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

Optimal. Leaf size=394 \[ -\frac{\left (31 a^2 c d^2 e^4-15 a^3 e^6-9 a c^2 d^4 e^2+9 c^3 d^6\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 a^2 d^3 e^2 x \left (c d^2-a e^2\right )^3}+\frac{2 \left (c d e x \left (-5 a^2 e^4+10 a c d^2 e^2+3 c^2 d^4\right )+9 a^2 c d^2 e^4-5 a^3 e^6+a c^2 d^4 e^2+3 c^3 d^6\right )}{3 a d^2 e x \left (c d^2-a e^2\right )^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{\left (5 a e^2+3 c d^2\right ) \tanh ^{-1}\left (\frac{x \left (a e^2+c d^2\right )+2 a d e}{2 \sqrt{a} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{2 a^{5/2} d^{7/2} e^{5/2}}-\frac{2 e (a e+c d x)}{3 d x \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \]

[Out]

(-2*e*(a*e + c*d*x))/(3*d*(c*d^2 - a*e^2)*x*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) + (2*(3*c^3*d^6 + a
*c^2*d^4*e^2 + 9*a^2*c*d^2*e^4 - 5*a^3*e^6 + c*d*e*(3*c^2*d^4 + 10*a*c*d^2*e^2 - 5*a^2*e^4)*x))/(3*a*d^2*e*(c*
d^2 - a*e^2)^3*x*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - ((9*c^3*d^6 - 9*a*c^2*d^4*e^2 + 31*a^2*c*d^2*e
^4 - 15*a^3*e^6)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(3*a^2*d^3*e^2*(c*d^2 - a*e^2)^3*x) + ((3*c*d^2
+ 5*a*e^2)*ArcTanh[(2*a*d*e + (c*d^2 + a*e^2)*x)/(2*Sqrt[a]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c
*d*e*x^2])])/(2*a^(5/2)*d^(7/2)*e^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.590436, antiderivative size = 394, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {851, 822, 806, 724, 206} \[ -\frac{\left (31 a^2 c d^2 e^4-15 a^3 e^6-9 a c^2 d^4 e^2+9 c^3 d^6\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 a^2 d^3 e^2 x \left (c d^2-a e^2\right )^3}+\frac{2 \left (c d e x \left (-5 a^2 e^4+10 a c d^2 e^2+3 c^2 d^4\right )+9 a^2 c d^2 e^4-5 a^3 e^6+a c^2 d^4 e^2+3 c^3 d^6\right )}{3 a d^2 e x \left (c d^2-a e^2\right )^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{\left (5 a e^2+3 c d^2\right ) \tanh ^{-1}\left (\frac{x \left (a e^2+c d^2\right )+2 a d e}{2 \sqrt{a} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{2 a^{5/2} d^{7/2} e^{5/2}}-\frac{2 e (a e+c d x)}{3 d x \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*(a*e + c*d*x))/(3*d*(c*d^2 - a*e^2)*x*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) + (2*(3*c^3*d^6 + a
*c^2*d^4*e^2 + 9*a^2*c*d^2*e^4 - 5*a^3*e^6 + c*d*e*(3*c^2*d^4 + 10*a*c*d^2*e^2 - 5*a^2*e^4)*x))/(3*a*d^2*e*(c*
d^2 - a*e^2)^3*x*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - ((9*c^3*d^6 - 9*a*c^2*d^4*e^2 + 31*a^2*c*d^2*e
^4 - 15*a^3*e^6)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(3*a^2*d^3*e^2*(c*d^2 - a*e^2)^3*x) + ((3*c*d^2
+ 5*a*e^2)*ArcTanh[(2*a*d*e + (c*d^2 + a*e^2)*x)/(2*Sqrt[a]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c
*d*e*x^2])])/(2*a^(5/2)*d^(7/2)*e^(5/2))

Rule 851

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Int[((f + g*x)^n*(a + b*x + c*x^2)^(m + p))/(a/d + (c*x)/e)^m, x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] &&
NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[m, 0] && In
tegerQ[n] && (LtQ[n, 0] || GtQ[p, 0])

