### 3.575 $$\int \frac{1}{(d+e x)^3 (a+c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=223 $\frac{c d e \sqrt{a+c x^2} \left (2 c d^2-13 a e^2\right )}{2 a (d+e x) \left (a e^2+c d^2\right )^3}+\frac{e \sqrt{a+c x^2} \left (2 c d^2-3 a e^2\right )}{2 a (d+e x)^2 \left (a e^2+c d^2\right )^2}+\frac{a e+c d x}{a \sqrt{a+c x^2} (d+e x)^2 \left (a e^2+c d^2\right )}-\frac{3 c e^2 \left (4 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{2 \left (a e^2+c d^2\right )^{7/2}}$

[Out]

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

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Rubi [A]  time = 0.198464, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.263, Rules used = {741, 835, 807, 725, 206} $\frac{c d e \sqrt{a+c x^2} \left (2 c d^2-13 a e^2\right )}{2 a (d+e x) \left (a e^2+c d^2\right )^3}+\frac{e \sqrt{a+c x^2} \left (2 c d^2-3 a e^2\right )}{2 a (d+e x)^2 \left (a e^2+c d^2\right )^2}+\frac{a e+c d x}{a \sqrt{a+c x^2} (d+e x)^2 \left (a e^2+c d^2\right )}-\frac{3 c e^2 \left (4 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{2 \left (a e^2+c d^2\right )^{7/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((d + e*x)^3*(a + c*x^2)^(3/2)),x]

[Out]

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

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 807

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

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

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{1}{(d+e x)^3 \left (a+c x^2\right )^{3/2}} \, dx &=\frac{a e+c d x}{a \left (c d^2+a e^2\right ) (d+e x)^2 \sqrt{a+c x^2}}-\frac{\int \frac{-3 a e^2-2 c d e x}{(d+e x)^3 \sqrt{a+c x^2}} \, dx}{a \left (c d^2+a e^2\right )}\\ &=\frac{a e+c d x}{a \left (c d^2+a e^2\right ) (d+e x)^2 \sqrt{a+c x^2}}+\frac{e \left (2 c d^2-3 a e^2\right ) \sqrt{a+c x^2}}{2 a \left (c d^2+a e^2\right )^2 (d+e x)^2}+\frac{\int \frac{10 a c d e^2+c e \left (2 c d^2-3 a e^2\right ) x}{(d+e x)^2 \sqrt{a+c x^2}} \, dx}{2 a \left (c d^2+a e^2\right )^2}\\ &=\frac{a e+c d x}{a \left (c d^2+a e^2\right ) (d+e x)^2 \sqrt{a+c x^2}}+\frac{e \left (2 c d^2-3 a e^2\right ) \sqrt{a+c x^2}}{2 a \left (c d^2+a e^2\right )^2 (d+e x)^2}+\frac{c d e \left (2 c d^2-13 a e^2\right ) \sqrt{a+c x^2}}{2 a \left (c d^2+a e^2\right )^3 (d+e x)}+\frac{\left (3 c e^2 \left (4 c d^2-a e^2\right )\right ) \int \frac{1}{(d+e x) \sqrt{a+c x^2}} \, dx}{2 \left (c d^2+a e^2\right )^3}\\ &=\frac{a e+c d x}{a \left (c d^2+a e^2\right ) (d+e x)^2 \sqrt{a+c x^2}}+\frac{e \left (2 c d^2-3 a e^2\right ) \sqrt{a+c x^2}}{2 a \left (c d^2+a e^2\right )^2 (d+e x)^2}+\frac{c d e \left (2 c d^2-13 a e^2\right ) \sqrt{a+c x^2}}{2 a \left (c d^2+a e^2\right )^3 (d+e x)}-\frac{\left (3 c e^2 \left (4 c d^2-a e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c d^2+a e^2-x^2} \, dx,x,\frac{a e-c d x}{\sqrt{a+c x^2}}\right )}{2 \left (c d^2+a e^2\right )^3}\\ &=\frac{a e+c d x}{a \left (c d^2+a e^2\right ) (d+e x)^2 \sqrt{a+c x^2}}+\frac{e \left (2 c d^2-3 a e^2\right ) \sqrt{a+c x^2}}{2 a \left (c d^2+a e^2\right )^2 (d+e x)^2}+\frac{c d e \left (2 c d^2-13 a e^2\right ) \sqrt{a+c x^2}}{2 a \left (c d^2+a e^2\right )^3 (d+e x)}-\frac{3 c e^2 \left (4 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{c d^2+a e^2} \sqrt{a+c x^2}}\right )}{2 \left (c d^2+a e^2\right )^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.399432, size = 240, normalized size = 1.08 $\frac{1}{2} \left (\frac{-a^2 c e^3 \left (10 d^2+11 d e x+3 e^2 x^2\right )-a^3 e^5+a c^2 d e \left (6 d^2 e x+6 d^3-14 d e^2 x^2-13 e^3 x^3\right )+2 c^3 d^3 x (d+e x)^2}{a \sqrt{a+c x^2} (d+e x)^2 \left (a e^2+c d^2\right )^3}+\frac{3 c e^2 \left (a e^2-4 c d^2\right ) \log \left (\sqrt{a+c x^2} \sqrt{a e^2+c d^2}+a e-c d x\right )}{\left (a e^2+c d^2\right )^{7/2}}+\frac{3 c e^2 \left (4 c d^2-a e^2\right ) \log (d+e x)}{\left (a e^2+c d^2\right )^{7/2}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((d + e*x)^3*(a + c*x^2)^(3/2)),x]

