3.1821 \(\int \frac{A+B x}{\sqrt{d+e x} (a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=209 \[ -\frac{e^2 (-a B e-5 A b e+6 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{3/2} (b d-a e)^{7/2}}+\frac{e \sqrt{d+e x} (-a B e-5 A b e+6 b B d)}{8 b (a+b x) (b d-a e)^3}-\frac{\sqrt{d+e x} (-a B e-5 A b e+6 b B d)}{12 b (a+b x)^2 (b d-a e)^2}-\frac{\sqrt{d+e x} (A b-a B)}{3 b (a+b x)^3 (b d-a e)} \]

[Out]

-((A*b - a*B)*Sqrt[d + e*x])/(3*b*(b*d - a*e)*(a + b*x)^3) - ((6*b*B*d - 5*A*b*e - a*B*e)*Sqrt[d + e*x])/(12*b
*(b*d - a*e)^2*(a + b*x)^2) + (e*(6*b*B*d - 5*A*b*e - a*B*e)*Sqrt[d + e*x])/(8*b*(b*d - a*e)^3*(a + b*x)) - (e
^2*(6*b*B*d - 5*A*b*e - a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(3/2)*(b*d - a*e)^(7/2))

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Rubi [A]  time = 0.201498, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {27, 78, 51, 63, 208} \[ -\frac{e^2 (-a B e-5 A b e+6 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{3/2} (b d-a e)^{7/2}}+\frac{e \sqrt{d+e x} (-a B e-5 A b e+6 b B d)}{8 b (a+b x) (b d-a e)^3}-\frac{\sqrt{d+e x} (-a B e-5 A b e+6 b B d)}{12 b (a+b x)^2 (b d-a e)^2}-\frac{\sqrt{d+e x} (A b-a B)}{3 b (a+b x)^3 (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A*b - a*B)*Sqrt[d + e*x])/(3*b*(b*d - a*e)*(a + b*x)^3) - ((6*b*B*d - 5*A*b*e - a*B*e)*Sqrt[d + e*x])/(12*b
*(b*d - a*e)^2*(a + b*x)^2) + (e*(6*b*B*d - 5*A*b*e - a*B*e)*Sqrt[d + e*x])/(8*b*(b*d - a*e)^3*(a + b*x)) - (e
^2*(6*b*B*d - 5*A*b*e - a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(3/2)*(b*d - a*e)^(7/2))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 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}{\sqrt{d+e x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{A+B x}{(a+b x)^4 \sqrt{d+e x}} \, dx\\ &=-\frac{(A b-a B) \sqrt{d+e x}}{3 b (b d-a e) (a+b x)^3}+\frac{(6 b B d-5 A b e-a B e) \int \frac{1}{(a+b x)^3 \sqrt{d+e x}} \, dx}{6 b (b d-a e)}\\ &=-\frac{(A b-a B) \sqrt{d+e x}}{3 b (b d-a e) (a+b x)^3}-\frac{(6 b B d-5 A b e-a B e) \sqrt{d+e x}}{12 b (b d-a e)^2 (a+b x)^2}-\frac{(e (6 b B d-5 A b e-a B e)) \int \frac{1}{(a+b x)^2 \sqrt{d+e x}} \, dx}{8 b (b d-a e)^2}\\ &=-\frac{(A b-a B) \sqrt{d+e x}}{3 b (b d-a e) (a+b x)^3}-\frac{(6 b B d-5 A b e-a B e) \sqrt{d+e x}}{12 b (b d-a e)^2 (a+b x)^2}+\frac{e (6 b B d-5 A b e-a B e) \sqrt{d+e x}}{8 b (b d-a e)^3 (a+b x)}+\frac{\left (e^2 (6 b B d-5 A b e-a B e)\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{16 b (b d-a e)^3}\\ &=-\frac{(A b-a B) \sqrt{d+e x}}{3 b (b d-a e) (a+b x)^3}-\frac{(6 b B d-5 A b e-a B e) \sqrt{d+e x}}{12 b (b d-a e)^2 (a+b x)^2}+\frac{e (6 b B d-5 A b e-a B e) \sqrt{d+e x}}{8 b (b d-a e)^3 (a+b x)}+\frac{(e (6 b B d-5 A b e-a B e)) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{8 b (b d-a e)^3}\\ &=-\frac{(A b-a B) \sqrt{d+e x}}{3 b (b d-a e) (a+b x)^3}-\frac{(6 b B d-5 A b e-a B e) \sqrt{d+e x}}{12 b (b d-a e)^2 (a+b x)^2}+\frac{e (6 b B d-5 A b e-a B e) \sqrt{d+e x}}{8 b (b d-a e)^3 (a+b x)}-\frac{e^2 (6 b B d-5 A b e-a B e) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{3/2} (b d-a e)^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0637855, size = 97, normalized size = 0.46 \[ \frac{\sqrt{d+e x} \left (\frac{e^2 (a B e+5 A b e-6 b B d) \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};\frac{b (d+e x)}{b d-a e}\right )}{(b d-a e)^3}+\frac{a B-A b}{(a+b x)^3}\right )}{3 b (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*((-(A*b) + a*B)/(a + b*x)^3 + (e^2*(-6*b*B*d + 5*A*b*e + a*B*e)*Hypergeometric2F1[1/2, 3, 3/2,
(b*(d + e*x))/(b*d - a*e)])/(b*d - a*e)^3))/(3*b*(b*d - a*e))

