3.2229 \(\int \frac{(a+b x)^{5/2} (A+B x)}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=256 \[ \frac{b^2 \sqrt{a+b x} \sqrt{d+e x} (-5 a B e-2 A b e+7 b B d)}{e^4 (b d-a e)}-\frac{b^{3/2} (-5 a B e-2 A b e+7 b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{e^{9/2}}-\frac{2 (a+b x)^{5/2} (-5 a B e-2 A b e+7 b B d)}{15 e^2 (d+e x)^{3/2} (b d-a e)}-\frac{2 b (a+b x)^{3/2} (-5 a B e-2 A b e+7 b B d)}{3 e^3 \sqrt{d+e x} (b d-a e)}-\frac{2 (a+b x)^{7/2} (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)} \]

[Out]

(-2*(B*d - A*e)*(a + b*x)^(7/2))/(5*e*(b*d - a*e)*(d + e*x)^(5/2)) - (2*(7*b*B*d - 2*A*b*e - 5*a*B*e)*(a + b*x
)^(5/2))/(15*e^2*(b*d - a*e)*(d + e*x)^(3/2)) - (2*b*(7*b*B*d - 2*A*b*e - 5*a*B*e)*(a + b*x)^(3/2))/(3*e^3*(b*
d - a*e)*Sqrt[d + e*x]) + (b^2*(7*b*B*d - 2*A*b*e - 5*a*B*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/(e^4*(b*d - a*e)) -
(b^(3/2)*(7*b*B*d - 2*A*b*e - 5*a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/e^(9/2)

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Rubi [A]  time = 0.195276, antiderivative size = 256, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {78, 47, 50, 63, 217, 206} \[ \frac{b^2 \sqrt{a+b x} \sqrt{d+e x} (-5 a B e-2 A b e+7 b B d)}{e^4 (b d-a e)}-\frac{b^{3/2} (-5 a B e-2 A b e+7 b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{e^{9/2}}-\frac{2 (a+b x)^{5/2} (-5 a B e-2 A b e+7 b B d)}{15 e^2 (d+e x)^{3/2} (b d-a e)}-\frac{2 b (a+b x)^{3/2} (-5 a B e-2 A b e+7 b B d)}{3 e^3 \sqrt{d+e x} (b d-a e)}-\frac{2 (a+b x)^{7/2} (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/(d + e*x)^(7/2),x]

[Out]

(-2*(B*d - A*e)*(a + b*x)^(7/2))/(5*e*(b*d - a*e)*(d + e*x)^(5/2)) - (2*(7*b*B*d - 2*A*b*e - 5*a*B*e)*(a + b*x
)^(5/2))/(15*e^2*(b*d - a*e)*(d + e*x)^(3/2)) - (2*b*(7*b*B*d - 2*A*b*e - 5*a*B*e)*(a + b*x)^(3/2))/(3*e^3*(b*
d - a*e)*Sqrt[d + e*x]) + (b^2*(7*b*B*d - 2*A*b*e - 5*a*B*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/(e^4*(b*d - a*e)) -
(b^(3/2)*(7*b*B*d - 2*A*b*e - 5*a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/e^(9/2)

