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

Optimal. Leaf size=159 \[ 2 d e^2 (1-p) x \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )-\frac{3 e \left (d^2-e^2 x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;1-\frac{e^2 x^2}{d^2}\right )}{2 (p+1)}-\frac{e \left (d^2-e^2 x^2\right )^{p+1}}{2 (p+1)}-\frac{d \left (d^2-e^2 x^2\right )^{p+1}}{x} \]

[Out]

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

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Rubi [A]  time = 0.323462, antiderivative size = 159, 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 \[ 2 d e^2 (1-p) x \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )-\frac{3 e \left (d^2-e^2 x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;1-\frac{e^2 x^2}{d^2}\right )}{2 (p+1)}-\frac{e \left (d^2-e^2 x^2\right )^{p+1}}{2 (p+1)}-\frac{d \left (d^2-e^2 x^2\right )^{p+1}}{x} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 51.8829, size = 158, normalized size = 0.99 \[ - \frac{d^{3} \left (1 - \frac{e^{2} x^{2}}{d^{2}}\right )^{- p} \left (d^{2} - e^{2} x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, - \frac{1}{2} \\ \frac{1}{2} \end{matrix}\middle |{\frac{e^{2} x^{2}}{d^{2}}} \right )}}{x} + 3 d e^{2} x \left (1 - \frac{e^{2} x^{2}}{d^{2}}\right )^{- p} \left (d^{2} - e^{2} x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{1}{2} \\ \frac{3}{2} \end{matrix}\middle |{\frac{e^{2} x^{2}}{d^{2}}} \right )} - \frac{3 e \left (d^{2} - e^{2} x^{2}\right )^{p + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, p + 1 \\ p + 2 \end{matrix}\middle |{1 - \frac{e^{2} x^{2}}{d^{2}}} \right )}}{2 \left (p + 1\right )} - \frac{e \left (d^{2} - e^{2} x^{2}\right )^{p + 1}}{2 \left (p + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.362842, size = 264, normalized size = 1.66 \[ \frac{\left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (e x \left (6 d e p (p+1) x \left (1-\frac{d^2}{e^2 x^2}\right )^p \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )+3 d^2 (p+1) \left (1-\frac{e^2 x^2}{d^2}\right )^p \, _2F_1\left (-p,-p;1-p;\frac{d^2}{e^2 x^2}\right )-p \left (d^2 \left (\left (1-\frac{e^2 x^2}{d^2}\right )^p-1\right )-e^2 x^2 \left (1-\frac{e^2 x^2}{d^2}\right )^p\right ) \left (1-\frac{d^2}{e^2 x^2}\right )^p\right )-2 d^3 p (p+1) \left (1-\frac{d^2}{e^2 x^2}\right )^p \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )\right )}{2 p (p+1) x} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [F]  time = 0.047, size = 0, normalized size = 0. \[ \int{\frac{ \left ( ex+d \right ) ^{3} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{p}}{{x}^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 14.0684, size = 177, normalized size = 1.11 \[ - \frac{d^{3} d^{2 p}{{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - p \\ \frac{1}{2} \end{matrix}\middle |{\frac{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{x} - \frac{3 d^{2} e e^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (- p\right ){{}_{2}F_{1}\left (\begin{matrix} - p, - p \\ - p + 1 \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 \Gamma \left (- p + 1\right )} + 3 d d^{2 p} e^{2} x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} + e^{3} \left (\begin{cases} \frac{x^{2} \left (d^{2}\right )^{p}}{2} & \text{for}\: e^{2} = 0 \\- \frac{\begin{cases} \frac{\left (d^{2} - e^{2} x^{2}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (d^{2} - e^{2} x^{2} \right )} & \text{otherwise} \end{cases}}{2 e^{2}} & \text{otherwise} \end{cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-d**3*d**(2*p)*hyper((-1/2, -p), (1/2,), e**2*x**2*exp_polar(2*I*pi)/d**2)/x - 3
*d**2*e*e**(2*p)*x**(2*p)*exp(I*pi*p)*gamma(-p)*hyper((-p, -p), (-p + 1,), d**2/
(e**2*x**2))/(2*gamma(-p + 1)) + 3*d*d**(2*p)*e**2*x*hyper((1/2, -p), (3/2,), e*
*2*x**2*exp_polar(2*I*pi)/d**2) + e**3*Piecewise((x**2*(d**2)**p/2, Eq(e**2, 0))
, (-Piecewise(((d**2 - e**2*x**2)**(p + 1)/(p + 1), Ne(p, -1)), (log(d**2 - e**2
*x**2), True))/(2*e**2), True))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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