3.543 \(\int \frac{x^2 (e+f x)^n}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=236 \[ \frac{\left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-b f+\sqrt{b^2-4 a c} f}\right )}{c (n+1) \left (2 c e-f \left (b-\sqrt{b^2-4 a c}\right )\right )}+\frac{\left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (\sqrt{b^2-4 a c}+b\right )\right )}+\frac{(e+f x)^{n+1}}{c f (n+1)} \]

[Out]

(e + f*x)^(1 + n)/(c*f*(1 + n)) + ((b - (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*(e + f*
x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(2*c*e - b*f + Sqr
t[b^2 - 4*a*c]*f)])/(c*(2*c*e - (b - Sqrt[b^2 - 4*a*c])*f)*(1 + n)) + ((b + (b^2
- 2*a*c)/Sqrt[b^2 - 4*a*c])*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n
, (2*c*(e + f*x))/(2*c*e - (b + Sqrt[b^2 - 4*a*c])*f)])/(c*(2*c*e - (b + Sqrt[b^
2 - 4*a*c])*f)*(1 + n))

_______________________________________________________________________________________

Rubi [A]  time = 0.707815, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087 \[ \frac{\left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-b f+\sqrt{b^2-4 a c} f}\right )}{c (n+1) \left (2 c e-f \left (b-\sqrt{b^2-4 a c}\right )\right )}+\frac{\left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (\sqrt{b^2-4 a c}+b\right )\right )}+\frac{(e+f x)^{n+1}}{c f (n+1)} \]

Antiderivative was successfully verified.

[In]  Int[(x^2*(e + f*x)^n)/(a + b*x + c*x^2),x]

[Out]

(e + f*x)^(1 + n)/(c*f*(1 + n)) + ((b - (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*(e + f*
x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(2*c*e - b*f + Sqr
t[b^2 - 4*a*c]*f)])/(c*(2*c*e - (b - Sqrt[b^2 - 4*a*c])*f)*(1 + n)) + ((b + (b^2
- 2*a*c)/Sqrt[b^2 - 4*a*c])*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n
, (2*c*(e + f*x))/(2*c*e - (b + Sqrt[b^2 - 4*a*c])*f)])/(c*(2*c*e - (b + Sqrt[b^
2 - 4*a*c])*f)*(1 + n))

_______________________________________________________________________________________

Rubi in Sympy [A]  time = 66.5773, size = 228, normalized size = 0.97 \[ \frac{\left (e + f x\right )^{n + 1} \left (- 2 a c + b^{2} + b \sqrt{- 4 a c + b^{2}}\right ){{}_{2}F_{1}\left (\begin{matrix} 1, n + 1 \\ n + 2 \end{matrix}\middle |{\frac{c \left (- 2 e - 2 f x\right )}{b f - 2 c e + f \sqrt{- 4 a c + b^{2}}}} \right )}}{c \left (n + 1\right ) \sqrt{- 4 a c + b^{2}} \left (2 c e - f \left (b + \sqrt{- 4 a c + b^{2}}\right )\right )} - \frac{\left (e + f x\right )^{n + 1} \left (- 2 a c + b^{2} - b \sqrt{- 4 a c + b^{2}}\right ){{}_{2}F_{1}\left (\begin{matrix} 1, n + 1 \\ n + 2 \end{matrix}\middle |{\frac{c \left (- 2 e - 2 f x\right )}{b f - 2 c e - f \sqrt{- 4 a c + b^{2}}}} \right )}}{c \left (n + 1\right ) \sqrt{- 4 a c + b^{2}} \left (2 c e - f \left (b - \sqrt{- 4 a c + b^{2}}\right )\right )} + \frac{\left (e + f x\right )^{n + 1}}{c f \left (n + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**2*(f*x+e)**n/(c*x**2+b*x+a),x)

[Out]

(e + f*x)**(n + 1)*(-2*a*c + b**2 + b*sqrt(-4*a*c + b**2))*hyper((1, n + 1), (n
+ 2,), c*(-2*e - 2*f*x)/(b*f - 2*c*e + f*sqrt(-4*a*c + b**2)))/(c*(n + 1)*sqrt(-
4*a*c + b**2)*(2*c*e - f*(b + sqrt(-4*a*c + b**2)))) - (e + f*x)**(n + 1)*(-2*a*
c + b**2 - b*sqrt(-4*a*c + b**2))*hyper((1, n + 1), (n + 2,), c*(-2*e - 2*f*x)/(
b*f - 2*c*e - f*sqrt(-4*a*c + b**2)))/(c*(n + 1)*sqrt(-4*a*c + b**2)*(2*c*e - f*
(b - sqrt(-4*a*c + b**2)))) + (e + f*x)**(n + 1)/(c*f*(n + 1))

