3.2084 \(\int \frac{(a+b x) (d+e x)^{9/2}}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=171 \[ -\frac{21 e^2 (d+e x)^{5/2}}{32 b^3 (a+b x)^2}-\frac{105 e^3 (d+e x)^{3/2}}{64 b^4 (a+b x)}-\frac{315 e^4 \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{11/2}}-\frac{3 e (d+e x)^{7/2}}{8 b^2 (a+b x)^3}-\frac{(d+e x)^{9/2}}{4 b (a+b x)^4}+\frac{315 e^4 \sqrt{d+e x}}{64 b^5} \]

[Out]

(315*e^4*Sqrt[d + e*x])/(64*b^5) - (105*e^3*(d + e*x)^(3/2))/(64*b^4*(a + b*x)) - (21*e^2*(d + e*x)^(5/2))/(32
*b^3*(a + b*x)^2) - (3*e*(d + e*x)^(7/2))/(8*b^2*(a + b*x)^3) - (d + e*x)^(9/2)/(4*b*(a + b*x)^4) - (315*e^4*S
qrt[b*d - a*e]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(64*b^(11/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0814466, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {27, 47, 50, 63, 208} \[ -\frac{21 e^2 (d+e x)^{5/2}}{32 b^3 (a+b x)^2}-\frac{105 e^3 (d+e x)^{3/2}}{64 b^4 (a+b x)}-\frac{315 e^4 \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{11/2}}-\frac{3 e (d+e x)^{7/2}}{8 b^2 (a+b x)^3}-\frac{(d+e x)^{9/2}}{4 b (a+b x)^4}+\frac{315 e^4 \sqrt{d+e x}}{64 b^5} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(d + e*x)^(9/2))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

(315*e^4*Sqrt[d + e*x])/(64*b^5) - (105*e^3*(d + e*x)^(3/2))/(64*b^4*(a + b*x)) - (21*e^2*(d + e*x)^(5/2))/(32
*b^3*(a + b*x)^2) - (3*e*(d + e*x)^(7/2))/(8*b^2*(a + b*x)^3) - (d + e*x)^(9/2)/(4*b*(a + b*x)^4) - (315*e^4*S
qrt[b*d - a*e]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(64*b^(11/2))

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 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 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) (d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac{(d+e x)^{9/2}}{(a+b x)^5} \, dx\\ &=-\frac{(d+e x)^{9/2}}{4 b (a+b x)^4}+\frac{(9 e) \int \frac{(d+e x)^{7/2}}{(a+b x)^4} \, dx}{8 b}\\ &=-\frac{3 e (d+e x)^{7/2}}{8 b^2 (a+b x)^3}-\frac{(d+e x)^{9/2}}{4 b (a+b x)^4}+\frac{\left (21 e^2\right ) \int \frac{(d+e x)^{5/2}}{(a+b x)^3} \, dx}{16 b^2}\\ &=-\frac{21 e^2 (d+e x)^{5/2}}{32 b^3 (a+b x)^2}-\frac{3 e (d+e x)^{7/2}}{8 b^2 (a+b x)^3}-\frac{(d+e x)^{9/2}}{4 b (a+b x)^4}+\frac{\left (105 e^3\right ) \int \frac{(d+e x)^{3/2}}{(a+b x)^2} \, dx}{64 b^3}\\ &=-\frac{105 e^3 (d+e x)^{3/2}}{64 b^4 (a+b x)}-\frac{21 e^2 (d+e x)^{5/2}}{32 b^3 (a+b x)^2}-\frac{3 e (d+e x)^{7/2}}{8 b^2 (a+b x)^3}-\frac{(d+e x)^{9/2}}{4 b (a+b x)^4}+\frac{\left (315 e^4\right ) \int \frac{\sqrt{d+e x}}{a+b x} \, dx}{128 b^4}\\ &=\frac{315 e^4 \sqrt{d+e x}}{64 b^5}-\frac{105 e^3 (d+e x)^{3/2}}{64 b^4 (a+b x)}-\frac{21 e^2 (d+e x)^{5/2}}{32 b^3 (a+b x)^2}-\frac{3 e (d+e x)^{7/2}}{8 b^2 (a+b x)^3}-\frac{(d+e x)^{9/2}}{4 b (a+b x)^4}+\frac{\left (315 e^4 (b d-a e)\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{128 b^5}\\ &=\frac{315 e^4 \sqrt{d+e x}}{64 b^5}-\frac{105 e^3 (d+e x)^{3/2}}{64 b^4 (a+b x)}-\frac{21 e^2 (d+e x)^{5/2}}{32 b^3 (a+b x)^2}-\frac{3 e (d+e x)^{7/2}}{8 b^2 (a+b x)^3}-\frac{(d+e x)^{9/2}}{4 b (a+b x)^4}+\frac{\left (315 e^3 (b d-a e)\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{64 b^5}\\ &=\frac{315 e^4 \sqrt{d+e x}}{64 b^5}-\frac{105 e^3 (d+e x)^{3/2}}{64 b^4 (a+b x)}-\frac{21 e^2 (d+e x)^{5/2}}{32 b^3 (a+b x)^2}-\frac{3 e (d+e x)^{7/2}}{8 b^2 (a+b x)^3}-\frac{(d+e x)^{9/2}}{4 b (a+b x)^4}-\frac{315 e^4 \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0203705, size = 52, normalized size = 0.3 \[ \frac{2 e^4 (d+e x)^{11/2} \, _2F_1\left (5,\frac{11}{2};\frac{13}{2};-\frac{b (d+e x)}{a e-b d}\right )}{11 (a e-b d)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(d + e*x)^(9/2))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

