3.38 \(\int \sqrt{c+e x+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4} \, dx\)

Optimal. Leaf size=283 \[ -\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} (2 d x+e) \sqrt{c+d x^2+e x} \left (-16 a d^2+4 b c d-5 b e^2\right )}{64 d^3 \left (a+b x^2\right )}-\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \left (4 c d-e^2\right ) \left (-16 a d^2+4 b c d-5 b e^2\right ) \tanh ^{-1}\left (\frac{2 d x+e}{2 \sqrt{d} \sqrt{c+d x^2+e x}}\right )}{128 d^{7/2} \left (a+b x^2\right )}-\frac{5 b e \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2}}{24 d^2 \left (a+b x^2\right )}+\frac{b x \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2}}{4 d \left (a+b x^2\right )} \]

[Out]

-((4*b*c*d - 16*a*d^2 - 5*b*e^2)*(e + 2*d*x)*Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(64*d^3*(a
+ b*x^2)) - (5*b*e*(c + e*x + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(24*d^2*(a + b*x^2)) + (b*x*(c +
e*x + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*d*(a + b*x^2)) - ((4*c*d - e^2)*(4*b*c*d - 16*a*d^2 - 5
*b*e^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + e*x + d*x^2])])/(128*d^(7/2)*(
a + b*x^2))

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Rubi [A]  time = 0.340535, antiderivative size = 283, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {6744, 1661, 640, 612, 621, 206} \[ -\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} (2 d x+e) \sqrt{c+d x^2+e x} \left (-16 a d^2+4 b c d-5 b e^2\right )}{64 d^3 \left (a+b x^2\right )}-\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \left (4 c d-e^2\right ) \left (-16 a d^2+4 b c d-5 b e^2\right ) \tanh ^{-1}\left (\frac{2 d x+e}{2 \sqrt{d} \sqrt{c+d x^2+e x}}\right )}{128 d^{7/2} \left (a+b x^2\right )}-\frac{5 b e \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2}}{24 d^2 \left (a+b x^2\right )}+\frac{b x \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2}}{4 d \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

-((4*b*c*d - 16*a*d^2 - 5*b*e^2)*(e + 2*d*x)*Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(64*d^3*(a
+ b*x^2)) - (5*b*e*(c + e*x + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(24*d^2*(a + b*x^2)) + (b*x*(c +
e*x + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*d*(a + b*x^2)) - ((4*c*d - e^2)*(4*b*c*d - 16*a*d^2 - 5
*b*e^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + e*x + d*x^2])])/(128*d^(7/2)*(
a + b*x^2))

Rule 6744

Int[(u_)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[Sqrt[a + b*x^n + c*x^(2*n)]/((4
*c)^(p - 1/2)*(b + 2*c*x^n)), Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] &
& EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 1661

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[(e*x^(q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*(q + 2*p + 1)), x] + Dist[1/(c*(q + 2*p + 1)), Int
[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + p)*x^(q - 1) - c*e*(q +
2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 640

