3.1581 \(\int \frac{(b+2 c x) (d+e x)^4}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=202 \[ \frac{e^2 \sqrt{a+b x+c x^2} \left (-2 c e (8 a e+27 b d)+15 b^2 e^2+10 c e x (2 c d-b e)+64 c^2 d^2\right )}{3 c^3}+\frac{e (2 c d-b e) \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 c^{7/2}}+\frac{8 e^2 (d+e x)^2 \sqrt{a+b x+c x^2}}{3 c}-\frac{2 (d+e x)^4}{\sqrt{a+b x+c x^2}} \]

[Out]

(-2*(d + e*x)^4)/Sqrt[a + b*x + c*x^2] + (8*e^2*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/(3*c) + (e^2*(64*c^2*d^2 +
15*b^2*e^2 - 2*c*e*(27*b*d + 8*a*e) + 10*c*e*(2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2])/(3*c^3) + (e*(2*c*d - b*e
)*(8*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(2*b*d + 3*a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*c
^(7/2))

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Rubi [A]  time = 0.206239, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {768, 742, 779, 621, 206} \[ \frac{e^2 \sqrt{a+b x+c x^2} \left (-2 c e (8 a e+27 b d)+15 b^2 e^2+10 c e x (2 c d-b e)+64 c^2 d^2\right )}{3 c^3}+\frac{e (2 c d-b e) \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 c^{7/2}}+\frac{8 e^2 (d+e x)^2 \sqrt{a+b x+c x^2}}{3 c}-\frac{2 (d+e x)^4}{\sqrt{a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((b + 2*c*x)*(d + e*x)^4)/(a + b*x + c*x^2)^(3/2),x]

[Out]

(-2*(d + e*x)^4)/Sqrt[a + b*x + c*x^2] + (8*e^2*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/(3*c) + (e^2*(64*c^2*d^2 +
15*b^2*e^2 - 2*c*e*(27*b*d + 8*a*e) + 10*c*e*(2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2])/(3*c^3) + (e*(2*c*d - b*e
)*(8*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(2*b*d + 3*a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*c
^(7/2))

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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{(b+2 c x) (d+e x)^4}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac{2 (d+e x)^4}{\sqrt{a+b x+c x^2}}+(8 e) \int \frac{(d+e x)^3}{\sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 (d+e x)^4}{\sqrt{a+b x+c x^2}}+\frac{8 e^2 (d+e x)^2 \sqrt{a+b x+c x^2}}{3 c}+\frac{(8 e) \int \frac{(d+e x) \left (\frac{1}{2} \left (6 c d^2-e (b d+4 a e)\right )+\frac{5}{2} e (2 c d-b e) x\right )}{\sqrt{a+b x+c x^2}} \, dx}{3 c}\\ &=-\frac{2 (d+e x)^4}{\sqrt{a+b x+c x^2}}+\frac{8 e^2 (d+e x)^2 \sqrt{a+b x+c x^2}}{3 c}+\frac{e^2 \left (64 c^2 d^2+15 b^2 e^2-2 c e (27 b d+8 a e)+10 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{3 c^3}+\frac{\left (e (2 c d-b e) \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{2 c^3}\\ &=-\frac{2 (d+e x)^4}{\sqrt{a+b x+c x^2}}+\frac{8 e^2 (d+e x)^2 \sqrt{a+b x+c x^2}}{3 c}+\frac{e^2 \left (64 c^2 d^2+15 b^2 e^2-2 c e (27 b d+8 a e)+10 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{3 c^3}+\frac{\left (e (2 c d-b e) \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{c^3}\\ &=-\frac{2 (d+e x)^4}{\sqrt{a+b x+c x^2}}+\frac{8 e^2 (d+e x)^2 \sqrt{a+b x+c x^2}}{3 c}+\frac{e^2 \left (64 c^2 d^2+15 b^2 e^2-2 c e (27 b d+8 a e)+10 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{3 c^3}+\frac{e (2 c d-b e) \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.34658, size = 248, normalized size = 1.23 \[ \frac{-c e^3 \left (16 a^2 e+a b (54 d+26 e x)+b^2 x (54 d-5 e x)\right )+15 b^2 e^4 (a+b x)+2 c^2 e^2 \left (2 a \left (18 d^2+9 d e x-2 e^2 x^2\right )-b x \left (-36 d^2+9 d e x+e^2 x^2\right )\right )+c^3 \left (36 d^2 e^2 x^2-24 d^3 e x-6 d^4+12 d e^3 x^3+2 e^4 x^4\right )}{3 c^3 \sqrt{a+x (b+c x)}}+\frac{e (2 c d-b e) \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{2 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((b + 2*c*x)*(d + e*x)^4)/(a + b*x + c*x^2)^(3/2),x]

