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

Optimal. Leaf size=166 \[ -\frac{e^2 (3-p) \left (d^2-e^2 x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;1-\frac{e^2 x^2}{d^2}\right )}{2 d (p+1)}-\frac{3 e \left (d^2-e^2 x^2\right )^{p+1}}{x}-\frac{d \left (d^2-e^2 x^2\right )^{p+1}}{2 x^2}-2 e^3 (3 p+1) x \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.367427, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24 \[ -\frac{e^2 (3-p) \left (d^2-e^2 x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;1-\frac{e^2 x^2}{d^2}\right )}{2 d (p+1)}-\frac{3 e \left (d^2-e^2 x^2\right )^{p+1}}{x}-\frac{d \left (d^2-e^2 x^2\right )^{p+1}}{2 x^2}-2 e^3 (3 p+1) x \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right ) \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 58.7051, size = 182, normalized size = 1.1 \[ - \frac{3 d^{2} e \left (1 - \frac{e^{2} x^{2}}{d^{2}}\right )^{- p} \left (d^{2} - e^{2} x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, - \frac{1}{2} \\ \frac{1}{2} \end{matrix}\middle |{\frac{e^{2} x^{2}}{d^{2}}} \right )}}{x} + e^{3} x \left (1 - \frac{e^{2} x^{2}}{d^{2}}\right )^{- p} \left (d^{2} - e^{2} x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{1}{2} \\ \frac{3}{2} \end{matrix}\middle |{\frac{e^{2} x^{2}}{d^{2}}} \right )} - \frac{3 e^{2} \left (d^{2} - e^{2} x^{2}\right )^{p + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, p + 1 \\ p + 2 \end{matrix}\middle |{1 - \frac{e^{2} x^{2}}{d^{2}}} \right )}}{2 d \left (p + 1\right )} - \frac{e^{2} \left (d^{2} - e^{2} x^{2}\right )^{p + 1}{{}_{2}F_{1}\left (\begin{matrix} 2, p + 1 \\ p + 2 \end{matrix}\middle |{1 - \frac{e^{2} x^{2}}{d^{2}}} \right )}}{2 d \left (p + 1\right )} \]

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**3,x)

[Out]

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

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Mathematica [A]  time = 0.290167, size = 201, normalized size = 1.21 \[ \frac{\left (d^2-e^2 x^2\right )^p \left (\frac{d \left (1-\frac{d^2}{e^2 x^2}\right )^{-p} \left (d^2 p \, _2F_1\left (1-p,-p;2-p;\frac{d^2}{e^2 x^2}\right )+3 e^2 (p-1) x^2 \, _2F_1\left (-p,-p;1-p;\frac{d^2}{e^2 x^2}\right )\right )}{(p-1) p}-6 d^2 e x \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};\frac{e^2 x^2}{d^2}\right )+2 e^3 x^3 \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};\frac{e^2 x^2}{d^2}\right )\right )}{2 x^2} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [F]  time = 0.058, size = 0, normalized size = 0. \[ \int{\frac{ \left ( ex+d \right ) ^{3} \left ( -{e}^{2}{x}^{2}+{d}^{2} \right ) ^{p}}{{x}^{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

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Sympy [A]  time = 18.313, size = 177, normalized size = 1.07 \[ - \frac{d^{3} e^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (- p + 1\right ){{}_{2}F_{1}\left (\begin{matrix} - p, - p + 1 \\ - p + 2 \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 x^{2} \Gamma \left (- p + 2\right )} - \frac{3 d^{2} d^{2 p} e{{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - p \\ \frac{1}{2} \end{matrix}\middle |{\frac{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{x} - \frac{3 d e^{2} e^{2 p} x^{2 p} e^{i \pi p} \Gamma \left (- p\right ){{}_{2}F_{1}\left (\begin{matrix} - p, - p \\ - p + 1 \end{matrix}\middle |{\frac{d^{2}}{e^{2} x^{2}}} \right )}}{2 \Gamma \left (- p + 1\right )} + d^{2 p} e^{3} x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-d**3*e**(2*p)*x**(2*p)*exp(I*pi*p)*gamma(-p + 1)*hyper((-p, -p + 1), (-p + 2,),
d**2/(e**2*x**2))/(2*x**2*gamma(-p + 2)) - 3*d**2*d**(2*p)*e*hyper((-1/2, -p),
(1/2,), e**2*x**2*exp_polar(2*I*pi)/d**2)/x - 3*d*e**2*e**(2*p)*x**(2*p)*exp(I*p
i*p)*gamma(-p)*hyper((-p, -p), (-p + 1,), d**2/(e**2*x**2))/(2*gamma(-p + 1)) +
d**(2*p)*e**3*x*hyper((1/2, -p), (3/2,), e**2*x**2*exp_polar(2*I*pi)/d**2)

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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