3.844 \(\int \frac{\sqrt{d+e x} (15 d^2+20 d e x+8 e^2 x^2)}{\sqrt{a+b x}} \, dx\)

Optimal. Leaf size=176 \[ \frac{\sqrt{a+b x} \sqrt{d+e x} \left (5 a^2 e^2-13 a b d e+11 b^2 d^2\right )}{b^3}+\frac{(b d-a e) \left (5 a^2 e^2-13 a b d e+11 b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{b^{7/2} \sqrt{e}}+\frac{8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac{2 \sqrt{a+b x} (d+e x)^{3/2} (4 b d-3 a e)}{b^2} \]

[Out]

((11*b^2*d^2 - 13*a*b*d*e + 5*a^2*e^2)*Sqrt[a + b*x]*Sqrt[d + e*x])/b^3 + (2*(4*b*d - 3*a*e)*Sqrt[a + b*x]*(d
+ e*x)^(3/2))/b^2 + (8*e*(a + b*x)^(3/2)*(d + e*x)^(3/2))/(3*b^2) + ((b*d - a*e)*(11*b^2*d^2 - 13*a*b*d*e + 5*
a^2*e^2)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(b^(7/2)*Sqrt[e])

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Rubi [A]  time = 0.171327, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {951, 80, 50, 63, 217, 206} \[ \frac{\sqrt{a+b x} \sqrt{d+e x} \left (5 a^2 e^2-13 a b d e+11 b^2 d^2\right )}{b^3}+\frac{(b d-a e) \left (5 a^2 e^2-13 a b d e+11 b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{b^{7/2} \sqrt{e}}+\frac{8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac{2 \sqrt{a+b x} (d+e x)^{3/2} (4 b d-3 a e)}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[d + e*x]*(15*d^2 + 20*d*e*x + 8*e^2*x^2))/Sqrt[a + b*x],x]

[Out]

((11*b^2*d^2 - 13*a*b*d*e + 5*a^2*e^2)*Sqrt[a + b*x]*Sqrt[d + e*x])/b^3 + (2*(4*b*d - 3*a*e)*Sqrt[a + b*x]*(d
+ e*x)^(3/2))/b^2 + (8*e*(a + b*x)^(3/2)*(d + e*x)^(3/2))/(3*b^2) + ((b*d - a*e)*(11*b^2*d^2 - 13*a*b*d*e + 5*
a^2*e^2)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(b^(7/2)*Sqrt[e])

Rule 951

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Simp[(c^p*(d + e*x)^(m + 2*p)*(f + g*x)^(n + 1))/(g*e^(2*p)*(m + n + 2*p + 1)), x] + Dist[1/(g*e^(2*p)*(m +
n + 2*p + 1)), Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x + c*x^2)^p - c^p*
(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x
] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*
p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

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 \frac{\sqrt{d+e x} \left (15 d^2+20 d e x+8 e^2 x^2\right )}{\sqrt{a+b x}} \, dx &=\frac{8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac{\int \frac{\sqrt{d+e x} \left (3 e (3 b d-2 a e) (5 b d+2 a e)+12 b e^2 (4 b d-3 a e) x\right )}{\sqrt{a+b x}} \, dx}{3 b^2 e}\\ &=\frac{2 (4 b d-3 a e) \sqrt{a+b x} (d+e x)^{3/2}}{b^2}+\frac{8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac{\left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right ) \int \frac{\sqrt{d+e x}}{\sqrt{a+b x}} \, dx}{b^2}\\ &=\frac{\left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right ) \sqrt{a+b x} \sqrt{d+e x}}{b^3}+\frac{2 (4 b d-3 a e) \sqrt{a+b x} (d+e x)^{3/2}}{b^2}+\frac{8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac{\left ((b d-a e) \left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right )\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{d+e x}} \, dx}{2 b^3}\\ &=\frac{\left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right ) \sqrt{a+b x} \sqrt{d+e x}}{b^3}+\frac{2 (4 b d-3 a e) \sqrt{a+b x} (d+e x)^{3/2}}{b^2}+\frac{8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac{\left ((b d-a e) \left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right )\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 )}{b^4}\\ &=\frac{\left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right ) \sqrt{a+b x} \sqrt{d+e x}}{b^3}+\frac{2 (4 b d-3 a e) \sqrt{a+b x} (d+e x)^{3/2}}{b^2}+\frac{8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac{\left ((b d-a e) \left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right )\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 )}{b^4}\\ &=\frac{\left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right ) \sqrt{a+b x} \sqrt{d+e x}}{b^3}+\frac{2 (4 b d-3 a e) \sqrt{a+b x} (d+e x)^{3/2}}{b^2}+\frac{8 e (a+b x)^{3/2} (d+e x)^{3/2}}{3 b^2}+\frac{(b d-a e) \left (11 b^2 d^2-13 a b d e+5 a^2 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{b^{7/2} \sqrt{e}}\\ \end{align*}