[Out]

(15*b^2*e^4*(a + b*x) + c^3*(-6*d^4 - 24*d^3*e*x + 36*d^2*e^2*x^2 + 12*d*e^3*x^3 + 2*e^4*x^4) - c*e^3*(16*a^2*
e + b^2*x*(54*d - 5*e*x) + a*b*(54*d + 26*e*x)) + 2*c^2*e^2*(2*a*(18*d^2 + 9*d*e*x - 2*e^2*x^2) - b*x*(-36*d^2
+ 9*d*e*x + e^2*x^2)))/(3*c^3*Sqrt[a + x*(b + c*x)]) + (e*(2*c*d - b*e)*(8*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(2*b*d
+ 3*a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(2*c^(7/2))

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Maple [B]  time = 0.017, size = 1320, normalized size = 6.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*x+b)*(e*x+d)^4/(c*x^2+b*x+a)^(3/2),x)

[Out]

24*a/c*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*d^2*e^2-6/c^2*e^4*b*a*x/(c*x^2+b*x+a)^(1/2)-16/3/c^2*e^4*a^2*b^2/(4
*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-5/2/c^3*e^4*b^5/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-4*b/(4*a*c-b^2)/(c*x^2+b*x+a)^
(1/2)*x*c*d^4-6*b^4/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*d^2*e^2+12*b/c*x/(c*x^2+b*x+a)^(1/2)*d^2*e^2+12*a/c*x/
(c*x^2+b*x+a)^(1/2)*d*e^3+9/2*b^5/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*d*e^3-18*b/c^2*a/(c*x^2+b*x+a)^(1/2)*d*e
^3-9*b^2/c^2*x/(c*x^2+b*x+a)^(1/2)*d*e^3-6*b/c*x^2/(c*x^2+b*x+a)^(1/2)*d*e^3+19/3/c^3*e^4*b^4*a/(4*a*c-b^2)/(c
*x^2+b*x+a)^(1/2)-8*x/(c*x^2+b*x+a)^(1/2)*d^3*e-2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*d^4+4*x^3/(c*x^2+b*x+a)^
(1/2)*d*e^3+12*x^2/(c*x^2+b*x+a)^(1/2)*d^2*e^2-5/4/c^4*e^4*b^4/(c*x^2+b*x+a)^(1/2)-16/3/c^2*e^4*a^2/(c*x^2+b*x
+a)^(1/2)+8/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d^3*e-5/2/c^(7/2)*e^4*b^3*ln((1/2*b+c*x)/c^(1/
2)+(c*x^2+b*x+a)^(1/2))-2/(c*x^2+b*x+a)^(1/2)*d^4+2/3*e^4*x^4/(c*x^2+b*x+a)^(1/2)+5/3/c^2*e^4*b^2*x^2/(c*x^2+b
*x+a)^(1/2)+5/2/c^3*e^4*b^3*x/(c*x^2+b*x+a)^(1/2)-5/4/c^4*e^4*b^6/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-2/3/c*e^4*b*
x^3/(c*x^2+b*x+a)^(1/2)+2*b*d^4*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+9*b^2/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)
+(c*x^2+b*x+a)^(1/2))*d*e^3+6/c^(5/2)*e^4*b*a*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-6*b^2/c^2/(c*x^2+b*x
+a)^(1/2)*d^2*e^2+24*a/c/(c*x^2+b*x+a)^(1/2)*d^2*e^2+9/2*b^3/c^3/(c*x^2+b*x+a)^(1/2)*d*e^3+19/3/c^3*e^4*b^2*a/
(c*x^2+b*x+a)^(1/2)-8/3/c*e^4*a*x^2/(c*x^2+b*x+a)^(1/2)-12*b/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2
))*d^2*e^2-12*a/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d*e^3-32/3/c*e^4*a^2*b/(4*a*c-b^2)/(c*x^2+
