3.367 \(\int \frac{x^3 (d+e x^2)^{3/2}}{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=460 \[ \frac{\left (b c \left (e \left (2 d \sqrt{b^2-4 a c}-3 a e\right )+c d^2\right )+c \left (a e^2 \sqrt{b^2-4 a c}-c d \left (d \sqrt{b^2-4 a c}-4 a e\right )\right )-b^2 e \left (e \sqrt{b^2-4 a c}+2 c d\right )+b^3 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} c^{5/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\left (b c \left (c d^2-e \left (2 d \sqrt{b^2-4 a c}+3 a e\right )\right )-c \left (a e^2 \sqrt{b^2-4 a c}-c d \left (d \sqrt{b^2-4 a c}+4 a e\right )\right )-b^2 e \left (2 c d-e \sqrt{b^2-4 a c}\right )+b^3 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} c^{5/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{\sqrt{d+e x^2} (c d-b e)}{c^2}+\frac{\left (d+e x^2\right )^{3/2}}{3 c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 5.08444, antiderivative size = 460, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {1251, 824, 826, 1166, 208} \[ \frac{\left (b c \left (e \left (2 d \sqrt{b^2-4 a c}-3 a e\right )+c d^2\right )+c \left (a e^2 \sqrt{b^2-4 a c}-c d \left (d \sqrt{b^2-4 a c}-4 a e\right )\right )-b^2 e \left (e \sqrt{b^2-4 a c}+2 c d\right )+b^3 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} c^{5/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}-\frac{\left (b c \left (c d^2-e \left (2 d \sqrt{b^2-4 a c}+3 a e\right )\right )-c \left (a e^2 \sqrt{b^2-4 a c}-c d \left (d \sqrt{b^2-4 a c}+4 a e\right )\right )-b^2 e \left (2 c d-e \sqrt{b^2-4 a c}\right )+b^3 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} c^{5/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{\sqrt{d+e x^2} (c d-b e)}{c^2}+\frac{\left (d+e x^2\right )^{3/2}}{3 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(d + e*x^2)^(3/2))/(a + b*x^2 + c*x^4),x]

[Out]

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

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 824

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

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

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

Mathematica [A]  time = 0.988296, size = 457, normalized size = 0.99 \[ -\frac{\left (-b c \left (e \left (2 d \sqrt{b^2-4 a c}-3 a e\right )+c d^2\right )+c \left (c d \left (d \sqrt{b^2-4 a c}-4 a e\right )-a e^2 \sqrt{b^2-4 a c}\right )+b^2 e \left (e \sqrt{b^2-4 a c}+2 c d\right )+b^3 \left (-e^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )}{\sqrt{2} c^{5/2} \sqrt{b^2-4 a c} \sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}-\frac{\left (b c \left (c d^2-e \left (2 d \sqrt{b^2-4 a c}+3 a e\right )\right )+c \left (c d \left (d \sqrt{b^2-4 a c}+4 a e\right )-a e^2 \sqrt{b^2-4 a c}\right )+b^2 e \left (e \sqrt{b^2-4 a c}-2 c d\right )+b^3 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} c^{5/2} \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{\sqrt{d+e x^2} (c d-b e)}{c^2}+\frac{\left (d+e x^2\right )^{3/2}}{3 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(d + e*x^2)^(3/2))/(a + b*x^2 + c*x^4),x]

[Out]

