3.453 \(\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{x^4 (d+e x)} \, dx\)
Optimal. Leaf size=211 \[ \frac{\left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac{x \left (a e^2+c d^2\right )+2 a d e}{2 \sqrt{a} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{16 a^{3/2} d^{5/2} e^{3/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 d x^3}-\frac{\left (\frac{c}{a e}-\frac{e}{d^2}\right ) \left (x \left (a e^2+c d^2\right )+2 a d e\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 x^2} \]
[Out]
-((c/(a*e) - e/d^2)*(2*a*d*e + (c*d^2 + a*e^2)*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(8*x^2) - (a*d*
e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(3*d*x^3) + ((c*d^2 - a*e^2)^3*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])])/(16*a^(3/2)*d^(5/2)*e^(3/2))
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Rubi [A] time = 0.236498, antiderivative size = 211, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used =
{849, 806, 720, 724, 206} \[ \frac{\left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac{x \left (a e^2+c d^2\right )+2 a d e}{2 \sqrt{a} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{16 a^{3/2} d^{5/2} e^{3/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 d x^3}-\frac{\left (\frac{c}{a e}-\frac{e}{d^2}\right ) \left (x \left (a e^2+c d^2\right )+2 a d e\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 x^2} \]
Antiderivative was successfully verified.
[In]
Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(x^4*(d + e*x)),x]
[Out]
-((c/(a*e) - e/d^2)*(2*a*d*e + (c*d^2 + a*e^2)*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(8*x^2) - (a*d*
e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(3*d*x^3) + ((c*d^2 - a*e^2)^3*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])])/(16*a^(3/2)*d^(5/2)*e^(3/2))
Rule 849
Int[((x_)^(n_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Int[x^n*(a/d + (c*
x)/e)*(a + b*x + c*x^2)^(p - 1), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b
*d*e + a*e^2, 0] && !IntegerQ[p] && ( !IntegerQ[n] || !IntegerQ[2*p] || IGtQ[n, 2] || (GtQ[p, 0] && NeQ[n, 2
]))
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 720
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*
(d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^p)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(p*(b^2 -
4*a*c))/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m +
2*p + 2, 0] && GtQ[p, 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{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{x^4 (d+e x)} \, dx &=\int \frac{(a e+c d x) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{x^4} \, dx\\ &=-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 d x^3}-\frac{\left (-2 a c d^2 e+a e \left (c d^2+a e^2\right )\right ) \int \frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{x^3} \, dx}{2 a d e}\\ &=-\frac{\left (\frac{c}{a e}-\frac{e}{d^2}\right ) \left (2 a d e+\left (c d^2+a e^2\right ) x\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 x^2}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 d x^3}-\frac{\left (c d^2-a e^2\right )^3 \int \frac{1}{x \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{16 a d^2 e}\\ &=-\frac{\left (\frac{c}{a e}-\frac{e}{d^2}\right ) \left (2 a d e+\left (c d^2+a e^2\right ) x\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 x^2}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 d x^3}+\frac{\left (c d^2-a e^2\right )^3 \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 )}{8 a d^2 e}\\ &=-\frac{\left (\frac{c}{a e}-\frac{e}{d^2}\right ) \left (2 a d e+\left (c d^2+a e^2\right ) x\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 x^2}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 d x^3}+\frac{\left (c d^2-a e^2\right )^3 \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 )}{16 a^{3/2} d^{5/2} e^{3/2}}\\ \end{align*}
Mathematica [A] time = 0.283088, size = 188, normalized size = 0.89 \[ \frac{\sqrt{(d+e x) (a e+c d x)} \left (\frac{3 \left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a e+c d x}}{\sqrt{a} \sqrt{e} \sqrt{d+e x}}\right )}{\sqrt{d+e x} \sqrt{a e+c d x}}-\frac{\sqrt{a} \sqrt{d} \sqrt{e} \left (a^2 e^2 \left (8 d^2+2 d e x-3 e^2 x^2\right )+2 a c d^2 e x (7 d+4 e x)+3 c^2 d^4 x^2\right )}{x^3}\right )}{24 a^{3/2} d^{5/2} e^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(x^4*(d + e*x)),x]
[Out]
(Sqrt[(a*e + c*d*x)*(d + e*x)]*(-((Sqrt[a]*Sqrt[d]*Sqrt[e]*(3*c^2*d^4*x^2 + 2*a*c*d^2*e*x*(7*d + 4*e*x) + a^2*
e^2*(8*d^2 + 2*d*e*x - 3*e^2*x^2)))/x^3) + (3*(c*d^2 - a*e^2)^3*ArcTanh[(Sqrt[d]*Sqrt[a*e + c*d*x])/(Sqrt[a]*S
qrt[e]*Sqrt[d + e*x])])/(Sqrt[a*e + c*d*x]*Sqrt[d + e*x])))/(24*a^(3/2)*d^(5/2)*e^(3/2))
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Maple [B] time = 0.077, size = 1945, normalized size = 9.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/x^4/(e*x+d),x)
[Out]
-1/8*d^3/a^2/e^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*c^3+11/24/d^2/a*e*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(
3/2)*c+3/16*a*e^3/(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+1/24/a^3/e^3/x*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)*c^2-1/3/d^2/a/e/x^3*(a*d*e+(a*e^2+c*d^2)*x+