[Out]

((-(a^3*e^5) + 2*c^3*d^3*x*(d + e*x)^2 - a^2*c*e^3*(10*d^2 + 11*d*e*x + 3*e^2*x^2) + a*c^2*d*e*(6*d^3 + 6*d^2*
e*x - 14*d*e^2*x^2 - 13*e^3*x^3))/(a*(c*d^2 + a*e^2)^3*(d + e*x)^2*Sqrt[a + c*x^2]) + (3*c*e^2*(4*c*d^2 - a*e^
2)*Log[d + e*x])/(c*d^2 + a*e^2)^(7/2) + (3*c*e^2*(-4*c*d^2 + a*e^2)*Log[a*e - c*d*x + Sqrt[c*d^2 + a*e^2]*Sqr
t[a + c*x^2]])/(c*d^2 + a*e^2)^(7/2))/2

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Maple [B]  time = 0.193, size = 681, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/e/(a*e^2+c*d^2)/(d/e+x)^2/(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)-5/2*c*d/(a*e^2+c*d^2)^2/(
d/e+x)/(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)+15/2*e*c^2*d^2/(a*e^2+c*d^2)^3/(c*(d/e+x)^2-2*c*d
/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)+15/2*c^3*d^3/(a*e^2+c*d^2)^3/a/(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/
e^2)^(1/2)*x-15/2*e*c^2*d^2/(a*e^2+c*d^2)^3/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2*c*d/e*(d/e+x)+
2*((a*e^2+c*d^2)/e^2)^(1/2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))/(d/e+x))-13/2*c^2*d/(a*e^2+
c*d^2)^2/a/(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)*x-3/2*e/(a*e^2+c*d^2)^2*c/(c*(d/e+x)^2-2*c*d/
e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2)+3/2*e/(a*e^2+c*d^2)^2*c/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2
*c*d/e*(d/e+x)+2*((a*e^2+c*d^2)/e^2)^(1/2)*(c*(d/e+x)^2-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))/(d/e+x))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 7.5163, size = 3078, normalized size = 13.8 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(3*(4*a^2*c^2*d^4*e^2 - a^3*c*d^2*e^4 + (4*a*c^3*d^2*e^4 - a^2*c^2*e^6)*x^4 + 2*(4*a*c^3*d^3*e^3 - a^2*c^
2*d*e^5)*x^3 + (4*a*c^3*d^4*e^2 + 3*a^2*c^2*d^2*e^4 - a^3*c*e^6)*x^2 + 2*(4*a^2*c^2*d^3*e^3 - a^3*c*d*e^5)*x)*
sqrt(c*d^2 + a*e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 - 2*sqrt(c*d^2 + a*e^2)
*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*(6*a*c^3*d^6*e - 4*a^2*c^2*d^4*e^3 - 11*a^3*c*d
^2*e^5 - a^4*e^7 + (2*c^4*d^5*e^2 - 11*a*c^3*d^3*e^4 - 13*a^2*c^2*d*e^6)*x^3 + (4*c^4*d^6*e - 10*a*c^3*d^4*e^3
- 17*a^2*c^2*d^2*e^5 - 3*a^3*c*e^7)*x^2 + (2*c^4*d^7 + 8*a*c^3*d^5*e^2 - 5*a^2*c^2*d^3*e^4 - 11*a^3*c*d*e^6)*