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 47

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

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 \frac{(a+b x)^{5/2} (A+B x)}{(d+e x)^{7/2}} \, dx &=-\frac{2 (B d-A e) (a+b x)^{7/2}}{5 e (b d-a e) (d+e x)^{5/2}}+\frac{(7 b B d-2 A b e-5 a B e) \int \frac{(a+b x)^{5/2}}{(d+e x)^{5/2}} \, dx}{5 e (b d-a e)}\\ &=-\frac{2 (B d-A e) (a+b x)^{7/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac{2 (7 b B d-2 A b e-5 a B e) (a+b x)^{5/2}}{15 e^2 (b d-a e) (d+e x)^{3/2}}+\frac{(b (7 b B d-2 A b e-5 a B e)) \int \frac{(a+b x)^{3/2}}{(d+e x)^{3/2}} \, dx}{3 e^2 (b d-a e)}\\ &=-\frac{2 (B d-A e) (a+b x)^{7/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac{2 (7 b B d-2 A b e-5 a B e) (a+b x)^{5/2}}{15 e^2 (b d-a e) (d+e x)^{3/2}}-\frac{2 b (7 b B d-2 A b e-5 a B e) (a+b x)^{3/2}}{3 e^3 (b d-a e) \sqrt{d+e x}}+\frac{\left (b^2 (7 b B d-2 A b e-5 a B e)\right ) \int \frac{\sqrt{a+b x}}{\sqrt{d+e x}} \, dx}{e^3 (b d-a e)}\\ &=-\frac{2 (B d-A e) (a+b x)^{7/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac{2 (7 b B d-2 A b e-5 a B e) (a+b x)^{5/2}}{15 e^2 (b d-a e) (d+e x)^{3/2}}-\frac{2 b (7 b B d-2 A b e-5 a B e) (a+b x)^{3/2}}{3 e^3 (b d-a e) \sqrt{d+e x}}+\frac{b^2 (7 b B d-2 A b e-5 a B e) \sqrt{a+b x} \sqrt{d+e x}}{e^4 (b d-a e)}-\frac{\left (b^2 (7 b B d-2 A b e-5 a B e)\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{d+e x}} \, dx}{2 e^4}\\ &=-\frac{2 (B d-A e) (a+b x)^{7/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac{2 (7 b B d-2 A b e-5 a B e) (a+b x)^{5/2}}{15 e^2 (b d-a e) (d+e x)^{3/2}}-\frac{2 b (7 b B d-2 A b e-5 a B e) (a+b x)^{3/2}}{3 e^3 (b d-a e) \sqrt{d+e x}}+\frac{b^2 (7 b B d-2 A b e-5 a B e) \sqrt{a+b x} \sqrt{d+e x}}{e^4 (b d-a e)}-\frac{(b (7 b B d-2 A b e-5 a B e)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d-\frac{a e}{b}+\frac{e x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{e^4}\\ &=-\frac{2 (B d-A e) (a+b x)^{7/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac{2 (7 b B d-2 A b e-5 a B e) (a+b x)^{5/2}}{15 e^2 (b d-a e) (d+e x)^{3/2}}-\frac{2 b (7 b B d-2 A b e-5 a B e) (a+b x)^{3/2}}{3 e^3 (b d-a e) \sqrt{d+e x}}+\frac{b^2 (7 b B d-2 A b e-5 a B e) \sqrt{a+b x} \sqrt{d+e x}}{e^4 (b d-a e)}-\frac{(b (7 b B d-2 A b e-5 a B e)) \operatorname{Subst}\left (\int \frac{1}{1-\frac{e x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{d+e x}}\right )}{e^4}\\ &=-\frac{2 (B d-A e) (a+b x)^{7/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac{2 (7 b B d-2 A b e-5 a B e) (a+b x)^{5/2}}{15 e^2 (b d-a e) (d+e x)^{3/2}}-\frac{2 b (7 b B d-2 A b e-5 a B e) (a+b x)^{3/2}}{3 e^3 (b d-a e) \sqrt{d+e x}}+\frac{b^2 (7 b B d-2 A b e-5 a B e) \sqrt{a+b x} \sqrt{d+e x}}{e^4 (b d-a e)}-\frac{b^{3/2} (7 b B d-2 A b e-5 a B e) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{e^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.187546, size = 113, normalized size = 0.44 \[ \frac{2 (a+b x)^{7/2} \left (-\frac{\left (\frac{b (d+e x)}{b d-a e}\right )^{5/2} (-5 a B e-2 A b e+7 b B d) \, _2F_1\left (\frac{5}{2},\frac{7}{2};\frac{9}{2};\frac{e (a+b x)}{a e-b d}\right )}{b}-7 A e+7 B d\right )}{35 e (d+e x)^{5/2} (a e-b d)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/(d + e*x)^(7/2),x]

[Out]

(2*(a + b*x)^(7/2)*(7*B*d - 7*A*e - ((7*b*B*d - 2*A*b*e - 5*a*B*e)*((b*(d + e*x))/(b*d - a*e))^(5/2)*Hypergeom
etric2F1[5/2, 7/2, 9/2, (e*(a + b*x))/(-(b*d) + a*e)])/b))/(35*e*(-(b*d) + a*e)*(d + e*x)^(5/2))