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

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

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 \sqrt{c+e x+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4} \, dx &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \int \left (2 a b+2 b^2 x^2\right ) \sqrt{c+e x+d x^2} \, dx}{2 a b+2 b^2 x^2}\\ &=\frac{b x \left (c+e x+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 d \left (a+b x^2\right )}+\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \int \left (-2 b (b c-4 a d)-5 b^2 e x\right ) \sqrt{c+e x+d x^2} \, dx}{4 d \left (2 a b+2 b^2 x^2\right )}\\ &=-\frac{5 b e \left (c+e x+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{24 d^2 \left (a+b x^2\right )}+\frac{b x \left (c+e x+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 d \left (a+b x^2\right )}+\frac{\left (\left (-4 b d (b c-4 a d)+5 b^2 e^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}\right ) \int \sqrt{c+e x+d x^2} \, dx}{8 d^2 \left (2 a b+2 b^2 x^2\right )}\\ &=-\frac{\left (4 b c d-16 a d^2-5 b e^2\right ) (e+2 d x) \sqrt{c+e x+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{64 d^3 \left (a+b x^2\right )}-\frac{5 b e \left (c+e x+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{24 d^2 \left (a+b x^2\right )}+\frac{b x \left (c+e x+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 d \left (a+b x^2\right )}+\frac{\left (\left (4 c d-e^2\right ) \left (-4 b d (b c-4 a d)+5 b^2 e^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}\right ) \int \frac{1}{\sqrt{c+e x+d x^2}} \, dx}{64 d^3 \left (2 a b+2 b^2 x^2\right )}\\ &=-\frac{\left (4 b c d-16 a d^2-5 b e^2\right ) (e+2 d x) \sqrt{c+e x+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{64 d^3 \left (a+b x^2\right )}-\frac{5 b e \left (c+e x+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{24 d^2 \left (a+b x^2\right )}+\frac{b x \left (c+e x+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 d \left (a+b x^2\right )}+\frac{\left (\left (4 c d-e^2\right ) \left (-4 b d (b c-4 a d)+5 b^2 e^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}\right ) \operatorname{Subst}\left (\int \frac{1}{4 d-x^2} \, dx,x,\frac{e+2 d x}{\sqrt{c+e x+d x^2}}\right )}{32 d^3 \left (2 a b+2 b^2 x^2\right )}\\ &=-\frac{\left (4 b c d-16 a d^2-5 b e^2\right ) (e+2 d x) \sqrt{c+e x+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{64 d^3 \left (a+b x^2\right )}-\frac{5 b e \left (c+e x+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{24 d^2 \left (a+b x^2\right )}+\frac{b x \left (c+e x+d x^2\right )^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 d \left (a+b x^2\right )}-\frac{\left (4 c d-e^2\right ) \left (4 b c d-16 a d^2-5 b e^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac{e+2 d x}{2 \sqrt{d} \sqrt{c+e x+d x^2}}\right )}{128 d^{7/2} \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.231036, size = 168, normalized size = 0.59 \[ \frac{\sqrt{\left (a+b x^2\right )^2} \left (2 \sqrt{d} \sqrt{c+x (d x+e)} \left (48 a d^2 (2 d x+e)+b \left (4 c d (6 d x-13 e)+8 d^2 e x^2+48 d^3 x^3-10 d e^2 x+15 e^3\right )\right )-3 \left (4 c d-e^2\right ) \left (-16 a d^2+4 b c d-5 b e^2\right ) \tanh ^{-1}\left (\frac{2 d x+e}{2 \sqrt{d} \sqrt{c+x (d x+e)}}\right )\right )}{384 d^{7/2} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

(Sqrt[(a + b*x^2)^2]*(2*Sqrt[d]*Sqrt[c + x*(e + d*x)]*(48*a*d^2*(e + 2*d*x) + b*(15*e^3 - 10*d*e^2*x + 8*d^2*e
*x^2 + 48*d^3*x^3 + 4*c*d*(-13*e + 6*d*x))) - 3*(4*c*d - e^2)*(4*b*c*d - 16*a*d^2 - 5*b*e^2)*ArcTanh[(e + 2*d*
x)/(2*Sqrt[d]*Sqrt[c + x*(e + d*x)])]))/(384*d^(7/2)*(a + b*x^2))