Rule 822

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

Rule 806

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

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 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{1}{x^2 (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=\int \frac{a e+c d x}{x^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx\\ &=-\frac{2 e (a e+c d x)}{3 d \left (c d^2-a e^2\right ) x \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac{2 \int \frac{-\frac{1}{2} a e \left (3 c d^2-5 a e^2\right ) \left (c d^2-a e^2\right )+3 a c d e^2 \left (c d^2-a e^2\right ) x}{x^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{3 a d e \left (c d^2-a e^2\right )^2}\\ &=-\frac{2 e (a e+c d x)}{3 d \left (c d^2-a e^2\right ) x \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac{2 \left (3 c^3 d^6+a c^2 d^4 e^2+9 a^2 c d^2 e^4-5 a^3 e^6+c d e \left (3 c^2 d^4+10 a c d^2 e^2-5 a^2 e^4\right ) x\right )}{3 a d^2 e \left (c d^2-a e^2\right )^3 x \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{4 \int \frac{\frac{1}{4} a e \left (c d^2-a e^2\right ) \left (9 c^3 d^6-9 a c^2 d^4 e^2+31 a^2 c d^2 e^4-15 a^3 e^6\right )+\frac{1}{2} a c d e^2 \left (c d^2-a e^2\right ) \left (3 c^2 d^4+10 a c d^2 e^2-5 a^2 e^4\right ) x}{x^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{3 a^2 d^2 e^2 \left (c d^2-a e^2\right )^4}\\ &=-\frac{2 e (a e+c d x)}{3 d \left (c d^2-a e^2\right ) x \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac{2 \left (3 c^3 d^6+a c^2 d^4 e^2+9 a^2 c d^2 e^4-5 a^3 e^6+c d e \left (3 c^2 d^4+10 a c d^2 e^2-5 a^2 e^4\right ) x\right )}{3 a d^2 e \left (c d^2-a e^2\right )^3 x \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{\left (9 c^3 d^6-9 a c^2 d^4 e^2+31 a^2 c d^2 e^4-15 a^3 e^6\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 a^2 d^3 e^2 \left (c d^2-a e^2\right )^3 x}-\frac{\left (3 c d^2+5 a e^2\right ) \int \frac{1}{x \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{2 a^2 d^3 e^2}\\ &=-\frac{2 e (a e+c d x)}{3 d \left (c d^2-a e^2\right ) x \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac{2 \left (3 c^3 d^6+a c^2 d^4 e^2+9 a^2 c d^2 e^4-5 a^3 e^6+c d e \left (3 c^2 d^4+10 a c d^2 e^2-5 a^2 e^4\right ) x\right )}{3 a d^2 e \left (c d^2-a e^2\right )^3 x \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{\left (9 c^3 d^6-9 a c^2 d^4 e^2+31 a^2 c d^2 e^4-15 a^3 e^6\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 a^2 d^3 e^2 \left (c d^2-a e^2\right )^3 x}+\frac{\left (3 c d^2+5 a e^2\right ) \operatorname{Subst}\left (\int \frac{1}{4 a d e-x^2} \, dx,x,\frac{2 a d e-\left (-c d^2-a e^2\right ) x}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{a^2 d^3 e^2}\\ &=-\frac{2 e (a e+c d x)}{3 d \left (c d^2-a e^2\right ) x \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac{2 \left (3 c^3 d^6+a c^2 d^4 e^2+9 a^2 c d^2 e^4-5 a^3 e^6+c d e \left (3 c^2 d^4+10 a c d^2 e^2-5 a^2 e^4\right ) x\right )}{3 a d^2 e \left (c d^2-a e^2\right )^3 x \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{\left (9 c^3 d^6-9 a c^2 d^4 e^2+31 a^2 c d^2 e^4-15 a^3 