3.343 \(\int \frac {x^5}{(d+e x^2) (a+b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=167 \[ \frac {d^2 \tanh ^{-1}\left (\frac {-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt {a+b x^2+c x^4} \sqrt {a e^2-b d e+c d^2}}\right )}{2 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac {x^2 \left (-a b e-2 a c d+b^2 d\right )+a (b d-2 a e)}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4} \left (a e^2-b d e+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1251, 1646, 12, 724, 206} \[ \frac {d^2 \tanh ^{-1}\left (\frac {-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt {a+b x^2+c x^4} \sqrt {a e^2-b d e+c d^2}}\right )}{2 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac {x^2 \left (-a b e-2 a c d+b^2 d\right )+a (b d-2 a e)}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4} \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^5/((d + e*x^2)*(a + b*x^2 + c*x^4)^(3/2)),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 1646

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomi
alQuotient[(d + e*x)^m*Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2,
x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2
*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(d
+ e*x)^m*(a + b*x + c*x^2)^(p + 1)*ExpandToSum[((p + 1)*(b^2 - 4*a*c)*Q)/(d + e*x)^m - ((2*p + 3)*(2*c*f - b*
g))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.62, size = 204, normalized size = 1.22 \[ \frac {1}{2} \left (\frac {2 \left (-2 a^2 e+a b \left (d-e x^2\right )-2 a c d x^2+b^2 d x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4} \left (e (b d-a e)-c d^2\right )}-\frac {d^2 \log \left (2 \sqrt {a+b x^2+c x^4} \sqrt {a e^2-b d e+c d^2}+2 a e-b d+b e x^2-2 c d x^2\right )}{\left (e (a e-b d)+c d^2\right )^{3/2}}+\frac {d^2 \log \left (d+e x^2\right )}{\left (e (a e-b d)+c d^2\right )^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/((d + e*x^2)*(a + b*x^2 + c*x^4)^(3/2)),x]

[Out]

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

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fricas [B]  time = 1.97, size = 1381, normalized size = 8.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(e*x^2+d)/(c*x^4+b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(((b^2*c - 4*a*c^2)*d^2*x^4 + (b^3 - 4*a*b*c)*d^2*x^2 + (a*b^2 - 4*a^2*c)*d^2)*sqrt(c*d^2 - b*d*e + a*e^2
)*log(-((8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*x^4 - 8*a*b*d*e + 8*a^2*e^2 + (b^2 + 4*a*c)*d^2 + 2*(4*b*c
*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x^2 + 4*sqrt(c*x^4 + b*x^2 + a)*sqrt(c*d^2 - b*d*e + a*e^2)*((2*c*d -
b*e)*x^2 + b*d - 2*a*e))/(e^2*x^4 + 2*d*e*x^2 + d^2)) - 4*(a*b*c*d^3 + 3*a^2*b*d*e^2 - 2*a^3*e^3 - (a*b^2 + 2*
a^2*c)*d^2*e - (a^2*b*e^3 - (b^2*c - 2*a*c^2)*d^3 + (b^3 - a*b*c)*d^2*e - 2*(a*b^2 - a^2*c)*d*e^2)*x^2)*sqrt(c
*x^4 + b*x^2 + a))/((a*b^2*c^2 - 4*a^2*c^3)*d^4 - 2*(a*b^3*c - 4*a^2*b*c^2)*d^3*e + (a*b^4 - 2*a^2*b^2*c - 8*a
^3*c^2)*d^2*e^2 - 2*(a^2*b^3 - 4*a^3*b*c)*d*e^3 + (a^3*b^2 - 4*a^4*c)*e^4 + ((b^2*c^3 - 4*a*c^4)*d^4 - 2*(b^3*
c^2 - 4*a*b*c^3)*d^3*e + (b^4*c - 2*a*b^2*c^2 - 8*a^2*c^3)*d^2*e^2 - 2*(a*b^3*c - 4*a^2*b*c^2)*d*e^3 + (a^2*b^
2*c - 4*a^3*c^2)*e^4)*x^4 + ((b^3*c^2 - 4*a*b*c^3)*d^4 - 2*(b^4*c - 4*a*b^2*c^2)*d^3*e + (b^5 - 2*a*b^3*c - 8*
a^2*b*c^2)*d^2*e^2 - 2*(a*b^4 - 4*a^2*b^2*c)*d*e^3 + (a^2*b^3 - 4*a^3*b*c)*e^4)*x^2), 1/2*(((b^2*c - 4*a*c^2)*
d^2*x^4 + (b^3 - 4*a*b*c)*d^2*x^2 + (a*b^2 - 4*a^2*c)*d^2)*sqrt(-c*d^2 + b*d*e - a*e^2)*arctan(-1/2*sqrt(c*x^4
+ b*x^2 + a)*sqrt(-c*d^2 + b*d*e - a*e^2)*((2*c*d - b*e)*x^2 + b*d - 2*a*e)/((c^2*d^2 - b*c*d*e + a*c*e^2)*x^
4 + a*c*d^2 - a*b*d*e + a^2*e^2 + (b*c*d^2 - b^2*d*e + a*b*e^2)*x^2)) - 2*(a*b*c*d^3 + 3*a^2*b*d*e^2 - 2*a^3*e
^3 - (a*b^2 + 2*a^2*c)*d^2*e - (a^2*b*e^3 - (b^2*c - 2*a*c^2)*d^3 + (b^3 - a*b*c)*d^2*e - 2*(a*b^2 - a^2*c)*d*
e^2)*x^2)*sqrt(c*x^4 + b*x^2 + a))/((a*b^2*c^2 - 4*a^2*c^3)*d^4 - 2*(a*b^3*c - 4*a^2*b*c^2)*d^3*e + (a*b^4 - 2
*a^2*b^2*c - 8*a^3*c^2)*d^2*e^2 - 2*(a^2*b^3 - 4*a^3*b*c)*d*e^3 + (a^3*b^2 - 4*a^4*c)*e^4 + ((b^2*c^3 - 4*a*c^
4)*d^4 - 2*(b^3*c^2 - 4*a*b*c^3)*d^3*e + (b^4*c - 2*a*b^2*c^2 - 8*a^2*c^3)*d^2*e^2 - 2*(a*b^3*c - 4*a^2*b*c^2)
*d*e^3 + (a^2*b^2*c - 4*a^3*c^2)*e^4)*x^4 + ((b^3*c^2 - 4*a*b*c^3)*d^4 - 2*(b^4*c - 4*a*b^2*c^2)*d^3*e + (b^5
- 2*a*b^3*c - 8*a^2*b*c^2)*d^2*e^2 - 2*(a*b^4 - 4*a^2*b^2*c)*d*e^3 + (a^2*b^3 - 4*a^3*b*c)*e^4)*x^2)]

