3.8 \(\int \frac{A+B x+C x^2+D x^3}{(a+b x)^4 \sqrt{c+d x}} \, dx\)

Optimal. Leaf size=375 \[ -\frac{\sqrt{c+d x} \left (a^2 b d (30 c D+C d)-11 a^3 d^2 D-a b^2 \left (-B d^2+24 c^2 D+4 c C d\right )+b^3 \left (5 A d^2-6 B c d+8 c^2 C\right )\right )}{8 b^3 (a+b x) (b c-a d)^3}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right ) \left (a^2 b d^2 (C d-18 c D)+5 a^3 d^3 D-a b^2 d \left (-B d^2-24 c^2 D+4 c C d\right )+b^3 \left (5 A d^3-6 B c d^2+8 c^2 C d-16 c^3 D\right )\right )}{8 b^{7/2} (b c-a d)^{7/2}}-\frac{\sqrt{c+d x} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^3 (a+b x)^3 (b c-a d)}-\frac{\sqrt{c+d x} \left (a^2 b (18 c D+7 C d)-13 a^3 d D-a b^2 (B d+12 c C)+b^3 (6 B c-5 A d)\right )}{12 b^3 (a+b x)^2 (b c-a d)^2} \]

[Out]

-((A*b^3 - a*(b^2*B - a*b*C + a^2*D))*Sqrt[c + d*x])/(3*b^3*(b*c - a*d)*(a + b*x)^3) - ((b^3*(6*B*c - 5*A*d) -
a*b^2*(12*c*C + B*d) - 13*a^3*d*D + a^2*b*(7*C*d + 18*c*D))*Sqrt[c + d*x])/(12*b^3*(b*c - a*d)^2*(a + b*x)^2)
- ((b^3*(8*c^2*C - 6*B*c*d + 5*A*d^2) - 11*a^3*d^2*D + a^2*b*d*(C*d + 30*c*D) - a*b^2*(4*c*C*d - B*d^2 + 24*c
^2*D))*Sqrt[c + d*x])/(8*b^3*(b*c - a*d)^3*(a + b*x)) + ((5*a^3*d^3*D + a^2*b*d^2*(C*d - 18*c*D) - a*b^2*d*(4*
c*C*d - B*d^2 - 24*c^2*D) + b^3*(8*c^2*C*d - 6*B*c*d^2 + 5*A*d^3 - 16*c^3*D))*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/
Sqrt[b*c - a*d]])/(8*b^(7/2)*(b*c - a*d)^(7/2))

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Rubi [A]  time = 0.821268, antiderivative size = 375, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {1621, 897, 1157, 385, 208} \[ -\frac{\sqrt{c+d x} \left (a^2 b d (30 c D+C d)-11 a^3 d^2 D-a b^2 \left (-B d^2+24 c^2 D+4 c C d\right )+b^3 \left (5 A d^2-6 B c d+8 c^2 C\right )\right )}{8 b^3 (a+b x) (b c-a d)^3}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right ) \left (a^2 b d^2 (C d-18 c D)+5 a^3 d^3 D-a b^2 d \left (-B d^2-24 c^2 D+4 c C d\right )+b^3 \left (5 A d^3-6 B c d^2+8 c^2 C d-16 c^3 D\right )\right )}{8 b^{7/2} (b c-a d)^{7/2}}-\frac{\sqrt{c+d x} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^3 (a+b x)^3 (b c-a d)}-\frac{\sqrt{c+d x} \left (a^2 b (18 c D+7 C d)-13 a^3 d D-a b^2 (B d+12 c C)+b^3 (6 B c-5 A d)\right )}{12 b^3 (a+b x)^2 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2 + D*x^3)/((a + b*x)^4*Sqrt[c + d*x]),x]

[Out]

-((A*b^3 - a*(b^2*B - a*b*C + a^2*D))*Sqrt[c + d*x])/(3*b^3*(b*c - a*d)*(a + b*x)^3) - ((b^3*(6*B*c - 5*A*d) -
a*b^2*(12*c*C + B*d) - 13*a^3*d*D + a^2*b*(7*C*d + 18*c*D))*Sqrt[c + d*x])/(12*b^3*(b*c - a*d)^2*(a + b*x)^2)
- ((b^3*(8*c^2*C - 6*B*c*d + 5*A*d^2) - 11*a^3*d^2*D + a^2*b*d*(C*d + 30*c*D) - a*b^2*(4*c*C*d - B*d^2 + 24*c
^2*D))*Sqrt[c + d*x])/(8*b^3*(b*c - a*d)^3*(a + b*x)) + ((5*a^3*d^3*D + a^2*b*d^2*(C*d - 18*c*D) - a*b^2*d*(4*
c*C*d - B*d^2 - 24*c^2*D) + b^3*(8*c^2*C*d - 6*B*c*d^2 + 5*A*d^3 - 16*c^3*D))*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/
Sqrt[b*c - a*d]])/(8*b^(7/2)*(b*c - a*d)^(7/2))

