\(\int (d+e x)^2 (1-\frac {e^2 x^2}{d^2})^p \, dx\) [961]

Optimal result

Integrand size = 23, antiderivative size = 57 \[ \int (d+e x)^2 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=-\frac {2^{2+p} d^3 \left (\frac {d-e x}{d}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-2-p,1+p,2+p,\frac {d-e x}{2 d}\right )}{e (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {690, 71} \[ \int (d+e x)^2 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=-\frac {d^3 2^{p+2} \left (\frac {d-e x}{d}\right )^{p+1} \operatorname {Hypergeometric2F1}\left (-p-2,p+1,p+2,\frac {d-e x}{2 d}\right )}{e (p+1)} \]

[In]

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

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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]))

Rule 690

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/e)*x)^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]))

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.51 \[ \int (d+e x)^2 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=-\frac {d^3 \left (1-\frac {e^2 x^2}{d^2}\right )^{1+p}}{e (1+p)}+d^2 x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )+\frac {1}{3} e^2 x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {e^2 x^2}{d^2}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.53 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.32

method	result	size
meijerg	\(\frac {e^{2} x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {3}{2},-p ;\frac {5}{2};\frac {e^{2} x^{2}}{d^{2}}\right )}{3}+e d \,x^{2} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (1,-p ;2;\frac {e^{2} x^{2}}{d^{2}}\right )+d^{2} x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},-p ;\frac {3}{2};\frac {e^{2} x^{2}}{d^{2}}\right )\)	\(75\)

[In]

int((e*x+d)^2*(1-e^2*x^2/d^2)^p,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int (d+e x)^2 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (-\frac {e^{2} x^{2}}{d^{2}} + 1\right )}^{p} \,d x } \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.74 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.04 \[ \int (d+e x)^2 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=d^{2} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} + 2 d e \left (\begin {cases} \frac {x^{2}}{2} & \text {for}\: e^{2} = 0 \\- \frac {d^{2} \left (\begin {cases} \frac {\left (1 - \frac {e^{2} x^{2}}{d^{2}}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (1 - \frac {e^{2} x^{2}}{d^{2}} \right )} & \text {otherwise} \end {cases}\right )}{2 e^{2}} & \text {otherwise} \end {cases}\right ) + \frac {e^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{3} \]

[In]

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

[Out]

d**2*x*hyper((1/2, -p), (3/2,), e**2*x**2*exp_polar(2*I*pi)/d**2) + 2*d*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), T
rue)) + e**2*x**3*hyper((3/2, -p), (5/2,), e**2*x**2*exp_polar(2*I*pi)/d**2)/3

Maxima [F]

\[ \int (d+e x)^2 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (-\frac {e^{2} x^{2}}{d^{2}} + 1\right )}^{p} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (d+e x)^2 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (-\frac {e^{2} x^{2}}{d^{2}} + 1\right )}^{p} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (d+e x)^2 \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx=\int {\left (1-\frac {e^2\,x^2}{d^2}\right )}^p\,{\left (d+e\,x\right )}^2 \,d x \]

[In]

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

[Out]

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