e^6\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 a^2 d^3 e^2 \left (c d^2-a e^2\right )^3 x}+\frac{\left (3 c d^2+5 a e^2\right ) \tanh ^{-1}\left (\frac{2 a d e+\left (c d^2+a e^2\right ) x}{2 \sqrt{a} \sqrt{d} \sqrt{e} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{2 a^{5/2} d^{7/2} e^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.589288, size = 370, normalized size = 0.94 \[ \frac{(a e+c d x) \left (\sqrt{a} d^{3/2} \sqrt{e} x \left (a e^2-c d^2\right ) \left (5 a^2 e^5-6 a c d^2 e^3+9 c^2 d^4 e\right ) (a e+c d x)+x (d+e x) \sqrt{a e+c d x} \left (\sqrt{a} \sqrt{d} \sqrt{e} \left (-31 a^2 c d^2 e^5+15 a^3 e^7+9 a c^2 d^4 e^3-9 c^3 d^6 e\right ) \sqrt{a e+c d x}+3 \sqrt{d+e x} \left (5 a e^2+3 c d^2\right ) \left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a e+c d x}}{\sqrt{a} \sqrt{e} \sqrt{d+e x}}\right )\right )+3 a^{3/2} d^{5/2} e^{3/2} \left (a e^2-c d^2\right )^3+3 \sqrt{a} c d^{7/2} \sqrt{e} x \left (c d^2-a e^2\right )^2 \left (a e^2-3 c d^2\right )\right )}{3 a^{5/2} d^{7/2} e^{5/2} x \left (c d^2-a e^2\right )^3 ((d+e x) (a e+c d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*e + c*d*x)*(3*a^(3/2)*d^(5/2)*e^(3/2)*(-(c*d^2) + a*e^2)^3 + 3*Sqrt[a]*c*d^(7/2)*Sqrt[e]*(c*d^2 - a*e^2)^2
*(-3*c*d^2 + a*e^2)*x + Sqrt[a]*d^(3/2)*Sqrt[e]*(-(c*d^2) + a*e^2)*(9*c^2*d^4*e - 6*a*c*d^2*e^3 + 5*a^2*e^5)*x
*(a*e + c*d*x) + x*Sqrt[a*e + c*d*x]*(d + e*x)*(Sqrt[a]*Sqrt[d]*Sqrt[e]*(-9*c^3*d^6*e + 9*a*c^2*d^4*e^3 - 31*a
^2*c*d^2*e^5 + 15*a^3*e^7)*Sqrt[a*e + c*d*x] + 3*(c*d^2 - a*e^2)^3*(3*c*d^2 + 5*a*e^2)*Sqrt[d + e*x]*ArcTanh[(
Sqrt[d]*Sqrt[a*e + c*d*x])/(Sqrt[a]*Sqrt[e]*Sqrt[d + e*x])])))/(3*a^(5/2)*d^(7/2)*e^(5/2)*(c*d^2 - a*e^2)^3*x*
((a*e + c*d*x)*(d + e*x))^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.065, size = 912, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2/d^3/a/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+5*e^3/d^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*
d^2)*x+c*d*e*x^2)^(1/2)*x*c+5/2*e^4/d^3*a/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(
1/2)+5/2*e^2/d/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*c+3/2*d/a/(-a^2*e^4+2*
a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*c^2+5/2/d^3/a/(a*d*e)^(1/2)*ln((2*a*d*e+(a*e^2+c*
d^2)*x+2*(a*d*e)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/x)-2/3*e/d^2/(a*e^2-c*d^2)/(d/e+x)/(c*d*e*(d/e
+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)+16/3*e^3*c^2/(a*e^2-c*d^2)^3/(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*
x+8/3*e^4/d*c/(a*e^2-c*d^2)^3/(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*a+8/3*e^2*d*c^2/(a*e^2-c*d^2)^3/(c
*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)-1/d^2/a/e/x/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-3/2/d/a^2/e^2/
(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*c+3*d^2/a^2/e/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+
c*d*e*x^2)^(1/2)*x*c^3+3/2*d^3/a^2/e^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2
)*c^3+3/2/d/a^2/e^2/(a*d*e)^(1/2)*ln((2*a*d*e+(a*e^2+c*d^2)*x+2*(a*d*e)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2