Rule 1621

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> With[{Qx = PolynomialQuotient[Px,
a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[(R*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/((m + 1)*
(b*c - a*d)), x] + Dist[1/((m + 1)*(b*c - a*d)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*ExpandToSum[(m + 1)*(b*c -
a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[m, -1] && GtQ[Expo
n[Px, x], 2]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 385

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

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{A+B x+C x^2+D x^3}{(a+b x)^4 \sqrt{c+d x}} \, dx &=-\frac{\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \sqrt{c+d x}}{3 b^3 (b c-a d) (a+b x)^3}-\frac{\int \frac{-\frac{b^3 (6 B c-5 A d)-a b^2 (6 c C+B d)-a^3 d D+a^2 b (C d+6 c D)}{2 b^3}-\frac{3 (b c-a d) (b C-a D) x}{b^2}-3 \left (c-\frac{a d}{b}\right ) D x^2}{(a+b x)^3 \sqrt{c+d x}} \, dx}{3 (b c-a d)}\\ &=-\frac{\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \sqrt{c+d x}}{3 b^3 (b c-a d) (a+b x)^3}-\frac{2 \operatorname{Subst}\left (\int \frac{\frac{-3 c^2 \left (c-\frac{a d}{b}\right ) D+\frac{3 c d (b c-a d) (b C-a D)}{b^2}-\frac{d^2 \left (b^3 (6 B c-5 A d)-a b^2 (6 c C+B d)-a^3 d D+a^2 b (C d+6 c D)\right )}{2 b^3}}{d^2}-\frac{\left (-6 c \left (c-\frac{a d}{b}\right ) D+\frac{3 d (b c-a d) (b C-a D)}{b^2}\right ) x^2}{d^2}-\frac{3 \left (c-\frac{a d}{b}\right ) D x^4}{d^2}}{\left (\frac{-b c+a d}{d}+\frac{b x^2}{d}\right )^3} \, dx,x,\sqrt{c+d x}\right )}{3 d (b c-a d)}\\ &=-\frac{\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \sqrt{c+d x}}{3 b^3 (b c-a d) (a+b x)^3}-\frac{\left (b^3 (6 B c-5 A d)-a b^2 (12 c C+B d)-13 a^3 d D+a^2 b (7 C d+18 c D)\right ) \sqrt{c+d x}}{12 b^3 (b c-a d)^2 (a+b x)^2}-\frac{\operatorname{Subst}\left (\int \frac{\frac{3 \left (a b^2 d^2 (4 c C-B d)+3 a^3 d^3 D-a^2 b d^2 (C d+6 c D)-b^3 \left (8 c^2 C d-6 B c d^2+5 A d^3-8 c^3 D\right )\right )}{2 b^3 d^2}-\frac{12 (b c-a d)^2 D x^2}{b^2 d^2}}{\left (\frac{-b c+a d}{d}+\frac{b x^2}{d}\right )^2} \, dx,x,\sqrt{c+d x}\right )}{6 (b c-a d)^2}\\ &=-\frac{\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \sqrt{c+d x}}{3 b^3 (b c-a d) (a+b x)^3}-\frac{\left (b^3 (6 B c-5 A d)-a b^2 (12 c C+B d)-13 a^3 d D+a^2 b (7 C d+18 c D)\right ) \sqrt{c+d x}}{12 b^3 (b c-a d)^2 (a+b x)^2}-\frac{\left (b^3 \left (8 c^2 C-6 B c d+5 A d^2\right )-11 a^3 d^2 D+a^2 b d (C d+30 c D)-a b^2 \left (4 c C d-B d^2+24 c^2 D\right )\right ) \sqrt{c+d x}}{8 b^3 (b c-a d)^3 (a+b x)}-\frac{\left (5 a^3 d^3 D+a^2 b d^2 (C d-18 c D)-a b^2 d \left (4 c C d-B d^2-24 c^2 D\right )+b^3 \left (8 c^2 C d-6 B c d^2+5 A d^3-16 c^3 D\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{-b c+a d}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{8 b^3 d (b c-a d)^3}\\ &=-\frac{\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \sqrt{c+d x}}{3 b^3 (b c-a d) (a+b x)^3}-\frac{\left (b^3 (6 B c-5 A d)-a b^2 (12 c C+B d)-13 a^3 d D+a^2 b (7 C d+18 c D)\right ) \sqrt{c+d x}}{12 b^3 (b c-a d)^2 (a+b x)^2}-\frac{\left (b^3 \left (8 c^2 C-6 B c d+5 A d^2\right )-11 a^3 d^2 D+a^2 b d (C d+30 c D)-a b^2 \left (4 c C d-B d^2+24 c^2 D\right )\right ) \sqrt{c+d x}}{8 b^3 (b c-a d)^3 (a+b x)}+\frac{\left (5 a^3 d^3 D+a^2 b d^2 (C d-18 c D)-a b^2 d \left (4 c C d-B d^2-24 c^2 D\right )+b^3 \left (8 c^2 C d-6 B c d^2+5 A d^3-16 c^3 D\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{8 b^{7/2} (b c-a d)^{7/2}}\\ \end{align*}

