3.2201 \(\int \sqrt{a+b x} (A+B x) (d+e x)^{3/2} \, dx\)

Optimal. Leaf size=250 \[ -\frac{\sqrt{a+b x} \sqrt{d+e x} (b d-a e)^2 (5 a B e-8 A b e+3 b B d)}{64 b^3 e^2}+\frac{(b d-a e)^3 (5 a B e-8 A b e+3 b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{64 b^{7/2} e^{5/2}}-\frac{(a+b x)^{3/2} \sqrt{d+e x} (b d-a e) (5 a B e-8 A b e+3 b B d)}{32 b^3 e}-\frac{(a+b x)^{3/2} (d+e x)^{3/2} (5 a B e-8 A b e+3 b B d)}{24 b^2 e}+\frac{B (a+b x)^{3/2} (d+e x)^{5/2}}{4 b e} \]

[Out]

-((b*d - a*e)^2*(3*b*B*d - 8*A*b*e + 5*a*B*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/(64*b^3*e^2) - ((b*d - a*e)*(3*b*B*
d - 8*A*b*e + 5*a*B*e)*(a + b*x)^(3/2)*Sqrt[d + e*x])/(32*b^3*e) - ((3*b*B*d - 8*A*b*e + 5*a*B*e)*(a + b*x)^(3
/2)*(d + e*x)^(3/2))/(24*b^2*e) + (B*(a + b*x)^(3/2)*(d + e*x)^(5/2))/(4*b*e) + ((b*d - a*e)^3*(3*b*B*d - 8*A*
b*e + 5*a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(64*b^(7/2)*e^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.200071, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {80, 50, 63, 217, 206} \[ -\frac{\sqrt{a+b x} \sqrt{d+e x} (b d-a e)^2 (5 a B e-8 A b e+3 b B d)}{64 b^3 e^2}+\frac{(b d-a e)^3 (5 a B e-8 A b e+3 b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{64 b^{7/2} e^{5/2}}-\frac{(a+b x)^{3/2} \sqrt{d+e x} (b d-a e) (5 a B e-8 A b e+3 b B d)}{32 b^3 e}-\frac{(a+b x)^{3/2} (d+e x)^{3/2} (5 a B e-8 A b e+3 b B d)}{24 b^2 e}+\frac{B (a+b x)^{3/2} (d+e x)^{5/2}}{4 b e} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x]*(A + B*x)*(d + e*x)^(3/2),x]

[Out]

-((b*d - a*e)^2*(3*b*B*d - 8*A*b*e + 5*a*B*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/(64*b^3*e^2) - ((b*d - a*e)*(3*b*B*
d - 8*A*b*e + 5*a*B*e)*(a + b*x)^(3/2)*Sqrt[d + e*x])/(32*b^3*e) - ((3*b*B*d - 8*A*b*e + 5*a*B*e)*(a + b*x)^(3
/2)*(d + e*x)^(3/2))/(24*b^2*e) + (B*(a + b*x)^(3/2)*(d + e*x)^(5/2))/(4*b*e) + ((b*d - a*e)^3*(3*b*B*d - 8*A*
b*e + 5*a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(64*b^(7/2)*e^(5/2))