b*x+a)^(1/2)*x+48*a*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*d^2*e^2-18*b^3/c^2*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*d
*e^3+9*b^4/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*d*e^3+38/3/c^2*e^4*b^3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-12
*b^3/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*d^2*e^2-36*b^2/c*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*d*e^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((2*c*x+b)*(e*x+d)^4/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 5.25393, size = 2055, normalized size = 10.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^4/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[1/12*(3*(16*a*c^3*d^3*e - 24*a*b*c^2*d^2*e^2 + 6*(3*a*b^2*c - 4*a^2*c^2)*d*e^3 - (5*a*b^3 - 12*a^2*b*c)*e^4 +
(16*c^4*d^3*e - 24*b*c^3*d^2*e^2 + 6*(3*b^2*c^2 - 4*a*c^3)*d*e^3 - (5*b^3*c - 12*a*b*c^2)*e^4)*x^2 + (16*b*c^
3*d^3*e - 24*b^2*c^2*d^2*e^2 + 6*(3*b^3*c - 4*a*b*c^2)*d*e^3 - (5*b^4 - 12*a*b^2*c)*e^4)*x)*sqrt(c)*log(-8*c^2
*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(2*c^4*e^4*x^4 - 6*c^4*d^4 + 7
2*a*c^3*d^2*e^2 - 54*a*b*c^2*d*e^3 + (15*a*b^2*c - 16*a^2*c^2)*e^4 + 2*(6*c^4*d*e^3 - b*c^3*e^4)*x^3 + (36*c^4
*d^2*e^2 - 18*b*c^3*d*e^3 + (5*b^2*c^2 - 8*a*c^3)*e^4)*x^2 - (24*c^4*d^3*e - 72*b*c^3*d^2*e^2 + 18*(3*b^2*c^2
- 2*a*c^3)*d*e^3 - (15*b^3*c - 26*a*b*c^2)*e^4)*x)*sqrt(c*x^2 + b*x + a))/(c^5*x^2 + b*c^4*x + a*c^4), -1/6*(3
*(16*a*c^3*d^3*e - 24*a*b*c^2*d^2*e^2 + 6*(3*a*b^2*c - 4*a^2*c^2)*d*e^3 - (5*a*b^3 - 12*a^2*b*c)*e^4 + (16*c^4
*d^3*e - 24*b*c^3*d^2*e^2 + 6*(3*b^2*c^2 - 4*a*c^3)*d*e^3 - (5*b^3*c - 12*a*b*c^2)*e^4)*x^2 + (16*b*c^3*d^3*e
- 24*b^2*c^2*d^2*e^2 + 6*(3*b^3*c - 4*a*b*c^2)*d*e^3 - (5*b^4 - 12*a*b^2*c)*e^4)*x)*sqrt(-c)*arctan(1/2*sqrt(c
*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 2*(2*c^4*e^4*x^4 - 6*c^4*d^4 + 72*a*c^3*d^2*e^
2 - 54*a*b*c^2*d*e^3 + (15*a*b^2*c - 16*a^2*c^2)*e^4 + 2*(6*c^4*d*e^3 - b*c^3*e^4)*x^3 + (36*c^4*d^2*e^2 - 18*
b*c^3*d*e^3 + (5*b^2*c^2 - 8*a*c^3)*e^4)*x^2 - (24*c^4*d^3*e - 72*b*c^3*d^2*e^2 + 18*(3*b^2*c^2 - 2*a*c^3)*d*e
^3 - (15*b^3*c - 26*a*b*c^2)*e^4)*x)*sqrt(c*x^2 + b*x + a))/(c^5*x^2 + b*c^4*x + a*c^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b + 2 c x\right ) \left (d + e x\right )^{4}}{\left (a + b x + c x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)**4/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral((b + 2*c*x)*(d + e*x)**4/(a + b*x + c*x**2)**(3/2), x)

