### 3.960 $$\int (d+e x)^3 (1-\frac{e^2 x^2}{d^2})^p \, dx$$

Optimal. Leaf size=57 $-\frac{d^4 2^{p+3} \left (\frac{d-e x}{d}\right )^{p+1} \, _2F_1\left (-p-3,p+1;p+2;\frac{d-e x}{2 d}\right )}{e (p+1)}$

[Out]

-((2^(3 + p)*d^4*((d - e*x)/d)^(1 + p)*Hypergeometric2F1[-3 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/(e*(1 + p)))

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Rubi [A]  time = 0.0386394, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.087, Rules used = {676, 69} $-\frac{d^4 2^{p+3} \left (\frac{d-e x}{d}\right )^{p+1} \, _2F_1\left (-p-3,p+1;p+2;\frac{d-e x}{2 d}\right )}{e (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((2^(3 + p)*d^4*((d - e*x)/d)^(1 + p)*Hypergeometric2F1[-3 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/(e*(1 + p)))

Rule 676

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a^(p + 1)*d^(m - 1)*((d - 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]) && GtQ[a, 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 (d+e x)^3 \left (1-\frac{e^2 x^2}{d^2}\right )^p \, dx &=\left (d^2 \left (\frac{d-e x}{d}\right )^{1+p} \left (\frac{1}{d}-\frac{e x}{d^2}\right )^{-1-p}\right ) \int \left (\frac{1}{d}-\frac{e x}{d^2}\right )^p \left (1+\frac{e x}{d}\right )^{3+p} \, dx\\ &=-\frac{2^{3+p} d^4 \left (\frac{d-e x}{d}\right )^{1+p} \, _2F_1\left (-3-p,1+p;2+p;\frac{d-e x}{2 d}\right )}{e (1+p)}\\ \end{align*}

Mathematica [B]  time = 0.112035, size = 116, normalized size = 2.04 $d^3 x \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )+d e^2 x^3 \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};\frac{e^2 x^2}{d^2}\right )-\frac{2 d^4 \left (1-\frac{e^2 x^2}{d^2}\right )^{p+1}}{e (p+1)}+\frac{d^4 \left (1-\frac{e^2 x^2}{d^2}\right )^{p+2}}{2 e (p+2)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*d^4*(1 - (e^2*x^2)/d^2)^(1 + p))/(e*(1 + p)) + (d^4*(1 - (e^2*x^2)/d^2)^(2 + p))/(2*e*(2 + p)) + d^3*x*Hyp
ergeometric2F1[1/2, -p, 3/2, (e^2*x^2)/d^2] + d*e^2*x^3*Hypergeometric2F1[3/2, -p, 5/2, (e^2*x^2)/d^2]

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Maple [A]  time = 0.616, size = 104, normalized size = 1.8 \begin{align*}{\frac{{e}^{3}{x}^{4}}{4}{\mbox{$_2$F$_1$}(2,-p;\,3;\,{\frac{{e}^{2}{x}^{2}}{{d}^{2}}})}}+d{e}^{2}{x}^{3}{\mbox{$_2$F$_1$}({\frac{3}{2}},-p;\,{\frac{5}{2}};\,{\frac{{e}^{2}{x}^{2}}{{d}^{2}}})}+{\frac{3\,{d}^{2}e{x}^{2}}{2}{\mbox{$_2$F$_1$}(1,-p;\,2;\,{\frac{{e}^{2}{x}^{2}}{{d}^{2}}})}}+{d}^{3}x{\mbox{$_2$F$_1$}({\frac{1}{2}},-p;\,{\frac{3}{2}};\,{\frac{{e}^{2}{x}^{2}}{{d}^{2}}})} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*e^3*x^4*hypergeom([2,-p],[3],e^2*x^2/d^2)+d*e^2*x^3*hypergeom([3/2,-p],[5/2],e^2*x^2/d^2)+3/2*e*d^2*x^2*hy
pergeom([1,-p],[2],e^2*x^2/d^2)+d^3*x*hypergeom([1/2,-p],[3/2],e^2*x^2/d^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}^{3}{\left (-\frac{e^{2} x^{2}}{d^{2}} + 1\right )}^{p}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(1-e^2*x^2/d^2)^p,x, algorithm="maxima")