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Maple [B]  time = 0.023, size = 1092, 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((b*x+a)^(5/2)*(B*x+A)/(e*x+d)^(7/2),x)

[Out]

1/30*(75*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x^3*a*b^2*e^4-105*B*ln(
1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x^3*b^3*d*e^3-315*B*ln(1/2*(2*b*x*e+2
*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x^2*b^3*d^2*e^2-92*A*x^2*b^2*e^3*((b*x+a)*(e*x+d))^
(1/2)*(b*e)^(1/2)-315*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x*b^3*d^3*
e+75*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b^2*d^3*e-20*B*x*a^2*e^3*
((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-8*B*a^2*d*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-140*B*x^2*a*b*e^3*((b*x+
a)*(e*x+d))^(1/2)*(b*e)^(1/2)+322*B*x^2*b^2*d*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-44*A*x*a*b*e^3*((b*x+a)*
(e*x+d))^(1/2)*(b*e)^(1/2)+490*B*x*b^2*d^2*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+30*B*x^3*b^2*e^3*(b*e)^(1/2)*
((b*x+a)*(e*x+d))^(1/2)+30*A*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^3*d
^3*e+30*A*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x^3*b^3*e^4-12*A*a^2*e^3
*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-140*A*x*b^2*d*e^2*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)-20*A*a*b*d*e^2*(b*e
)^(1/2)*((b*x+a)*(e*x+d))^(1/2)-60*A*b^2*d^2*e*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)+90*A*ln(1/2*(2*b*x*e+2*((b*
x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x*b^3*d^2*e^2-80*B*a*b*d^2*e*((b*x+a)*(e*x+d))^(1/2)*(b*
e)^(1/2)+90*A*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x^2*b^3*d*e^3-196*B*
x*a*b*d*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-105*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*
e+b*d)/(b*e)^(1/2))*b^3*d^4+210*B*b^2*d^3*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+225*B*ln(1/2*(2*b*x*e+2*((b*x+a)
*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x^2*a*b^2*d*e^3+225*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1
/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x*a*b^2*d^2*e^2)*(b*x+a)^(1/2)/((b*x+a)*(e*x+d))^(1/2)/(b*e)^(1/2)/(e*x+
d)^(5/2)/e^4

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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((b*x+a)^(5/2)*(B*x+A)/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 40.6799, size = 1825, normalized size = 7.13 \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)^(5/2)*(B*x+A)/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

[-1/60*(15*(7*B*b^2*d^4 - (5*B*a*b + 2*A*b^2)*d^3*e + (7*B*b^2*d*e^3 - (5*B*a*b + 2*A*b^2)*e^4)*x^3 + 3*(7*B*b
^2*d^2*e^2 - (5*B*a*b + 2*A*b^2)*d*e^3)*x^2 + 3*(7*B*b^2*d^3*e - (5*B*a*b + 2*A*b^2)*d^2*e^2)*x)*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^2*x + b*d*e + a*e^2)*sqrt(b*x + a)*sqrt(e*x + d)*sqr
t(b/e) + 8*(b^2*d*e + a*b*e^2)*x) - 4*(15*B*b^2*e^3*x^3 + 105*B*b^2*d^3 - 6*A*a^2*e^3 - 10*(4*B*a*b + 3*A*b^2)
*d^2*e - 2*(2*B*a^2 + 5*A*a*b)*d*e^2 + (161*B*b^2*d*e^2 - 2*(35*B*a*b + 23*A*b^2)*e^3)*x^2 + (245*B*b^2*d^2*e
- 14*(7*B*a*b + 5*A*b^2)*d*e^2 - 2*(5*B*a^2 + 11*A*a*b)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(e^7*x^3 + 3*d*e^
6*x^2 + 3*d^2*e^5*x + d^3*e^4), 1/30*(15*(7*B*b^2*d^4 - (5*B*a*b + 2*A*b^2)*d^3*e + (7*B*b^2*d*e^3 - (5*B*a*b
+ 2*A*b^2)*e^4)*x^3 + 3*(7*B*b^2*d^2*e^2 - (5*B*a*b + 2*A*b^2)*d*e^3)*x^2 + 3*(7*B*b^2*d^3*e - (5*B*a*b + 2*A*
b^2)*d^2*e^2)*x)*sqrt(-b/e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(b*x + a)*sqrt(e*x + d)*sqrt(-b/e)/(b^2*e*x^2
+ a*b*d + (b^2*d + a*b*e)*x)) + 2*(15*B*b^2*e^3*x^3 + 105*B*b^2*d^3 - 6*A*a^2*e^3 - 10*(4*B*a*b + 3*A*b^2)*d^
2*e - 2*(2*B*a^2 + 5*A*a*b)*d*e^2 + (161*B*b^2*d*e^2 - 2*(35*B*a*b + 23*A*b^2)*e^3)*x^2 + (245*B*b^2*d^2*e - 1
4*(7*B*a*b + 5*A*b^2)*d*e^2 - 2*(5*B*a^2 + 11*A*a*b)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(e^7*x^3 + 3*d*e^6*x
^2 + 3*d^2*e^5*x + d^3*e^4)]

