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

Optimal. Leaf size=89 \[ \frac {1}{3} e x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )-\frac {d \left (d^2-e^2 x^2\right )^{p+1}}{2 e^2 (p+1)} \]

[Out]

-1/2*d*(-e^2*x^2+d^2)^(1+p)/e^2/(1+p)+1/3*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 = 0.03, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {764, 261, 365, 364} \[ \frac {1}{3} e x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )-\frac {d \left (d^2-e^2 x^2\right )^{p+1}}{2 e^2 (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x*(d + e*x)*(d^2 - e^2*x^2)^p,x]

[Out]

-(d*(d^2 - e^2*x^2)^(1 + p))/(2*e^2*(1 + p)) + (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 261

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

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 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]

Rubi steps

\begin {align*} \int x (d+e x) \left (d^2-e^2 x^2\right )^p \, dx &=d \int x \left (d^2-e^2 x^2\right )^p \, dx+e \int x^2 \left (d^2-e^2 x^2\right )^p \, dx\\ &=-\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{2 e^2 (1+p)}+\left (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 \left (d^2-e^2 x^2\right )^{1+p}}{2 e^2 (1+p)}+\frac {1}{3} 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 = 0.04, size = 89, normalized size = 1.00 \[ \frac {1}{3} e x^3 \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )-\frac {d \left (d^2-e^2 x^2\right )^{p+1}}{2 e^2 (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e x^{2} + d x\right )} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(d^2 - e^2*x^2)^p*(d + e*x),x)

[Out]

int(x*(d^2 - e^2*x^2)^p*(d + e*x), x)

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sympy [A]  time = 3.52, size = 85, normalized size = 0.96 \[ d \left (\begin {cases} \frac {x^{2} \left (d^{2}\right )^{p}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\begin {cases} \frac {\left (d^{2} - e^{2} x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (d^{2} - e^{2} x^{2} \right )} & \text {otherwise} \end {cases}}{2 e^{2}} & \text {otherwise} \end {cases}\right ) + \frac {d^{2 p} e x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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