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

Optimal. Leaf size=119 \[ \frac{1}{3} d x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};\frac{e^2 x^2}{d^2}\right )-\frac{d^2 \left (d^2-e^2 x^2\right )^{p+1}}{2 e^3 (p+1)}+\frac{\left (d^2-e^2 x^2\right )^{p+2}}{2 e^3 (p+2)} \]

[Out]

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

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Rubi [A]  time = 0.0671999, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {764, 365, 364, 266, 43} \[ \frac{1}{3} d x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};\frac{e^2 x^2}{d^2}\right )-\frac{d^2 \left (d^2-e^2 x^2\right )^{p+1}}{2 e^3 (p+1)}+\frac{\left (d^2-e^2 x^2\right )^{p+2}}{2 e^3 (p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 764

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

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

Mathematica [A]  time = 0.0806659, size = 103, normalized size = 0.87 \[ \frac{1}{6} \left (d^2-e^2 x^2\right )^p \left (2 d x^3 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};\frac{e^2 x^2}{d^2}\right )-\frac{3 \left (d^2-e^2 x^2\right ) \left (d^2+e^2 (p+1) x^2\right )}{e^3 (p+1) (p+2)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d^2 - e^2*x^2)^p*((-3*(d^2 - e^2*x^2)*(d^2 + e^2*(1 + p)*x^2))/(e^3*(1 + p)*(2 + p)) + (2*d*x^3*Hypergeometr
ic2F1[3/2, -p, 5/2, (e^2*x^2)/d^2])/(1 - (e^2*x^2)/d^2)^p))/6

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Maple [F]  time = 0.454, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( ex+d \right ) \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)*(-e^2*x^2+d^2)^p,x)

[Out]

int(x^2*(e*x+d)*(-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 )}{\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)*(-e^2*x^2+d^2)^p,x, algorithm="maxima")

[Out]

integrate((e*x + d)*(-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 x^{3} + d 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)*(-e^2*x^2+d^2)^p,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 5.01855, size = 382, normalized size = 3.21 \begin{align*} \frac{d 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} + 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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*d**(2*p)*x**3*hyper((3/2, -p), (5/2,), e**2*x**2*exp_polar(2*I*pi)/d**2)/3 + 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))

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

[Out]

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