(2*e^4*(d + e*x)^(11/2)*Hypergeometric2F1[5, 11/2, 13/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(11*(-(b*d) + a*e)^
5)

________________________________________________________________________________________

Maple [B]  time = 0.02, size = 497, normalized size = 2.9 \begin{align*} 2\,{\frac{{e}^{4}\sqrt{ex+d}}{{b}^{5}}}+{\frac{325\,{e}^{5}a}{64\,{b}^{2} \left ( bex+ae \right ) ^{4}} \left ( ex+d \right ) ^{{\frac{7}{2}}}}-{\frac{325\,{e}^{4}d}{64\,b \left ( bex+ae \right ) ^{4}} \left ( ex+d \right ) ^{{\frac{7}{2}}}}+{\frac{765\,{e}^{6}{a}^{2}}{64\,{b}^{3} \left ( bex+ae \right ) ^{4}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{765\,{e}^{5}ad}{32\,{b}^{2} \left ( bex+ae \right ) ^{4}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}+{\frac{765\,{e}^{4}{d}^{2}}{64\,b \left ( bex+ae \right ) ^{4}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}+{\frac{643\,{e}^{7}{a}^{3}}{64\,{b}^{4} \left ( bex+ae \right ) ^{4}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{1929\,{e}^{6}{a}^{2}d}{64\,{b}^{3} \left ( bex+ae \right ) ^{4}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{1929\,{e}^{5}a{d}^{2}}{64\,{b}^{2} \left ( bex+ae \right ) ^{4}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{643\,{e}^{4}{d}^{3}}{64\,b \left ( bex+ae \right ) ^{4}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{187\,{e}^{8}{a}^{4}}{64\,{b}^{5} \left ( bex+ae \right ) ^{4}}\sqrt{ex+d}}-{\frac{187\,{e}^{7}d{a}^{3}}{16\,{b}^{4} \left ( bex+ae \right ) ^{4}}\sqrt{ex+d}}+{\frac{561\,{e}^{6}{a}^{2}{d}^{2}}{32\,{b}^{3} \left ( bex+ae \right ) ^{4}}\sqrt{ex+d}}-{\frac{187\,{e}^{5}a{d}^{3}}{16\,{b}^{2} \left ( bex+ae \right ) ^{4}}\sqrt{ex+d}}+{\frac{187\,{e}^{4}{d}^{4}}{64\,b \left ( bex+ae \right ) ^{4}}\sqrt{ex+d}}-{\frac{315\,{e}^{5}a}{64\,{b}^{5}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}}+{\frac{315\,{e}^{4}d}{64\,{b}^{4}}\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)*(e*x+d)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^3,x)

[Out]