Mathematica [A]  time = 0.472013, size = 163, normalized size = 0.93 \[ \frac{\sqrt{d+e x} \left (\sqrt{a+b x} \left (15 a^2 e^2-a b e (49 d+10 e x)+b^2 \left (57 d^2+32 d e x+8 e^2 x^2\right )\right )+\frac{3 \sqrt{b d-a e} \left (5 a^2 e^2-13 a b d e+11 b^2 d^2\right ) \sinh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b d-a e}}\right )}{\sqrt{e} \sqrt{\frac{b (d+e x)}{b d-a e}}}\right )}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[d + e*x]*(15*d^2 + 20*d*e*x + 8*e^2*x^2))/Sqrt[a + b*x],x]

[Out]

(Sqrt[d + e*x]*(Sqrt[a + b*x]*(15*a^2*e^2 - a*b*e*(49*d + 10*e*x) + b^2*(57*d^2 + 32*d*e*x + 8*e^2*x^2)) + (3*
Sqrt[b*d - a*e]*(11*b^2*d^2 - 13*a*b*d*e + 5*a^2*e^2)*ArcSinh[(Sqrt[e]*Sqrt[a + b*x])/Sqrt[b*d - a*e]])/(Sqrt[
e]*Sqrt[(b*(d + e*x))/(b*d - a*e)])))/(3*b^3)

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Maple [B]  time = 0.318, size = 392, normalized size = 2.2 \begin{align*} -{\frac{1}{6\,{b}^{3}}\sqrt{ex+d}\sqrt{bx+a} \left ( -16\,{x}^{2}{b}^{2}{e}^{2}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+15\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{3}{e}^{3}-54\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{2}bd{e}^{2}+72\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) a{b}^{2}{d}^{2}e-33\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){b}^{3}{d}^{3}+20\,\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }xab{e}^{2}-64\,\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }x{b}^{2}de-30\,\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }{a}^{2}{e}^{2}+98\,\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }abde-114\,\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }{b}^{2}{d}^{2} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }}}{\frac{1}{\sqrt{be}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(1/2)*(8*e^2*x^2+20*d*e*x+15*d^2)/(b*x+a)^(1/2),x)

[Out]

-1/6*(e*x+d)^(1/2)*(b*x+a)^(1/2)*(-16*x^2*b^2*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+15*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^3-54*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^2+72*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*e-33*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+20*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*x*a*b*e^2-64*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*x*b^2*d*e-30*(b*e)^(
1/2)*((b*x+a)*(e*x+d))^(1/2)*a^2*e^2+98*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*a*b*d*e-114*(b*e)^(1/2)*((b*x+a)*(
e*x+d))^(1/2)*b^2*d^2)/b^3/((b*x+a)*(e*x+d))^(1/2)/(b*e)^(1/2)

