/2)/(a*d-b*c)/c/((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)^(3/2)
*x+1/b^2*d*(-a*b)^(1/2)/(a*d-b*c)/c^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-1/2/b/(a*d-b*c)^2*(-a*b)^(1/2)/c/((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*d+1/2/b/(a*d-b*c)^2*(-a*b)^(1/2)/c/((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*d-5/4*a*d/(a*d-b*c)^3/((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)-5/4*a*d/(a*d-b*c)^3/((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)-1/6/b/(a*d-b*c)/((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)^(3/2)-1/2/(a*d-b*c)^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)))+1/4/b^2*(-a*b)^(1/2)/(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)^(3/2)+5/12/b*a*d/(a*d-b*c)^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)^(3/2)+5/4*a*d/(a*d-b*c)^3/(-(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*a*d/(a*d-b*c)^3/(-(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)))-1/6/b/(a*d-b*c)/((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)^(3/2)-1/2/(a*d-b*c)^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*(-a*b)^(1/2)*d^2*a/(a*d-b*c)^3/c
/((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-1/4/b^2*(-a*b)^(1/2)/(
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)^(
3/2)+5/12/b*a*d/(a*d-b*c)^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)^(3/
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(x^3/(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.75511, size = 2022, normalized size = 11.89 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[-1/24*(3*((2*b^2*c*d^2 + 3*a*b*d^3)*x^6 + 2*a*b*c^3 + 3*a^2*c^2*d + (4*b^2*c^2*d + 8*a*b*c*d^2 + 3*a^2*d^3)*x
^4 + (2*b^2*c^3 + 7*a*b*c^2*d + 6*a^2*c*d^2)*x^2)*sqrt(b/(b*c - a*d))*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d
+ a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)*x^2 + 4*(2*b^2*c^2 - 3*a*b*c*d + a^2*d^2 + (b^2*c*d - a*b*d^2)*x^2)*sqr
t(d*x^2 + c)*sqrt(b/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*(3*(2*b^2*c*d + 3*a*b*d^2)*x^4 + 11*a*b*c^2
+ 4*a^2*c*d + 2*(4*b^2*c^2 + 8*a*b*c*d + 3*a^2*d^2)*x^2)*sqrt(d*x^2 + c))/(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^
3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^6 + (2*b^4*c^4*d -
5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^4 + (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2
+ 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x^2), 1/12*(3*((2*b^2*c*d^2 + 3*a*b*d^3)*x^6 + 2*a*b*c^3 + 3*a^2*c^2*d + (4*
b^2*c^2*d + 8*a*b*c*d^2 + 3*a^2*d^3)*x^4 + (2*b^2*c^3 + 7*a*b*c^2*d + 6*a^2*c*d^2)*x^2)*sqrt(-b/(b*c - a*d))*a
rctan(1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-b/(b*c - a*d))/(b*d*x^2 + b*c)) + 2*(3*(2*b^2*c*d + 3*
a*b*d^2)*x^4 + 11*a*b*c^2 + 4*a^2*c*d + 2*(4*b^2*c^2 + 8*a*b*c*d + 3*a^2*d^2)*x^2)*sqrt(d*x^2 + c))/(a*b^3*c^5
- 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*
d^5)*x^6 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^4 + (b^4*c^5 - a*b^3*
c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x^2)]

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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(x**3/(b*x**2+a)**2/(d*x**2+c)**(5/2),x)

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.18173, size = 351, normalized size = 2.06 \begin{align*} \frac{\frac{3 \, \sqrt{d x^{2} + c} a b d^{2}}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}{\left ({\left (d x^{2} + c\right )} b - b c + a d\right )}} + \frac{3 \,{\left (2 \, b^{2} c d + 3 \, a b d^{2}\right )} \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{-b^{2} c + a b d}} + \frac{2 \,{\left (3 \,{\left (d x^{2} + c\right )} b c d + b c^{2} d + 3 \,{\left (d x^{2} + c\right )} a d^{2} - a c d^{2}\right )}}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}{\left (d x^{2} + c\right )}^{\frac{3}{2}}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

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