2*e^4*(e*x+d)^(1/2)/b^5+325/64*e^5/b^2/(b*e*x+a*e)^4*(e*x+d)^(7/2)*a-325/64*e^4/b/(b*e*x+a*e)^4*(e*x+d)^(7/2)*
d+765/64*e^6/b^3/(b*e*x+a*e)^4*(e*x+d)^(5/2)*a^2-765/32*e^5/b^2/(b*e*x+a*e)^4*(e*x+d)^(5/2)*a*d+765/64*e^4/b/(
b*e*x+a*e)^4*(e*x+d)^(5/2)*d^2+643/64*e^7/b^4/(b*e*x+a*e)^4*(e*x+d)^(3/2)*a^3-1929/64*e^6/b^3/(b*e*x+a*e)^4*(e
*x+d)^(3/2)*a^2*d+1929/64*e^5/b^2/(b*e*x+a*e)^4*(e*x+d)^(3/2)*a*d^2-643/64*e^4/b/(b*e*x+a*e)^4*(e*x+d)^(3/2)*d
^3+187/64*e^8/b^5/(b*e*x+a*e)^4*(e*x+d)^(1/2)*a^4-187/16*e^7/b^4/(b*e*x+a*e)^4*(e*x+d)^(1/2)*d*a^3+561/32*e^6/
b^3/(b*e*x+a*e)^4*(e*x+d)^(1/2)*a^2*d^2-187/16*e^5/b^2/(b*e*x+a*e)^4*(e*x+d)^(1/2)*a*d^3+187/64*e^4/b/(b*e*x+a
*e)^4*(e*x+d)^(1/2)*d^4-315/64*e^5/b^5/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a+315/6
4*e^4/b^4/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*d

________________________________________________________________________________________

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)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.07907, size = 1453, normalized size = 8.5 \begin{align*} \left [\frac{315 \,{\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \sqrt{\frac{b d - a e}{b}} \log \left (\frac{b e x + 2 \, b d - a e - 2 \, \sqrt{e x + d} b \sqrt{\frac{b d - a e}{b}}}{b x + a}\right ) + 2 \,{\left (128 \, b^{4} e^{4} x^{4} - 16 \, b^{4} d^{4} - 24 \, a b^{3} d^{3} e - 42 \, a^{2} b^{2} d^{2} e^{2} - 105 \, a^{3} b d e^{3} + 315 \, a^{4} e^{4} -{\left (325 \, b^{4} d e^{3} - 837 \, a b^{3} e^{4}\right )} x^{3} - 3 \,{\left (70 \, b^{4} d^{2} e^{2} + 185 \, a b^{3} d e^{3} - 511 \, a^{2} b^{2} e^{4}\right )} x^{2} -{\left (88 \, b^{4} d^{3} e + 156 \, a b^{3} d^{2} e^{2} + 399 \, a^{2} b^{2} d e^{3} - 1155 \, a^{3} b e^{4}\right )} x\right )} \sqrt{e x + d}}{128 \,{\left (b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}\right )}}, -\frac{315 \,{\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \sqrt{-\frac{b d - a e}{b}} \arctan \left (-\frac{\sqrt{e x + d} b \sqrt{-\frac{b d - a e}{b}}}{b d - a e}\right ) -{\left (128 \, b^{4} e^{4} x^{4} - 16 \, b^{4} d^{4} - 24 \, a b^{3} d^{3} e - 42 \, a^{2} b^{2} d^{2} e^{2} - 105 \, a^{3} b d e^{3} + 315 \, a^{4} e^{4} -{\left (325 \, b^{4} d e^{3} - 837 \, a b^{3} e^{4}\right )} x^{3} - 3 \,{\left (70 \, b^{4} d^{2} e^{2} + 185 \, a b^{3} d e^{3} - 511 \, a^{2} b^{2} e^{4}\right )} x^{2} -{\left (88 \, b^{4} d^{3} e + 156 \, a b^{3} d^{2} e^{2} + 399 \, a^{2} b^{2} d e^{3} - 1155 \, a^{3} b e^{4}\right )} x\right )} \sqrt{e x + d}}{64 \,{\left (b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")

[Out]