_______________________________________________________________________________________

Mathematica [A]  time = 1.90567, size = 346, normalized size = 1.47 \[ \frac{2^{-n-1} (e+f x)^n \left (-\left (b \sqrt{f^2 \left (b^2-4 a c\right )}+2 a c f+b^2 (-f)\right ) \left (\frac{c (e+f x)}{-\sqrt{f^2 \left (b^2-4 a c\right )}+b f+2 c f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;\frac{2 c e-b f+\sqrt{\left (b^2-4 a c\right ) f^2}}{-b f-2 c x f+\sqrt{\left (b^2-4 a c\right ) f^2}}\right )-\left (b \sqrt{f^2 \left (b^2-4 a c\right )}-2 a c f+b^2 f\right ) \left (\frac{c (e+f x)}{\sqrt{f^2 \left (b^2-4 a c\right )}+b f+2 c f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;\frac{-2 c e+b f+\sqrt{\left (b^2-4 a c\right ) f^2}}{b f+2 c x f+\sqrt{\left (b^2-4 a c\right ) f^2}}\right )+\frac{c 2^{n+1} n (e+f x) \sqrt{f^2 \left (b^2-4 a c\right )}}{f (n+1)}\right )}{c^2 n \sqrt{f^2 \left (b^2-4 a c\right )}} \]

Antiderivative was successfully verified.

[In]  Integrate[(x^2*(e + f*x)^n)/(a + b*x + c*x^2),x]

[Out]

(2^(-1 - n)*(e + f*x)^n*((2^(1 + n)*c*Sqrt[(b^2 - 4*a*c)*f^2]*n*(e + f*x))/(f*(1
+ n)) - ((-(b^2*f) + 2*a*c*f + b*Sqrt[(b^2 - 4*a*c)*f^2])*Hypergeometric2F1[-n,
-n, 1 - n, (2*c*e - b*f + Sqrt[(b^2 - 4*a*c)*f^2])/(-(b*f) + Sqrt[(b^2 - 4*a*c)
*f^2] - 2*c*f*x)])/((c*(e + f*x))/(b*f - Sqrt[(b^2 - 4*a*c)*f^2] + 2*c*f*x))^n -
((b^2*f - 2*a*c*f + b*Sqrt[(b^2 - 4*a*c)*f^2])*Hypergeometric2F1[-n, -n, 1 - n,
(-2*c*e + b*f + Sqrt[(b^2 - 4*a*c)*f^2])/(b*f + Sqrt[(b^2 - 4*a*c)*f^2] + 2*c*f
*x)])/((c*(e + f*x))/(b*f + Sqrt[(b^2 - 4*a*c)*f^2] + 2*c*f*x))^n))/(c^2*Sqrt[(b
^2 - 4*a*c)*f^2]*n)

_______________________________________________________________________________________

Maple [F]  time = 0.136, size = 0, normalized size = 0. \[ \int{\frac{{x}^{2} \left ( fx+e \right ) ^{n}}{c{x}^{2}+bx+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x^2*(f*x+e)^n/(c*x^2+b*x+a),x)

[Out]

int(x^2*(f*x+e)^n/(c*x^2+b*x+a),x)

_______________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((f*x + e)^n*x^2/(c*x^2 + b*x + a),x, algorithm="maxima")

[Out]

integrate((f*x + e)^n*x^2/(c*x^2 + b*x + a), x)

_______________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((f*x + e)^n*x^2/(c*x^2 + b*x + a),x, algorithm="fricas")

[Out]

integral((f*x + e)^n*x^2/(c*x^2 + b*x + a), x)

_______________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x**2*(f*x+e)**n/(c*x**2+b*x+a),x)

[Out]

Timed out

_______________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((f*x + e)^n*x^2/(c*x^2 + b*x + a),x, algorithm="giac")

[Out]

integrate((f*x + e)^n*x^2/(c*x^2 + b*x + a), x)