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giac [B]  time = 0.72, size = 458, normalized size = 2.74 \[ \frac {d^{2} \arctan \left (-\frac {{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} + b d e - a e^{2}}}\right )}{{\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {-c d^{2} + b d e - a e^{2}}} - \frac {\frac {{\left (b^{2} c d^{3} - 2 \, a c^{2} d^{3} - b^{3} d^{2} e + a b c d^{2} e + 2 \, a b^{2} d e^{2} - 2 \, a^{2} c d e^{2} - a^{2} b e^{3}\right )} x^{2}}{b^{2} c^{2} d^{4} - 4 \, a c^{3} d^{4} - 2 \, b^{3} c d^{3} e + 8 \, a b c^{2} d^{3} e + b^{4} d^{2} e^{2} - 2 \, a b^{2} c d^{2} e^{2} - 8 \, a^{2} c^{2} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + 8 \, a^{2} b c d e^{3} + a^{2} b^{2} e^{4} - 4 \, a^{3} c e^{4}} + \frac {a b c d^{3} - a b^{2} d^{2} e - 2 \, a^{2} c d^{2} e + 3 \, a^{2} b d e^{2} - 2 \, a^{3} e^{3}}{b^{2} c^{2} d^{4} - 4 \, a c^{3} d^{4} - 2 \, b^{3} c d^{3} e + 8 \, a b c^{2} d^{3} e + b^{4} d^{2} e^{2} - 2 \, a b^{2} c d^{2} e^{2} - 8 \, a^{2} c^{2} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + 8 \, a^{2} b c d e^{3} + a^{2} b^{2} e^{4} - 4 \, a^{3} c e^{4}}}{\sqrt {c x^{4} + b x^{2} + a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(e*x^2+d)/(c*x^4+b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

