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

Optimal. Leaf size=156 \[ \frac {2 (p+2) x^3 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {3}{2},2-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )}{3 d^2 (2 p+1)}-\frac {x^3 \left (d^2-e^2 x^2\right )^{p-1}}{2 p+1}-\frac {d \left (d^2-e^2 x^2\right )^p}{e^3 p}-\frac {d^3 \left (d^2-e^2 x^2\right )^{p-1}}{e^3 (1-p)} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {852, 1652, 459, 365, 364, 12, 266, 43} \[ \frac {2 (p+2) x^3 \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac {3}{2},2-p;\frac {5}{2};\frac {e^2 x^2}{d^2}\right )}{3 d^2 (2 p+1)}-\frac {x^3 \left (d^2-e^2 x^2\right )^{p-1}}{2 p+1}-\frac {d^3 \left (d^2-e^2 x^2\right )^{p-1}}{e^3 (1-p)}-\frac {d \left (d^2-e^2 x^2\right )^p}{e^3 p} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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

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])

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 459

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 177, normalized size = 1.13 \[ \frac {2^{p-2} \left (\frac {e x}{d}+1\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \left (4 e (p+1) x \left (\frac {e x}{2 d}+\frac {1}{2}\right )^p \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right )+(d-e x) \left (1-\frac {e^2 x^2}{d^2}\right )^p \left (4 \, _2F_1\left (1-p,p+1;p+2;\frac {d-e x}{2 d}\right )-\, _2F_1\left (2-p,p+1;p+2;\frac {d-e x}{2 d}\right )\right )\right )}{e^3 (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(-2 + p)*(d^2 - e^2*x^2)^p*(4*e*(1 + p)*x*(1/2 + (e*x)/(2*d))^p*Hypergeometric2F1[1/2, -p, 3/2, (e^2*x^2)/d
^2] + (d - e*x)*(1 - (e^2*x^2)/d^2)^p*(4*Hypergeometric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)] - Hypergeomet
ric2F1[2 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])))/(e^3*(1 + p)*(1 + (e*x)/d)^p*(1 - (e^2*x^2)/d^2)^p)

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fricas [F]  time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{p} x^{2}}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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