3.2246 \(\int \frac{(A+B x) (d+e x)^{3/2}}{(a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=199 \[ \frac{\sqrt{a+b x} (d+e x)^{3/2} (-5 a B e+4 A b e+b B d)}{2 b^2 (b d-a e)}+\frac{3 \sqrt{a+b x} \sqrt{d+e x} (-5 a B e+4 A b e+b B d)}{4 b^3}+\frac{3 (b d-a e) (-5 a B e+4 A b e+b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{4 b^{7/2} \sqrt{e}}-\frac{2 (d+e x)^{5/2} (A b-a B)}{b \sqrt{a+b x} (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.150539, antiderivative size = 199, 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 = {78, 50, 63, 217, 206} \[ \frac{\sqrt{a+b x} (d+e x)^{3/2} (-5 a B e+4 A b e+b B d)}{2 b^2 (b d-a e)}+\frac{3 \sqrt{a+b x} \sqrt{d+e x} (-5 a B e+4 A b e+b B d)}{4 b^3}+\frac{3 (b d-a e) (-5 a B e+4 A b e+b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{4 b^{7/2} \sqrt{e}}-\frac{2 (d+e x)^{5/2} (A b-a B)}{b \sqrt{a+b x} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.52072, size = 161, normalized size = 0.81 \[ \frac{\sqrt{d+e x} \left (\frac{B \left (-15 a^2 e+a b (13 d-5 e x)+b^2 x (5 d+2 e x)\right )+4 A b (3 a e-2 b d+b e x)}{\sqrt{a+b x}}+\frac{3 \sqrt{b d-a e} (-5 a B e+4 A b e+b B d) \sinh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b d-a e}}\right )}{\sqrt{e} \sqrt{\frac{b (d+e x)}{b d-a e}}}\right )}{4 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*((4*A*b*(-2*b*d + 3*a*e + b*e*x) + B*(-15*a^2*e + a*b*(13*d - 5*e*x) + b^2*x*(5*d + 2*e*x)))/Sq
rt[a + b*x] + (3*Sqrt[b*d - a*e]*(b*B*d + 4*A*b*e - 5*a*B*e)*ArcSinh[(Sqrt[e]*Sqrt[a + b*x])/Sqrt[b*d - a*e]])
/(Sqrt[e]*Sqrt[(b*(d + e*x))/(b*d - a*e)])))/(4*b^3)

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Maple [B]  time = 0.022, size = 740, normalized size = 3.7 \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)^(3/2),x)

[Out]

