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

Optimal. Leaf size=137 \[ -\frac{e \left (d^2-e^2 x^2\right )^{p-1} \, _2F_1\left (1,p-1;p;1-\frac{e^2 x^2}{d^2}\right )}{d (1-p)}+\frac{2 e^2 (2-p) x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{1}{2},2-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^4}-\frac{\left (d^2-e^2 x^2\right )^{p-1}}{x} \]

[Out]

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

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Rubi [A]  time = 0.161907, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {852, 1807, 764, 266, 65, 246, 245} \[ -\frac{e \left (d^2-e^2 x^2\right )^{p-1} \, _2F_1\left (1,p-1;p;1-\frac{e^2 x^2}{d^2}\right )}{d (1-p)}+\frac{2 e^2 (2-p) x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{1}{2},2-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^4}-\frac{\left (d^2-e^2 x^2\right )^{p-1}}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d^2 - e^2*x^2)^(-1 + p)/x) + (2*e^2*(2 - p)*x*(d^2 - e^2*x^2)^p*Hypergeometric2F1[1/2, 2 - p, 3/2, (e^2*x^2
)/d^2])/(d^4*(1 - (e^2*x^2)/d^2)^p) - (e*(d^2 - e^2*x^2)^(-1 + p)*Hypergeometric2F1[1, -1 + p, p, 1 - (e^2*x^2
)/d^2])/(d*(1 - 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 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 764

Int[(x_)^(m_.)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[f, Int[x^m*(a + c*x^2)^p, x]
, x] + Dist[g, Int[x^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && IntegerQ[m] &&  !IntegerQ[2
*p]

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 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

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

Rubi steps

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

Mathematica [C]  time = 0.112843, size = 82, normalized size = 0.6 \[ \frac{\left (1-\frac{d^2}{e^2 x^2}\right )^{-p} (d-e x)^p (d+e x)^p F_1\left (3-2 p;-p,2-p;4-2 p;\frac{d}{e x},-\frac{d}{e x}\right )}{e^2 (2 p-3) x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((d - e*x)^p*(d + e*x)^p*AppellF1[3 - 2*p, -p, 2 - p, 4 - 2*p, d/(e*x), -(d/(e*x))])/(e^2*(-3 + 2*p)*(1 - d^2/
(e^2*x^2))^p*x^3)

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Maple [F]  time = 0.656, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{{x}^{2} \left ( ex+d \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-e^2*x^2 + d^2)^p/((e*x + d)^2*x^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}}{e^{2} x^{4} + 2 \, d e x^{3} + d^{2} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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