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

Optimal. Leaf size=210 \[ -\frac{e^2 (27-2 p) \left (d^2-e^2 x^2\right )^{p-3}}{3 x}+\frac{2 d e \left (d^2-e^2 x^2\right )^{p-3}}{x^2}-\frac{d^2 \left (d^2-e^2 x^2\right )^{p-3}}{3 x^3}-\frac{2 e^3 (5-p) \left (d^2-e^2 x^2\right )^{p-3} \, _2F_1\left (1,p-3;p-2;1-\frac{e^2 x^2}{d^2}\right )}{d (3-p)}+\frac{4 e^4 \left (p^2-17 p+48\right ) x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{1}{2},4-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{3 d^8} \]

[Out]

-(d^2*(d^2 - e^2*x^2)^(-3 + p))/(3*x^3) + (2*d*e*(d^2 - e^2*x^2)^(-3 + p))/x^2 -
(e^2*(27 - 2*p)*(d^2 - e^2*x^2)^(-3 + p))/(3*x) + (4*e^4*(48 - 17*p + p^2)*x*(d
^2 - e^2*x^2)^p*Hypergeometric2F1[1/2, 4 - p, 3/2, (e^2*x^2)/d^2])/(3*d^8*(1 - (
e^2*x^2)/d^2)^p) - (2*e^3*(5 - p)*(d^2 - e^2*x^2)^(-3 + p)*Hypergeometric2F1[1,
-3 + p, -2 + p, 1 - (e^2*x^2)/d^2])/(d*(3 - p))

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Rubi [A]  time = 0.665328, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28 \[ -\frac{e^2 (27-2 p) \left (d^2-e^2 x^2\right )^{p-3}}{3 x}+\frac{2 d e \left (d^2-e^2 x^2\right )^{p-3}}{x^2}-\frac{d^2 \left (d^2-e^2 x^2\right )^{p-3}}{3 x^3}-\frac{2 e^3 (5-p) \left (d^2-e^2 x^2\right )^{p-3} \, _2F_1\left (1,p-3;p-2;1-\frac{e^2 x^2}{d^2}\right )}{d (3-p)}+\frac{4 e^4 \left (p^2-17 p+48\right ) x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{1}{2},4-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{3 d^8} \]

Antiderivative was successfully verified.

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

[Out]

-(d^2*(d^2 - e^2*x^2)^(-3 + p))/(3*x^3) + (2*d*e*(d^2 - e^2*x^2)^(-3 + p))/x^2 -
(e^2*(27 - 2*p)*(d^2 - e^2*x^2)^(-3 + p))/(3*x) + (4*e^4*(48 - 17*p + p^2)*x*(d
^2 - e^2*x^2)^p*Hypergeometric2F1[1/2, 4 - p, 3/2, (e^2*x^2)/d^2])/(3*d^8*(1 - (
e^2*x^2)/d^2)^p) - (2*e^3*(5 - p)*(d^2 - e^2*x^2)^(-3 + p)*Hypergeometric2F1[1,
-3 + p, -2 + p, 1 - (e^2*x^2)/d^2])/(d*(3 - p))

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Rubi in Sympy [A]  time = 96.312, size = 240, normalized size = 1.14 \[ - \frac{2 e^{3} \left (d^{2} - e^{2} x^{2}\right )^{p - 3}{{}_{2}F_{1}\left (\begin{matrix} 1, p - 3 \\ p - 2 \end{matrix}\middle |{1 - \frac{e^{2} x^{2}}{d^{2}}} \right )}}{d \left (- p + 3\right )} - \frac{2 e^{3} \left (d^{2} - e^{2} x^{2}\right )^{p - 3}{{}_{2}F_{1}\left (\begin{matrix} 2, p - 3 \\ p - 2 \end{matrix}\middle |{1 - \frac{e^{2} x^{2}}{d^{2}}} \right )}}{d \left (- p + 3\right )} - \frac{\left (1 - \frac{e^{2} x^{2}}{d^{2}}\right )^{- p} \left (d^{2} - e^{2} x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p + 4, - \frac{3}{2} \\ - \frac{1}{2} \end{matrix}\middle |{\frac{e^{2} x^{2}}{d^{2}}} \right )}}{3 d^{4} x^{3}} - \frac{6 e^{2} \left (1 - \frac{e^{2} x^{2}}{d^{2}}\right )^{- p} \left (d^{2} - e^{2} x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p + 4, - \frac{1}{2} \\ \frac{1}{2} \end{matrix}\middle |{\frac{e^{2} x^{2}}{d^{2}}} \right )}}{d^{6} x} + \frac{e^{4} x \left (1 - \frac{e^{2} x^{2}}{d^{2}}\right )^{- p} \left (d^{2} - e^{2} x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p + 4, \frac{1}{2} \\ \frac{3}{2} \end{matrix}\middle |{\frac{e^{2} x^{2}}{d^{2}}} \right )}}{d^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((-e**2*x**2+d**2)**p/x**4/(e*x+d)**4,x)

