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

Optimal. Leaf size=118 \[ \frac {2}{3} d e x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )-\frac {d^2 \left (d^2-e^2 x^2\right )^{p+1}}{e^2 (p+1)}+\frac {\left (d^2-e^2 x^2\right )^{p+2}}{2 e^2 (p+2)} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {1652, 444, 43, 12, 365, 364} \[ \frac {2}{3} d e x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )-\frac {d^2 \left (d^2-e^2 x^2\right )^{p+1}}{e^2 (p+1)}+\frac {\left (d^2-e^2 x^2\right )^{p+2}}{2 e^2 (p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d^2*(d^2 - e^2*x^2)^(1 + p))/(e^2*(1 + p))) + (d^2 - e^2*x^2)^(2 + p)/(2*e^2*(2 + p)) + (2*d*e*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 12

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

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])

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 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 444

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 110, normalized size = 0.93 \[ \frac {\left (d^2-e^2 x^2\right )^p \left (4 d e^3 x^3 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )-\frac {3 \left (d^2-e^2 x^2\right ) \left (d^2 (p+3)+e^2 (p+1) x^2\right )}{(p+1) (p+2)}\right )}{6 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(d + e*x)^2*(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*(3 + p) + e^2*(1 + p)*x^2))/((1 + p)*(2 + p)) + (4*d*e^3*x^3*Hype
rgeometric2F1[3/2, -p, 5/2, (e^2*x^2)/d^2])/(1 - (e^2*x^2)/d^2)^p))/(6*e^2)

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e^{2} x^{3} + 2 \, d e x^{2} + d^{2} x\right )} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{2} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(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, x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{2} x \left (-e^{2} x^{2}+d^{2}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p + 1} d^{2}}{2 \, e^{2} {\left (p + 1\right )}} + \int {\left (e^{2} x^{3} + 2 \, d e x^{2}\right )} e^{\left (p \log \left (e x + d\right ) + p \log \left (-e x + d\right )\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-e^2*x^2 + d^2)^(p + 1)*d^2/(e^2*(p + 1)) + integrate((e^2*x^3 + 2*d*e*x^2)*e^(p*log(e*x + d) + p*log(-e
*x + d)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\left (d^2-e^2\,x^2\right )}^p\,{\left (d+e\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 5.92, size = 440, normalized size = 3.73 \[ d^{2} \left (\begin {cases} \frac {x^{2} \left (d^{2}\right )^{p}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\begin {cases} \frac {\left (d^{2} - e^{2} x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (d^{2} - e^{2} x^{2} \right )} & \text {otherwise} \end {cases}}{2 e^{2}} & \text {otherwise} \end {cases}\right ) + \frac {2 d d^{2 p} e x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{3} + e^{2} \left (\begin {cases} \frac {x^{4} \left (d^{2}\right )^{p}}{4} & \text {for}\: e = 0 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac {d^{2}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (- \frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac {e^{2} x^{2} \log {\left (\frac {d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} & \text {for}\: p = -2 \\- \frac {d^{2} \log {\left (- \frac {d}{e} + x \right )}}{2 e^{4}} - \frac {d^{2} \log {\left (\frac {d}{e} + x \right )}}{2 e^{4}} - \frac {x^{2}}{2 e^{2}} & \text {for}\: p = -1 \\- \frac {d^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} - \frac {d^{2} e^{2} p x^{2} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} p x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac {e^{4} x^{4} \left (d^{2} - e^{2} x^{2}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*Piecewise((x**2*(d**2)**p/2, Eq(e**2, 0)), (-Piecewise(((d**2 - e**2*x**2)**(p + 1)/(p + 1), Ne(p, -1)),
(log(d**2 - e**2*x**2), True))/(2*e**2), True)) + 2*d*d**(2*p)*e*x**3*hyper((3/2, -p), (5/2,), e**2*x**2*exp_p
olar(2*I*pi)/d**2)/3 + e**2*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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