)^(1/2))/x)*c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 128.576, size = 3671, normalized size = 9.32 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*((3*c^5*d^9*e^2 - 4*a*c^4*d^7*e^4 - 6*a^2*c^3*d^5*e^6 + 12*a^3*c^2*d^3*e^8 - 5*a^4*c*d*e^10)*x^4 + (6
*c^5*d^10*e - 5*a*c^4*d^8*e^3 - 16*a^2*c^3*d^6*e^5 + 18*a^3*c^2*d^4*e^7 + 2*a^4*c*d^2*e^9 - 5*a^5*e^11)*x^3 +
(3*c^5*d^11 + 2*a*c^4*d^9*e^2 - 14*a^2*c^3*d^7*e^4 + 19*a^4*c*d^3*e^8 - 10*a^5*d*e^10)*x^2 + (3*a*c^4*d^10*e -
4*a^2*c^3*d^8*e^3 - 6*a^3*c^2*d^6*e^5 + 12*a^4*c*d^4*e^7 - 5*a^5*d^2*e^9)*x)*sqrt(a*d*e)*log((8*a^2*d^2*e^2 +
(c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4)*x^2 + 4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*a*d*e + (c*d^2 + a
*e^2)*x)*sqrt(a*d*e) + 8*(a*c*d^3*e + a^2*d*e^3)*x)/x^2) - 4*(3*a^2*c^3*d^9*e^2 - 9*a^3*c^2*d^7*e^4 + 9*a^4*c*
d^5*e^6 - 3*a^5*d^3*e^8 + (9*a*c^4*d^8*e^3 - 9*a^2*c^3*d^6*e^5 + 31*a^3*c^2*d^4*e^7 - 15*a^4*c*d^2*e^9)*x^3 +
(18*a*c^4*d^9*e^2 - 15*a^2*c^3*d^7*e^4 + 33*a^3*c^2*d^5*e^6 + 11*a^4*c*d^3*e^8 - 15*a^5*d*e^10)*x^2 + (9*a*c^4
*d^10*e - 3*a^2*c^3*d^8*e^3 - 9*a^3*c^2*d^6*e^5 + 39*a^4*c*d^4*e^7 - 20*a^5*d^2*e^9)*x)*sqrt(c*d*e*x^2 + a*d*e
+ (c*d^2 + a*e^2)*x))/((a^3*c^4*d^11*e^5 - 3*a^4*c^3*d^9*e^7 + 3*a^5*c^2*d^7*e^9 - a^6*c*d^5*e^11)*x^4 + (2*a
^3*c^4*d^12*e^4 - 5*a^4*c^3*d^10*e^6 + 3*a^5*c^2*d^8*e^8 + a^6*c*d^6*e^10 - a^7*d^4*e^12)*x^3 + (a^3*c^4*d^13*
e^3 - a^4*c^3*d^11*e^5 - 3*a^5*c^2*d^9*e^7 + 5*a^6*c*d^7*e^9 - 2*a^7*d^5*e^11)*x^2 + (a^4*c^3*d^12*e^4 - 3*a^5
*c^2*d^10*e^6 + 3*a^6*c*d^8*e^8 - a^7*d^6*e^10)*x), -1/6*(3*((3*c^5*d^9*e^2 - 4*a*c^4*d^7*e^4 - 6*a^2*c^3*d^5*
e^6 + 12*a^3*c^2*d^3*e^8 - 5*a^4*c*d*e^10)*x^4 + (6*c^5*d^10*e - 5*a*c^4*d^8*e^3 - 16*a^2*c^3*d^6*e^5 + 18*a^3
*c^2*d^4*e^7 + 2*a^4*c*d^2*e^9 - 5*a^5*e^11)*x^3 + (3*c^5*d^11 + 2*a*c^4*d^9*e^2 - 14*a^2*c^3*d^7*e^4 + 19*a^4
*c*d^3*e^8 - 10*a^5*d*e^10)*x^2 + (3*a*c^4*d^10*e - 4*a^2*c^3*d^8*e^3 - 6*a^3*c^2*d^6*e^5 + 12*a^4*c*d^4*e^7 -
5*a^5*d^2*e^9)*x)*sqrt(-a*d*e)*arctan(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*a*d*e + (c*d^2 + a*e
^2)*x)*sqrt(-a*d*e)/(a*c*d^2*e^2*x^2 + a^2*d^2*e^2 + (a*c*d^3*e + a^2*d*e^3)*x)) + 2*(3*a^2*c^3*d^9*e^2 - 9*a^
3*c^2*d^7*e^4 + 9*a^4*c*d^5*e^6 - 3*a^5*d^3*e^8 + (9*a*c^4*d^8*e^3 - 9*a^2*c^3*d^6*e^5 + 31*a^3*c^2*d^4*e^7 -
15*a^4*c*d^2*e^9)*x^3 + (18*a*c^4*d^9*e^2 - 15*a^2*c^3*d^7*e^4 + 33*a^3*c^2*d^5*e^6 + 11*a^4*c*d^3*e^8 - 15*a^
5*d*e^10)*x^2 + (9*a*c^4*d^10*e - 3*a^2*c^3*d^8*e^3 - 9*a^3*c^2*d^6*e^5 + 39*a^4*c*d^4*e^7 - 20*a^5*d^2*e^9)*x
)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/((a^3*c^4*d^11*e^5 - 3*a^4*c^3*d^9*e^7 + 3*a^5*c^2*d^7*e^9 - a^
6*c*d^5*e^11)*x^4 + (2*a^3*c^4*d^12*e^4 - 5*a^4*c^3*d^10*e^6 + 3*a^5*c^2*d^8*e^8 + a^6*c*d^6*e^10 - a^7*d^4*e^
12)*x^3 + (a^3*c^4*d^13*e^3 - a^4*c^3*d^11*e^5 - 3*a^5*c^2*d^9*e^7 + 5*a^6*c*d^7*e^9 - 2*a^7*d^5*e^11)*x^2 + (
a^4*c^3*d^12*e^4 - 3*a^5*c^2*d^10*e^6 + 3*a^6*c*d^8*e^8 - a^7*d^6*e^10)*x)]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \left [\mathit{undef}, \mathit{undef}, \mathit{undef}, 1\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[undef, undef, undef, 1]