3.3.44 \(\int x (d+e x) (d^2-e^2 x^2)^p \, dx\) [244]

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

[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 = 0.02, 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 = {778, 267, 372, 371} \begin {gather*} \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)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[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)

Rule 267

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 778

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 = 0.13, size = 89, normalized size = 1.00 \begin {gather*} -\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 {gather*}

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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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int x \left (e x +d \right ) \left (-e^{2} x^{2}+d^{2}\right )^{p}\, dx\]

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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

[Out]

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

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Sympy [A]
time = 1.45, size = 85, normalized size = 0.96 \begin {gather*} 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} \end {gather*}

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

[Out]

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

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

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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