3.87 \(\int \frac{1}{(a+b x^2)^{3/2} (c+d x^2)^2} \, dx\)

Optimal. Leaf size=143 \[ -\frac{d (4 b c-a d) \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{2 c^{3/2} (b c-a d)^{5/2}}+\frac{b x (a d+2 b c)}{2 a c \sqrt{a+b x^2} (b c-a d)^2}-\frac{d x}{2 c \sqrt{a+b x^2} \left (c+d x^2\right ) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.109018, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {414, 527, 12, 377, 208} \[ -\frac{d (4 b c-a d) \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{2 c^{3/2} (b c-a d)^{5/2}}+\frac{b x (a d+2 b c)}{2 a c \sqrt{a+b x^2} (b c-a d)^2}-\frac{d x}{2 c \sqrt{a+b x^2} \left (c+d x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 414

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

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -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 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{1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2} \, dx &=-\frac{d x}{2 c (b c-a d) \sqrt{a+b x^2} \left (c+d x^2\right )}+\frac{\int \frac{2 b c-a d-2 b d x^2}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )} \, dx}{2 c (b c-a d)}\\ &=\frac{b (2 b c+a d) x}{2 a c (b c-a d)^2 \sqrt{a+b x^2}}-\frac{d x}{2 c (b c-a d) \sqrt{a+b x^2} \left (c+d x^2\right )}-\frac{\int \frac{a d (4 b c-a d)}{\sqrt{a+b x^2} \left (c+d x^2\right )} \, dx}{2 a c (b c-a d)^2}\\ &=\frac{b (2 b c+a d) x}{2 a c (b c-a d)^2 \sqrt{a+b x^2}}-\frac{d x}{2 c (b c-a d) \sqrt{a+b x^2} \left (c+d x^2\right )}-\frac{(d (4 b c-a d)) \int \frac{1}{\sqrt{a+b x^2} \left (c+d x^2\right )} \, dx}{2 c (b c-a d)^2}\\ &=\frac{b (2 b c+a d) x}{2 a c (b c-a d)^2 \sqrt{a+b x^2}}-\frac{d x}{2 c (b c-a d) \sqrt{a+b x^2} \left (c+d x^2\right )}-\frac{(d (4 b c-a d)) \operatorname{Subst}\left (\int \frac{1}{c-(b c-a d) x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{2 c (b c-a d)^2}\\ &=\frac{b (2 b c+a d) x}{2 a c (b c-a d)^2 \sqrt{a+b x^2}}-\frac{d x}{2 c (b c-a d) \sqrt{a+b x^2} \left (c+d x^2\right )}-\frac{d (4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{c} \sqrt{a+b x^2}}\right )}{2 c^{3/2} (b c-a d)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 2.52783, size = 758, normalized size = 5.3 \[ \frac{x \left (\frac{24 d^2 x^4 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{7/2} \text{HypergeometricPFQ}\left (\left \{2,2,\frac{5}{2}\right \},\left \{1,\frac{9}{2}\right \},\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )}{c^2}+\frac{48 d x^2 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{7/2} \text{HypergeometricPFQ}\left (\left \{2,2,\frac{5}{2}\right \},\left \{1,\frac{9}{2}\right \},\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )}{c}+24 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{7/2} \text{HypergeometricPFQ}\left (\left \{2,2,\frac{5}{2}\right \},\left \{1,\frac{9}{2}\right \},\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )+\frac{280 d^2 x^4 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{3/2}}{c^2}-\frac{2310 d^2 x^4 \sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}}{c^2}+\frac{1050 d^2 x^6 (a d-b c) \tanh ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )}{c^3 \left (a+b x^2\right )}+\frac{2310 d^2 x^4 \tanh ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )}{c^2}+\frac{2310 d x^4 (a d-b c) \tanh ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )}{c^2 \left (a+b x^2\right )}+\frac{560 d x^2 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{3/2}}{c}-\frac{5250 d x^2 \sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}}{c}+70 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{3/2}-2625 \sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}+\frac{5250 d x^2 \tanh ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )}{c}-\frac{945 x^2 (b c-a d) \tanh ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )}{c \left (a+b x^2\right )}+2625 \tanh ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )\right )}{210 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(-2625*Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))] - (5250*d*x^2*Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))])/c - (
2310*d^2*x^4*Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))])/c^2 + 70*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(3/2) + (56
0*d*x^2*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(3/2))/c + (280*d^2*x^4*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(3/2))
/c^2 + 2625*ArcTanh[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]] + (5250*d*x^2*ArcTanh[Sqrt[((b*c - a*d)*x^2)/(c*(
a + b*x^2))]])/c + (2310*d^2*x^4*ArcTanh[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]])/c^2 - (945*(b*c - a*d)*x^2*
ArcTanh[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]])/(c*(a + b*x^2)) + (2310*d*(-(b*c) + a*d)*x^4*ArcTanh[Sqrt[((
b*c - a*d)*x^2)/(c*(a + b*x^2))]])/(c^2*(a + b*x^2)) + (1050*d^2*(-(b*c) + a*d)*x^6*ArcTanh[Sqrt[((b*c - a*d)*
x^2)/(c*(a + b*x^2))]])/(c^3*(a + b*x^2)) + 24*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(7/2)*HypergeometricPFQ[{2,
2, 5/2}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + (48*d*x^2*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(7/2)*H
ypergeometricPFQ[{2, 2, 5/2}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^2))])/c + (24*d^2*x^4*(((b*c - a*d)*x^2)
/(c*(a + b*x^2)))^(7/2)*HypergeometricPFQ[{2, 2, 5/2}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^2))])/c^2))/(21
0*c*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(5/2)*(a + b*x^2)^(3/2)*(c + d*x^2))

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Maple [B]  time = 0.017, size = 1439, normalized size = 10.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)^(3/2)/(d*x^2+c)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{3}{2}}{\left (d x^{2} + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(3/2)/(d*x^2+c)^2,x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(3/2)*(d*x^2 + c)^2), x)

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Fricas [B]  time = 4.48744, size = 1706, normalized size = 11.93 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(3/2)/(d*x^2+c)^2,x, algorithm="fricas")

[Out]

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

[Out]

Exception raised: ValueError

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Giac [B]  time = 5.45968, size = 429, normalized size = 3. \begin{align*} \frac{b^{2} x}{{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt{b x^{2} + a}} + \frac{{\left (4 \, b^{\frac{3}{2}} c d - a \sqrt{b} d^{2}\right )} \arctan \left (\frac{{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} d + 2 \, b c - a d}{2 \, \sqrt{-b^{2} c^{2} + a b c d}}\right )}{2 \,{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt{-b^{2} c^{2} + a b c d}} + \frac{2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} b^{\frac{3}{2}} c d -{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a \sqrt{b} d^{2} + a^{2} \sqrt{b} d^{2}}{{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )}{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} d + 4 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} b c - 2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a d + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(3/2)/(d*x^2+c)^2,x, algorithm="giac")

[Out]

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