-1/8*(e*x+d)^(1/2)*(12*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*a*b^2*e
^2-12*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*e-15*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^2*b*e^2+18*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*e-3*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^2-4*B*x^2*b^2*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+12*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))*a^2*b*e^2-12*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))*a*b^2*d*e-8*A*x*b^2*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-15*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^3*e^2+18*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^2*b*d*e-3*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^2+10*B*x*a*b*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-10*B*x*b^2*d*((b*x+a)
*(e*x+d))^(1/2)*(b*e)^(1/2)-24*A*a*b*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+16*A*b^2*d*((b*x+a)*(e*x+d))^(1/2)*
(b*e)^(1/2)+30*B*a^2*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-26*B*a*b*d*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/((b
*x+a)*(e*x+d))^(1/2)/(b*e)^(1/2)/(b*x+a)^(1/2)/b^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((B*x+A)*(e*x+d)^(3/2)/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.22691, size = 1297, normalized size = 6.52 \begin{align*} \left [\frac{3 \,{\left (B a b^{2} d^{2} - 2 \,{\left (3 \, B a^{2} b - 2 \, A a b^{2}\right )} d e +{\left (5 \, B a^{3} - 4 \, A a^{2} b\right )} e^{2} +{\left (B b^{3} d^{2} - 2 \,{\left (3 \, B a b^{2} - 2 \, A b^{3}\right )} d e +{\left (5 \, B a^{2} b - 4 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt{b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \,{\left (2 \, b e x + b d + a e\right )} \sqrt{b e} \sqrt{b x + a} \sqrt{e x + d} + 8 \,{\left (b^{2} d e + a b e^{2}\right )} x\right ) + 4 \,{\left (2 \, B b^{3} e^{2} x^{2} +{\left (13 \, B a b^{2} - 8 \, A b^{3}\right )} d e - 3 \,{\left (5 \, B a^{2} b - 4 \, A a b^{2}\right )} e^{2} +{\left (5 \, B b^{3} d e -{\left (5 \, B a b^{2} - 4 \, A b^{3}\right )} e^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{16 \,{\left (b^{5} e x + a b^{4} e\right )}}, -\frac{3 \,{\left (B a b^{2} d^{2} - 2 \,{\left (3 \, B a^{2} b - 2 \, A a b^{2}\right )} d e +{\left (5 \, B a^{3} - 4 \, A a^{2} b\right )} e^{2} +{\left (B b^{3} d^{2} - 2 \,{\left (3 \, B a b^{2} - 2 \, A b^{3}\right )} d e +{\left (5 \, B a^{2} b - 4 \, A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt{-b e} \arctan \left (\frac{{\left (2 \, b e x + b d + a e\right )} \sqrt{-b e} \sqrt{b x + a} \sqrt{e x + d}}{2 \,{\left (b^{2} e^{2} x^{2} + a b d e +{\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) - 2 \,{\left (2 \, B b^{3} e^{2} x^{2} +{\left (13 \, B a b^{2} - 8 \, A b^{3}\right )} d e - 3 \,{\left (5 \, B a^{2} b - 4 \, A a b^{2}\right )} e^{2} +{\left (5 \, B b^{3} d e -{\left (5 \, B a b^{2} - 4 \, A b^{3}\right )} e^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{8 \,{\left (b^{5} e x + a b^{4} e\right )}}\right ] \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)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(3*(B*a*b^2*d^2 - 2*(3*B*a^2*b - 2*A*a*b^2)*d*e + (5*B*a^3 - 4*A*a^2*b)*e^2 + (B*b^3*d^2 - 2*(3*B*a*b^2
- 2*A*b^3)*d*e + (5*B*a^2*b - 4*A*a*b^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*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*(2*B*b^3*e^2*x^2
+ (13*B*a*b^2 - 8*A*b^3)*d*e - 3*(5*B*a^2*b - 4*A*a*b^2)*e^2 + (5*B*b^3*d*e - (5*B*a*b^2 - 4*A*b^3)*e^2)*x)*s
qrt(b*x + a)*sqrt(e*x + d))/(b^5*e*x + a*b^4*e), -1/8*(3*(B*a*b^2*d^2 - 2*(3*B*a^2*b - 2*A*a*b^2)*d*e + (5*B*a
^3 - 4*A*a^2*b)*e^2 + (B*b^3*d^2 - 2*(3*B*a*b^2 - 2*A*b^3)*d*e + (5*B*a^2*b - 4*A*a*b^2)*e^2)*x)*sqrt(-b*e)*ar
ctan(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*(2*B*b^3*e^2*x^2 + (13*B*a*b^2 - 8*A*b^3)*d*e - 3*(5*B*a^2*b - 4*A*a*b^2)*e^2 + (5*B*b^3*d*e - (5
*B*a*b^2 - 4*A*b^3)*e^2)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b^5*e*x + a*b^4*e)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (d + e x\right )^{\frac{3}{2}}}{\left (a + b x\right )^{\frac{3}{2}}}\, dx \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)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.95466, size = 455, normalized size = 2.29 \begin{align*} \frac{1}{4} \, \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \sqrt{b x + a}{\left (\frac{2 \,{\left (b x + a\right )} B{\left | b \right |} e}{b^{5}} + \frac{{\left (5 \, B b^{10} d{\left | b \right |} e^{2} - 9 \, B a b^{9}{\left | b \right |} e^{3} + 4 \, A b^{10}{\left | b \right |} e^{3}\right )} e^{\left (-2\right )}}{b^{14}}\right )} - \frac{3 \,{\left (B b^{\frac{5}{2}} d^{2}{\left | b \right |} e^{\frac{1}{2}} - 6 \, B a b^{\frac{3}{2}} d{\left | b \right |} e^{\frac{3}{2}} + 4 \, A b^{\frac{5}{2}} d{\left | b \right |} e^{\frac{3}{2}} + 5 \, B a^{2} \sqrt{b}{\left | b \right |} e^{\frac{5}{2}} - 4 \, A a b^{\frac{3}{2}}{\left | b \right |} e^{\frac{5}{2}}\right )} e^{\left (-1\right )} \log \left ({\left (\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} - \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e}\right )}^{2}\right )}{8 \, b^{5}} + \frac{4 \,{\left (B a b^{\frac{5}{2}} d^{2}{\left | b \right |} e^{\frac{1}{2}} - A b^{\frac{7}{2}} d^{2}{\left | b \right |} e^{\frac{1}{2}} - 2 \, B a^{2} b^{\frac{3}{2}} d{\left | b \right |} e^{\frac{3}{2}} + 2 \, A a b^{\frac{5}{2}} d{\left | b \right |} e^{\frac{3}{2}} + B a^{3} \sqrt{b}{\left | b \right |} e^{\frac{5}{2}} - A a^{2} b^{\frac{3}{2}}{\left | b \right |} e^{\frac{5}{2}}\right )}}{{\left (b^{2} d - a b e -{\left (\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} - \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e}\right )}^{2}\right )} b^{4}} \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)^(3/2),x, algorithm="giac")

[Out]

1/4*sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*B*abs(b)*e/b^5 + (5*B*b^10*d*abs(b)*e^2 - 9
*B*a*b^9*abs(b)*e^3 + 4*A*b^10*abs(b)*e^3)*e^(-2)/b^14) - 3/8*(B*b^(5/2)*d^2*abs(b)*e^(1/2) - 6*B*a*b^(3/2)*d*
abs(b)*e^(3/2) + 4*A*b^(5/2)*d*abs(b)*e^(3/2) + 5*B*a^2*sqrt(b)*abs(b)*e^(5/2) - 4*A*a*b^(3/2)*abs(b)*e^(5/2))
*e^(-1)*log((sqrt(b*x + a)*sqrt(b)*e^(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2)/b^5 + 4*(B*a*b^(5/2)*d^2*
abs(b)*e^(1/2) - A*b^(7/2)*d^2*abs(b)*e^(1/2) - 2*B*a^2*b^(3/2)*d*abs(b)*e^(3/2) + 2*A*a*b^(5/2)*d*abs(b)*e^(3
/2) + B*a^3*sqrt(b)*abs(b)*e^(5/2) - A*a^2*b^(3/2)*abs(b)*e^(5/2))/((b^2*d - a*b*e - (sqrt(b*x + a)*sqrt(b)*e^
(1/2) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2)*b^4)