3.5.25 \(\int \frac {x^2 (a+b x^2)^p}{(d+e x)^3} \, dx\) [425]

Optimal. Leaf size=396 \[ -\frac {d^2 \left (a+b x^2\right )^{1+p}}{2 e \left (b d^2+a e^2\right ) (d+e x)^2}+\frac {d \left (2 a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right )^2 (d+e x)}+\frac {\left (a^2 e^4+a b d^2 e^2 (2+5 p)+b^2 d^4 \left (1+3 p+2 p^2\right )\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1}{2};-p,1;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d e^2 \left (b d^2+a e^2\right )^2}-\frac {b d (1+2 p) \left (2 a e^2+b d^2 (1+p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e^2 \left (b d^2+a e^2\right )^2}-\frac {\left (a^2 e^4+a b d^2 e^2 (2+5 p)+b^2 d^4 \left (1+3 p+2 p^2\right )\right ) \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 e \left (b d^2+a e^2\right )^3 (1+p)} \]

[Out]

-1/2*d^2*(b*x^2+a)^(1+p)/e/(a*e^2+b*d^2)/(e*x+d)^2+d*(2*a*e^2+b*d^2*(1+p))*(b*x^2+a)^(1+p)/e/(a*e^2+b*d^2)^2/(
e*x+d)+(a^2*e^4+a*b*d^2*e^2*(2+5*p)+b^2*d^4*(2*p^2+3*p+1))*x*(b*x^2+a)^p*AppellF1(1/2,1,-p,3/2,e^2*x^2/d^2,-b*
x^2/a)/d/e^2/(a*e^2+b*d^2)^2/((1+b*x^2/a)^p)-b*d*(1+2*p)*(2*a*e^2+b*d^2*(1+p))*x*(b*x^2+a)^p*hypergeom([1/2, -
p],[3/2],-b*x^2/a)/e^2/(a*e^2+b*d^2)^2/((1+b*x^2/a)^p)-1/2*(a^2*e^4+a*b*d^2*e^2*(2+5*p)+b^2*d^4*(2*p^2+3*p+1))
*(b*x^2+a)^(1+p)*hypergeom([1, 1+p],[2+p],e^2*(b*x^2+a)/(a*e^2+b*d^2))/e/(a*e^2+b*d^2)^3/(1+p)

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Rubi [A]
time = 0.37, antiderivative size = 396, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1665, 849, 858, 252, 251, 771, 441, 440, 455, 70} \begin {gather*} \frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a^2 e^4+a b d^2 e^2 (5 p+2)+b^2 d^4 \left (2 p^2+3 p+1\right )\right ) F_1\left (\frac {1}{2};-p,1;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d e^2 \left (a e^2+b d^2\right )^2}-\frac {\left (a+b x^2\right )^{p+1} \left (a^2 e^4+a b d^2 e^2 (5 p+2)+b^2 d^4 \left (2 p^2+3 p+1\right )\right ) \, _2F_1\left (1,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 e (p+1) \left (a e^2+b d^2\right )^3}-\frac {b d (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a e^2+b d^2 (p+1)\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e^2 \left (a e^2+b d^2\right )^2}+\frac {d \left (a+b x^2\right )^{p+1} \left (2 a e^2+b d^2 (p+1)\right )}{e (d+e x) \left (a e^2+b d^2\right )^2}-\frac {d^2 \left (a+b x^2\right )^{p+1}}{2 e (d+e x)^2 \left (a e^2+b d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(d^2*(a + b*x^2)^(1 + p))/(e*(b*d^2 + a*e^2)*(d + e*x)^2) + (d*(2*a*e^2 + b*d^2*(1 + p))*(a + b*x^2)^(1 +
p))/(e*(b*d^2 + a*e^2)^2*(d + e*x)) + ((a^2*e^4 + a*b*d^2*e^2*(2 + 5*p) + b^2*d^4*(1 + 3*p + 2*p^2))*x*(a + b
*x^2)^p*AppellF1[1/2, -p, 1, 3/2, -((b*x^2)/a), (e^2*x^2)/d^2])/(d*e^2*(b*d^2 + a*e^2)^2*(1 + (b*x^2)/a)^p) -
(b*d*(1 + 2*p)*(2*a*e^2 + b*d^2*(1 + p))*x*(a + b*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(e^2*(
b*d^2 + a*e^2)^2*(1 + (b*x^2)/a)^p) - ((a^2*e^4 + a*b*d^2*e^2*(2 + 5*p) + b^2*d^4*(1 + 3*p + 2*p^2))*(a + b*x^
2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + a*e^2)])/(2*e*(b*d^2 + a*e^2)^3*(1 +
p))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 771

