3.288 \(\int \frac{x^2 (d^2-e^2 x^2)^p}{(d+e x)^3} \, dx\)

Optimal. Leaf size=157 \[ \frac{2^{p-3} (p+4) \left (d^2-e^2 x^2\right )^{p+1} \left (\frac{e x}{d}+1\right )^{-p-1} \, _2F_1\left (2-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{d^2 e^3 (2-p) p (p+1)}-\frac{\left (d^2-e^2 x^2\right )^{p+1}}{2 e^3 p (d+e x)^2}-\frac{d \left (d^2-e^2 x^2\right )^{p+1}}{2 e^3 (2-p) (d+e x)^3} \]

[Out]

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

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Rubi [A]  time = 0.183063, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {1639, 793, 678, 69} \[ \frac{2^{p-3} (p+4) \left (d^2-e^2 x^2\right )^{p+1} \left (\frac{e x}{d}+1\right )^{-p-1} \, _2F_1\left (2-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{d^2 e^3 (2-p) p (p+1)}-\frac{\left (d^2-e^2 x^2\right )^{p+1}}{2 e^3 p (d+e x)^2}-\frac{d \left (d^2-e^2 x^2\right )^{p+1}}{2 e^3 (2-p) (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1639

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 793

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

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 \frac{x^2 \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx &=-\frac{\left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 p (d+e x)^2}-\frac{\int \frac{\left (2 d^2 e^2+2 d e^3 (1+p) x\right ) \left (d^2-e^2 x^2\right )^p}{(d+e x)^3} \, dx}{2 e^4 p}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 (2-p) (d+e x)^3}-\frac{\left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 p (d+e x)^2}-\frac{(d (4+p)) \int \frac{\left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx}{2 e^2 (2-p) p}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 (2-p) (d+e x)^3}-\frac{\left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 p (d+e x)^2}-\frac{\left ((4+p) (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}{2 d^2 e^2 (2-p) p}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 (2-p) (d+e x)^3}-\frac{\left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 p (d+e x)^2}+\frac{2^{-3+p} (4+p) \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 )}{d^2 e^3 (2-p) p (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.133383, size = 130, normalized size = 0.83 \[ -\frac{2^{p-3} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (4 \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )-4 \, _2F_1\left (2-p,p+1;p+2;\frac{d-e x}{2 d}\right )+\, _2F_1\left (3-p,p+1;p+2;\frac{d-e x}{2 d}\right )\right )}{d e^3 (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^(-3 + p)*(d - e*x)*(d^2 - e^2*x^2)^p*(4*Hypergeometric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)] - 4*Hyper
geometric2F1[2 - p, 1 + p, 2 + p, (d - e*x)/(2*d)] + Hypergeometric2F1[3 - p, 1 + p, 2 + p, (d - e*x)/(2*d)]))
/(d*e^3*(1 + p)*(1 + (e*x)/d)^p))

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Maple [F]  time = 0.668, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{2} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{ \left ( ex+d \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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