x)*sqrt(c*x^2 + a))/(a^2*c^4*d^10 + 4*a^3*c^3*d^8*e^2 + 6*a^4*c^2*d^6*e^4 + 4*a^5*c*d^4*e^6 + a^6*d^2*e^8 + (a
*c^5*d^8*e^2 + 4*a^2*c^4*d^6*e^4 + 6*a^3*c^3*d^4*e^6 + 4*a^4*c^2*d^2*e^8 + a^5*c*e^10)*x^4 + 2*(a*c^5*d^9*e +
4*a^2*c^4*d^7*e^3 + 6*a^3*c^3*d^5*e^5 + 4*a^4*c^2*d^3*e^7 + a^5*c*d*e^9)*x^3 + (a*c^5*d^10 + 5*a^2*c^4*d^8*e^2
+ 10*a^3*c^3*d^6*e^4 + 10*a^4*c^2*d^4*e^6 + 5*a^5*c*d^2*e^8 + a^6*e^10)*x^2 + 2*(a^2*c^4*d^9*e + 4*a^3*c^3*d^
7*e^3 + 6*a^4*c^2*d^5*e^5 + 4*a^5*c*d^3*e^7 + a^6*d*e^9)*x), -1/2*(3*(4*a^2*c^2*d^4*e^2 - a^3*c*d^2*e^4 + (4*a
*c^3*d^2*e^4 - a^2*c^2*e^6)*x^4 + 2*(4*a*c^3*d^3*e^3 - a^2*c^2*d*e^5)*x^3 + (4*a*c^3*d^4*e^2 + 3*a^2*c^2*d^2*e
^4 - a^3*c*e^6)*x^2 + 2*(4*a^2*c^2*d^3*e^3 - a^3*c*d*e^5)*x)*sqrt(-c*d^2 - a*e^2)*arctan(sqrt(-c*d^2 - a*e^2)*
(c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) - (6*a*c^3*d^6*e - 4*a^2*c^2*d^4*
e^3 - 11*a^3*c*d^2*e^5 - a^4*e^7 + (2*c^4*d^5*e^2 - 11*a*c^3*d^3*e^4 - 13*a^2*c^2*d*e^6)*x^3 + (4*c^4*d^6*e -
10*a*c^3*d^4*e^3 - 17*a^2*c^2*d^2*e^5 - 3*a^3*c*e^7)*x^2 + (2*c^4*d^7 + 8*a*c^3*d^5*e^2 - 5*a^2*c^2*d^3*e^4 -
11*a^3*c*d*e^6)*x)*sqrt(c*x^2 + a))/(a^2*c^4*d^10 + 4*a^3*c^3*d^8*e^2 + 6*a^4*c^2*d^6*e^4 + 4*a^5*c*d^4*e^6 +
a^6*d^2*e^8 + (a*c^5*d^8*e^2 + 4*a^2*c^4*d^6*e^4 + 6*a^3*c^3*d^4*e^6 + 4*a^4*c^2*d^2*e^8 + a^5*c*e^10)*x^4 + 2
*(a*c^5*d^9*e + 4*a^2*c^4*d^7*e^3 + 6*a^3*c^3*d^5*e^5 + 4*a^4*c^2*d^3*e^7 + a^5*c*d*e^9)*x^3 + (a*c^5*d^10 + 5
*a^2*c^4*d^8*e^2 + 10*a^3*c^3*d^6*e^4 + 10*a^4*c^2*d^4*e^6 + 5*a^5*c*d^2*e^8 + a^6*e^10)*x^2 + 2*(a^2*c^4*d^9*
e + 4*a^3*c^3*d^7*e^3 + 6*a^4*c^2*d^5*e^5 + 4*a^5*c*d^3*e^7 + a^6*d*e^9)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + c x^{2}\right )^{\frac{3}{2}} \left (d + e x\right )^{3}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.69494, size = 876, normalized size = 3.93 \begin{align*} \frac{\frac{{\left (c^{6} d^{9} - 6 \, a^{2} c^{4} d^{5} e^{4} - 8 \, a^{3} c^{3} d^{3} e^{6} - 3 \, a^{4} c^{2} d e^{8}\right )} x}{a c^{6} d^{12} + 6 \, a^{2} c^{5} d^{10} e^{2} + 15 \, a^{3} c^{4} d^{8} e^{4} + 20 \, a^{4} c^{3} d^{6} e^{6} + 15 \, a^{5} c^{2} d^{4} e^{8} + 6 \, a^{6} c d^{2} e^{10} + a^{7} e^{12}} + \frac{3 \, a c^{5} d^{8} e + 8 \, a^{2} c^{4} d^{6} e^{3} + 6 \, a^{3} c^{3} d^{4} e^{5} - a^{5} c