3.3.98 \(\int \frac {x^2 (d^2-e^2 x^2)^p}{(d+e x)^4} \, dx\) [298]

Optimal. Leaf size=163 \[ -\frac {d \left (d^2-e^2 x^2\right )^{1+p}}{2 e^3 (3-p) (d+e x)^4}+\frac {\left (d^2-e^2 x^2\right )^{1+p}}{e^3 (1-2 p) (d+e x)^3}-\frac {2^{-3+p} (7+p) \left (1+\frac {e x}{d}\right )^{-1-p} \left (d^2-e^2 x^2\right )^{1+p} \, _2F_1\left (3-p,1+p;2+p;\frac {d-e x}{2 d}\right )}{d^3 e^3 (1-2 p) (3-p) (1+p)} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1653, 807, 692, 71} \begin {gather*} \frac {\left (d^2-e^2 x^2\right )^{p+1}}{e^3 (1-2 p) (d+e x)^3}-\frac {d \left (d^2-e^2 x^2\right )^{p+1}}{2 e^3 (3-p) (d+e x)^4}-\frac {2^{p-3} (p+7) \left (d^2-e^2 x^2\right )^{p+1} \left (\frac {e x}{d}+1\right )^{-p-1} \, _2F_1\left (3-p,p+1;p+2;\frac {d-e x}{2 d}\right )}{d^3 e^3 (1-2 p) (3-p) (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(d^2 - e^2*x^2)^p)/(d + e*x)^4,x]

[Out]

-1/2*(d*(d^2 - e^2*x^2)^(1 + p))/(e^3*(3 - p)*(d + e*x)^4) + (d^2 - e^2*x^2)^(1 + p)/(e^3*(1 - 2*p)*(d + e*x)^
3) - (2^(-3 + p)*(7 + p)*(1 + (e*x)/d)^(-1 - p)*(d^2 - e^2*x^2)^(1 + p)*Hypergeometric2F1[3 - p, 1 + p, 2 + p,
(d - e*x)/(2*d)])/(d^3*e^3*(1 - 2*p)*(3 - p)*(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 692

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/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]) &&  !(IGtQ[m, 0] && (
IntegerQ[3*p] || IntegerQ[4*p]))

Rule 807

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 1653

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]

Rubi steps

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

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Mathematica [A]
time = 0.36, size = 130, normalized size = 0.80 \begin {gather*} -\frac {2^{-4+p} (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (4 \, _2F_1\left (2-p,1+p;2+p;\frac {d-e x}{2 d}\right )-4 \, _2F_1\left (3-p,1+p;2+p;\frac {d-e x}{2 d}\right )+\, _2F_1\left (4-p,1+p;2+p;\frac {d-e x}{2 d}\right )\right )}{d^2 e^3 (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(d^2 - e^2*x^2)^p)/(d + e*x)^4,x]

[Out]

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

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{p}}{\left (e x +d \right )^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(-e^2*x^2+d^2)^p/(e*x+d)^4,x)

[Out]

int(x^2*(-e^2*x^2+d^2)^p/(e*x+d)^4,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-e^2*x^2+d^2)^p/(e*x+d)^4,x, algorithm="maxima")

[Out]

integrate((-x^2*e^2 + d^2)^p*x^2/(x*e + d)^4, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-e^2*x^2+d^2)^p/(e*x+d)^4,x, algorithm="fricas")

[Out]

integral((-x^2*e^2 + d^2)^p*x^2/(x^4*e^4 + 4*d*x^3*e^3 + 6*d^2*x^2*e^2 + 4*d^3*x*e + d^4), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{\left (d + e x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(-e**2*x**2+d**2)**p/(e*x+d)**4,x)

[Out]

Integral(x**2*(-(-d + e*x)*(d + e*x))**p/(d + e*x)**4, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-e^2*x^2+d^2)^p/(e*x+d)^4,x, algorithm="giac")

[Out]

integrate((-x^2*e^2 + d^2)^p*x^2/(x*e + d)^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,{\left (d^2-e^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(d^2 - e^2*x^2)^p)/(d + e*x)^4,x)

[Out]

int((x^2*(d^2 - e^2*x^2)^p)/(d + e*x)^4, x)

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