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(1-e^2*x^2/d^2)^p,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 6.26666, size = 479, normalized size = 8.4 \begin{align*} d^{3} x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} + 3 d^{2} e \left (\begin{cases} \frac{x^{2}}{2} & \text{for}\: e^{2} = 0 \\- \frac{d^{2} \left (\begin{cases} \frac{\left (1 - \frac{e^{2} x^{2}}{d^{2}}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (1 - \frac{e^{2} x^{2}}{d^{2}} \right )} & \text{otherwise} \end{cases}\right )}{2 e^{2}} & \text{otherwise} \end{cases}\right ) + d e^{2} 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 )} + e^{3} \left (\begin{cases} \frac{x^{4}}{4} & \text{for}\: e = 0 \\- \frac{d^{6} \log{\left (- \frac{d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac{d^{6} \log{\left (\frac{d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} - \frac{d^{6}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac{d^{4} e^{2} x^{2} \log{\left (- \frac{d}{e} + x \right )}}{- 2 d^{2} e^{4} + 2 e^{6} x^{2}} + \frac{d^{4} 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^{4} \log{\left (- \frac{d}{e} + x \right )}}{2 e^{4}} - \frac{d^{4} \log{\left (\frac{d}{e} + x \right )}}{2 e^{4}} - \frac{d^{2} x^{2}}{2 e^{2}} & \text{for}\: p = -1 \\- \frac{d^{4} \left (1 - \frac{e^{2} x^{2}}{d^{2}}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} - \frac{d^{2} e^{2} p x^{2} \left (1 - \frac{e^{2} x^{2}}{d^{2}}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac{e^{4} p x^{4} \left (1 - \frac{e^{2} x^{2}}{d^{2}}\right )^{p}}{2 e^{4} p^{2} + 6 e^{4} p + 4 e^{4}} + \frac{e^{4} x^{4} \left (1 - \frac{e^{2} x^{2}}{d^{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((e*x+d)**3*(1-e**2*x**2/d**2)**p,x)

[Out]

d**3*x*hyper((1/2, -p), (3/2,), e**2*x**2*exp_polar(2*I*pi)/d**2) + 3*d**2*e*Piecewise((x**2/2, Eq(e**2, 0)),
(-d**2*Piecewise(((1 - e**2*x**2/d**2)**(p + 1)/(p + 1), Ne(p, -1)), (log(1 - e**2*x**2/d**2), True))/(2*e**2)
, True)) + d*e**2*x**3*hyper((3/2, -p), (5/2,), e**2*x**2*exp_polar(2*I*pi)/d**2) + e**3*Piecewise((x**4/4, Eq
(e, 0)), (-d**6*log(-d/e + x)/(-2*d**2*e**4 + 2*e**6*x**2) - d**6*log(d/e + x)/(-2*d**2*e**4 + 2*e**6*x**2) -
d**6/(-2*d**2*e**4 + 2*e**6*x**2) + d**4*e**2*x**2*log(-d/e + x)/(-2*d**2*e**4 + 2*e**6*x**2) + d**4*e**2*x**2
*log(d/e + x)/(-2*d**2*e**4 + 2*e**6*x**2), Eq(p, -2)), (-d**4*log(-d/e + x)/(2*e**4) - d**4*log(d/e + x)/(2*e
**4) - d**2*x**2/(2*e**2), Eq(p, -1)), (-d**4*(1 - e**2*x**2/d**2)**p/(2*e**4*p**2 + 6*e**4*p + 4*e**4) - d**2
*e**2*p*x**2*(1 - e**2*x**2/d**2)**p/(2*e**4*p**2 + 6*e**4*p + 4*e**4) + e**4*p*x**4*(1 - e**2*x**2/d**2)**p/(
2*e**4*p**2 + 6*e**4*p + 4*e**4) + e**4*x**4*(1 - e**2*x**2/d**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 )}^{3}{\left (-\frac{e^{2} x^{2}}{d^{2}} + 1\right )}^{p}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(1-e^2*x^2/d^2)^p,x, algorithm="giac")

[Out]

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