e^{9}}{a c^{6} d^{12} + 6 \, a^{2} c^{5} d^{10} e^{2} + 15 \, a^{3} c^{4} d^{8} e^{4} + 20 \, a^{4} c^{3} d^{6} e^{6} + 15 \, a^{5} c^{2} d^{4} e^{8} + 6 \, a^{6} c d^{2} e^{10} + a^{7} e^{12}}}{\sqrt{c x^{2} + a}} + \frac{3 \,{\left (4 \, c^{2} d^{2} e^{2} - a c e^{4}\right )} \arctan \left (-\frac{{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} e + \sqrt{c} d}{\sqrt{-c d^{2} - a e^{2}}}\right )}{{\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^{2} - a e^{2}}} - \frac{14 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} c^{\frac{5}{2}} d^{3} e^{2} + 6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} c^{2} d^{2} e^{3} - 22 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} a c^{2} d^{2} e^{3} - 7 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} a c^{\frac{3}{2}} d e^{4} -{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} a c e^{5} + 7 \, a^{2} c^{\frac{3}{2}} d e^{4} -{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} a^{2} c e^{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 )}{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} e + 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} \sqrt{c} d - a e\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c^6*d^9 - 6*a^2*c^4*d^5*e^4 - 8*a^3*c^3*d^3*e^6 - 3*a^4*c^2*d*e^8)*x/(a*c^6*d^12 + 6*a^2*c^5*d^10*e^2 + 15*a
^3*c^4*d^8*e^4 + 20*a^4*c^3*d^6*e^6 + 15*a^5*c^2*d^4*e^8 + 6*a^6*c*d^2*e^10 + a^7*e^12) + (3*a*c^5*d^8*e + 8*a
^2*c^4*d^6*e^3 + 6*a^3*c^3*d^4*e^5 - a^5*c*e^9)/(a*c^6*d^12 + 6*a^2*c^5*d^10*e^2 + 15*a^3*c^4*d^8*e^4 + 20*a^4
*c^3*d^6*e^6 + 15*a^5*c^2*d^4*e^8 + 6*a^6*c*d^2*e^10 + a^7*e^12))/sqrt(c*x^2 + a) + 3*(4*c^2*d^2*e^2 - a*c*e^4
)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))/((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^2 - a*e^2)) - (14*(sqrt(c)*x - sqrt(c*x^2 + a))^2*c^(5/2)*d^3*e^2 + 6*(sqrt
(c)*x - sqrt(c*x^2 + a))^3*c^2*d^2*e^3 - 22*(sqrt(c)*x - sqrt(c*x^2 + a))*a*c^2*d^2*e^3 - 7*(sqrt(c)*x - sqrt(
c*x^2 + a))^2*a*c^(3/2)*d*e^4 - (sqrt(c)*x - sqrt(c*x^2 + a))^3*a*c*e^5 + 7*a^2*c^(3/2)*d*e^4 - (sqrt(c)*x - s
qrt(c*x^2 + a))*a^2*c*e^5)/((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)*x - sqrt(c*x^2 +
a))^2*e + 2*(sqrt(c)*x - sqrt(c*x^2 + a))*sqrt(c)*d - a*e)^2)