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

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

[Out]

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

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Rubi [A]  time = 0.325915, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4 \[ -\frac{d^3 \left (a+b x^2\right )^{p+1}}{a x}+\frac{d x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 a e^2+b d^2 (2 p+1)\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{a}-\frac{3 d^2 e \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a (p+1)}+\frac{e^3 \left (a+b x^2\right )^{p+1}}{2 b (p+1)} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 29.5236, size = 139, normalized size = 0.87 \[ - \frac{d^{3} \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, - \frac{1}{2} \\ \frac{1}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{x} + 3 d e^{2} x \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{1}{2} \\ \frac{3}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )} + \frac{e^{3} \left (a + b x^{2}\right )^{p + 1}}{2 b \left (p + 1\right )} - \frac{3 d^{2} e \left (a + b x^{2}\right )^{p + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, p + 1 \\ p + 2 \end{matrix}\middle |{1 + \frac{b x^{2}}{a}} \right )}}{2 a \left (p + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.955778, size = 174, normalized size = 1.09 \[ \frac{1}{2} \left (a+b x^2\right )^p \left (e \left (\frac{3 d^2 \left (\frac{a}{b x^2}+1\right )^{-p} \, _2F_1\left (-p,-p;1-p;-\frac{a}{b x^2}\right )}{p}+6 d e x \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )+\frac{e^2 \left (-a \left (\frac{b x^2}{a}+1\right )^{-p}+a+b x^2\right )}{b p+b}\right )-\frac{2 d^3 \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )}{x}\right ) \]

Antiderivative was successfully verified.

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

[Out]

((a + b*x^2)^p*((-2*d^3*Hypergeometric2F1[-1/2, -p, 1/2, -((b*x^2)/a)])/(x*(1 +
(b*x^2)/a)^p) + e*((e^2*(a + b*x^2 - a/(1 + (b*x^2)/a)^p))/(b + b*p) + (6*d*e*x*
Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p + (3*d^2*Hyperg
eometric2F1[-p, -p, 1 - p, -(a/(b*x^2))])/(p*(1 + a/(b*x^2))^p))))/2

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Maple [F]  time = 0.087, size = 0, normalized size = 0. \[ \int{\frac{ \left ( ex+d \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{p}}{{x}^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

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Sympy [A]  time = 69.6426, size = 143, normalized size = 0.9 \[ - \frac{a^{p} d^{3}{{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - p \\ \frac{1}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{x} + 3 a^{p} d e^{2} x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )} - \frac{3 b^{p} d^{2} e x^{2 p} \Gamma \left (- p\right ){{}_{2}F_{1}\left (\begin{matrix} - p, - p \\ - p + 1 \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (- p + 1\right )} + e^{3} \left (\begin{cases} \frac{a^{p} x^{2}}{2} & \text{for}\: b = 0 \\\frac{\begin{cases} \frac{\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (a + b x^{2} \right )} & \text{otherwise} \end{cases}}{2 b} & \text{otherwise} \end{cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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