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Maple [A]  time = 0.008, size = 373, normalized size = 1.3 \begin{align*}{\frac{1}{384\,b{x}^{2}+384\,a}\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}} \left ( 96\,{d}^{7/2} \left ({x}^{2}d+ex+c \right ) ^{3/2}xb-80\,{d}^{5/2} \left ({x}^{2}d+ex+c \right ) ^{3/2}be+192\,{d}^{9/2}\sqrt{{x}^{2}d+ex+c}xa-48\,{d}^{7/2}\sqrt{{x}^{2}d+ex+c}xbc+60\,{d}^{5/2}\sqrt{{x}^{2}d+ex+c}xb{e}^{2}+96\,{d}^{7/2}\sqrt{{x}^{2}d+ex+c}ae-24\,{d}^{5/2}\sqrt{{x}^{2}d+ex+c}bce+30\,{d}^{3/2}\sqrt{{x}^{2}d+ex+c}b{e}^{3}+192\,\ln \left ( 1/2\,{\frac{2\,\sqrt{{x}^{2}d+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) ac{d}^{4}-48\,\ln \left ( 1/2\,{\frac{2\,\sqrt{{x}^{2}d+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) a{d}^{3}{e}^{2}-48\,\ln \left ( 1/2\,{\frac{2\,\sqrt{{x}^{2}d+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) b{c}^{2}{d}^{3}+72\,\ln \left ( 1/2\,{\frac{2\,\sqrt{{x}^{2}d+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) bc{d}^{2}{e}^{2}-15\,\ln \left ( 1/2\,{\frac{2\,\sqrt{{x}^{2}d+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) bd{e}^{4} \right ){d}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*((b*x^2+a)^2)^(1/2)*(96*d^(7/2)*(d*x^2+e*x+c)^(3/2)*x*b-80*d^(5/2)*(d*x^2+e*x+c)^(3/2)*b*e+192*d^(9/2)*(
d*x^2+e*x+c)^(1/2)*x*a-48*d^(7/2)*(d*x^2+e*x+c)^(1/2)*x*b*c+60*d^(5/2)*(d*x^2+e*x+c)^(1/2)*x*b*e^2+96*d^(7/2)*
(d*x^2+e*x+c)^(1/2)*a*e-24*d^(5/2)*(d*x^2+e*x+c)^(1/2)*b*c*e+30*d^(3/2)*(d*x^2+e*x+c)^(1/2)*b*e^3+192*ln(1/2*(
2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2))*a*c*d^4-48*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1
/2))*a*d^3*e^2-48*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2))*b*c^2*d^3+72*ln(1/2*(2*(d*x^2+e*x+c)
^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2))*b*c*d^2*e^2-15*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2))*b*d*e^
4)/(b*x^2+a)/d^(9/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*x^2 + e*x + c)*sqrt((b*x^2 + a)^2), x)

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Fricas [A]  time = 1.92417, size = 857, normalized size = 3.03 \begin{align*} \left [\frac{3 \,{\left (16 \, b c^{2} d^{2} - 64 \, a c d^{3} + 5 \, b e^{4} - 8 \,{\left (3 \, b c d - 2 \, a d^{2}\right )} e^{2}\right )} \sqrt{d} \log \left (8 \, d^{2} x^{2} + 8 \, d e x - 4 \, \sqrt{d x^{2} + e x + c}{\left (2 \, d x + e\right )} \sqrt{d} + 4 \, c d + e^{2}\right ) + 4 \,{\left (48 \, b d^{4} x^{3} + 8 \, b d^{3} e x^{2} + 15 \, b d e^{3} - 4 \,{\left (13 \, b c d^{2} - 12 \, a d^{3}\right )} e + 2 \,{\left (12 \, b c d^{3} + 48 \, a d^{4} - 5 \, b d^{2} e^{2}\right )} x\right )} \sqrt{d x^{2} + e x + c}}{768 \, d^{4}}, \frac{3 \,{\left (16 \, b c^{2} d^{2} - 64 \, a c d^{3} + 5 \, b e^{4} - 8 \,{\left (3 \, b c d - 2 \, a d^{2}\right )} e^{2}\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{d x^{2} + e x + c}{\left (2 \, d x + e\right )} \sqrt{-d}}{2 \,{\left (d^{2} x^{2} + d e x + c d\right )}}\right ) + 2 \,{\left (48 \, b d^{4} x^{3} + 8 \, b d^{3} e x^{2} + 15 \, b d e^{3} - 4 \,{\left (13 \, b c d^{2} - 12 \, a d^{3}\right )} e + 2 \,{\left (12 \, b c d^{3} + 48 \, a d^{4} - 5 \, b d^{2} e^{2}\right )} x\right )} \sqrt{d x^{2} + e x + c}}{384 \, d^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/768*(3*(16*b*c^2*d^2 - 64*a*c*d^3 + 5*b*e^4 - 8*(3*b*c*d - 2*a*d^2)*e^2)*sqrt(d)*log(8*d^2*x^2 + 8*d*e*x -
4*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(d) + 4*c*d + e^2) + 4*(48*b*d^4*x^3 + 8*b*d^3*e*x^2 + 15*b*d*e^3 - 4*
(13*b*c*d^2 - 12*a*d^3)*e + 2*(12*b*c*d^3 + 48*a*d^4 - 5*b*d^2*e^2)*x)*sqrt(d*x^2 + e*x + c))/d^4, 1/384*(3*(1
6*b*c^2*d^2 - 64*a*c*d^3 + 5*b*e^4 - 8*(3*b*c*d - 2*a*d^2)*e^2)*sqrt(-d)*arctan(1/2*sqrt(d*x^2 + e*x + c)*(2*d
*x + e)*sqrt(-d)/(d^2*x^2 + d*e*x + c*d)) + 2*(48*b*d^4*x^3 + 8*b*d^3*e*x^2 + 15*b*d*e^3 - 4*(13*b*c*d^2 - 12*
a*d^3)*e + 2*(12*b*c*d^3 + 48*a*d^4 - 5*b*d^2*e^2)*x)*sqrt(d*x^2 + e*x + c))/d^4]