Mathematica [A]  time = 1.45928, size = 493, normalized size = 1.31 \[ \frac{5 d (a+b x) \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right ) \left (3 d^2 (a+b x)^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )+\sqrt{b} \sqrt{c+d x} \sqrt{b c-a d} (-5 a d+2 b c-3 b d x)\right )-8 \sqrt{b} \sqrt{c+d x} (b c-a d)^{5/2} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )-12 \sqrt{b} (a+b x) \sqrt{c+d x} (b c-a d)^{5/2} \left (3 a^2 D-2 a b C+b^2 B\right )+18 d (a+b x)^2 (b c-a d) \left (3 a^2 D-2 a b C+b^2 B\right ) \left (\sqrt{b} \sqrt{c+d x} \sqrt{b c-a d}-d (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )\right )-24 \sqrt{b} (a+b x)^2 \sqrt{c+d x} (b c-a d)^{5/2} (b C-3 a D)+24 d (a+b x)^3 (b c-a d)^2 (b C-3 a D) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )-48 D (a+b x)^3 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{24 b^{7/2} (a+b x)^3 (b c-a d)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x + C*x^2 + D*x^3)/((a + b*x)^4*Sqrt[c + d*x]),x]

[Out]

(-8*Sqrt[b]*(b*c - a*d)^(5/2)*(A*b^3 - a*(b^2*B - a*b*C + a^2*D))*Sqrt[c + d*x] - 12*Sqrt[b]*(b*c - a*d)^(5/2)
*(b^2*B - 2*a*b*C + 3*a^2*D)*(a + b*x)*Sqrt[c + d*x] - 24*Sqrt[b]*(b*c - a*d)^(5/2)*(b*C - 3*a*D)*(a + b*x)^2*
Sqrt[c + d*x] - 48*(b*c - a*d)^3*D*(a + b*x)^3*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]] + 24*d*(b*c -
a*d)^2*(b*C - 3*a*D)*(a + b*x)^3*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]] + 18*d*(b*c - a*d)*(b^2*B -
2*a*b*C + 3*a^2*D)*(a + b*x)^2*(Sqrt[b]*Sqrt[b*c - a*d]*Sqrt[c + d*x] - d*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[c +
d*x])/Sqrt[b*c - a*d]]) + 5*d*(A*b^3 - a*(b^2*B - a*b*C + a^2*D))*(a + b*x)*(Sqrt[b]*Sqrt[b*c - a*d]*Sqrt[c +
d*x]*(2*b*c - 5*a*d - 3*b*d*x) + 3*d^2*(a + b*x)^2*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]]))/(24*b^(7
/2)*(b*c - a*d)^(7/2)*(a + b*x)^3)

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Maple [B]  time = 0.023, size = 1186, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((D*x^3+C*x^2+B*x+A)/(b*x+a)^4/(d*x+c)^(1/2),x)

[Out]

2*(1/16*d*(5*A*b^3*d^2+B*a*b^2*d^2-6*B*b^3*c*d+C*a^2*b*d^2-4*C*a*b^2*c*d+8*C*b^3*c^2-11*D*a^3*d^2+30*D*a^2*b*c
*d-24*D*a*b^2*c^2)/b/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)*(d*x+c)^(5/2)+1/6*(5*A*b^3*d^2+B*a*b^2*d^2-
6*B*b^3*c*d-C*a^2*b*d^2+6*C*b^3*c^2-5*D*a^3*d^2+18*D*a^2*b*c*d-18*D*a*b^2*c^2)/b^2*d/(a^2*d^2-2*a*b*c*d+b^2*c^
2)*(d*x+c)^(3/2)+1/16*(11*A*b^3*d^2-B*a*b^2*d^2-10*B*b^3*c*d-C*a^2*b*d^2+4*C*a*b^2*c*d+8*C*b^3*c^2-5*D*a^3*d^2
+18*D*a^2*b*c*d-24*D*a*b^2*c^2)/b^3*d/(a*d-b*c)*(d*x+c)^(1/2))/(b*(d*x+c)+a*d-b*c)^3+5/8/(a^3*d^3-3*a^2*b*c*d^
2+3*a*b^2*c^2*d-b^3*c^3)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*A*d^3+1/8/b/(a^3*d^3-
3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*a*B*d^3-3
/4/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2
))*B*c*d^2+1/8/b^2/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((
a*d-b*c)*b)^(1/2))*a^2*C*d^3-1/2/b/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/((a*d-b*c)*b)^(1/2)*arctan(b*
(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*C*a*c*d^2+1/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/((a*d-b*c)*b)^(1/
2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*C*c^2*d+5/8/b^3/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/(
(a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*a^3*d^3*D-9/4/b^2/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^
2*c^2*d-b^3*c^3)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*D*a^2*c*d^2+3/b/(a^3*d^3-3*a^
2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*D*a*c^2*d-2/(
a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*D
*c^3