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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((e*x+d)^(1/2)*(8*e^2*x^2+20*d*e*x+15*d^2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.7625, size = 948, normalized size = 5.39 \begin{align*} \left [-\frac{3 \,{\left (11 \, b^{3} d^{3} - 24 \, a b^{2} d^{2} e + 18 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3}\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 (8 \, b^{3} e^{3} x^{2} + 57 \, b^{3} d^{2} e - 49 \, a b^{2} d e^{2} + 15 \, a^{2} b e^{3} + 2 \,{\left (16 \, b^{3} d e^{2} - 5 \, a b^{2} e^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{12 \, b^{4} e}, -\frac{3 \,{\left (11 \, b^{3} d^{3} - 24 \, a b^{2} d^{2} e + 18 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3}\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 (8 \, b^{3} e^{3} x^{2} + 57 \, b^{3} d^{2} e - 49 \, a b^{2} d e^{2} + 15 \, a^{2} b e^{3} + 2 \,{\left (16 \, b^{3} d e^{2} - 5 \, a b^{2} e^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{6 \, b^{4} e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)*(8*e^2*x^2+20*d*e*x+15*d^2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(11*b^3*d^3 - 24*a*b^2*d^2*e + 18*a^2*b*d*e^2 - 5*a^3*e^3)*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*(8*b^3*e^3*x^2 + 57*b^3*d^2*e - 49*a*b^2*d*e^2 + 15*a^2*b*e^3 + 2*(16*b^3*d*e^2 - 5*a*b^2*e^3)*x)*sqrt(b*x
+ a)*sqrt(e*x + d))/(b^4*e), -1/6*(3*(11*b^3*d^3 - 24*a*b^2*d^2*e + 18*a^2*b*d*e^2 - 5*a^3*e^3)*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*(8*b^3*e^3*x^2 + 57*b^3*d^2*e - 49*a*b^2*d*e^2 + 15*a^2*b*e^3 + 2*(16*b^3*d*e^2 - 5*a*b^2*e^3)*x)
*sqrt(b*x + a)*sqrt(e*x + d))/(b^4*e)]

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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((e*x+d)**(1/2)*(8*e**2*x**2+20*d*e*x+15*d**2)/(b*x+a)**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 1.32129, size = 602, normalized size = 3.42 \begin{align*} -\frac{\frac{180 \,{\left (\frac{{\left (b^{2} d - a b e\right )} e^{\left (-\frac{1}{2}\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 |}\right )}{\sqrt{b}} - \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \sqrt{b x + a}\right )} d^{2}{\left | b \right |}}{b^{2}} - \frac{4 \,{\left (\sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \sqrt{b x + a}{\left (2 \,{\left (b x + a\right )}{\left (\frac{4 \,{\left (b x + a\right )}}{b^{2}} + \frac{{\left (b^{6} d e^{3} - 13 \, a b^{5} e^{4}\right )} e^{\left (-4\right )}}{b^{7}}\right )} - \frac{3 \,{\left (b^{7} d^{2} e^{2} + 2 \, a b^{6} d e^{3} - 11 \, a^{2} b^{5} e^{4}\right )} e^{\left (-4\right )}}{b^{7}}\right )} - \frac{3 \,{\left (b^{3} d^{3} + a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3}\right )} e^{\left (-\frac{5}{2}\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 |}\right )}{b^{\frac{3}{2}}}\right )}{\left | b \right |} e^{2}}{b^{2}} - \frac{5 \,{\left (\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 )} e^{\left (-2\right )}}{b^{4}} + \frac{{\left (b d e - 5 \, a e^{2}\right )} e^{\left (-4\right )}}{b^{4}}\right )} + \frac{{\left (b^{2} d^{2} + 2 \, a b d e - 3 \, a^{2} e^{2}\right )} e^{\left (-\frac{7}{2}\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 |}\right )}{b^{\frac{7}{2}}}\right )} d{\left | b \right |} e}{b^{3}}}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)*(8*e^2*x^2+20*d*e*x+15*d^2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

-1/12*(180*((b^2*d - a*b*e)*e^(-1/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b
*e)))/sqrt(b) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a))*d^2*abs(b)/b^2 - 4*(sqrt(b^2*d + (b*x + a)*
b*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/b^2 + (b^6*d*e^3 - 13*a*b^5*e^4)*e^(-4)/b^7) - 3*(b^7*d^2
*e^2 + 2*a*b^6*d*e^3 - 11*a^2*b^5*e^4)*e^(-4)/b^7) - 3*(b^3*d^3 + a*b^2*d^2*e + 3*a^2*b*d*e^2 - 5*a^3*e^3)*e^(
-5/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/b^(3/2))*abs(b)*e^2/b^2 -
5*(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 - 5*a*e^2)*e^(-4)/b^4)
+ (b^2*d^2 + 2*a*b*d*e - 3*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))*d*abs(b)*e/b^3)/b