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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((b*x+a)**(5/2)*(B*x+A)/(e*x+d)**(7/2),x)

[Out]

Timed out

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Giac [B]  time = 2.3631, size = 902, normalized size = 3.52 \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)^(5/2)*(B*x+A)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

(7*B*b^2*d*abs(b) - 5*B*a*b*abs(b)*e - 2*A*b^2*abs(b)*e)*e^(-9/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqr
t(b^2*d + (b*x + a)*b*e - a*b*e)))/sqrt(b) + 1/15*(((b*x + a)*(15*(B*b^9*d^2*abs(b)*e^6 - 2*B*a*b^8*d*abs(b)*e
^7 + B*a^2*b^7*abs(b)*e^8)*(b*x + a)/(b^6*d^2*e^7 - 2*a*b^5*d*e^8 + a^2*b^4*e^9) + 23*(7*B*b^10*d^3*abs(b)*e^5
- 19*B*a*b^9*d^2*abs(b)*e^6 - 2*A*b^10*d^2*abs(b)*e^6 + 17*B*a^2*b^8*d*abs(b)*e^7 + 4*A*a*b^9*d*abs(b)*e^7 -
5*B*a^3*b^7*abs(b)*e^8 - 2*A*a^2*b^8*abs(b)*e^8)/(b^6*d^2*e^7 - 2*a*b^5*d*e^8 + a^2*b^4*e^9)) + 35*(7*B*b^11*d
^4*abs(b)*e^4 - 26*B*a*b^10*d^3*abs(b)*e^5 - 2*A*b^11*d^3*abs(b)*e^5 + 36*B*a^2*b^9*d^2*abs(b)*e^6 + 6*A*a*b^1
0*d^2*abs(b)*e^6 - 22*B*a^3*b^8*d*abs(b)*e^7 - 6*A*a^2*b^9*d*abs(b)*e^7 + 5*B*a^4*b^7*abs(b)*e^8 + 2*A*a^3*b^8
*abs(b)*e^8)/(b^6*d^2*e^7 - 2*a*b^5*d*e^8 + a^2*b^4*e^9))*(b*x + a) + 15*(7*B*b^12*d^5*abs(b)*e^3 - 33*B*a*b^1
1*d^4*abs(b)*e^4 - 2*A*b^12*d^4*abs(b)*e^4 + 62*B*a^2*b^10*d^3*abs(b)*e^5 + 8*A*a*b^11*d^3*abs(b)*e^5 - 58*B*a
^3*b^9*d^2*abs(b)*e^6 - 12*A*a^2*b^10*d^2*abs(b)*e^6 + 27*B*a^4*b^8*d*abs(b)*e^7 + 8*A*a^3*b^9*d*abs(b)*e^7 -
5*B*a^5*b^7*abs(b)*e^8 - 2*A*a^4*b^8*abs(b)*e^8)/(b^6*d^2*e^7 - 2*a*b^5*d*e^8 + a^2*b^4*e^9))*sqrt(b*x + a)/(b
^2*d + (b*x + a)*b*e - a*b*e)^(5/2)