3.262 \(\int x (d+e x)^3 (d^2-e^2 x^2)^p \, dx\)

Optimal. Leaf size=116 \[ -\frac{3 d^3 2^{p+3} \left (d^2-e^2 x^2\right )^{p+1} \left (\frac{e x}{d}+1\right )^{-p-1} \, _2F_1\left (-p-3,p+1;p+2;\frac{d-e x}{2 d}\right )}{e^2 (p+1) (2 p+5)}-\frac{(d+e x)^3 \left (d^2-e^2 x^2\right )^{p+1}}{e^2 (2 p+5)} \]

[Out]

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

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Rubi [A]  time = 0.0663772, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {795, 678, 69} \[ -\frac{3 d^3 2^{p+3} \left (d^2-e^2 x^2\right )^{p+1} \left (\frac{e x}{d}+1\right )^{-p-1} \, _2F_1\left (-p-3,p+1;p+2;\frac{d-e x}{2 d}\right )}{e^2 (p+1) (2 p+5)}-\frac{(d+e x)^3 \left (d^2-e^2 x^2\right )^{p+1}}{e^2 (2 p+5)} \]

Antiderivative was successfully verified.

[In]

Int[x*(d + e*x)^3*(d^2 - e^2*x^2)^p,x]

[Out]

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

Rule 795

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(g*(d + e*x)^m
*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)), Int[(d +
e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && NeQ[m + 2*p +
2, 0] && NeQ[m, 2]

Rule 678

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^(m - 1)*(a + c*x^2)^(p + 1))/((1
+ (e*x)/d)^(p + 1)*(a/d + (c*x)/e)^(p + 1)), Int[(1 + (e*x)/d)^(m + p)*(a/d + (c*x)/e)^p, x], x] /; FreeQ[{a,
c, d, e, m}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (
IntegerQ[3*p] || IntegerQ[4*p]))

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

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

Mathematica [A]  time = 0.290134, size = 159, normalized size = 1.37 \[ \frac{\left (d^2-e^2 x^2\right )^p \left (10 d^2 e^3 x^3 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};\frac{e^2 x^2}{d^2}\right )+2 e^5 x^5 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};\frac{e^2 x^2}{d^2}\right )-\frac{5 d \left (d^2-e^2 x^2\right ) \left (d^2 (p+5)+3 e^2 (p+1) x^2\right )}{(p+1) (p+2)}\right )}{10 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(d + e*x)^3*(d^2 - e^2*x^2)^p,x]

[Out]

((d^2 - e^2*x^2)^p*((-5*d*(d^2 - e^2*x^2)*(d^2*(5 + p) + 3*e^2*(1 + p)*x^2))/((1 + p)*(2 + p)) + (10*d^2*e^3*x
^3*Hypergeometric2F1[3/2, -p, 5/2, (e^2*x^2)/d^2])/(1 - (e^2*x^2)/d^2)^p + (2*e^5*x^5*Hypergeometric2F1[5/2, -
p, 7/2, (e^2*x^2)/d^2])/(1 - (e^2*x^2)/d^2)^p))/(10*e^2)

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Maple [F]  time = 0.593, size = 0, normalized size = 0. \begin{align*} \int x \left ( ex+d \right ) ^{3} \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*(e*x+d)^3*(-e^2*x^2+d^2)^p,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

[Out]

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

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Sympy [A]  time = 7.5621, size = 479, normalized size = 4.13 \begin{align*} d^{3} \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 ) + d^{2} 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 d 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 ) + \frac{d^{2 p} e^{3} 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*(e*x+d)**3*(-e**2*x**2+d**2)**p,x)

[Out]

d**3*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)) + d**2*d**(2*p)*e*x**3*hyper((3/2, -p), (5/2,), e**2*x**2*exp_
polar(2*I*pi)/d**2) + 3*d*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*lo
g(-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**3*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 )}^{3}{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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