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

Optimal. Leaf size=71 \[ -\frac {d 2^{p+2} \left (\frac {e x}{d}+1\right )^{-p-1} \left (d^2-e^2 x^2\right )^{p+1} \, _2F_1\left (-p-2,p+1;p+2;\frac {d-e x}{2 d}\right )}{e (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {678, 69} \[ -\frac {d 2^{p+2} \left (\frac {e x}{d}+1\right )^{-p-1} \left (d^2-e^2 x^2\right )^{p+1} \, _2F_1\left (-p-2,p+1;p+2;\frac {d-e x}{2 d}\right )}{e (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2*(d^2 - e^2*x^2)^p,x]

[Out]

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

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]))

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 134, normalized size = 1.89 \[ \frac {\left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (3 d^2 e (p+1) x \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )-3 d \left (d^2-e^2 x^2\right ) \left (1-\frac {e^2 x^2}{d^2}\right )^p+e^3 (p+1) x^3 \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )\right )}{3 e (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} {\left (-e^{2} x^{2} + d^{2}\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(-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)^p, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d^2 - e^2*x^2)^p*(d + e*x)^2,x)

[Out]

int((d^2 - e^2*x^2)^p*(d + e*x)^2, x)

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sympy [A]  time = 4.95, size = 124, normalized size = 1.75 \[ d^{2} d^{2 p} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} + 2 d e \left (\begin {cases} \frac {x^{2} \left (d^{2}\right )^{p}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\begin {cases} \frac {\left (d^{2} - e^{2} x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (d^{2} - e^{2} x^{2} \right )} & \text {otherwise} \end {cases}}{2 e^{2}} & \text {otherwise} \end {cases}\right ) + \frac {d^{2 p} e^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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