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Maple [B]  time = 0.02, size = 679, normalized size = 3.3 \begin{align*}{\frac{5\,A{b}^{2}{e}^{3}}{8\, \left ( bex+ae \right ) ^{3} \left ({a}^{3}{e}^{3}-3\,{a}^{2}bd{e}^{2}+3\,a{b}^{2}{d}^{2}e-{b}^{3}{d}^{3} \right ) } \left ( ex+d \right ) ^{{\frac{5}{2}}}}+{\frac{ab{e}^{3}B}{8\, \left ( bex+ae \right ) ^{3} \left ({a}^{3}{e}^{3}-3\,{a}^{2}bd{e}^{2}+3\,a{b}^{2}{d}^{2}e-{b}^{3}{d}^{3} \right ) } \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{3\,B{e}^{2}{b}^{2}d}{4\, \left ( bex+ae \right ) ^{3} \left ({a}^{3}{e}^{3}-3\,{a}^{2}bd{e}^{2}+3\,a{b}^{2}{d}^{2}e-{b}^{3}{d}^{3} \right ) } \left ( ex+d \right ) ^{{\frac{5}{2}}}}+{\frac{5\,{e}^{3}Ab}{3\, \left ( bex+ae \right ) ^{3} \left ({a}^{2}{e}^{2}-2\,abde+{b}^{2}{d}^{2} \right ) } \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{{e}^{3}aB}{3\, \left ( bex+ae \right ) ^{3} \left ({a}^{2}{e}^{2}-2\,abde+{b}^{2}{d}^{2} \right ) } \left ( ex+d \right ) ^{{\frac{3}{2}}}}-2\,{\frac{{e}^{2} \left ( ex+d \right ) ^{3/2}Bbd}{ \left ( bex+ae \right ) ^{3} \left ({a}^{2}{e}^{2}-2\,abde+{b}^{2}{d}^{2} \right ) }}+{\frac{11\,{e}^{3}A}{8\, \left ( bex+ae \right ) ^{3} \left ( ae-bd \right ) }\sqrt{ex+d}}-{\frac{{e}^{3}aB}{8\, \left ( bex+ae \right ) ^{3}b \left ( ae-bd \right ) }\sqrt{ex+d}}-{\frac{5\,B{e}^{2}d}{4\, \left ( bex+ae \right ) ^{3} \left ( ae-bd \right ) }\sqrt{ex+d}}+{\frac{5\,{e}^{3}A}{8\,{a}^{3}{e}^{3}-24\,{a}^{2}bd{e}^{2}+24\,a{b}^{2}{d}^{2}e-8\,{b}^{3}{d}^{3}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}}+{\frac{{e}^{3}aB}{8\,b \left ({a}^{3}{e}^{3}-3\,{a}^{2}bd{e}^{2}+3\,a{b}^{2}{d}^{2}e-{b}^{3}{d}^{3} \right ) }\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}}-{\frac{3\,B{e}^{2}d}{4\,{a}^{3}{e}^{3}-12\,{a}^{2}bd{e}^{2}+12\,a{b}^{2}{d}^{2}e-4\,{b}^{3}{d}^{3}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/8*e^3/(b*e*x+a*e)^3*b^2/(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)*(e*x+d)^(5/2)*A+1/8*e^3/(b*e*x+a*e)^3*
b/(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)*(e*x+d)^(5/2)*a*B-3/4*e^2/(b*e*x+a*e)^3*b^2/(a^3*e^3-3*a^2*b*d
*e^2+3*a*b^2*d^2*e-b^3*d^3)*(e*x+d)^(5/2)*B*d+5/3*e^3/(b*e*x+a*e)^3/(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^(3/2)*
A*b+1/3*e^3/(b*e*x+a*e)^3/(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^(3/2)*a*B-2*e^2/(b*e*x+a*e)^3/(a^2*e^2-2*a*b*d*e
+b^2*d^2)*(e*x+d)^(3/2)*B*b*d+11/8*e^3/(b*e*x+a*e)^3/(a*e-b*d)*(e*x+d)^(1/2)*A-1/8*e^3/(b*e*x+a*e)^3/b/(a*e-b*
d)*(e*x+d)^(1/2)*a*B-5/4*e^2/(b*e*x+a*e)^3/(a*e-b*d)*(e*x+d)^(1/2)*B*d+5/8*e^3/(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*
d^2*e-b^3*d^3)/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*A+1/8*e^3/b/(a^3*e^3-3*a^2*b*d*
e^2+3*a*b^2*d^2*e-b^3*d^3)/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a*B-3/4*e^2/(a^3*e^
3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*B*d