[1/128*(315*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*sqrt((b*d - a*e)/b)*
log((b*e*x + 2*b*d - a*e - 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) + 2*(128*b^4*e^4*x^4 - 16*b^4*d^4
- 24*a*b^3*d^3*e - 42*a^2*b^2*d^2*e^2 - 105*a^3*b*d*e^3 + 315*a^4*e^4 - (325*b^4*d*e^3 - 837*a*b^3*e^4)*x^3 -
3*(70*b^4*d^2*e^2 + 185*a*b^3*d*e^3 - 511*a^2*b^2*e^4)*x^2 - (88*b^4*d^3*e + 156*a*b^3*d^2*e^2 + 399*a^2*b^2*
d*e^3 - 1155*a^3*b*e^4)*x)*sqrt(e*x + d))/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*b^5), -1/
64*(315*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*sqrt(-(b*d - a*e)/b)*arc
tan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (128*b^4*e^4*x^4 - 16*b^4*d^4 - 24*a*b^3*d^3*e - 42*a
^2*b^2*d^2*e^2 - 105*a^3*b*d*e^3 + 315*a^4*e^4 - (325*b^4*d*e^3 - 837*a*b^3*e^4)*x^3 - 3*(70*b^4*d^2*e^2 + 185
*a*b^3*d*e^3 - 511*a^2*b^2*e^4)*x^2 - (88*b^4*d^3*e + 156*a*b^3*d^2*e^2 + 399*a^2*b^2*d*e^3 - 1155*a^3*b*e^4)*
x)*sqrt(e*x + d))/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*b^5)]

________________________________________________________________________________________

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)**(9/2)/(b**2*x**2+2*a*b*x+a**2)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.21711, size = 463, normalized size = 2.71 \begin{align*} \frac{315 \,{\left (b d e^{4} - a e^{5}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{64 \, \sqrt{-b^{2} d + a b e} b^{5}} + \frac{2 \, \sqrt{x e + d} e^{4}}{b^{5}} - \frac{325 \,{\left (x e + d\right )}^{\frac{7}{2}} b^{4} d e^{4} - 765 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{4} d^{2} e^{4} + 643 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{4} d^{3} e^{4} - 187 \, \sqrt{x e + d} b^{4} d^{4} e^{4} - 325 \,{\left (x e + d\right )}^{\frac{7}{2}} a b^{3} e^{5} + 1530 \,{\left (x e + d\right )}^{\frac{5}{2}} a b^{3} d e^{5} - 1929 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{3} d^{2} e^{5} + 748 \, \sqrt{x e + d} a b^{3} d^{3} e^{5} - 765 \,{\left (x e + d\right )}^{\frac{5}{2}} a^{2} b^{2} e^{6} + 1929 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b^{2} d e^{6} - 1122 \, \sqrt{x e + d} a^{2} b^{2} d^{2} e^{6} - 643 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{3} b e^{7} + 748 \, \sqrt{x e + d} a^{3} b d e^{7} - 187 \, \sqrt{x e + d} a^{4} e^{8}}{64 \,{\left ({\left (x e + d\right )} b - b d + a e\right )}^{4} b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")

[Out]

315/64*(b*d*e^4 - a*e^5)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^5) + 2*sqrt(x*e
+ d)*e^4/b^5 - 1/64*(325*(x*e + d)^(7/2)*b^4*d*e^4 - 765*(x*e + d)^(5/2)*b^4*d^2*e^4 + 643*(x*e + d)^(3/2)*b^4
*d^3*e^4 - 187*sqrt(x*e + d)*b^4*d^4*e^4 - 325*(x*e + d)^(7/2)*a*b^3*e^5 + 1530*(x*e + d)^(5/2)*a*b^3*d*e^5 -
1929*(x*e + d)^(3/2)*a*b^3*d^2*e^5 + 748*sqrt(x*e + d)*a*b^3*d^3*e^5 - 765*(x*e + d)^(5/2)*a^2*b^2*e^6 + 1929*
(x*e + d)^(3/2)*a^2*b^2*d*e^6 - 1122*sqrt(x*e + d)*a^2*b^2*d^2*e^6 - 643*(x*e + d)^(3/2)*a^3*b*e^7 + 748*sqrt(
x*e + d)*a^3*b*d*e^7 - 187*sqrt(x*e + d)*a^4*e^8)/(((x*e + d)*b - b*d + a*e)^4*b^5)