Rule 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sqrt{a+b x} (A+B x) (d+e x)^{3/2} \, dx &=\frac{B (a+b x)^{3/2} (d+e x)^{5/2}}{4 b e}+\frac{\left (4 A b e-B \left (\frac{3 b d}{2}+\frac{5 a e}{2}\right )\right ) \int \sqrt{a+b x} (d+e x)^{3/2} \, dx}{4 b e}\\ &=-\frac{(3 b B d-8 A b e+5 a B e) (a+b x)^{3/2} (d+e x)^{3/2}}{24 b^2 e}+\frac{B (a+b x)^{3/2} (d+e x)^{5/2}}{4 b e}-\frac{((b d-a e) (3 b B d-8 A b e+5 a B e)) \int \sqrt{a+b x} \sqrt{d+e x} \, dx}{16 b^2 e}\\ &=-\frac{(b d-a e) (3 b B d-8 A b e+5 a B e) (a+b x)^{3/2} \sqrt{d+e x}}{32 b^3 e}-\frac{(3 b B d-8 A b e+5 a B e) (a+b x)^{3/2} (d+e x)^{3/2}}{24 b^2 e}+\frac{B (a+b x)^{3/2} (d+e x)^{5/2}}{4 b e}-\frac{\left ((b d-a e)^2 (3 b B d-8 A b e+5 a B e)\right ) \int \frac{\sqrt{a+b x}}{\sqrt{d+e x}} \, dx}{64 b^3 e}\\ &=-\frac{(b d-a e)^2 (3 b B d-8 A b e+5 a B e) \sqrt{a+b x} \sqrt{d+e x}}{64 b^3 e^2}-\frac{(b d-a e) (3 b B d-8 A b e+5 a B e) (a+b x)^{3/2} \sqrt{d+e x}}{32 b^3 e}-\frac{(3 b B d-8 A b e+5 a B e) (a+b x)^{3/2} (d+e x)^{3/2}}{24 b^2 e}+\frac{B (a+b x)^{3/2} (d+e x)^{5/2}}{4 b e}+\frac{\left ((b d-a e)^3 (3 b B d-8 A b e+5 a B e)\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{d+e x}} \, dx}{128 b^3 e^2}\\ &=-\frac{(b d-a e)^2 (3 b B d-8 A b e+5 a B e) \sqrt{a+b x} \sqrt{d+e x}}{64 b^3 e^2}-\frac{(b d-a e) (3 b B d-8 A b e+5 a B e) (a+b x)^{3/2} \sqrt{d+e x}}{32 b^3 e}-\frac{(3 b B d-8 A b e+5 a B e) (a+b x)^{3/2} (d+e x)^{3/2}}{24 b^2 e}+\frac{B (a+b x)^{3/2} (d+e x)^{5/2}}{4 b e}+\frac{\left ((b d-a e)^3 (3 b B d-8 A b e+5 a B e)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d-\frac{a e}{b}+\frac{e x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{64 b^4 e^2}\\ &=-\frac{(b d-a e)^2 (3 b B d-8 A b e+5 a B e) \sqrt{a+b x} \sqrt{d+e x}}{64 b^3 e^2}-\frac{(b d-a e) (3 b B d-8 A b e+5 a B e) (a+b x)^{3/2} \sqrt{d+e x}}{32 b^3 e}-\frac{(3 b B d-8 A b e+5 a B e) (a+b x)^{3/2} (d+e x)^{3/2}}{24 b^2 e}+\frac{B (a+b x)^{3/2} (d+e x)^{5/2}}{4 b e}+\frac{\left ((b d-a e)^3 (3 b B d-8 A b e+5 a B e)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{e x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{d+e x}}\right )}{64 b^4 e^2}\\ &=-\frac{(b d-a e)^2 (3 b B d-8 A b e+5 a B e) \sqrt{a+b x} \sqrt{d+e x}}{64 b^3 e^2}-\frac{(b d-a e) (3 b B d-8 A b e+5 a B e) (a+b x)^{3/2} \sqrt{d+e x}}{32 b^3 e}-\frac{(3 b B d-8 A b e+5 a B e) (a+b x)^{3/2} (d+e x)^{3/2}}{24 b^2 e}+\frac{B (a+b x)^{3/2} (d+e x)^{5/2}}{4 b e}+\frac{(b d-a e)^3 (3 b B d-8 A b e+5 a B e) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{64 b^{7/2} e^{5/2}}\\ \end{align*}

Mathematica [A]  time = 1.93152, size = 266, normalized size = 1.06 \[ \frac{\sqrt{b d-a e} \left (\frac{b (d+e x)}{b d-a e}\right )^{3/2} (-5 a B e+8 A b e-3 b B d) \left (2 b^4 e^2 (a+b x)^2 (b d-a e)^{3/2} \sqrt{\frac{b (d+e x)}{b d-a e}} (-3 a e+7 b d+4 b e x)+3 b^4 e (a+b x) (b d-a e)^{7/2} \sqrt{\frac{b (d+e x)}{b d-a e}}-3 b^4 \sqrt{e} \sqrt{a+b x} (b d-a e)^4 \sinh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b d-a e}}\right )\right )+48 b^8 B e^2 (a+b x)^2 (d+e x)^4}{192 b^9 e^3 \sqrt{a+b x} (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x]*(A + B*x)*(d + e*x)^(3/2),x]

