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

Optimal. Leaf size=108 \[ \frac{e \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},1-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{d^2 x}-\frac{e^2 \left (d^2-e^2 x^2\right )^p \, _2F_1\left (2,p;p+1;1-\frac{e^2 x^2}{d^2}\right )}{2 d^3 p} \]

[Out]

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

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Rubi [A]  time = 0.217953, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24 \[ \frac{e \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},1-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{d^2 x}-\frac{e^2 \left (d^2-e^2 x^2\right )^p \, _2F_1\left (2,p;p+1;1-\frac{e^2 x^2}{d^2}\right )}{2 d^3 p} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 35.4839, size = 87, normalized size = 0.81 \[ \frac{e \left (1 - \frac{e^{2} x^{2}}{d^{2}}\right )^{- p} \left (d^{2} - e^{2} x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p + 1, - \frac{1}{2} \\ \frac{1}{2} \end{matrix}\middle |{\frac{e^{2} x^{2}}{d^{2}}} \right )}}{d^{2} x} - \frac{e^{2} \left (d^{2} - e^{2} x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} 2, p \\ p + 1 \end{matrix}\middle |{1 - \frac{e^{2} x^{2}}{d^{2}}} \right )}}{2 d^{3} p} \]

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),x)

[Out]

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

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Mathematica [B]  time = 1.23429, size = 219, normalized size = 2.03 \[ \frac{\left (d^2-e^2 x^2\right )^p \left (\frac{2 d^2 e \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )}{x}+\left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \left (e^2 \left (\frac{(d-e x) \left (2-\frac{2 d^2}{e^2 x^2}\right )^p \left (\frac{e x}{d}+1\right )^{-p} \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{p+1}+\frac{d \, _2F_1\left (-p,-p;1-p;\frac{d^2}{e^2 x^2}\right )}{p}\right )+\frac{d^3 \, _2F_1\left (1-p,-p;2-p;\frac{d^2}{e^2 x^2}\right )}{(p-1) x^2}\right )\right )}{2 d^4} \]

Antiderivative was successfully verified.

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

[Out]

((d^2 - e^2*x^2)^p*((2*d^2*e*Hypergeometric2F1[-1/2, -p, 1/2, (e^2*x^2)/d^2])/(x
*(1 - (e^2*x^2)/d^2)^p) + ((d^3*Hypergeometric2F1[1 - p, -p, 2 - p, d^2/(e^2*x^2
)])/((-1 + p)*x^2) + e^2*(((2 - (2*d^2)/(e^2*x^2))^p*(d - e*x)*Hypergeometric2F1
[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/((1 + p)*(1 + (e*x)/d)^p) + (d*Hypergeom
etric2F1[-p, -p, 1 - p, d^2/(e^2*x^2)])/p))/(1 - d^2/(e^2*x^2))^p))/(2*d^4)