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p, (d/(d
^2 - e^2*x^2) - e*(x/(d^2 - e^2*x^2)))^(-m), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&
!IntegerQ[p] && ILtQ[m, 0]

Rule 849

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*
(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/((m + 1)*(c*d^2 + a*e^2))), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 858

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

Rule 1665

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, d
+ e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1
)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p*ExpandToSum[(m
+ 1)*(c*d^2 + a*e^2)*Q + c*d*R*(m + 1) - c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e, p}, x] && Po
lyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+b x^2\right )^p}{(d+e x)^3} \, dx &=-\frac {d^2 \left (a+b x^2\right )^{1+p}}{2 e \left (b d^2+a e^2\right ) (d+e x)^2}-\frac {\int \frac {\left (2 a d-\frac {2 \left (a e^2+b d^2 (1+p)\right ) x}{e}\right ) \left (a+b x^2\right )^p}{(d+e x)^2} \, dx}{2 \left (b d^2+a e^2\right )}\\ &=-\frac {d^2 \left (a+b x^2\right )^{1+p}}{2 e \left (b d^2+a e^2\right ) (d+e x)^2}+\frac {d \left (2 a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right )^2 (d+e x)}+\frac {\int \frac {\left (2 a \left (a e^2+b d^2 p\right )-\frac {2 b d (1+2 p) \left (2 a e^2+b d^2 (1+p)\right ) x}{e}\right ) \left (a+b x^2\right )^p}{d+e x} \, dx}{2 \left (b d^2+a e^2\right )^2}\\ &=-\frac {d^2 \left (a+b x^2\right )^{1+p}}{2 e \left (b d^2+a e^2\right ) (d+e x)^2}+\frac {d \left (2 a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right )^2 (d+e x)}-\frac {\left (b d (1+2 p) \left (2 a e^2+b d^2 (1+p)\right )\right ) \int \left (a+b x^2\right )^p \, dx}{e^2 \left (b d^2+a e^2\right )^2}+\frac {\left (a^2 e^4+a b d^2 e^2 (2+5 p)+b^2 d^4 \left (1+3 p+2 p^2\right )\right ) \int \frac {\left (a+b x^2\right )^p}{d+e x} \, dx}{e^2 \left (b d^2+a e^2\right )^2}\\ &=-\frac {d^2 \left (a+b x^2\right )^{1+p}}{2 e \left (b d^2+a e^2\right ) (d+e x)^2}+\frac {d \left (2 a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right )^2 (d+e x)}+\frac {\left (a^2 e^4+a b d^2 e^2 (2+5 p)+b^2 d^4 \left (1+3 p+2 p^2\right )\right ) \int \left (\frac {d \left (a+b x^2\right )^p}{d^2-e^2 x^2}+\frac {e x \left (a+b x^2\right )^p}{-d^2+e^2 x^2}\right ) \, dx}{e^2 \left (b d^2+a e^2\right )^2}-\frac {\left (b d (1+2 p) \left (2 a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \left (1+\frac {b x^2}{a}\right )^p \, dx}{e^2 \left (b d^2+a e^2\right )^2}\\ &=-\frac {d^2 \left (a+b x^2\right )^{1+p}}{2 e \left (b d^2+a e^2\right ) (d+e x)^2}+\frac {d \left (2 a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right )^2 (d+e x)}-\frac {b d (1+2 p) \left (2 a e^2+b d^2 (1+p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e^2 \left (b d^2+a e^2\right )^2}+\frac {\left (d \left (a^2 e^4+a b d^2 e^2 (2+5 p)+b^2 d^4 \left (1+3 p+2 p^2\right )\right )\right ) \int \frac {\left (a+b x^2\right )^p}{d^2-e^2 x^2} \, dx}{e^2 \left (b d^2+a e^2\right )^2}+\frac {\left (a^2 e^4+a b d^2 e^2 (2+5 p)+b^2 d^4 \left (1+3 p+2 p^2\right )\right ) \int \frac {x \left (a+b x^2\right )^p}{-d^2+e^2 x^2} \, dx}{e \left (b d^2+a e^2\right )^2}\\ &=-\frac {d^2 \left (a+b x^2\right )^{1+p}}{2 e \left (b d^2+a e^2\right ) (d+e x)^2}+\frac {d \left (2 a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right )^2 (d+e x)}-\frac {b d (1+2 p) \left (2 a e^2+b d^2 (1+p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e^2 \left (b d^2+a e^2\right )^2}+\frac {\left (a^2 e^4+a b d^2 e^2 (2+5 p)+b^2 d^4 \left (1+3 p+2 p^2\right )\right ) \text {Subst}\left (\int \frac {(a+b x)^p}{-d^2+e^2 x} \, dx,x,x^2\right )}{2 e \left (b d^2+a e^2\right )^2}+\frac {\left (d \left (a^2 e^4+a b d^2 e^2 (2+5 p)+b^2 d^4 \left (1+3 p+2 p^2\right )\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \frac {\left (1+\frac {b x^2}{a}\right )^p}{d^2-e^2 x^2} \, dx}{e^2 \left (b d^2+a e^2\right )^2}\\ &=-\frac {d^2 \left (a+b x^2\right )^{1+p}}{2 e \left (b d^2+a e^2\right ) (d+e x)^2}+\frac {d \left (2 a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right )^2 (d+e x)}+\frac {\left (a^2 e^4+a b d^2 e^2 (2+5 p)+b^2 d^4 \left (1+3 p+2 p^2\right )\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1}{2};-p,1;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d e^2 \left (b d^2+a e^2\right )^2}-\frac {b d (1+2 p) \left (2 a e^2+b d^2 (1+p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e^2 \left (b d^2+a e^2\right )^2}-\frac {\left (a^2 e^4+a b d^2 e^2 (2+5 p)+b^2 d^4 \left (1+3 p+2 p^2\right )\right ) \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 e \left (b d^2+a e^2\right )^3 (1+p)}\\ \end {align*}

