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

Optimal. Leaf size=171 \[ -\frac{d^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a (p+1)}-\frac{e x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (a e^2-3 b d^2 (2 p+3)\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{b (2 p+3)}+\frac{3 d e^2 \left (a+b x^2\right )^{p+1}}{2 b (p+1)}+\frac{e^3 x \left (a+b x^2\right )^{p+1}}{b (2 p+3)} \]

[Out]

(3*d*e^2*(a + b*x^2)^(1 + p))/(2*b*(1 + p)) + (e^3*x*(a + b*x^2)^(1 + p))/(b*(3
+ 2*p)) - (e*(a*e^2 - 3*b*d^2*(3 + 2*p))*x*(a + b*x^2)^p*Hypergeometric2F1[1/2,
-p, 3/2, -((b*x^2)/a)])/(b*(3 + 2*p)*(1 + (b*x^2)/a)^p) - (d^3*(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 = 0.271101, antiderivative size = 165, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35 \[ -\frac{d^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a (p+1)}+e x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 d^2-\frac{a e^2}{2 b p+3 b}\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )+\frac{3 d e^2 \left (a+b x^2\right )^{p+1}}{2 b (p+1)}+\frac{e^3 x \left (a+b x^2\right )^{p+1}}{b (2 p+3)} \]

Antiderivative was successfully verified.

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

[Out]

(3*d*e^2*(a + b*x^2)^(1 + p))/(2*b*(1 + p)) + (e^3*x*(a + b*x^2)^(1 + p))/(b*(3
+ 2*p)) + (e*(3*d^2 - (a*e^2)/(3*b + 2*b*p))*x*(a + b*x^2)^p*Hypergeometric2F1[1
/2, -p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p - (d^3*(a + b*x^2)^(1 + p)*Hyperge
ometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a])/(2*a*(1 + p))

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Rubi in Sympy [A]  time = 29.7222, size = 141, normalized size = 0.82 \[ 3 d^{2} e x \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{1}{2} \\ \frac{3}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )} + \frac{e^{3} x^{3} \left (1 + \frac{b x^{2}}{a}\right )^{- p} \left (a + b x^{2}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{3}{2} \\ \frac{5}{2} \end{matrix}\middle |{- \frac{b x^{2}}{a}} \right )}}{3} + \frac{3 d e^{2} \left (a + b x^{2}\right )^{p + 1}}{2 b \left (p + 1\right )} - \frac{d^{3} \left (a + b x^{2}\right )^{p + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, p + 1 \\ p + 2 \end{matrix}\middle |{1 + \frac{b x^{2}}{a}} \right )}}{2 a \left (p + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

3*d**2*e*x*(1 + b*x**2/a)**(-p)*(a + b*x**2)**p*hyper((-p, 1/2), (3/2,), -b*x**2
/a) + e**3*x**3*(1 + b*x**2/a)**(-p)*(a + b*x**2)**p*hyper((-p, 3/2), (5/2,), -b
*x**2/a)/3 + 3*d*e**2*(a + b*x**2)**(p + 1)/(2*b*(p + 1)) - d**3*(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 = 0.838119, size = 175, normalized size = 1.02 \[ \frac{1}{6} \left (a+b x^2\right )^p \left (\frac{3 d^3 \left (\frac{a}{b x^2}+1\right )^{-p} \, _2F_1\left (-p,-p;1-p;-\frac{a}{b x^2}\right )}{p}+18 d^2 e x \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )+\frac{9 d e^2 \left (-a \left (\frac{b x^2}{a}+1\right )^{-p}+a+b x^2\right )}{b p+b}+2 e^3 x^3 \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )\right ) \]

Antiderivative was successfully verified.

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

[Out]

((a + b*x^2)^p*((9*d*e^2*(a + b*x^2 - a/(1 + (b*x^2)/a)^p))/(b + b*p) + (18*d^2*
e*x*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p + (2*e^3*x^
3*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p + (3*d^3*Hype
rgeometric2F1[-p, -p, 1 - p, -(a/(b*x^2))])/(p*(1 + a/(b*x^2))^p)))/6

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \[ \int{\frac{ \left ( ex+d \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{p}}{x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((e*x+d)^3*(b*x^2+a)^p/x,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

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Sympy [A]  time = 63.7909, size = 144, normalized size = 0.84 \[ 3 a^{p} d^{2} e x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )} + \frac{a^{p} e^{3} x^{3}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, - p \\ \frac{5}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{3} - \frac{b^{p} d^{3} x^{2 p} \Gamma \left (- p\right ){{}_{2}F_{1}\left (\begin{matrix} - p, - p \\ - p + 1 \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (- p + 1\right )} + 3 d e^{2} \left (\begin{cases} \frac{a^{p} x^{2}}{2} & \text{for}\: b = 0 \\\frac{\begin{cases} \frac{\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (a + b x^{2} \right )} & \text{otherwise} \end{cases}}{2 b} & \text{otherwise} \end{cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

3*a**p*d**2*e*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) + a**p*e**3*x
**3*hyper((3/2, -p), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 - b**p*d**3*x**(2*p)*ga
mma(-p)*hyper((-p, -p), (-p + 1,), a*exp_polar(I*pi)/(b*x**2))/(2*gamma(-p + 1))
+ 3*d*e**2*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. \[ \int \frac{{\left (e x + d\right )}^{3}{\left (b x^{2} + a\right )}^{p}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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