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

[Out]

Timed out

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Giac [A]  time = 1.15152, size = 358, normalized size = 1.27 \begin{align*} \frac{1}{192} \, \sqrt{d x^{2} + x e + c}{\left (2 \,{\left (4 \,{\left (6 \, b x \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{b e \mathrm{sgn}\left (b x^{2} + a\right )}{d}\right )} x + \frac{12 \, b c d^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + 48 \, a d^{3} \mathrm{sgn}\left (b x^{2} + a\right ) - 5 \, b d e^{2} \mathrm{sgn}\left (b x^{2} + a\right )}{d^{3}}\right )} x - \frac{52 \, b c d e \mathrm{sgn}\left (b x^{2} + a\right ) - 48 \, a d^{2} e \mathrm{sgn}\left (b x^{2} + a\right ) - 15 \, b e^{3} \mathrm{sgn}\left (b x^{2} + a\right )}{d^{3}}\right )} + \frac{{\left (16 \, b c^{2} d^{2} \mathrm{sgn}\left (b x^{2} + a\right ) - 64 \, a c d^{3} \mathrm{sgn}\left (b x^{2} + a\right ) - 24 \, b c d e^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + 16 \, a d^{2} e^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + 5 \, b e^{4} \mathrm{sgn}\left (b x^{2} + a\right )\right )} \log \left ({\left | -2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + x e + c}\right )} \sqrt{d} - e \right |}\right )}{128 \, d^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(d*x^2 + x*e + c)*(2*(4*(6*b*x*sgn(b*x^2 + a) + b*e*sgn(b*x^2 + a)/d)*x + (12*b*c*d^2*sgn(b*x^2 + a)
+ 48*a*d^3*sgn(b*x^2 + a) - 5*b*d*e^2*sgn(b*x^2 + a))/d^3)*x - (52*b*c*d*e*sgn(b*x^2 + a) - 48*a*d^2*e*sgn(b*
x^2 + a) - 15*b*e^3*sgn(b*x^2 + a))/d^3) + 1/128*(16*b*c^2*d^2*sgn(b*x^2 + a) - 64*a*c*d^3*sgn(b*x^2 + a) - 24
*b*c*d*e^2*sgn(b*x^2 + a) + 16*a*d^2*e^2*sgn(b*x^2 + a) + 5*b*e^4*sgn(b*x^2 + a))*log(abs(-2*(sqrt(d)*x - sqrt
(d*x^2 + x*e + c))*sqrt(d) - e))/d^(7/2)