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Mathematica [A]
time = 0.42, size = 290, normalized size = 0.73 \begin {gather*} \frac {\left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (a+b x^2\right )^p \left (-\frac {4 d F_1\left (1-2 p;-p,-p;2-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{(-1+2 p) (d+e x)}+\frac {d^2 F_1\left (2-2 p;-p,-p;3-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{(-1+p) (d+e x)^2}+\frac {F_1\left (-2 p;-p,-p;1-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{p}\right )}{2 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^p*((-4*d*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, (d - Sqrt[-(a/b)]*e)/(d + e*x), (d + Sqrt[-(a/b)]*e)/
(d + e*x)])/((-1 + 2*p)*(d + e*x)) + (d^2*AppellF1[2 - 2*p, -p, -p, 3 - 2*p, (d - Sqrt[-(a/b)]*e)/(d + e*x), (
d + Sqrt[-(a/b)]*e)/(d + e*x)])/((-1 + p)*(d + e*x)^2) + AppellF1[-2*p, -p, -p, 1 - 2*p, (d - Sqrt[-(a/b)]*e)/
(d + e*x), (d + Sqrt[-(a/b)]*e)/(d + e*x)]/p))/(2*e^3*((e*(-Sqrt[-(a/b)] + x))/(d + e*x))^p*((e*(Sqrt[-(a/b)]
+ x))/(d + e*x))^p)

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {x^{2} \left (b \,x^{2}+a \right )^{p}}{\left (e x +d \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^p*x^2/(x^3*e^3 + 3*d*x^2*e^2 + 3*d^2*x*e + d^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \left (a + b x^{2}\right )^{p}}{\left (d + e x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^2\,{\left (b\,x^2+a\right )}^p}{{\left (d+e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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