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]

-1/2*a^(3/2)*arctanh(1/2*(b*x^2+2*a)/a^(1/2)/(c*x^4+b*x^2+a)^(1/2))/d-1/2*(a*e^2-b*d*e+c*d^2)^(3/2)*arctanh(1/
2*(b*d-2*a*e+(-b*e+2*c*d)*x^2)/(a*e^2-b*d*e+c*d^2)^(1/2)/(c*x^4+b*x^2+a)^(1/2))/d/e^3+1/4*a*b*arctanh(1/2*(2*c
*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/d/c^(1/2)+1/16*(8*c^2*d^3+b*e^2*(-4*a*e+3*b*d)-12*c*d*e*(-a*e+b*d))*arc
tanh(1/2*(2*c*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/d/e^3/c^(1/2)+1/2*a*(c*x^4+b*x^2+a)^(1/2)/d-1/8*(4*c*d^2-e
*(-4*a*e+5*b*d)-2*c*d*e*x^2)*(c*x^4+b*x^2+a)^(1/2)/d/e^2

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Rubi [A]  time = 0.57, antiderivative size = 350, normalized size of antiderivative = 1.00, 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 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 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 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 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.53, 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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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r: Bad Argument Type

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maple [B]  time = 0.04, size = 1270, normalized size = 3.63 \[ \text {result too large to display} \]

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]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}}}{{\left (e x^{2} + d\right )} x}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{x\,\left (e\,x^2+d\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}{x \left (d + e x^{2}\right )}\, dx \]

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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