### 3.1946 $$\int \frac{(d+e x)^3}{\sqrt{a d e+(c d^2+a e^2) x+c d e x^2}} \, dx$$

Optimal. Leaf size=255 $\frac{5 \left (c d^2-a e^2\right )^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 c^3 d^3}+\frac{5 (d+e x) \left (c d^2-a e^2\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{12 c^2 d^2}+\frac{5 \left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{16 c^{7/2} d^{7/2} \sqrt{e}}+\frac{(d+e x)^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d}$

[Out]

(5*(c*d^2 - a*e^2)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(8*c^3*d^3) + (5*(c*d^2 - a*e^2)*(d + e*x)*S
qrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(12*c^2*d^2) + ((d + e*x)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e
*x^2])/(3*c*d) + (5*(c*d^2 - a*e^2)^3*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*
e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(16*c^(7/2)*d^(7/2)*Sqrt[e])

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Rubi [A]  time = 0.196533, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.108, Rules used = {670, 640, 621, 206} $\frac{5 \left (c d^2-a e^2\right )^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 c^3 d^3}+\frac{5 (d+e x) \left (c d^2-a e^2\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{12 c^2 d^2}+\frac{5 \left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{16 c^{7/2} d^{7/2} \sqrt{e}}+\frac{(d+e x)^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(5*(c*d^2 - a*e^2)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(8*c^3*d^3) + (5*(c*d^2 - a*e^2)*(d + e*x)*S
qrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(12*c^2*d^2) + ((d + e*x)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e
*x^2])/(3*c*d) + (5*(c*d^2 - a*e^2)^3*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*
e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(16*c^(7/2)*d^(7/2)*Sqrt[e])

Rule 670

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[((m + p)*(2*c*d - b*e))/(c*(m + 2*p + 1)), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

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 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{(d+e x)^3}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx &=\frac{(d+e x)^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c d}+\frac{\left (5 \left (d^2-\frac{a e^2}{c}\right )\right ) \int \frac{(d+e x)^2}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{6 d}\\ &=\frac{5 \left (c d^2-a e^2\right ) (d+e x) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{12 c^2 d^2}+\frac{(d+e x)^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c d}+\frac{\left (5 \left (d^2-\frac{a e^2}{c}\right )^2\right ) \int \frac{d+e x}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{8 d^2}\\ &=\frac{5 \left (c d^2-a e^2\right )^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 c^3 d^3}+\frac{5 \left (c d^2-a e^2\right ) (d+e x) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{12 c^2 d^2}+\frac{(d+e x)^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c d}+\frac{\left (5 \left (d^2-\frac{a e^2}{c}\right )^3\right ) \int \frac{1}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{16 d^3}\\ &=\frac{5 \left (c d^2-a e^2\right )^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 c^3 d^3}+\frac{5 \left (c d^2-a e^2\right ) (d+e x) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{12 c^2 d^2}+\frac{(d+e x)^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c d}+\frac{\left (5 \left (d^2-\frac{a e^2}{c}\right )^3\right ) \operatorname{Subst}\left (\int \frac{1}{4 c d e-x^2} \, dx,x,\frac{c d^2+a e^2+2 c d e x}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{8 d^3}\\ &=\frac{5 \left (c d^2-a e^2\right )^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 c^3 d^3}+\frac{5 \left (c d^2-a e^2\right ) (d+e x) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{12 c^2 d^2}+\frac{(d+e x)^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c d}+\frac{5 \left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac{c d^2+a e^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{16 c^{7/2} d^{7/2} \sqrt{e}}\\ \end{align*}

