3.398 \(\int \frac{(d+e x)^2 \left (a+b x^2\right )^p}{x^3} \, dx\)

Optimal. Leaf size=127 \[ -\frac{\left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 p\right ) \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a^2 (p+1)}-\frac{d^2 \left (a+b x^2\right )^{p+1}}{2 a x^2}-\frac{2 d e \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.238728, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3 \[ -\frac{\left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 p\right ) \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a^2 (p+1)}-\frac{d^2 \left (a+b x^2\right )^{p+1}}{2 a x^2}-\frac{2 d e \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )}{x} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 25.6477, size = 112, normalized size = 0.88 \[ - \frac{2 d e \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, - \frac{1}{2} \\ \frac{1}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{x} - \frac{e^{2} \left (a + b x^{2}\right )^{p + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, p + 1 \\ p + 2 \end{matrix}\middle |{1 + \frac{b x^{2}}{a}} \right )}}{2 a \left (p + 1\right )} + \frac{b d^{2} \left (a + b x^{2}\right )^{p + 1}{{}_{2}F_{1}\left (\begin{matrix} 2, p + 1 \\ p + 2 \end{matrix}\middle |{1 + \frac{b x^{2}}{a}} \right )}}{2 a^{2} \left (p + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.275336, size = 138, normalized size = 1.09 \[ \frac{\left (a+b x^2\right )^p \left (\frac{\left (\frac{a}{b x^2}+1\right )^{-p} \left (d^2 p \, _2F_1\left (1-p,-p;2-p;-\frac{a}{b x^2}\right )+e^2 (p-1) x^2 \, _2F_1\left (-p,-p;1-p;-\frac{a}{b x^2}\right )\right )}{(p-1) p}-4 d e x \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )\right )}{2 x^2} \]

Antiderivative was successfully verified.

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

[Out]

((a + b*x^2)^p*((-4*d*e*x*Hypergeometric2F1[-1/2, -p, 1/2, -((b*x^2)/a)])/(1 + (
b*x^2)/a)^p + (d^2*p*Hypergeometric2F1[1 - p, -p, 2 - p, -(a/(b*x^2))] + e^2*(-1
+ p)*x^2*Hypergeometric2F1[-p, -p, 1 - p, -(a/(b*x^2))])/((-1 + p)*p*(1 + a/(b*
x^2))^p)))/(2*x^2)

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Maple [F]  time = 0.142, size = 0, normalized size = 0. \[ \int{\frac{ \left ( ex+d \right ) ^{2} \left ( b{x}^{2}+a \right ) ^{p}}{{x}^{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 74.213, size = 119, normalized size = 0.94 \[ - \frac{2 a^{p} d e{{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - p \\ \frac{1}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{x} - \frac{b^{p} d^{2} x^{2 p} \Gamma \left (- p + 1\right ){{}_{2}F_{1}\left (\begin{matrix} - p, - p + 1 \\ - p + 2 \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{2 x^{2} \Gamma \left (- p + 2\right )} - \frac{b^{p} e^{2} x^{2 p} \Gamma \left (- p\right ){{}_{2}F_{1}\left (\begin{matrix} - p, - p \\ - p + 1 \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (- p + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-2*a**p*d*e*hyper((-1/2, -p), (1/2,), b*x**2*exp_polar(I*pi)/a)/x - b**p*d**2*x*
*(2*p)*gamma(-p + 1)*hyper((-p, -p + 1), (-p + 2,), a*exp_polar(I*pi)/(b*x**2))/
(2*x**2*gamma(-p + 2)) - b**p*e**2*x**(2*p)*gamma(-p)*hyper((-p, -p), (-p + 1,),
a*exp_polar(I*pi)/(b*x**2))/(2*gamma(-p + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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