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

Optimal. Leaf size=150 \[ -\frac{2 e x^5 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{5}{2},2-p;\frac{7}{2};\frac{e^2 x^2}{d^2}\right )}{5 d^3}+\frac{d^4 \left (d^2-e^2 x^2\right )^{p-1}}{e^4 (1-p)}+\frac{3 d^2 \left (d^2-e^2 x^2\right )^p}{2 e^4 p}-\frac{\left (d^2-e^2 x^2\right )^{p+1}}{2 e^4 (p+1)} \]

[Out]

(d^4*(d^2 - e^2*x^2)^(-1 + p))/(e^4*(1 - p)) + (3*d^2*(d^2 - e^2*x^2)^p)/(2*e^4*p) - (d^2 - e^2*x^2)^(1 + p)/(
2*e^4*(1 + p)) - (2*e*x^5*(d^2 - e^2*x^2)^p*Hypergeometric2F1[5/2, 2 - p, 7/2, (e^2*x^2)/d^2])/(5*d^3*(1 - (e^
2*x^2)/d^2)^p)

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Rubi [A]  time = 0.164881, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {852, 1652, 446, 77, 12, 365, 364} \[ -\frac{2 e x^5 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{5}{2},2-p;\frac{7}{2};\frac{e^2 x^2}{d^2}\right )}{5 d^3}+\frac{d^4 \left (d^2-e^2 x^2\right )^{p-1}}{e^4 (1-p)}+\frac{3 d^2 \left (d^2-e^2 x^2\right )^p}{2 e^4 p}-\frac{\left (d^2-e^2 x^2\right )^{p+1}}{2 e^4 (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^4*(d^2 - e^2*x^2)^(-1 + p))/(e^4*(1 - p)) + (3*d^2*(d^2 - e^2*x^2)^p)/(2*e^4*p) - (d^2 - e^2*x^2)^(1 + p)/(
2*e^4*(1 + p)) - (2*e*x^5*(d^2 - e^2*x^2)^p*Hypergeometric2F1[5/2, 2 - p, 7/2, (e^2*x^2)/d^2])/(5*d^3*(1 - (e^
2*x^2)/d^2)^p)

Rule 852

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

Rule 1652

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[x^m*Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2)^p, x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2] && IGtQ[m, -2] &&  !
IntegerQ[2*p]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{x^3 \left (d^2-e^2 x^2\right )^p}{(d+e x)^2} \, dx &=\int x^3 (d-e x)^2 \left (d^2-e^2 x^2\right )^{-2+p} \, dx\\ &=\int -2 d e x^4 \left (d^2-e^2 x^2\right )^{-2+p} \, dx+\int x^3 \left (d^2-e^2 x^2\right )^{-2+p} \left (d^2+e^2 x^2\right ) \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int x \left (d^2-e^2 x\right )^{-2+p} \left (d^2+e^2 x\right ) \, dx,x,x^2\right )-(2 d e) \int x^4 \left (d^2-e^2 x^2\right )^{-2+p} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{2 d^4 \left (d^2-e^2 x\right )^{-2+p}}{e^2}-\frac{3 d^2 \left (d^2-e^2 x\right )^{-1+p}}{e^2}+\frac{\left (d^2-e^2 x\right )^p}{e^2}\right ) \, dx,x,x^2\right )-\frac{\left (2 e \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p}\right ) \int x^4 \left (1-\frac{e^2 x^2}{d^2}\right )^{-2+p} \, dx}{d^3}\\ &=\frac{d^4 \left (d^2-e^2 x^2\right )^{-1+p}}{e^4 (1-p)}+\frac{3 d^2 \left (d^2-e^2 x^2\right )^p}{2 e^4 p}-\frac{\left (d^2-e^2 x^2\right )^{1+p}}{2 e^4 (1+p)}-\frac{2 e x^5 \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{5}{2},2-p;\frac{7}{2};\frac{e^2 x^2}{d^2}\right )}{5 d^3}\\ \end{align*}

Mathematica [B]  time = 0.313731, size = 332, normalized size = 2.21 \[ \frac{2^{p-2} \left (\frac{e x}{d}+1\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (-8 d e (p+1) x \left (\frac{e x}{2 d}+\frac{1}{2}\right )^p \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )-6 d (d-e x) \left (1-\frac{e^2 x^2}{d^2}\right )^p \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )+d^2 \left (1-\frac{e^2 x^2}{d^2}\right )^p \, _2F_1\left (2-p,p+1;p+2;\frac{d-e x}{2 d}\right )-d e x \left (1-\frac{e^2 x^2}{d^2}\right )^p \, _2F_1\left (2-p,p+1;p+2;\frac{d-e x}{2 d}\right )-2 d^2 \left (1-\frac{e^2 x^2}{d^2}\right )^p \left (\frac{e x}{2 d}+\frac{1}{2}\right )^p+2 e^2 x^2 \left (1-\frac{e^2 x^2}{d^2}\right )^p \left (\frac{e x}{2 d}+\frac{1}{2}\right )^p+2 d^2 \left (\frac{e x}{2 d}+\frac{1}{2}\right )^p\right )}{e^4 (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(-2 + p)*(d^2 - e^2*x^2)^p*(2*d^2*(1/2 + (e*x)/(2*d))^p - 2*d^2*(1/2 + (e*x)/(2*d))^p*(1 - (e^2*x^2)/d^2)^p
+ 2*e^2*x^2*(1/2 + (e*x)/(2*d))^p*(1 - (e^2*x^2)/d^2)^p - 8*d*e*(1 + p)*x*(1/2 + (e*x)/(2*d))^p*Hypergeometri
c2F1[1/2, -p, 3/2, (e^2*x^2)/d^2] - 6*d*(d - e*x)*(1 - (e^2*x^2)/d^2)^p*Hypergeometric2F1[1 - p, 1 + p, 2 + p,
(d - e*x)/(2*d)] + d^2*(1 - (e^2*x^2)/d^2)^p*Hypergeometric2F1[2 - p, 1 + p, 2 + p, (d - e*x)/(2*d)] - d*e*x*
(1 - (e^2*x^2)/d^2)^p*Hypergeometric2F1[2 - p, 1 + p, 2 + p, (d - e*x)/(2*d)]))/(e^4*(1 + p)*(1 + (e*x)/d)^p*(
1 - (e^2*x^2)/d^2)^p)

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Maple [F]  time = 0.705, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{ \left ( ex+d \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^3*(-e^2*x^2+d^2)^p/(e*x+d)^2,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^{3}}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-e^2*x^2 + d^2)^p*x^3/(e*x + d)^2, 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^{3}}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*(-(-d + e*x)*(d + e*x))**p/(d + e*x)**2, 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^{3}}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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