Mathematica [A]  time = 0.585563, size = 214, normalized size = 0.84 $\frac{\sqrt{(d+e x) (a e+c d x)} \left (\sqrt{c} \sqrt{d} \left (15 a^2 e^4-10 a c d e^2 (4 d+e x)+c^2 d^2 \left (33 d^2+26 d e x+8 e^2 x^2\right )\right )+\frac{15 \sqrt{c d} \left (c d^2-a e^2\right )^{5/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{e} \sqrt{a e+c d x}}{\sqrt{c d} \sqrt{c d^2-a e^2}}\right )}{\sqrt{e} \sqrt{a e+c d x} \sqrt{\frac{c d (d+e x)}{c d^2-a e^2}}}\right )}{24 c^{7/2} d^{7/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[(a*e + c*d*x)*(d + e*x)]*(Sqrt[c]*Sqrt[d]*(15*a^2*e^4 - 10*a*c*d*e^2*(4*d + e*x) + c^2*d^2*(33*d^2 + 26*
d*e*x + 8*e^2*x^2)) + (15*Sqrt[c*d]*(c*d^2 - a*e^2)^(5/2)*ArcSinh[(Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d*x])/
(Sqrt[c*d]*Sqrt[c*d^2 - a*e^2])])/(Sqrt[e]*Sqrt[a*e + c*d*x]*Sqrt[(c*d*(d + e*x))/(c*d^2 - a*e^2)])))/(24*c^(7
/2)*d^(7/2))

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Maple [B]  time = 0.056, size = 513, normalized size = 2. \begin{align*}{\frac{{e}^{2}{x}^{2}}{3\,cd}\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}}}-{\frac{5\,a{e}^{3}x}{12\,{c}^{2}{d}^{2}}\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}}}+{\frac{13\,ex}{12\,c}\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}}}+{\frac{5\,{a}^{2}{e}^{4}}{8\,{c}^{3}{d}^{3}}\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}}}-{\frac{5\,a{e}^{2}}{3\,{c}^{2}d}\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}}}+{\frac{11\,d}{8\,c}\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}}}-{\frac{5\,{a}^{3}{e}^{6}}{16\,{c}^{3}{d}^{3}}\ln \left ({ \left ({\frac{a{e}^{2}}{2}}+{\frac{c{d}^{2}}{2}}+cdex \right ){\frac{1}{\sqrt{dec}}}}+\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}} \right ){\frac{1}{\sqrt{dec}}}}+{\frac{15\,{a}^{2}{e}^{4}}{16\,{c}^{2}d}\ln \left ({ \left ({\frac{a{e}^{2}}{2}}+{\frac{c{d}^{2}}{2}}+cdex \right ){\frac{1}{\sqrt{dec}}}}+\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}} \right ){\frac{1}{\sqrt{dec}}}}-{\frac{15\,ad{e}^{2}}{16\,c}\ln \left ({ \left ({\frac{a{e}^{2}}{2}}+{\frac{c{d}^{2}}{2}}+cdex \right ){\frac{1}{\sqrt{dec}}}}+\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}} \right ){\frac{1}{\sqrt{dec}}}}+{\frac{5\,{d}^{3}}{16}\ln \left ({ \left ({\frac{a{e}^{2}}{2}}+{\frac{c{d}^{2}}{2}}+cdex \right ){\frac{1}{\sqrt{dec}}}}+\sqrt{ade+ \left ( a{e}^{2}+c{d}^{2} \right ) x+cde{x}^{2}} \right ){\frac{1}{\sqrt{dec}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x)

[Out]