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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((D*x^3+C*x^2+B*x+A)/(b*x+a)^4/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^4/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)**4/(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 2.3845, size = 1318, normalized size = 3.51 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^4/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

1/8*(16*D*b^3*c^3 - 24*D*a*b^2*c^2*d - 8*C*b^3*c^2*d + 18*D*a^2*b*c*d^2 + 4*C*a*b^2*c*d^2 + 6*B*b^3*c*d^2 - 5*
D*a^3*d^3 - C*a^2*b*d^3 - B*a*b^2*d^3 - 5*A*b^3*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^6*c^3 -
3*a*b^5*c^2*d + 3*a^2*b^4*c*d^2 - a^3*b^3*d^3)*sqrt(-b^2*c + a*b*d)) + 1/24*(72*(d*x + c)^(5/2)*D*a*b^4*c^2*d
- 24*(d*x + c)^(5/2)*C*b^5*c^2*d - 144*(d*x + c)^(3/2)*D*a*b^4*c^3*d + 48*(d*x + c)^(3/2)*C*b^5*c^3*d + 72*sqr
t(d*x + c)*D*a*b^4*c^4*d - 24*sqrt(d*x + c)*C*b^5*c^4*d - 90*(d*x + c)^(5/2)*D*a^2*b^3*c*d^2 + 12*(d*x + c)^(5
/2)*C*a*b^4*c*d^2 + 18*(d*x + c)^(5/2)*B*b^5*c*d^2 + 288*(d*x + c)^(3/2)*D*a^2*b^3*c^2*d^2 - 48*(d*x + c)^(3/2
)*C*a*b^4*c^2*d^2 - 48*(d*x + c)^(3/2)*B*b^5*c^2*d^2 - 198*sqrt(d*x + c)*D*a^2*b^3*c^3*d^2 + 36*sqrt(d*x + c)*
C*a*b^4*c^3*d^2 + 30*sqrt(d*x + c)*B*b^5*c^3*d^2 + 33*(d*x + c)^(5/2)*D*a^3*b^2*d^3 - 3*(d*x + c)^(5/2)*C*a^2*
b^3*d^3 - 3*(d*x + c)^(5/2)*B*a*b^4*d^3 - 15*(d*x + c)^(5/2)*A*b^5*d^3 - 184*(d*x + c)^(3/2)*D*a^3*b^2*c*d^3 -
8*(d*x + c)^(3/2)*C*a^2*b^3*c*d^3 + 56*(d*x + c)^(3/2)*B*a*b^4*c*d^3 + 40*(d*x + c)^(3/2)*A*b^5*c*d^3 + 195*s
qrt(d*x + c)*D*a^3*b^2*c^2*d^3 + 3*sqrt(d*x + c)*C*a^2*b^3*c^2*d^3 - 57*sqrt(d*x + c)*B*a*b^4*c^2*d^3 - 33*sqr
t(d*x + c)*A*b^5*c^2*d^3 + 40*(d*x + c)^(3/2)*D*a^4*b*d^4 + 8*(d*x + c)^(3/2)*C*a^3*b^2*d^4 - 8*(d*x + c)^(3/2
)*B*a^2*b^3*d^4 - 40*(d*x + c)^(3/2)*A*a*b^4*d^4 - 84*sqrt(d*x + c)*D*a^4*b*c*d^4 - 18*sqrt(d*x + c)*C*a^3*b^2
*c*d^4 + 24*sqrt(d*x + c)*B*a^2*b^3*c*d^4 + 66*sqrt(d*x + c)*A*a*b^4*c*d^4 + 15*sqrt(d*x + c)*D*a^5*d^5 + 3*sq
rt(d*x + c)*C*a^4*b*d^5 + 3*sqrt(d*x + c)*B*a^3*b^2*d^5 - 33*sqrt(d*x + c)*A*a^2*b^3*d^5)/((b^6*c^3 - 3*a*b^5*
c^2*d + 3*a^2*b^4*c*d^2 - a^3*b^3*d^3)*((d*x + c)*b - b*c + a*d)^3)