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

Optimal. Leaf size=173 \[ \frac{e^2 (6-p) \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac{e^2 x^2}{d^2}\right )}{2 d (2-p)}-\frac{2 e^3 (8-3 p) x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{1}{2},3-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^6}+\frac{3 e \left (d^2-e^2 x^2\right )^{p-2}}{x}-\frac{d \left (d^2-e^2 x^2\right )^{p-2}}{2 x^2} \]

[Out]

-(d*(d^2 - e^2*x^2)^(-2 + p))/(2*x^2) + (3*e*(d^2 - e^2*x^2)^(-2 + p))/x - (2*e^3*(8 - 3*p)*x*(d^2 - e^2*x^2)^
p*Hypergeometric2F1[1/2, 3 - p, 3/2, (e^2*x^2)/d^2])/(d^6*(1 - (e^2*x^2)/d^2)^p) + (e^2*(6 - p)*(d^2 - e^2*x^2
)^(-2 + p)*Hypergeometric2F1[1, -2 + p, -1 + p, 1 - (e^2*x^2)/d^2])/(2*d*(2 - p))

________________________________________________________________________________________

Rubi [A]  time = 0.265752, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{e^2 (6-p) \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac{e^2 x^2}{d^2}\right )}{2 d (2-p)}-\frac{2 e^3 (8-3 p) x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{1}{2},3-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^6}+\frac{3 e \left (d^2-e^2 x^2\right )^{p-2}}{x}-\frac{d \left (d^2-e^2 x^2\right )^{p-2}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(d^2 - e^2*x^2)^(-2 + p))/(2*x^2) + (3*e*(d^2 - e^2*x^2)^(-2 + p))/x - (2*e^3*(8 - 3*p)*x*(d^2 - e^2*x^2)^
p*Hypergeometric2F1[1/2, 3 - p, 3/2, (e^2*x^2)/d^2])/(d^6*(1 - (e^2*x^2)/d^2)^p) + (e^2*(6 - p)*(d^2 - e^2*x^2
)^(-2 + p)*Hypergeometric2F1[1, -2 + p, -1 + p, 1 - (e^2*x^2)/d^2])/(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 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^3 (d+e x)^3} \, dx &=\int \frac{(d-e x)^3 \left (d^2-e^2 x^2\right )^{-3+p}}{x^3} \, dx\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}-\frac{\int \frac{\left (d^2-e^2 x^2\right )^{-3+p} \left (6 d^4 e-2 d^3 e^2 (6-p) x+2 d^2 e^3 x^2\right )}{x^2} \, dx}{2 d^2}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}+\frac{3 e \left (d^2-e^2 x^2\right )^{-2+p}}{x}+\frac{\int \frac{\left (2 d^5 e^2 (6-p)-4 d^4 e^3 (8-3 p) x\right ) \left (d^2-e^2 x^2\right )^{-3+p}}{x} \, dx}{2 d^4}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}+\frac{3 e \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\left (2 e^3 (8-3 p)\right ) \int \left (d^2-e^2 x^2\right )^{-3+p} \, dx+\left (d e^2 (6-p)\right ) \int \frac{\left (d^2-e^2 x^2\right )^{-3+p}}{x} \, dx\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}+\frac{3 e \left (d^2-e^2 x^2\right )^{-2+p}}{x}+\frac{1}{2} \left (d e^2 (6-p)\right ) \operatorname{Subst}\left (\int \frac{\left (d^2-e^2 x\right )^{-3+p}}{x} \, dx,x,x^2\right )-\frac{\left (2 e^3 (8-3 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^6}\\ &=-\frac{d \left (d^2-e^2 x^2\right )^{-2+p}}{2 x^2}+\frac{3 e \left (d^2-e^2 x^2\right )^{-2+p}}{x}-\frac{2 e^3 (8-3 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^6}+\frac{e^2 (6-p) \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 )}{2 d (2-p)}\\ \end{align*}

Mathematica [A]  time = 0.680982, size = 341, normalized size = 1.97 \[ \frac{\left (d^2-e^2 x^2\right )^p \left (\frac{4 d^3 \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (1-p,-p;2-p;\frac{d^2}{e^2 x^2}\right )}{(p-1) x^2}+\frac{24 d e^2 \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (-p,-p;1-p;\frac{d^2}{e^2 x^2}\right )}{p}+\frac{24 d^2 e \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x}+\frac{3 e^2 2^{p+3} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{3 e^2 2^{p+1} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (2-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{e^2 2^p (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (3-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}\right )}{8 d^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d^2 - e^2*x^2)^p*((24*d^2*e*Hypergeometric2F1[-1/2, -p, 1/2, (e^2*x^2)/d^2])/(x*(1 - (e^2*x^2)/d^2)^p) + (4*
d^3*Hypergeometric2F1[1 - p, -p, 2 - p, d^2/(e^2*x^2)])/((-1 + p)*(1 - d^2/(e^2*x^2))^p*x^2) + (3*2^(3 + p)*e^
2*(d - e*x)*Hypergeometric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) + (3*2^(1 + p)*
e^2*(d - e*x)*Hypergeometric2F1[2 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) + (2^p*e^2*(d
- e*x)*Hypergeometric2F1[3 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) + (24*d*e^2*Hyperge
ometric2F1[-p, -p, 1 - p, d^2/(e^2*x^2)])/(p*(1 - d^2/(e^2*x^2))^p)))/(8*d^6)

________________________________________________________________________________________

Maple [F]  time = 0.68, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{{x}^{3} \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^3/(e*x+d)^3,x)

[Out]

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

________________________________________________________________________________________

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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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^{6} + 3 \, d e^{2} x^{5} + 3 \, d^{2} e x^{4} + d^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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^{3} \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**3/(e*x+d)**3,x)

[Out]

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

________________________________________________________________________________________

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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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