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

Optimal. Leaf size=199 \[ \frac{3 e^{3/2} (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{8 \sqrt{b} d^{7/2}}+\frac{b e \left (c+d x^2\right )^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{4 d^3}-\frac{e \left (c+d x^2\right ) (9 b c-5 a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{8 d^3}-\frac{c e (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{d^3} \]

[Out]

-((c*(b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/d^3) - ((9*b*c - 5*a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^
2)]*(c + d*x^2))/(8*d^3) + (b*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)^2)/(4*d^3) + (3*(b*c - a*d)*(5*b
*c - a*d)*e^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(8*Sqrt[b]*d^(7/2))

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Rubi [A]  time = 0.220538, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {1960, 455, 1157, 388, 208} \[ \frac{3 e^{3/2} (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{8 \sqrt{b} d^{7/2}}+\frac{b e \left (c+d x^2\right )^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{4 d^3}-\frac{e \left (c+d x^2\right ) (9 b c-5 a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{8 d^3}-\frac{c e (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*(b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/d^3) - ((9*b*c - 5*a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^
2)]*(c + d*x^2))/(8*d^3) + (b*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)^2)/(4*d^3) + (3*(b*c - a*d)*(5*b
*c - a*d)*e^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(8*Sqrt[b]*d^(7/2))

Rule 1960

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Den
ominator[p]}, Dist[(q*e*(b*c - a*d))/n, Subst[Int[(x^(q*(p + 1) - 1)*(-(a*e) + c*x^q)^(Simplify[(m + 1)/n] - 1
))/(b*e - d*x^q)^(Simplify[(m + 1)/n] + 1), x], x, ((e*(a + b*x^n))/(c + d*x^n))^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, m, n}, x] && FractionQ[p] && IntegerQ[Simplify[(m + 1)/n]]

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 388

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

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

Mathematica [A]  time = 0.621445, size = 191, normalized size = 0.96 \[ \frac{e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (3 \sqrt{b c-a d} \left (a^2 d^2-6 a b c d+5 b^2 c^2\right ) \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )+b \sqrt{d} \sqrt{a+b x^2} \left (a d \left (13 c+5 d x^2\right )+b \left (-15 c^2-5 c d x^2+2 d^2 x^4\right )\right )\right )}{8 b d^{7/2} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.013, size = 679, normalized size = 3.4 \begin{align*}{\frac{d{x}^{2}+c}{16\,{d}^{3} \left ( b{x}^{2}+a \right ) } \left ( 4\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}{x}^{4}b{d}^{2}+3\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}{a}^{2}{d}^{3}-18\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}abc{d}^{2}+15\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}{b}^{2}{c}^{2}d+10\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}{x}^{2}a{d}^{2}-10\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}{x}^{2}bcd+3\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}c{d}^{2}-18\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) ab{c}^{2}d+15\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{2}{c}^{3}+10\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}acd-14\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}b{c}^{2}+16\,\sqrt{bd}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }acd-16\,\sqrt{bd}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }b{c}^{2} \right ) \left ({\frac{e \left ( b{x}^{2}+a \right ) }{d{x}^{2}+c}} \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^4*b*d^2+3*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x
^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*a^2*d^3-18*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*
c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*a*b*c*d^2+15*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^
(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*b^2*c^2*d+10*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^2*a
*d^2-10*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^2*b*c*d+3*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x
^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*c*d^2-18*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)
^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b*c^2*d+15*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*
(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^2*c^3+10*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a*c*d-14*(b*d*x^4
+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b*c^2+16*(b*d)^(1/2)*((d*x^2+c)*(b*x^2+a))^(1/2)*a*c*d-16*(b*d)^(1/2)*
((d*x^2+c)*(b*x^2+a))^(1/2)*b*c^2)/d^3*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(3/2)/(b*d)^(1/2)/((d*x^2+c)*(b*x^2+a
))^(1/2)/(b*x^2+a)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.89172, size = 886, normalized size = 4.45 \begin{align*} \left [\frac{3 \,{\left (5 \, b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2}\right )} e \sqrt{\frac{e}{b d}} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} e x^{2} +{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e + 4 \,{\left (2 \, b^{2} d^{3} x^{4} + b^{2} c^{2} d + a b c d^{2} +{\left (3 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{\frac{e}{b d}}\right ) + 4 \,{\left (2 \, b d^{2} e x^{4} - 5 \,{\left (b c d - a d^{2}\right )} e x^{2} -{\left (15 \, b c^{2} - 13 \, a c d\right )} e\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{32 \, d^{3}}, -\frac{3 \,{\left (5 \, b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2}\right )} e \sqrt{-\frac{e}{b d}} \arctan \left (\frac{{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{-\frac{e}{b d}}}{2 \,{\left (b e x^{2} + a e\right )}}\right ) - 2 \,{\left (2 \, b d^{2} e x^{4} - 5 \,{\left (b c d - a d^{2}\right )} e x^{2} -{\left (15 \, b c^{2} - 13 \, a c d\right )} e\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{16 \, d^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(3*(5*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*e*sqrt(e/(b*d))*log(8*b^2*d^2*e*x^4 + 8*(b^2*c*d + a*b*d^2)*e*x^2 +
(b^2*c^2 + 6*a*b*c*d + a^2*d^2)*e + 4*(2*b^2*d^3*x^4 + b^2*c^2*d + a*b*c*d^2 + (3*b^2*c*d^2 + a*b*d^3)*x^2)*s
qrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(e/(b*d))) + 4*(2*b*d^2*e*x^4 - 5*(b*c*d - a*d^2)*e*x^2 - (15*b*c^2 - 13*
a*c*d)*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/d^3, -1/16*(3*(5*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*e*sqrt(-e/(b*d))*
arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(-e/(b*d))/(b*e*x^2 + a*e)) - 2*(2*b*
d^2*e*x^4 - 5*(b*c*d - a*d^2)*e*x^2 - (15*b*c^2 - 13*a*c*d)*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/d^3]