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.45626, size = 2726, normalized size = 13.04 \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)/(b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

[1/48*(3*(6*B*a^3*b*d*e^2 - (B*a^4 + 5*A*a^3*b)*e^3 + (6*B*b^4*d*e^2 - (B*a*b^3 + 5*A*b^4)*e^3)*x^3 + 3*(6*B*a
*b^3*d*e^2 - (B*a^2*b^2 + 5*A*a*b^3)*e^3)*x^2 + 3*(6*B*a^2*b^2*d*e^2 - (B*a^3*b + 5*A*a^2*b^2)*e^3)*x)*sqrt(b^
2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(4*(B*a*b^4 + 2*A*
b^5)*d^3 - 2*(10*B*a^2*b^3 + 17*A*a*b^4)*d^2*e + (13*B*a^3*b^2 + 59*A*a^2*b^3)*d*e^2 + 3*(B*a^4*b - 11*A*a^3*b
^2)*e^3 - 3*(6*B*b^5*d^2*e - (7*B*a*b^4 + 5*A*b^5)*d*e^2 + (B*a^2*b^3 + 5*A*a*b^4)*e^3)*x^2 + 2*(6*B*b^5*d^3 -
(31*B*a*b^4 + 5*A*b^5)*d^2*e + (29*B*a^2*b^3 + 25*A*a*b^4)*d*e^2 - 4*(B*a^3*b^2 + 5*A*a^2*b^3)*e^3)*x)*sqrt(e
*x + d))/(a^3*b^6*d^4 - 4*a^4*b^5*d^3*e + 6*a^5*b^4*d^2*e^2 - 4*a^6*b^3*d*e^3 + a^7*b^2*e^4 + (b^9*d^4 - 4*a*b
^8*d^3*e + 6*a^2*b^7*d^2*e^2 - 4*a^3*b^6*d*e^3 + a^4*b^5*e^4)*x^3 + 3*(a*b^8*d^4 - 4*a^2*b^7*d^3*e + 6*a^3*b^6
*d^2*e^2 - 4*a^4*b^5*d*e^3 + a^5*b^4*e^4)*x^2 + 3*(a^2*b^7*d^4 - 4*a^3*b^6*d^3*e + 6*a^4*b^5*d^2*e^2 - 4*a^5*b
^4*d*e^3 + a^6*b^3*e^4)*x), 1/24*(3*(6*B*a^3*b*d*e^2 - (B*a^4 + 5*A*a^3*b)*e^3 + (6*B*b^4*d*e^2 - (B*a*b^3 + 5
*A*b^4)*e^3)*x^3 + 3*(6*B*a*b^3*d*e^2 - (B*a^2*b^2 + 5*A*a*b^3)*e^3)*x^2 + 3*(6*B*a^2*b^2*d*e^2 - (B*a^3*b + 5
*A*a^2*b^2)*e^3)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (4*(B*a*b^
4 + 2*A*b^5)*d^3 - 2*(10*B*a^2*b^3 + 17*A*a*b^4)*d^2*e + (13*B*a^3*b^2 + 59*A*a^2*b^3)*d*e^2 + 3*(B*a^4*b - 11
*A*a^3*b^2)*e^3 - 3*(6*B*b^5*d^2*e - (7*B*a*b^4 + 5*A*b^5)*d*e^2 + (B*a^2*b^3 + 5*A*a*b^4)*e^3)*x^2 + 2*(6*B*b
^5*d^3 - (31*B*a*b^4 + 5*A*b^5)*d^2*e + (29*B*a^2*b^3 + 25*A*a*b^4)*d*e^2 - 4*(B*a^3*b^2 + 5*A*a^2*b^3)*e^3)*x
)*sqrt(e*x + d))/(a^3*b^6*d^4 - 4*a^4*b^5*d^3*e + 6*a^5*b^4*d^2*e^2 - 4*a^6*b^3*d*e^3 + a^7*b^2*e^4 + (b^9*d^4
- 4*a*b^8*d^3*e + 6*a^2*b^7*d^2*e^2 - 4*a^3*b^6*d*e^3 + a^4*b^5*e^4)*x^3 + 3*(a*b^8*d^4 - 4*a^2*b^7*d^3*e + 6
*a^3*b^6*d^2*e^2 - 4*a^4*b^5*d*e^3 + a^5*b^4*e^4)*x^2 + 3*(a^2*b^7*d^4 - 4*a^3*b^6*d^3*e + 6*a^4*b^5*d^2*e^2 -
4*a^5*b^4*d*e^3 + a^6*b^3*e^4)*x)]