[Out]

-2*e**3*(d**2 - e**2*x**2)**(p - 3)*hyper((1, p - 3), (p - 2,), 1 - e**2*x**2/d*
*2)/(d*(-p + 3)) - 2*e**3*(d**2 - e**2*x**2)**(p - 3)*hyper((2, p - 3), (p - 2,)
, 1 - e**2*x**2/d**2)/(d*(-p + 3)) - (1 - e**2*x**2/d**2)**(-p)*(d**2 - e**2*x**
2)**p*hyper((-p + 4, -3/2), (-1/2,), e**2*x**2/d**2)/(3*d**4*x**3) - 6*e**2*(1 -
e**2*x**2/d**2)**(-p)*(d**2 - e**2*x**2)**p*hyper((-p + 4, -1/2), (1/2,), e**2*
x**2/d**2)/(d**6*x) + e**4*x*(1 - e**2*x**2/d**2)**(-p)*(d**2 - e**2*x**2)**p*hy
per((-p + 4, 1/2), (3/2,), e**2*x**2/d**2)/d**8

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Mathematica [B]  time = 1.0128, size = 452, normalized size = 2.15 \[ \frac{\left (d^2-e^2 x^2\right )^p \left (-\frac{480 d^2 e^2 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x}-\frac{480 d e^3 \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (-p,-p;1-p;\frac{d^2}{e^2 x^2}\right )}{p}-\frac{16 d^4 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{3}{2},-p;-\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x^3}-\frac{96 d^3 e \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (1-p,-p;2-p;\frac{d^2}{e^2 x^2}\right )}{(p-1) x^2}+\frac{15 e^3 2^{p+5} (e x-d) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{15 e^3 2^{p+3} (e x-d) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (2-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{3 e^3 2^{p+3} (e x-d) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (3-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{3 e^3 2^p (e x-d) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (4-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}\right )}{48 d^8} \]

Antiderivative was successfully verified.

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

[Out]

((d^2 - e^2*x^2)^p*((-16*d^4*Hypergeometric2F1[-3/2, -p, -1/2, (e^2*x^2)/d^2])/(
x^3*(1 - (e^2*x^2)/d^2)^p) - (480*d^2*e^2*Hypergeometric2F1[-1/2, -p, 1/2, (e^2*
x^2)/d^2])/(x*(1 - (e^2*x^2)/d^2)^p) - (96*d^3*e*Hypergeometric2F1[1 - p, -p, 2
- p, d^2/(e^2*x^2)])/((-1 + p)*(1 - d^2/(e^2*x^2))^p*x^2) + (15*2^(5 + p)*e^3*(-
d + e*x)*Hypergeometric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 +
(e*x)/d)^p) + (15*2^(3 + p)*e^3*(-d + e*x)*Hypergeometric2F1[2 - p, 1 + p, 2 + p
, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) + (3*2^(3 + p)*e^3*(-d + e*x)*Hype
rgeometric2F1[3 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) +
(3*2^p*e^3*(-d + e*x)*Hypergeometric2F1[4 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/
((1 + p)*(1 + (e*x)/d)^p) - (480*d*e^3*Hypergeometric2F1[-p, -p, 1 - p, d^2/(e^2
*x^2)])/(p*(1 - d^2/(e^2*x^2))^p)))/(48*d^8)

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Maple [F]  time = 0.092, size = 0, normalized size = 0. \[ \int{\frac{ \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{p}}{{x}^{4} \left ( ex+d \right ) ^{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{e^{4} x^{8} + 4 \, d e^{3} x^{7} + 6 \, d^{2} e^{2} x^{6} + 4 \, d^{3} e x^{5} + d^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((-e^2*x^2 + d^2)^p/(e^4*x^8 + 4*d*e^3*x^7 + 6*d^2*e^2*x^6 + 4*d^3*e*x^5
+ d^4*x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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