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

Optimal. Leaf size=175 \[ \frac{\left (d^2-e^2 x^2\right )^{p-1} \, _2F_1\left (1,p-1;p;1-\frac{e^2 x^2}{d^2}\right )}{2 d (1-p)}-\frac{2 e (4-3 p) x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{1}{2},3-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^4 (3-2 p)}-\frac{e x \left (d^2-e^2 x^2\right )^{p-2}}{3-2 p}+\frac{2 d \left (d^2-e^2 x^2\right )^{p-2}}{2-p} \]

[Out]

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

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Rubi [A]  time = 0.155502, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {852, 1652, 446, 79, 65, 388, 246, 245} \[ \frac{\left (d^2-e^2 x^2\right )^{p-1} \, _2F_1\left (1,p-1;p;1-\frac{e^2 x^2}{d^2}\right )}{2 d (1-p)}-\frac{2 e (4-3 p) x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{1}{2},3-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^4 (3-2 p)}-\frac{e x \left (d^2-e^2 x^2\right )^{p-2}}{3-2 p}+\frac{2 d \left (d^2-e^2 x^2\right )^{p-2}}{2-p} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(d^2 - e^2*x^2)^(-2 + p))/(2 - p) - (e*x*(d^2 - e^2*x^2)^(-2 + p))/(3 - 2*p) - (2*e*(4 - 3*p)*x*(d^2 - e^
2*x^2)^p*Hypergeometric2F1[1/2, 3 - p, 3/2, (e^2*x^2)/d^2])/(d^4*(3 - 2*p)*(1 - (e^2*x^2)/d^2)^p) + ((d^2 - e^
2*x^2)^(-1 + p)*Hypergeometric2F1[1, -1 + p, p, 1 - (e^2*x^2)/d^2])/(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 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 (d+e x)^3} \, dx &=\int \frac{(d-e x)^3 \left (d^2-e^2 x^2\right )^{-3+p}}{x} \, dx\\ &=\int \frac{\left (d^2-e^2 x^2\right )^{-3+p} \left (d^3+3 d e^2 x^2\right )}{x} \, dx+\int \left (d^2-e^2 x^2\right )^{-3+p} \left (-3 d^2 e-e^3 x^2\right ) \, dx\\ &=-\frac{e x \left (d^2-e^2 x^2\right )^{-2+p}}{3-2 p}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (d^2-e^2 x\right )^{-3+p} \left (d^3+3 d e^2 x\right )}{x} \, dx,x,x^2\right )-\frac{\left (2 d^2 e (4-3 p)\right ) \int \left (d^2-e^2 x^2\right )^{-3+p} \, dx}{3-2 p}\\ &=\frac{2 d \left (d^2-e^2 x^2\right )^{-2+p}}{2-p}-\frac{e x \left (d^2-e^2 x^2\right )^{-2+p}}{3-2 p}+\frac{1}{2} d \operatorname{Subst}\left (\int \frac{\left (d^2-e^2 x\right )^{-2+p}}{x} \, dx,x,x^2\right )-\frac{\left (2 e (4-3 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 )^{-3+p} \, dx}{d^4 (3-2 p)}\\ &=\frac{2 d \left (d^2-e^2 x^2\right )^{-2+p}}{2-p}-\frac{e x \left (d^2-e^2 x^2\right )^{-2+p}}{3-2 p}-\frac{2 e (4-3 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},3-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^4 (3-2 p)}+\frac{\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 )}{2 d (1-p)}\\ \end{align*}

Mathematica [C]  time = 0.0971274, size = 82, normalized size = 0.47 \[ \frac{\left (1-\frac{d^2}{e^2 x^2}\right )^{-p} (d-e x)^p (d+e x)^p F_1\left (3-2 p;-p,3-p;4-2 p;\frac{d}{e x},-\frac{d}{e x}\right )}{e^3 (2 p-3) x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.65, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{x \left ( ex+d \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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