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Maple [F]  time = 0.091, size = 0, normalized size = 0. \[ \int{\frac{ \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{p}}{{x}^{3} \left ( ex+d \right ) }}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 22.4695, size = 498, normalized size = 4.61 \[ \begin{cases} - \frac{0^{p} d^{2 p}}{2 d x^{2}} + \frac{0^{p} d^{2 p} e}{d^{2} x} + \frac{0^{p} d^{2 p} e^{2} \log{\left (\frac{e^{2} x^{2}}{d^{2}} \right )}}{2 d^{3}} - \frac{0^{p} d^{2 p} e^{2} \log{\left (-1 + \frac{e^{2} x^{2}}{d^{2}} \right )}}{2 d^{3}} - \frac{0^{p} d^{2 p} e^{2} \operatorname{acoth}{\left (\frac{e x}{d} \right )}}{d^{3}} + \frac{d e^{2 p} p x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (- p + 2\right ){{}_{2}F_{1}\left (\begin{matrix} - p + 1, - p + 2 \\ - p + 3 \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 e^{2} x^{4} \Gamma \left (- p + 3\right ) \Gamma \left (p + 1\right )} - \frac{e^{2 p} p x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (- p + \frac{3}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} - p + 1, - p + \frac{3}{2} \\ - p + \frac{5}{2} \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 e x^{3} \Gamma \left (- p + \frac{5}{2}\right ) \Gamma \left (p + 1\right )} & \text{for}\: \left |{\frac{e^{2} x^{2}}{d^{2}}}\right | > 1 \\- \frac{0^{p} d^{2 p}}{2 d x^{2}} + \frac{0^{p} d^{2 p} e}{d^{2} x} + \frac{0^{p} d^{2 p} e^{2} \log{\left (\frac{e^{2} x^{2}}{d^{2}} \right )}}{2 d^{3}} - \frac{0^{p} d^{2 p} e^{2} \log{\left (1 - \frac{e^{2} x^{2}}{d^{2}} \right )}}{2 d^{3}} - \frac{0^{p} d^{2 p} e^{2} \operatorname{atanh}{\left (\frac{e x}{d} \right )}}{d^{3}} + \frac{d e^{2 p} p x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (- p + 2\right ){{}_{2}F_{1}\left (\begin{matrix} - p + 1, - p + 2 \\ - p + 3 \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 e^{2} x^{4} \Gamma \left (- p + 3\right ) \Gamma \left (p + 1\right )} - \frac{e^{2 p} p x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (- p + \frac{3}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} - p + 1, - p + \frac{3}{2} \\ - p + \frac{5}{2} \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 e x^{3} \Gamma \left (- p + \frac{5}{2}\right ) \Gamma \left (p + 1\right )} & \text{otherwise} \end{cases} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Piecewise((-0**p*d**(2*p)/(2*d*x**2) + 0**p*d**(2*p)*e/(d**2*x) + 0**p*d**(2*p)*
e**2*log(e**2*x**2/d**2)/(2*d**3) - 0**p*d**(2*p)*e**2*log(-1 + e**2*x**2/d**2)/
(2*d**3) - 0**p*d**(2*p)*e**2*acoth(e*x/d)/d**3 + d*e**(2*p)*p*x**(2*p)*exp(I*pi
*p)*gamma(p)*gamma(-p + 2)*hyper((-p + 1, -p + 2), (-p + 3,), d**2/(e**2*x**2))/
(2*e**2*x**4*gamma(-p + 3)*gamma(p + 1)) - e**(2*p)*p*x**(2*p)*exp(I*pi*p)*gamma
(p)*gamma(-p + 3/2)*hyper((-p + 1, -p + 3/2), (-p + 5/2,), d**2/(e**2*x**2))/(2*
e*x**3*gamma(-p + 5/2)*gamma(p + 1)), Abs(e**2*x**2/d**2) > 1), (-0**p*d**(2*p)/
(2*d*x**2) + 0**p*d**(2*p)*e/(d**2*x) + 0**p*d**(2*p)*e**2*log(e**2*x**2/d**2)/(
2*d**3) - 0**p*d**(2*p)*e**2*log(1 - e**2*x**2/d**2)/(2*d**3) - 0**p*d**(2*p)*e*
*2*atanh(e*x/d)/d**3 + d*e**(2*p)*p*x**(2*p)*exp(I*pi*p)*gamma(p)*gamma(-p + 2)*
hyper((-p + 1, -p + 2), (-p + 3,), d**2/(e**2*x**2))/(2*e**2*x**4*gamma(-p + 3)*
gamma(p + 1)) - e**(2*p)*p*x**(2*p)*exp(I*pi*p)*gamma(p)*gamma(-p + 3/2)*hyper((
-p + 1, -p + 3/2), (-p + 5/2,), d**2/(e**2*x**2))/(2*e*x**3*gamma(-p + 5/2)*gamm
a(p + 1)), True))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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