c*d*e*x^2)^(5/2)+1/3/d/a^2*c^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*x-3/16*d^2*e/(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^2-1/16*e^2*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)-1/16*e^5/d^2*a^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)-1/4*e^5/d^4*a*
(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x-1/8*e^6/d^5*a^2/c*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+1/3*e^3/d^
4*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(3/2)+3/8*e^3/d^4*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-3/16*e^6/d
^3*a^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)+3
/8*e^3/d^2*c*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x+1/4*e^5/d^4*a*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(
1/2)*x+1/8*e^6/d^5*a^2/c*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)+3/16*e^6/d^3*a^2*ln((1/2*a*e^2-1/2*c*d^
2+(d/e+x)*c*d*e)/(d*e*c)^(1/2)+(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2))/(d*e*c)^(1/2)-1/4*e^3/d^2*c*(c*d
*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*x+1/16*e^2*d*c^2*ln((1/2*a*e^2-1/2*c*d^2+(d/e+x)*c*d*e)/(d*e*c)^(1/2
)+(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2))/(d*e*c)^(1/2)-17/24/d^4/a*e/x*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^
2)^(5/2)-1/24*d^2/a^3/e^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*c^3+5/24/a^2/e*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*
x^2)^(3/2)*c^2+1/8*d/a*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*c^2+7/12/d^3/a/x^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e
*x^2)^(5/2)-1/8*e^2/d*c*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)+1/8*e^4/d^3*a*(a*d*e+(a*e^2+c*d^2)*x+c*d
*e*x^2)^(1/2)+1/16*e^8/d^5*a^3/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)+3/16*e^4/d*a*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)-1/16*e^8/d^5*a^3/c*ln((1/2*a*e^2-1/2*c*d^2+(d/e+x)*c*d*e)/(d*e*c)^(1/2)+(c*d*e
*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2))/(d*e*c)^(1/2)-3/16*e^4/d*a*c*ln((1/2*a*e^2-1/2*c*d^2+(d/e+x)*c*d*e)/(
d*e*c)^(1/2)+(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2))/(d*e*c)^(1/2)+1/16*d^4/a/e/(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^3+17/24/d^3/a*e^2*c*(a*d*e+(a
*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*x-1/24*d/a^3/e^2*c^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*x+1/12/d/a^2/e^2/x
^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)*c-1/3/d^2/a^2/e/x*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)*c-1/8*d^2
/a^2/e*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x*c^3
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/x^4/(e*x+d),x, algorithm="maxima")
[Out]
integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(3/2)/((e*x + d)*x^4), x)
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Fricas [A] time = 10.3689, size = 1168, normalized size = 5.54 \begin{align*} \left [-\frac{3 \,{\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{a d e} x^{3} \log \left (\frac{8 \, a^{2} d^{2} e^{2} +{\left (c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} x^{2} - 4 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (2 \, a d e +{\left (c d^{2} + a e^{2}\right )} x\right )} \sqrt{a d e} + 8 \,{\left (a c d^{3} e + a^{2} d e^{3}\right )} x}{x^{2}}\right ) + 4 \,{\left (8 \, a^{3} d^{3} e^{3} +{\left (3 \, a c^{2} d^{5} e + 8 \, a^{2} c d^{3} e^{3} - 3 \, a^{3} d e^{5}\right )} x^{2} + 2 \,{\left (7 \, a^{2} c d^{4} e^{2} + a^{3} 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 \, a^{2} d^{3} e^{2} x^{3}}, -\frac{3 \,{\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{-a d e} x^{3} \arctan \left (\frac{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (2 \, a d e +{\left (c d^{2} + a e^{2}\right )} x\right )} \sqrt{-a d e}}{2 \,{\left (a c d^{2} e^{2} x^{2} + a^{2} d^{2} e^{2} +{\left (a c d^{3} e + a^{2} d e^{3}\right )} x\right )}}\right ) + 2 \,{\left (8 \, a^{3} d^{3} e^{3} +{\left (3 \, a c^{2} d^{5} e + 8 \, a^{2} c d^{3} e^{3} - 3 \, a^{3} d e^{5}\right )} x^{2} + 2 \,{\left (7 \, a^{2} c d^{4} e^{2} + a^{3} 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 \, a^{2} d^{3} e^{2} x^{3}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/x^4/(e*x+d),x, algorithm="fricas")
[Out]
[-1/96*(3*(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(a*d*e)*x^3*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*(8*a^3*d^3*e^3 + (3*a*c^2*d^5*e + 8*a^2*c*d^3*e^3 - 3*a^3
*d*e^5)*x^2 + 2*(7*a^2*c*d^4*e^2 + a^3*d^2*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(a^2*d^3*e^2*x
^3), -1/48*(3*(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(-a*d*e)*x^3*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*(8*a^3*d^3*e^3 + (3*a*c^2*d^5*e + 8*a^2*c*d^3*e^3 - 3*a^3*d*e^5)*x^2 + 2*(7*a^2*
c*d^4*e^2 + a^3*d^2*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(a^2*d^3*e^2*x^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((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2)/x**4/(e*x+d),x)
[Out]
Timed out
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/x^4/(e*x+d),x, algorithm="giac")
[Out]
Exception raised: NotImplementedError