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

Optimal. Leaf size=190 \[ \frac{2 c (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{(n+1) \sqrt{b^2-4 a c} \left (2 c e-f \left (\sqrt{b^2-4 a c}+b\right )\right )}-\frac{2 c (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-b f+\sqrt{b^2-4 a c} f}\right )}{(n+1) \sqrt{b^2-4 a c} \left (2 c e-f \left (b-\sqrt{b^2-4 a c}\right )\right )} \]

[Out]

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

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Rubi [A]  time = 0.478302, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1 \[ \frac{2 c (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{(n+1) \sqrt{b^2-4 a c} \left (2 c e-f \left (\sqrt{b^2-4 a c}+b\right )\right )}-\frac{2 c (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-b f+\sqrt{b^2-4 a c} f}\right )}{(n+1) \sqrt{b^2-4 a c} \left (2 c e-f \left (b-\sqrt{b^2-4 a c}\right )\right )} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 58.5569, size = 172, normalized size = 0.91 \[ \frac{2 c \left (e + f x\right )^{n + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, n + 1 \\ n + 2 \end{matrix}\middle |{\frac{c \left (- 2 e - 2 f x\right )}{b f - 2 c e + f \sqrt{- 4 a c + b^{2}}}} \right )}}{\left (n + 1\right ) \sqrt{- 4 a c + b^{2}} \left (2 c e - f \left (b + \sqrt{- 4 a c + b^{2}}\right )\right )} - \frac{2 c \left (e + f x\right )^{n + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, n + 1 \\ n + 2 \end{matrix}\middle |{\frac{c \left (- 2 e - 2 f x\right )}{b f - 2 c e - f \sqrt{- 4 a c + b^{2}}}} \right )}}{\left (n + 1\right ) \sqrt{- 4 a c + b^{2}} \left (2 c e - f \left (b - \sqrt{- 4 a c + b^{2}}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.646766, size = 245, normalized size = 1.29 \[ \frac{f 2^{-n} (e+f x)^n \left (\left (\frac{c (e+f x)}{-\sqrt{f^2 \left (b^2-4 a c\right )}+b f+2 c f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;\frac{2 c e-b f+\sqrt{\left (b^2-4 a c\right ) f^2}}{-b f-2 c x f+\sqrt{\left (b^2-4 a c\right ) f^2}}\right )-\left (\frac{c (e+f x)}{\sqrt{f^2 \left (b^2-4 a c\right )}+b f+2 c f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;\frac{-2 c e+b f+\sqrt{\left (b^2-4 a c\right ) f^2}}{b f+2 c x f+\sqrt{\left (b^2-4 a c\right ) f^2}}\right )\right )}{n \sqrt{f^2 \left (b^2-4 a c\right )}} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [F]  time = 0.154, size = 0, normalized size = 0. \[ \int{\frac{ \left ( fx+e \right ) ^{n}}{c{x}^{2}+bx+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((f*x+e)^n/(c*x^2+b*x+a),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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