[Out]

(48*b^8*B*e^2*(a + b*x)^2*(d + e*x)^4 + Sqrt[b*d - a*e]*(-3*b*B*d + 8*A*b*e - 5*a*B*e)*((b*(d + e*x))/(b*d - a
*e))^(3/2)*(3*b^4*e*(b*d - a*e)^(7/2)*(a + b*x)*Sqrt[(b*(d + e*x))/(b*d - a*e)] + 2*b^4*e^2*(b*d - a*e)^(3/2)*
(a + b*x)^2*Sqrt[(b*(d + e*x))/(b*d - a*e)]*(7*b*d - 3*a*e + 4*b*e*x) - 3*b^4*Sqrt[e]*(b*d - a*e)^4*Sqrt[a + b
*x]*ArcSinh[(Sqrt[e]*Sqrt[a + b*x])/Sqrt[b*d - a*e]]))/(192*b^9*e^3*Sqrt[a + b*x]*(d + e*x)^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.014, size = 1150, normalized size = 4.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(3/2)*(b*x+a)^(1/2),x)

[Out]

1/384*(e*x+d)^(1/2)*(b*x+a)^(1/2)*(96*B*x^3*b^3*e^3*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)+16*B*x^2*a*b^2
*e^3*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)+144*B*x^2*b^3*d*e^2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/
2)+40*B*(b*e)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*x*a*b^2*d*e^2+30*B*(b*e)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(
1/2)*a^3*e^3+24*A*ln(1/2*(2*b*x*e+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^3*b*e^
4-24*A*ln(1/2*(2*b*x*e+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^4*d^3*e-18*B*(b*e
)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*b^3*d^3-15*B*ln(1/2*(2*b*x*e+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(
1/2)+a*e+b*d)/(b*e)^(1/2))*a^4*e^4+9*B*ln(1/2*(2*b*x*e+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)+a*e+b*d)/
(b*e)^(1/2))*b^4*d^4-20*B*(b*e)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*x*a^2*b*e^3+128*A*a*b^2*d*e^2*(b*e*x^2+a
*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)+32*A*(b*e)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*x*a*b^2*e^3+224*A*(b*e)^(1/
2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*x*b^3*d*e^2+12*B*(b*e)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*x*b^3*d^2*e-62
*B*(b*e)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*a^2*b*d*e^2+18*B*(b*e)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*a*
b^2*d^2*e+128*A*x^2*b^3*e^3*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)-72*A*ln(1/2*(2*b*x*e+2*(b*e*x^2+a*e*x+
b*d*x+a*d)^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*b^2*d*e^3+72*A*ln(1/2*(2*b*x*e+2*(b*e*x^2+a*e*x+b*d*x+a
*d)^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b^3*d^2*e^2+36*B*ln(1/2*(2*b*x*e+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/
2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^3*b*d*e^3-18*B*ln(1/2*(2*b*x*e+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^
(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*b^2*d^2*e^2-12*B*ln(1/2*(2*b*x*e+2*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*(b*e)^(1/2)
+a*e+b*d)/(b*e)^(1/2))*a*b^3*d^3*e-48*A*(b*e)^(1/2)*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*a^2*b*e^3+48*A*(b*e)^(1/2)
*(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)*b^3*d^2*e)/e^2/(b*e*x^2+a*e*x+b*d*x+a*d)^(1/2)/b^3/(b*e)^(1/2)

