3.252 \(\int x^2 (d+e x)^2 (d^2-e^2 x^2)^p \, dx\)

Optimal. Leaf size=155 \[ \frac{2 d^2 (p+4) x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};\frac{e^2 x^2}{d^2}\right )}{3 (2 p+5)}-\frac{x^3 \left (d^2-e^2 x^2\right )^{p+1}}{2 p+5}-\frac{d^3 \left (d^2-e^2 x^2\right )^{p+1}}{e^3 (p+1)}+\frac{d \left (d^2-e^2 x^2\right )^{p+2}}{e^3 (p+2)} \]

[Out]

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

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Rubi [A]  time = 0.138683, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {1652, 459, 365, 364, 12, 266, 43} \[ \frac{2 d^2 (p+4) x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};\frac{e^2 x^2}{d^2}\right )}{3 (2 p+5)}-\frac{x^3 \left (d^2-e^2 x^2\right )^{p+1}}{2 p+5}-\frac{d^3 \left (d^2-e^2 x^2\right )^{p+1}}{e^3 (p+1)}+\frac{d \left (d^2-e^2 x^2\right )^{p+2}}{e^3 (p+2)} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(d + e*x)^2*(d^2 - e^2*x^2)^p,x]

[Out]

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

Rule 1652

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[x^m*Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2)^p, x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2] && IGtQ[m, -2] &&  !
IntegerQ[2*p]

Rule 459

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

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

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

Rule 12

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.116613, size = 168, normalized size = 1.08 \[ \frac{\left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (5 d^2 e^3 \left (p^2+3 p+2\right ) x^3 \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};\frac{e^2 x^2}{d^2}\right )+3 e^5 \left (p^2+3 p+2\right ) x^5 \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};\frac{e^2 x^2}{d^2}\right )-15 d \left (d^2-e^2 x^2\right ) \left (d^2+e^2 (p+1) x^2\right ) \left (1-\frac{e^2 x^2}{d^2}\right )^p\right )}{15 e^3 (p+1) (p+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(d + e*x)^2*(d^2 - e^2*x^2)^p,x]

[Out]

((d^2 - e^2*x^2)^p*(-15*d*(d^2 - e^2*x^2)*(1 - (e^2*x^2)/d^2)^p*(d^2 + e^2*(1 + p)*x^2) + 5*d^2*e^3*(2 + 3*p +
p^2)*x^3*Hypergeometric2F1[3/2, -p, 5/2, (e^2*x^2)/d^2] + 3*e^5*(2 + 3*p + p^2)*x^5*Hypergeometric2F1[5/2, -p
, 7/2, (e^2*x^2)/d^2]))/(15*e^3*(1 + p)*(2 + p)*(1 - (e^2*x^2)/d^2)^p)

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Maple [F]  time = 0.653, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( ex+d \right ) ^{2} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(e*x+d)^2*(-e^2*x^2+d^2)^p,x)

[Out]

int(x^2*(e*x+d)^2*(-e^2*x^2+d^2)^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}^{2}{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x+d)^2*(-e^2*x^2+d^2)^p,x, algorithm="maxima")

[Out]

integrate((e*x + d)^2*(-e^2*x^2 + d^2)^p*x^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (e^{2} x^{4} + 2 \, d e x^{3} + d^{2} x^{2}\right )}{\left (-e^{2} x^{2} + d^{2}\right )}^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x+d)^2*(-e^2*x^2+d^2)^p,x, algorithm="fricas")

[Out]

integral((e^2*x^4 + 2*d*e*x^3 + d^2*x^2)*(-e^2*x^2 + d^2)^p, x)

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Sympy [B]  time = 6.48471, size = 425, normalized size = 2.74 \begin{align*} \frac{d^{2} d^{2 p} x^{3}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, - p \\ \frac{5}{2} \end{matrix}\middle |{\frac{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{3} + 2 d e \left (\begin{cases} \frac{x^{4} \left (d^{2}\right )^{p}}{4} & \text{for}\: e = 0 \\- \frac{d^{2} \log{\left (- \frac{d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac{d^{2} \log{\left (\frac{d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac{d^{2}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac{e^{2} x^{2} \log{\left (- \frac{d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac{e^{2} x^{2} \log{\left (\frac{d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} & \text{for}\: p = -2 \\- \frac{d^{2} \log{\left (- \frac{d}{e} + x \right )}}{2 e^{4}} - \frac{d^{2} \log{\left (\frac{d}{e} + x \right )}}{2 e^{4}} - \frac{x^{2}}{2 e^{2}} & \text{for}\: p = -1 \\- \frac{d^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} - \frac{d^{2} e^{2} p x^{2} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac{e^{4} p x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac{e^{4} x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} & \text{otherwise} \end{cases}\right ) + \frac{d^{2 p} e^{2} x^{5}{{}_{2}F_{1}\left (\begin{matrix} \frac{5}{2}, - p \\ \frac{7}{2} \end{matrix}\middle |{\frac{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(e*x+d)**2*(-e**2*x**2+d**2)**p,x)

[Out]

d**2*d**(2*p)*x**3*hyper((3/2, -p), (5/2,), e**2*x**2*exp_polar(2*I*pi)/d**2)/3 + 2*d*e*Piecewise((x**4*(d**2)
**p/4, Eq(e, 0)), (-d**2*log(-d/e + x)/(-2*d**2*e**4 + 2*e**6*x**2) - d**2*log(d/e + x)/(-2*d**2*e**4 + 2*e**6
*x**2) - d**2/(-2*d**2*e**4 + 2*e**6*x**2) + e**2*x**2*log(-d/e + x)/(-2*d**2*e**4 + 2*e**6*x**2) + e**2*x**2*
log(d/e + x)/(-2*d**2*e**4 + 2*e**6*x**2), Eq(p, -2)), (-d**2*log(-d/e + x)/(2*e**4) - d**2*log(d/e + x)/(2*e*
*4) - x**2/(2*e**2), Eq(p, -1)), (-d**4*(d**2 - e**2*x**2)**p/(2*e**4*p**2 + 6*e**4*p + 4*e**4) - d**2*e**2*p*
x**2*(d**2 - e**2*x**2)**p/(2*e**4*p**2 + 6*e**4*p + 4*e**4) + e**4*p*x**4*(d**2 - e**2*x**2)**p/(2*e**4*p**2
+ 6*e**4*p + 4*e**4) + e**4*x**4*(d**2 - e**2*x**2)**p/(2*e**4*p**2 + 6*e**4*p + 4*e**4), True)) + d**(2*p)*e*
*2*x**5*hyper((5/2, -p), (7/2,), e**2*x**2*exp_polar(2*I*pi)/d**2)/5

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}^{2}{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x+d)^2*(-e^2*x^2+d^2)^p,x, algorithm="giac")

[Out]

integrate((e*x + d)^2*(-e^2*x^2 + d^2)^p*x^2, x)