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Giac [B]  time = 1.5505, size = 736, normalized size = 3.64 \begin{align*} \frac{{\left ({\left (2 \,{\left (\frac{{\left (b^{2} c^{3} e^{4} - 4 \, a c^{4} e^{4}\right )} x}{b^{2} c^{3} - 4 \, a c^{4}} + \frac{6 \, b^{2} c^{3} d e^{3} - 24 \, a c^{4} d e^{3} - b^{3} c^{2} e^{4} + 4 \, a b c^{3} e^{4}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x + \frac{36 \, b^{2} c^{3} d^{2} e^{2} - 144 \, a c^{4} d^{2} e^{2} - 18 \, b^{3} c^{2} d e^{3} + 72 \, a b c^{3} d e^{3} + 5 \, b^{4} c e^{4} - 28 \, a b^{2} c^{2} e^{4} + 32 \, a^{2} c^{3} e^{4}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x - \frac{24 \, b^{2} c^{3} d^{3} e - 96 \, a c^{4} d^{3} e - 72 \, b^{3} c^{2} d^{2} e^{2} + 288 \, a b c^{3} d^{2} e^{2} + 54 \, b^{4} c d e^{3} - 252 \, a b^{2} c^{2} d e^{3} + 144 \, a^{2} c^{3} d e^{3} - 15 \, b^{5} e^{4} + 86 \, a b^{3} c e^{4} - 104 \, a^{2} b c^{2} e^{4}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x - \frac{6 \, b^{2} c^{3} d^{4} - 24 \, a c^{4} d^{4} - 72 \, a b^{2} c^{2} d^{2} e^{2} + 288 \, a^{2} c^{3} d^{2} e^{2} + 54 \, a b^{3} c d e^{3} - 216 \, a^{2} b c^{2} d e^{3} - 15 \, a b^{4} e^{4} + 76 \, a^{2} b^{2} c e^{4} - 64 \, a^{3} c^{2} e^{4}}{b^{2} c^{3} - 4 \, a c^{4}}}{3 \, \sqrt{c x^{2} + b x + a}} - \frac{{\left (16 \, c^{3} d^{3} e - 24 \, b c^{2} d^{2} e^{2} + 18 \, b^{2} c d e^{3} - 24 \, a c^{2} d e^{3} - 5 \, b^{3} e^{4} + 12 \, a b c e^{4}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{2 \, c^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^4/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

1/3*(((2*((b^2*c^3*e^4 - 4*a*c^4*e^4)*x/(b^2*c^3 - 4*a*c^4) + (6*b^2*c^3*d*e^3 - 24*a*c^4*d*e^3 - b^3*c^2*e^4
+ 4*a*b*c^3*e^4)/(b^2*c^3 - 4*a*c^4))*x + (36*b^2*c^3*d^2*e^2 - 144*a*c^4*d^2*e^2 - 18*b^3*c^2*d*e^3 + 72*a*b*
c^3*d*e^3 + 5*b^4*c*e^4 - 28*a*b^2*c^2*e^4 + 32*a^2*c^3*e^4)/(b^2*c^3 - 4*a*c^4))*x - (24*b^2*c^3*d^3*e - 96*a
*c^4*d^3*e - 72*b^3*c^2*d^2*e^2 + 288*a*b*c^3*d^2*e^2 + 54*b^4*c*d*e^3 - 252*a*b^2*c^2*d*e^3 + 144*a^2*c^3*d*e
^3 - 15*b^5*e^4 + 86*a*b^3*c*e^4 - 104*a^2*b*c^2*e^4)/(b^2*c^3 - 4*a*c^4))*x - (6*b^2*c^3*d^4 - 24*a*c^4*d^4 -
72*a*b^2*c^2*d^2*e^2 + 288*a^2*c^3*d^2*e^2 + 54*a*b^3*c*d*e^3 - 216*a^2*b*c^2*d*e^3 - 15*a*b^4*e^4 + 76*a^2*b
^2*c*e^4 - 64*a^3*c^2*e^4)/(b^2*c^3 - 4*a*c^4))/sqrt(c*x^2 + b*x + a) - 1/2*(16*c^3*d^3*e - 24*b*c^2*d^2*e^2 +
18*b^2*c*d*e^3 - 24*a*c^2*d*e^3 - 5*b^3*e^4 + 12*a*b*c*e^4)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sq
rt(c) - b))/c^(7/2)