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

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Maple [C]  time = 0.03, size = 490, normalized size = 1.1 \begin{align*} -{\frac{{x}^{3}}{6\,c}{e}^{{\frac{3}{2}}}}+{\frac{e{x}^{2}}{8\,c}\sqrt{e{x}^{2}+d}}-{\frac{3\,dx}{4\,c}\sqrt{e}}+{\frac{1}{24\,c} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{bx}{2\,{c}^{2}}{e}^{{\frac{3}{2}}}}-{\frac{be}{2\,{c}^{2}}\sqrt{e{x}^{2}+d}}+{\frac{5\,d}{8\,c}\sqrt{e{x}^{2}+d}}-{\frac{deb}{2\,{c}^{2}} \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{-1}}+{\frac{5\,{d}^{2}}{8\,c} \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{-1}}+{\frac{{d}^{3}}{24\,c} \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{-3}}+{\frac{1}{4\,{c}^{2}}\sum _{{\it \_R}={\it RootOf} \left ( c{{\it \_Z}}^{8}+ \left ( 4\,be-4\,cd \right ){{\it \_Z}}^{6}+ \left ( 16\,a{e}^{2}-8\,deb+6\,c{d}^{2} \right ){{\it \_Z}}^{4}+ \left ( 4\,b{d}^{2}e-4\,c{d}^{3} \right ){{\it \_Z}}^{2}+c{d}^{4} \right ) }{\frac{ \left ( -ac{e}^{2}+{b}^{2}{e}^{2}-2\,bcde+{c}^{2}{d}^{2} \right ){{\it \_R}}^{6}+ \left ( 4\,ab{e}^{3}-5\,{e}^{2}dac-3\,{b}^{2}d{e}^{2}+6\,bc{d}^{2}e-3\,{c}^{2}{d}^{3} \right ){{\it \_R}}^{4}+d \left ( -4\,ab{e}^{3}+5\,{e}^{2}dac+3\,{b}^{2}d{e}^{2}-6\,bc{d}^{2}e+3\,{c}^{2}{d}^{3} \right ){{\it \_R}}^{2}+ac{d}^{3}{e}^{2}-{b}^{2}{d}^{3}{e}^{2}+2\,bc{d}^{4}e-{c}^{2}{d}^{5}}{{{\it \_R}}^{7}c+3\,{{\it \_R}}^{5}be-3\,{{\it \_R}}^{5}cd+8\,{{\it \_R}}^{3}a{e}^{2}-4\,{{\it \_R}}^{3}bde+3\,{{\it \_R}}^{3}c{d}^{2}+{\it \_R}\,b{d}^{2}e-{\it \_R}\,c{d}^{3}}\ln \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x)

[Out]

-1/6/c*e^(3/2)*x^3+1/8/c*e*(e*x^2+d)^(1/2)*x^2-3/4/c*e^(1/2)*x*d+1/24*(e*x^2+d)^(3/2)/c+1/2/c^2*e^(3/2)*x*b-1/
2/c^2*(e*x^2+d)^(1/2)*b*e+5/8/c*(e*x^2+d)^(1/2)*d-1/2/c^2*d/((e*x^2+d)^(1/2)-e^(1/2)*x)*b*e+5/8/c*d^2/((e*x^2+
d)^(1/2)-e^(1/2)*x)+1/24/c*d^3/((e*x^2+d)^(1/2)-e^(1/2)*x)^3+1/4/c^2*sum(((-a*c*e^2+b^2*e^2-2*b*c*d*e+c^2*d^2)
*_R^6+(4*a*b*e^3-5*a*c*d*e^2-3*b^2*d*e^2+6*b*c*d^2*e-3*c^2*d^3)*_R^4+d*(-4*a*b*e^3+5*a*c*d*e^2+3*b^2*d*e^2-6*b
*c*d^2*e+3*c^2*d^3)*_R^2+a*c*d^3*e^2-b^2*d^3*e^2+2*b*c*d^4*e-c^2*d^5)/(_R^7*c+3*_R^5*b*e-3*_R^5*c*d+8*_R^3*a*e
^2-4*_R^3*b*d*e+3*_R^3*c*d^2+_R*b*d^2*e-_R*c*d^3)*ln((e*x^2+d)^(1/2)-e^(1/2)*x-_R),_R=RootOf(c*_Z^8+(4*b*e-4*c
*d)*_Z^6+(16*a*e^2-8*b*d*e+6*c*d^2)*_Z^4+(4*b*d^2*e-4*c*d^3)*_Z^2+c*d^4))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}^{\frac{3}{2}} x^{3}}{c x^{4} + b x^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)^(3/2)*x^3/(c*x^4 + b*x^2 + a), x)

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Fricas [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*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

Timed out

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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*x**2+d)**(3/2)/(c*x**4+b*x**2+a),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [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*x^2+d)^(3/2)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

Timed out