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.35384, size = 1669, normalized size = 6.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/768*(3*(3*B*b^4*d^4 - 4*(B*a*b^3 + 2*A*b^4)*d^3*e - 6*(B*a^2*b^2 - 4*A*a*b^3)*d^2*e^2 + 12*(B*a^3*b - 2*A*a
^2*b^2)*d*e^3 - (5*B*a^4 - 8*A*a^3*b)*e^4)*sqrt(b*e)*log(8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*
b*e*x + b*d + a*e)*sqrt(b*e)*sqrt(b*x + a)*sqrt(e*x + d) + 8*(b^2*d*e + a*b*e^2)*x) + 4*(48*B*b^4*e^4*x^3 - 9*
B*b^4*d^3*e + 3*(3*B*a*b^3 + 8*A*b^4)*d^2*e^2 - (31*B*a^2*b^2 - 64*A*a*b^3)*d*e^3 + 3*(5*B*a^3*b - 8*A*a^2*b^2
)*e^4 + 8*(9*B*b^4*d*e^3 + (B*a*b^3 + 8*A*b^4)*e^4)*x^2 + 2*(3*B*b^4*d^2*e^2 + 2*(5*B*a*b^3 + 28*A*b^4)*d*e^3
- (5*B*a^2*b^2 - 8*A*a*b^3)*e^4)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b^4*e^3), -1/384*(3*(3*B*b^4*d^4 - 4*(B*a*b^
3 + 2*A*b^4)*d^3*e - 6*(B*a^2*b^2 - 4*A*a*b^3)*d^2*e^2 + 12*(B*a^3*b - 2*A*a^2*b^2)*d*e^3 - (5*B*a^4 - 8*A*a^3
*b)*e^4)*sqrt(-b*e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(-b*e)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*e^2*x^2 + a*b
*d*e + (b^2*d*e + a*b*e^2)*x)) - 2*(48*B*b^4*e^4*x^3 - 9*B*b^4*d^3*e + 3*(3*B*a*b^3 + 8*A*b^4)*d^2*e^2 - (31*B
*a^2*b^2 - 64*A*a*b^3)*d*e^3 + 3*(5*B*a^3*b - 8*A*a^2*b^2)*e^4 + 8*(9*B*b^4*d*e^3 + (B*a*b^3 + 8*A*b^4)*e^4)*x
^2 + 2*(3*B*b^4*d^2*e^2 + 2*(5*B*a*b^3 + 28*A*b^4)*d*e^3 - (5*B*a^2*b^2 - 8*A*a*b^3)*e^4)*x)*sqrt(b*x + a)*sqr
t(e*x + d))/(b^4*e^3)]

________________________________________________________________________________________

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((B*x+A)*(e*x+d)**(3/2)*(b*x+a)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 2.3457, size = 1058, normalized size = 4.23 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*(20*(sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*e^(-2)/b^4 + (b*d*e - a*e^2)*e^(-4)
/b^4) + (b^2*d^2 - 2*a*b*d*e + a^2*e^2)*e^(-7/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x +
a)*b*e - a*b*e)))/b^(7/2))*A*d*abs(b)/b^2 + 10*(sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*(2*(b*x + a)*(4*(b*x + a)*
(6*(b*x + a)/b^2 + (b^7*d*e^5 - 17*a*b^6*e^6)*e^(-6)/b^8) - (5*b^8*d^2*e^4 + 6*a*b^7*d*e^5 - 59*a^2*b^6*e^6)*e
^(-6)/b^8) + 3*(5*b^9*d^3*e^3 + a*b^8*d^2*e^4 - a^2*b^7*d*e^5 - 5*a^3*b^6*e^6)*e^(-6)/b^8)*sqrt(b*x + a) + 3*(
5*b^4*d^4 - 4*a*b^3*d^3*e - 2*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + 5*a^4*e^4)*e^(-7/2)*log(abs(-sqrt(b*x + a)*sqr
t(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/b^(3/2))*B*abs(b)*e/b^2 + (sqrt(b^2*d + (b*x + a)*b*e - a
*b*e)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)*e^(-2)/b^6 + (b*d*e^3 - 7*a*e^4)*e^(-6)/b^6) - 3*(b^2*d^2*e^2 -
a^2*e^4)*e^(-6)/b^6) - 3*(b^3*d^3 - a*b^2*d^2*e - a^2*b*d*e^2 + a^3*e^3)*e^(-9/2)*log(abs(-sqrt(b*x + a)*sqrt(
b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/b^(11/2))*B*d*abs(b)/b^3 + (sqrt(b^2*d + (b*x + a)*b*e - a*
b*e)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)*e^(-2)/b^6 + (b*d*e^3 - 7*a*e^4)*e^(-6)/b^6) - 3*(b^2*d^2*e^2 - a
^2*e^4)*e^(-6)/b^6) - 3*(b^3*d^3 - a*b^2*d^2*e - a^2*b*d*e^2 + a^3*e^3)*e^(-9/2)*log(abs(-sqrt(b*x + a)*sqrt(b
)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/b^(11/2))*A*abs(b)*e/b^3)/b