3.302 \(\int \frac{(d^2-e^2 x^2)^p}{x^2 (d+e x)^4} \, dx\)

Optimal. Leaf size=207 \[ \frac{4 e^2 \left (p^2-9 p+16\right ) x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{1}{2},4-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^6 (5-2 p)}-\frac{2 e \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac{e^2 x^2}{d^2}\right )}{d (2-p)}+\frac{e^2 x \left (d^2-e^2 x^2\right )^{p-3}}{5-2 p}-\frac{4 d e \left (d^2-e^2 x^2\right )^{p-3}}{3-p}-\frac{d^2 \left (d^2-e^2 x^2\right )^{p-3}}{x} \]

[Out]

(-4*d*e*(d^2 - e^2*x^2)^(-3 + p))/(3 - p) - (d^2*(d^2 - e^2*x^2)^(-3 + p))/x + (e^2*x*(d^2 - e^2*x^2)^(-3 + p)
)/(5 - 2*p) + (4*e^2*(16 - 9*p + p^2)*x*(d^2 - e^2*x^2)^p*Hypergeometric2F1[1/2, 4 - p, 3/2, (e^2*x^2)/d^2])/(
d^6*(5 - 2*p)*(1 - (e^2*x^2)/d^2)^p) - (2*e*(d^2 - e^2*x^2)^(-2 + p)*Hypergeometric2F1[1, -2 + p, -1 + p, 1 -
(e^2*x^2)/d^2])/(d*(2 - p))

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Rubi [A]  time = 0.277891, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {852, 1807, 1652, 446, 79, 65, 388, 246, 245} \[ \frac{4 e^2 \left (p^2-9 p+16\right ) x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{1}{2},4-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^6 (5-2 p)}-\frac{2 e \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac{e^2 x^2}{d^2}\right )}{d (2-p)}+\frac{e^2 x \left (d^2-e^2 x^2\right )^{p-3}}{5-2 p}-\frac{4 d e \left (d^2-e^2 x^2\right )^{p-3}}{3-p}-\frac{d^2 \left (d^2-e^2 x^2\right )^{p-3}}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*d*e*(d^2 - e^2*x^2)^(-3 + p))/(3 - p) - (d^2*(d^2 - e^2*x^2)^(-3 + p))/x + (e^2*x*(d^2 - e^2*x^2)^(-3 + p)
)/(5 - 2*p) + (4*e^2*(16 - 9*p + p^2)*x*(d^2 - e^2*x^2)^p*Hypergeometric2F1[1/2, 4 - p, 3/2, (e^2*x^2)/d^2])/(
d^6*(5 - 2*p)*(1 - (e^2*x^2)/d^2)^p) - (2*e*(d^2 - e^2*x^2)^(-2 + p)*Hypergeometric2F1[1, -2 + p, -1 + p, 1 -
(e^2*x^2)/d^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 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 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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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)^4} \, dx &=\int \frac{(d-e x)^4 \left (d^2-e^2 x^2\right )^{-4+p}}{x^2} \, dx\\ &=-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{x}-\frac{\int \frac{\left (d^2-e^2 x^2\right )^{-4+p} \left (4 d^5 e-d^4 e^2 (13-2 p) x+4 d^3 e^3 x^2-d^2 e^4 x^3\right )}{x} \, dx}{d^2}\\ &=-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{x}-\frac{\int \frac{\left (d^2-e^2 x^2\right )^{-4+p} \left (4 d^5 e+4 d^3 e^3 x^2\right )}{x} \, dx}{d^2}-\frac{\int \left (d^2-e^2 x^2\right )^{-4+p} \left (-d^4 e^2 (13-2 p)-d^2 e^4 x^2\right ) \, dx}{d^2}\\ &=-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac{e^2 x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}-\frac{\operatorname{Subst}\left (\int \frac{\left (d^2-e^2 x\right )^{-4+p} \left (4 d^5 e+4 d^3 e^3 x\right )}{x} \, dx,x,x^2\right )}{2 d^2}+\frac{\left (4 d^2 e^2 \left (16-9 p+p^2\right )\right ) \int \left (d^2-e^2 x^2\right )^{-4+p} \, dx}{5-2 p}\\ &=-\frac{4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{3-p}-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac{e^2 x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}-(2 d e) \operatorname{Subst}\left (\int \frac{\left (d^2-e^2 x\right )^{-3+p}}{x} \, dx,x,x^2\right )+\frac{\left (4 e^2 \left (16-9 p+p^2\right ) \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 )^{-4+p} \, dx}{d^6 (5-2 p)}\\ &=-\frac{4 d e \left (d^2-e^2 x^2\right )^{-3+p}}{3-p}-\frac{d^2 \left (d^2-e^2 x^2\right )^{-3+p}}{x}+\frac{e^2 x \left (d^2-e^2 x^2\right )^{-3+p}}{5-2 p}+\frac{4 e^2 \left (16-9 p+p^2\right ) 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},4-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^6 (5-2 p)}-\frac{2 e \left (d^2-e^2 x^2\right )^{-2+p} \, _2F_1\left (1,-2+p;-1+p;1-\frac{e^2 x^2}{d^2}\right )}{d (2-p)}\\ \end{align*}

Mathematica [C]  time = 0.190724, size = 82, normalized size = 0.4 \[ \frac{\left (1-\frac{d^2}{e^2 x^2}\right )^{-p} (d-e x)^p (d+e x)^p F_1\left (5-2 p;-p,4-p;6-2 p;\frac{d}{e x},-\frac{d}{e x}\right )}{e^4 (2 p-5) x^5} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((d - e*x)^p*(d + e*x)^p*AppellF1[5 - 2*p, -p, 4 - p, 6 - 2*p, d/(e*x), -(d/(e*x))])/(e^4*(-5 + 2*p)*(1 - d^2/
(e^2*x^2))^p*x^5)

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Maple [F]  time = 0.679, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{{x}^{2} \left ( ex+d \right ) ^{4}}}\, 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)^4,x)

[Out]

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

[Out]

integrate((-e^2*x^2 + d^2)^p/((e*x + d)^4*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^{4} x^{6} + 4 \, d e^{3} x^{5} + 6 \, d^{2} e^{2} x^{4} + 4 \, d^{3} e x^{3} + d^{4} 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)^4,x, algorithm="fricas")

[Out]

integral((-e^2*x^2 + d^2)^p/(e^4*x^6 + 4*d*e^3*x^5 + 6*d^2*e^2*x^4 + 4*d^3*e*x^3 + d^4*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 )^{4}}\, 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)**4,x)

[Out]

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

[Out]

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