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

Optimal. Leaf size=350 \[ -\frac{a^{3/2} \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{2 d}+\frac{\left (-12 c d e (b d-a e)+b e^2 (3 b d-4 a e)+8 c^2 d^3\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{16 \sqrt{c} d e^3}-\frac{\sqrt{a+b x^2+c x^4} \left (-e (5 b d-4 a e)+4 c d^2-2 c d e x^2\right )}{8 d e^2}-\frac{\left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt{a+b x^2+c x^4} \sqrt{a e^2-b d e+c d^2}}\right )}{2 d e^3}+\frac{a \sqrt{a+b x^2+c x^4}}{2 d}+\frac{a b \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{4 \sqrt{c} d} \]

[Out]

(a*Sqrt[a + b*x^2 + c*x^4])/(2*d) - ((4*c*d^2 - e*(5*b*d - 4*a*e) - 2*c*d*e*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*d
*e^2) - (a^(3/2)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(2*d) + (a*b*ArcTanh[(b + 2*c*x^2
)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(4*Sqrt[c]*d) + ((8*c^2*d^3 + b*e^2*(3*b*d - 4*a*e) - 12*c*d*e*(b*d -
a*e))*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(16*Sqrt[c]*d*e^3) - ((c*d^2 - b*d*e + a*e^2
)^(3/2)*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x^2)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x^2 + c*x^4])])/(2
*d*e^3)

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Rubi [A]  time = 0.573223, antiderivative size = 350, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {1251, 895, 734, 843, 621, 206, 724, 814} \[ -\frac{a^{3/2} \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{2 d}+\frac{\left (-12 c d e (b d-a e)+b e^2 (3 b d-4 a e)+8 c^2 d^3\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{16 \sqrt{c} d e^3}-\frac{\sqrt{a+b x^2+c x^4} \left (-e (5 b d-4 a e)+4 c d^2-2 c d e x^2\right )}{8 d e^2}-\frac{\left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt{a+b x^2+c x^4} \sqrt{a e^2-b d e+c d^2}}\right )}{2 d e^3}+\frac{a \sqrt{a+b x^2+c x^4}}{2 d}+\frac{a b \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{4 \sqrt{c} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a + b*x^2 + c*x^4])/(2*d) - ((4*c*d^2 - e*(5*b*d - 4*a*e) - 2*c*d*e*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*d
*e^2) - (a^(3/2)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(2*d) + (a*b*ArcTanh[(b + 2*c*x^2
)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(4*Sqrt[c]*d) + ((8*c^2*d^3 + b*e^2*(3*b*d - 4*a*e) - 12*c*d*e*(b*d -
a*e))*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(16*Sqrt[c]*d*e^3) - ((c*d^2 - b*d*e + a*e^2
)^(3/2)*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x^2)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x^2 + c*x^4])])/(2
*d*e^3)

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 895

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

Rule 734

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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]
&&  !IGtQ[m, 0]

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])

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 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*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] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps

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

Mathematica [A]  time = 0.521335, size = 251, normalized size = 0.72 \[ \frac{1}{16} \left (-\frac{8 a^{3/2} \tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{d}+\frac{\left (12 c e (a e-b d)+3 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{\sqrt{c} e^3}+\frac{2 \left (4 \left (e (a e-b d)+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{2 a e-b d+b e x^2-2 c d x^2}{2 \sqrt{a+b x^2+c x^4} \sqrt{e (a e-b d)+c d^2}}\right )+d e \sqrt{a+b x^2+c x^4} \left (5 b e-4 c d+2 c e x^2\right )\right )}{d e^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*a^(3/2)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/d + ((8*c^2*d^2 + 3*b^2*e^2 + 12*c*e*
(-(b*d) + a*e))*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(Sqrt[c]*e^3) + (2*(d*e*(-4*c*d +
5*b*e + 2*c*e*x^2)*Sqrt[a + b*x^2 + c*x^4] + 4*(c*d^2 + e*(-(b*d) + a*e))^(3/2)*ArcTanh[(-(b*d) + 2*a*e - 2*c*
d*x^2 + b*e*x^2)/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + b*x^2 + c*x^4])]))/(d*e^3))/16

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Maple [B]  time = 0.021, size = 1270, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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