3.303 \(\int \frac{\left (d^2-e^2 x^2\right )^p}{x^3 (d+e x)^4} \, dx\)

Optimal. Leaf size=211 \[ \frac{e^2 (10-p) \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac{e^2 x^2}{d^2}\right )}{2 d^2 (2-p)}+\frac{e^2 (11-p) \left (d^2-e^2 x^2\right )^{p-3}}{2 (3-p)}+\frac{4 d e \left (d^2-e^2 x^2\right )^{p-3}}{x}-\frac{d^2 \left (d^2-e^2 x^2\right )^{p-3}}{2 x^2}-\frac{8 e^3 (4-p) x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{1}{2},4-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^7} \]

[Out]

(e^2*(11 - p)*(d^2 - e^2*x^2)^(-3 + p))/(2*(3 - p)) - (d^2*(d^2 - e^2*x^2)^(-3 +
p))/(2*x^2) + (4*d*e*(d^2 - e^2*x^2)^(-3 + p))/x - (8*e^3*(4 - p)*x*(d^2 - e^2*
x^2)^p*Hypergeometric2F1[1/2, 4 - p, 3/2, (e^2*x^2)/d^2])/(d^7*(1 - (e^2*x^2)/d^
2)^p) + (e^2*(10 - 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*(2 - p))

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Rubi [A]  time = 0.634138, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36 \[ \frac{e^2 (10-p) \left (d^2-e^2 x^2\right )^{p-2} \, _2F_1\left (1,p-2;p-1;1-\frac{e^2 x^2}{d^2}\right )}{2 d^2 (2-p)}+\frac{e^2 (11-p) \left (d^2-e^2 x^2\right )^{p-3}}{2 (3-p)}+\frac{4 d e \left (d^2-e^2 x^2\right )^{p-3}}{x}-\frac{d^2 \left (d^2-e^2 x^2\right )^{p-3}}{2 x^2}-\frac{8 e^3 (4-p) x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (\frac{1}{2},4-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )}{d^7} \]

Antiderivative was successfully verified.

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

[Out]

(e^2*(11 - p)*(d^2 - e^2*x^2)^(-3 + p))/(2*(3 - p)) - (d^2*(d^2 - e^2*x^2)^(-3 +
p))/(2*x^2) + (4*d*e*(d^2 - e^2*x^2)^(-3 + p))/x - (8*e^3*(4 - p)*x*(d^2 - e^2*
x^2)^p*Hypergeometric2F1[1/2, 4 - p, 3/2, (e^2*x^2)/d^2])/(d^7*(1 - (e^2*x^2)/d^
2)^p) + (e^2*(10 - 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*(2 - p))

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Rubi in Sympy [A]  time = 88.1931, size = 204, normalized size = 0.97 \[ \frac{3 e^{2} \left (d^{2} - e^{2} x^{2}\right )^{p - 3}{{}_{2}F_{1}\left (\begin{matrix} 1, p - 3 \\ p - 2 \end{matrix}\middle |{1 - \frac{e^{2} x^{2}}{d^{2}}} \right )}}{- p + 3} + \frac{e^{2} \left (d^{2} - e^{2} x^{2}\right )^{p - 3}{{}_{2}F_{1}\left (\begin{matrix} 2, p - 3 \\ p - 2 \end{matrix}\middle |{1 - \frac{e^{2} x^{2}}{d^{2}}} \right )}}{2 \left (- p + 3\right )} + \frac{e^{2} \left (d^{2} - e^{2} x^{2}\right )^{p - 3}}{2 \left (- p + 3\right )} + \frac{4 e \left (1 - \frac{e^{2} x^{2}}{d^{2}}\right )^{- p} \left (d^{2} - e^{2} x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p + 4, - \frac{1}{2} \\ \frac{1}{2} \end{matrix}\middle |{\frac{e^{2} x^{2}}{d^{2}}} \right )}}{d^{5} x} - \frac{4 e^{3} x \left (1 - \frac{e^{2} x^{2}}{d^{2}}\right )^{- p} \left (d^{2} - e^{2} x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p + 4, \frac{1}{2} \\ \frac{3}{2} \end{matrix}\middle |{\frac{e^{2} x^{2}}{d^{2}}} \right )}}{d^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((-e**2*x**2+d**2)**p/x**3/(e*x+d)**4,x)

[Out]

3*e**2*(d**2 - e**2*x**2)**(p - 3)*hyper((1, p - 3), (p - 2,), 1 - e**2*x**2/d**
2)/(-p + 3) + e**2*(d**2 - e**2*x**2)**(p - 3)*hyper((2, p - 3), (p - 2,), 1 - e
**2*x**2/d**2)/(2*(-p + 3)) + e**2*(d**2 - e**2*x**2)**(p - 3)/(2*(-p + 3)) + 4*
e*(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**5*x) - 4*e**3*x*(1 - e**2*x**2/d**2)**(-p)*(d**2 - e**2*x**
2)**p*hyper((-p + 4, 1/2), (3/2,), e**2*x**2/d**2)/d**7

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Mathematica [A]  time = 2.03137, size = 399, normalized size = 1.89 \[ \frac{\left (d^2-e^2 x^2\right )^p \left (\frac{80 d e^2 \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (-p,-p;1-p;\frac{d^2}{e^2 x^2}\right )}{p}+\frac{64 d^2 e \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x}+\frac{8 d^3 \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \, _2F_1\left (1-p,-p;2-p;\frac{d^2}{e^2 x^2}\right )}{(p-1) x^2}+\frac{5 e^2 2^{p+4} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{3 e^2 2^{p+3} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (2-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{3 e^2 2^{p+1} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (3-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{e^2 2^p (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (4-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}\right )}{16 d^7} \]

Antiderivative was successfully verified.

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

[Out]

((d^2 - e^2*x^2)^p*((64*d^2*e*Hypergeometric2F1[-1/2, -p, 1/2, (e^2*x^2)/d^2])/(
x*(1 - (e^2*x^2)/d^2)^p) + (8*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) + (5*2^(4 + p)*e^2*(d - e*x)*Hypergeo
metric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) + (3*
2^(3 + 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) + (3*2^(1 + 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) + (2^p*e^2*(d - e*x)*
Hypergeometric2F1[4 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^
p) + (80*d*e^2*Hypergeometric2F1[-p, -p, 1 - p, d^2/(e^2*x^2)])/(p*(1 - d^2/(e^2
*x^2))^p)))/(16*d^7)

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Maple [F]  time = 0.127, size = 0, normalized size = 0. \[ \int{\frac{ \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{p}}{{x}^{3} \left ( ex+d \right ) ^{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((-e^2*x^2 + d^2)^p/((e*x + d)^4*x^3),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((-e^2*x^2 + d^2)^p/((e*x + d)^4*x^3),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p}}{x^{3} \left (d + e x\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((-e**2*x**2+d**2)**p/x**3/(e*x+d)**4,x)

[Out]

Integral((-(-d + e*x)*(d + e*x))**p/(x**3*(d + e*x)**4), x)

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{p}}{{\left (e x + d\right )}^{4} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((-e^2*x^2 + d^2)^p/((e*x + d)^4*x^3),x, algorithm="giac")

[Out]

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