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

[Out]

Timed out

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Giac [A]  time = 1.39202, size = 409, normalized size = 2.06 \begin{align*} \frac{1}{16} \,{\left (2 \, \sqrt{b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}{\left (\frac{2 \, b x^{2}}{d^{2}} - \frac{7 \, b^{2} c d^{5} - 5 \, a b d^{6}}{b d^{8}}\right )} - \frac{16 \,{\left (b^{2} c^{3} e - 2 \, a b c^{2} d e + a^{2} c d^{2} e\right )}}{{\left (\sqrt{b d} c e^{\frac{1}{2}} +{\left (\sqrt{b d} x^{2} e^{\frac{1}{2}} - \sqrt{b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )} d\right )} d^{3}} - \frac{3 \,{\left (5 \, \sqrt{b d} b^{2} c^{2} e^{\frac{1}{2}} - 6 \, \sqrt{b d} a b c d e^{\frac{1}{2}} + \sqrt{b d} a^{2} d^{2} e^{\frac{1}{2}}\right )} \log \left ({\left | -\sqrt{b d} b c e^{\frac{1}{2}} - \sqrt{b d} a d e^{\frac{1}{2}} - 2 \,{\left (\sqrt{b d} x^{2} e^{\frac{1}{2}} - \sqrt{b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )} b d \right |}\right )}{b d^{4}}\right )} e \mathrm{sgn}\left (d x^{2} + c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*sqrt(b*d*x^4*e + b*c*x^2*e + a*d*x^2*e + a*c*e)*(2*b*x^2/d^2 - (7*b^2*c*d^5 - 5*a*b*d^6)/(b*d^8)) - 16
*(b^2*c^3*e - 2*a*b*c^2*d*e + a^2*c*d^2*e)/((sqrt(b*d)*c*e^(1/2) + (sqrt(b*d)*x^2*e^(1/2) - sqrt(b*d*x^4*e + b
*c*x^2*e + a*d*x^2*e + a*c*e))*d)*d^3) - 3*(5*sqrt(b*d)*b^2*c^2*e^(1/2) - 6*sqrt(b*d)*a*b*c*d*e^(1/2) + sqrt(b
*d)*a^2*d^2*e^(1/2))*log(abs(-sqrt(b*d)*b*c*e^(1/2) - sqrt(b*d)*a*d*e^(1/2) - 2*(sqrt(b*d)*x^2*e^(1/2) - sqrt(
b*d*x^4*e + b*c*x^2*e + a*d*x^2*e + a*c*e))*b*d))/(b*d^4))*e*sgn(d*x^2 + c)