d^2*arctan(-((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e - a*e^2))/((c*d^2 - b*
d*e + a*e^2)*sqrt(-c*d^2 + b*d*e - a*e^2)) - ((b^2*c*d^3 - 2*a*c^2*d^3 - b^3*d^2*e + a*b*c*d^2*e + 2*a*b^2*d*e
^2 - 2*a^2*c*d*e^2 - a^2*b*e^3)*x^2/(b^2*c^2*d^4 - 4*a*c^3*d^4 - 2*b^3*c*d^3*e + 8*a*b*c^2*d^3*e + b^4*d^2*e^2
- 2*a*b^2*c*d^2*e^2 - 8*a^2*c^2*d^2*e^2 - 2*a*b^3*d*e^3 + 8*a^2*b*c*d*e^3 + a^2*b^2*e^4 - 4*a^3*c*e^4) + (a*b
*c*d^3 - a*b^2*d^2*e - 2*a^2*c*d^2*e + 3*a^2*b*d*e^2 - 2*a^3*e^3)/(b^2*c^2*d^4 - 4*a*c^3*d^4 - 2*b^3*c*d^3*e +
8*a*b*c^2*d^3*e + b^4*d^2*e^2 - 2*a*b^2*c*d^2*e^2 - 8*a^2*c^2*d^2*e^2 - 2*a*b^3*d*e^3 + 8*a^2*b*c*d*e^3 + a^2
*b^2*e^4 - 4*a^3*c*e^4))/sqrt(c*x^4 + b*x^2 + a)

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maple [B]  time = 0.02, size = 613, normalized size = 3.67 \[ \frac {2 c \,d^{2} \ln \left (\frac {\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {2 a \,e^{2}-2 d e b +2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x^{2}+\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}}{x^{2}+\frac {d}{e}}\right )}{\left (-b e +2 c d +\sqrt {-4 a c +b^{2}}\, e \right ) \left (b e -2 c d +\sqrt {-4 a c +b^{2}}\, e \right ) \sqrt {\frac {a \,e^{2}-d e b +c \,d^{2}}{e^{2}}}\, e}-\frac {b \,x^{2}}{\sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (4 a c -b^{2}\right ) e}-\frac {2 c d \,x^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, e^{2}}+\frac {2 \sqrt {\left (x^{2}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )^{2} c -\sqrt {-4 a c +b^{2}}\, \left (x^{2}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}\, c \,d^{2}}{\left (b e -2 c d +\sqrt {-4 a c +b^{2}}\, e \right ) \left (-4 a c +b^{2}\right ) \left (x^{2}+\frac {b}{2 c}+\frac {\sqrt {-4 a c +b^{2}}}{2 c}\right ) e^{2}}-\frac {2 \sqrt {\left (x^{2}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )^{2} c +\sqrt {-4 a c +b^{2}}\, \left (x^{2}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}\, c \,d^{2}}{\left (-b e +2 c d +\sqrt {-4 a c +b^{2}}\, e \right ) \left (-4 a c +b^{2}\right ) \left (x^{2}+\frac {b}{2 c}-\frac {\sqrt {-4 a c +b^{2}}}{2 c}\right ) e^{2}}-\frac {2 a}{\sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (4 a c -b^{2}\right ) e}-\frac {b d}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(e*x^2+d)/(c*x^4+b*x^2+a)^(3/2),x)

[Out]

-1/e/(c*x^4+b*x^2+a)^(1/2)/(4*a*c-b^2)*x^2*b-2/e/(c*x^4+b*x^2+a)^(1/2)/(4*a*c-b^2)*a-2/e^2*d/(4*a*c-b^2)/(c*x^
4+b*x^2+a)^(1/2)*c*x^2-1/e^2*d/(4*a*c-b^2)/(c*x^4+b*x^2+a)^(1/2)*b+2*d^2/e^2*c/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e
)/(-4*a*c+b^2)/(x^2+1/2*b/c+1/2*(-4*a*c+b^2)^(1/2)/c)*((x^2+1/2*(b+(-4*a*c+b^2)^(1/2))/c)^2*c-(-4*a*c+b^2)^(1/
2)*(x^2+1/2*(b+(-4*a*c+b^2)^(1/2))/c))^(1/2)+2*d^2/e*c/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)/(b*e-2*c*d+(-4*a*c+b^
2)^(1/2)*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))-2*d^2/
e^2*c/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-4*a*c+b^2)/(x^2+1/2*b/c-1/2*(-4*a*c+b^2)^(1/2)/c)*((x^2-1/2*(-b+(-4*
a*c+b^2)^(1/2))/c)^2*c+(-4*a*c+b^2)^(1/2)*(x^2-1/2*(-b+(-4*a*c+b^2)^(1/2))/c))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac {3}{2}} {\left (e x^{2} + d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(e*x^2+d)/(c*x^4+b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^5/((c*x^4 + b*x^2 + a)^(3/2)*(e*x^2 + d)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^5}{\left (e\,x^2+d\right )\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/((d + e*x^2)*(a + b*x^2 + c*x^4)^(3/2)),x)

[Out]

int(x^5/((d + e*x^2)*(a + b*x^2 + c*x^4)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{\left (d + e x^{2}\right ) \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(e*x**2+d)/(c*x**4+b*x**2+a)**(3/2),x)

[Out]

Integral(x**5/((d + e*x**2)*(a + b*x**2 + c*x**4)**(3/2)), x)

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