1/3*e^2*x^2/d/c*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-5/12*e^3/d^2/c^2*x*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(
1/2)*a+13/12*e/c*x*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+5/8*e^4/d^3/c^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(
1/2)*a^2-5/3*e^2/d/c^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a+11/8*d/c*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1
/2)-5/16*e^6/d^3/c^3*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(d*e*c)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(
d*e*c)^(1/2)*a^3+15/16*e^4/d/c^2*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(d*e*c)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x
^2)^(1/2))/(d*e*c)^(1/2)*a^2-15/16*e^2*d/c*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(d*e*c)^(1/2)+(a*d*e+(a*e^2+c*d^2)
*x+c*d*e*x^2)^(1/2))/(d*e*c)^(1/2)*a+5/16*d^3*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(d*e*c)^(1/2)+(a*d*e+(a*e^2+c*d
^2)*x+c*d*e*x^2)^(1/2))/(d*e*c)^(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)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.54175, size = 1137, normalized size = 4.46 \begin{align*} \left [\frac{15 \,{\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt{c d e} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} + 4 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt{c d e} + 8 \,{\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right ) + 4 \,{\left (8 \, c^{3} d^{3} e^{3} x^{2} + 33 \, c^{3} d^{5} e - 40 \, a c^{2} d^{3} e^{3} + 15 \, a^{2} c d e^{5} + 2 \,{\left (13 \, c^{3} d^{4} e^{2} - 5 \, a c^{2} d^{2} e^{4}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{96 \, c^{4} d^{4} e}, -\frac{15 \,{\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt{-c d e} \arctan \left (\frac{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt{-c d e}}{2 \,{\left (c^{2} d^{2} e^{2} x^{2} + a c d^{2} e^{2} +{\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right )}}\right ) - 2 \,{\left (8 \, c^{3} d^{3} e^{3} x^{2} + 33 \, c^{3} d^{5} e - 40 \, a c^{2} d^{3} e^{3} + 15 \, a^{2} c d e^{5} + 2 \,{\left (13 \, c^{3} d^{4} e^{2} - 5 \, a c^{2} d^{2} e^{4}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{48 \, c^{4} d^{4} e}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(15*(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*sqrt(c*d*e)*log(8*c^2*d^2*e^2*x^2 + c^2*d^4
+ 6*a*c*d^2*e^2 + a^2*e^4 + 4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(c*d
*e) + 8*(c^2*d^3*e + a*c*d*e^3)*x) + 4*(8*c^3*d^3*e^3*x^2 + 33*c^3*d^5*e - 40*a*c^2*d^3*e^3 + 15*a^2*c*d*e^5 +
2*(13*c^3*d^4*e^2 - 5*a*c^2*d^2*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^4*d^4*e), -1/48*(15*(
c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*sqrt(-c*d*e)*arctan(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2
+ a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(-c*d*e)/(c^2*d^2*e^2*x^2 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)
*x)) - 2*(8*c^3*d^3*e^3*x^2 + 33*c^3*d^5*e - 40*a*c^2*d^3*e^3 + 15*a^2*c*d*e^5 + 2*(13*c^3*d^4*e^2 - 5*a*c^2*d
^2*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^4*d^4*e)]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2),x)

[Out]

Integral((d + e*x)**3/sqrt((d + e*x)*(a*e + c*d*x)), x)

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Giac [A]  time = 1.39729, size = 316, normalized size = 1.24 \begin{align*} \frac{1}{24} \, \sqrt{c d x^{2} e + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (2 \, x{\left (\frac{4 \, x e^{2}}{c d} + \frac{{\left (13 \, c^{2} d^{3} e^{3} - 5 \, a c d e^{5}\right )} e^{\left (-2\right )}}{c^{3} d^{3}}\right )} + \frac{{\left (33 \, c^{2} d^{4} e^{2} - 40 \, a c d^{2} e^{4} + 15 \, a^{2} e^{6}\right )} e^{\left (-2\right )}}{c^{3} d^{3}}\right )} - \frac{5 \,{\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt{c d} e^{\left (-\frac{1}{2}\right )} \log \left ({\left | -\sqrt{c d} c d^{2} e^{\frac{1}{2}} - 2 \,{\left (\sqrt{c d} x e^{\frac{1}{2}} - \sqrt{c d x^{2} e + a d e +{\left (c d^{2} + a e^{2}\right )} x}\right )} c d e - \sqrt{c d} a e^{\frac{5}{2}} \right |}\right )}{16 \, c^{4} d^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="giac")

[Out]

1/24*sqrt(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)*(2*x*(4*x*e^2/(c*d) + (13*c^2*d^3*e^3 - 5*a*c*d*e^5)*e^(-2)/(
c^3*d^3)) + (33*c^2*d^4*e^2 - 40*a*c*d^2*e^4 + 15*a^2*e^6)*e^(-2)/(c^3*d^3)) - 5/16*(c^3*d^6 - 3*a*c^2*d^4*e^2
+ 3*a^2*c*d^2*e^4 - a^3*e^6)*sqrt(c*d)*e^(-1/2)*log(abs(-sqrt(c*d)*c*d^2*e^(1/2) - 2*(sqrt(c*d)*x*e^(1/2) - s
qrt(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x))*c*d*e - sqrt(c*d)*a*e^(5/2)))/(c^4*d^4)