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

[Out]

Timed out

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Giac [B]  time = 1.183, size = 579, normalized size = 2.77 \begin{align*} \frac{{\left (6 \, B b d e^{2} - B a e^{3} - 5 \, A b e^{3}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{8 \,{\left (b^{4} d^{3} - 3 \, a b^{3} d^{2} e + 3 \, a^{2} b^{2} d e^{2} - a^{3} b e^{3}\right )} \sqrt{-b^{2} d + a b e}} + \frac{18 \,{\left (x e + d\right )}^{\frac{5}{2}} B b^{3} d e^{2} - 48 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{3} d^{2} e^{2} + 30 \, \sqrt{x e + d} B b^{3} d^{3} e^{2} - 3 \,{\left (x e + d\right )}^{\frac{5}{2}} B a b^{2} e^{3} - 15 \,{\left (x e + d\right )}^{\frac{5}{2}} A b^{3} e^{3} + 56 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b^{2} d e^{3} + 40 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{3} d e^{3} - 57 \, \sqrt{x e + d} B a b^{2} d^{2} e^{3} - 33 \, \sqrt{x e + d} A b^{3} d^{2} e^{3} - 8 \,{\left (x e + d\right )}^{\frac{3}{2}} B a^{2} b e^{4} - 40 \,{\left (x e + d\right )}^{\frac{3}{2}} A a b^{2} e^{4} + 24 \, \sqrt{x e + d} B a^{2} b d e^{4} + 66 \, \sqrt{x e + d} A a b^{2} d e^{4} + 3 \, \sqrt{x e + d} B a^{3} e^{5} - 33 \, \sqrt{x e + d} A a^{2} b e^{5}}{24 \,{\left (b^{4} d^{3} - 3 \, a b^{3} d^{2} e + 3 \, a^{2} b^{2} d e^{2} - a^{3} b e^{3}\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(6*B*b*d*e^2 - B*a*e^3 - 5*A*b*e^3)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^4*d^3 - 3*a*b^3*d^2*e
+ 3*a^2*b^2*d*e^2 - a^3*b*e^3)*sqrt(-b^2*d + a*b*e)) + 1/24*(18*(x*e + d)^(5/2)*B*b^3*d*e^2 - 48*(x*e + d)^(3
/2)*B*b^3*d^2*e^2 + 30*sqrt(x*e + d)*B*b^3*d^3*e^2 - 3*(x*e + d)^(5/2)*B*a*b^2*e^3 - 15*(x*e + d)^(5/2)*A*b^3*
e^3 + 56*(x*e + d)^(3/2)*B*a*b^2*d*e^3 + 40*(x*e + d)^(3/2)*A*b^3*d*e^3 - 57*sqrt(x*e + d)*B*a*b^2*d^2*e^3 - 3
3*sqrt(x*e + d)*A*b^3*d^2*e^3 - 8*(x*e + d)^(3/2)*B*a^2*b*e^4 - 40*(x*e + d)^(3/2)*A*a*b^2*e^4 + 24*sqrt(x*e +
d)*B*a^2*b*d*e^4 + 66*sqrt(x*e + d)*A*a*b^2*d*e^4 + 3*sqrt(x*e + d)*B*a^3*e^5 - 33*sqrt(x*e + d)*A*a^2*b*e^5)
/((b^4*d^3 - 3*a*b^3*d^2*e + 3*a^2*b^2*d*e^2 - a^3*b*e^3)*((x*e + d)*b - b*d + a*e)^3)