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

Optimal. Leaf size=295 \[ -\frac{c \left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-b f+\sqrt{b^2-4 a c} f}\right )}{a^2 (n+1) \left (2 c e-f \left (b-\sqrt{b^2-4 a c}\right )\right )}-\frac{c \left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{a^2 (n+1) \left (2 c e-f \left (\sqrt{b^2-4 a c}+b\right )\right )}+\frac{b (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{f x}{e}+1\right )}{a^2 e (n+1)}+\frac{f (e+f x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{f x}{e}+1\right )}{a e^2 (n+1)} \]

[Out]

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

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Rubi [A]  time = 1.00389, antiderivative size = 295, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217 \[ -\frac{c \left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-b f+\sqrt{b^2-4 a c} f}\right )}{a^2 (n+1) \left (2 c e-f \left (b-\sqrt{b^2-4 a c}\right )\right )}-\frac{c \left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{a^2 (n+1) \left (2 c e-f \left (\sqrt{b^2-4 a c}+b\right )\right )}+\frac{b (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{f x}{e}+1\right )}{a^2 e (n+1)}+\frac{f (e+f x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{f x}{e}+1\right )}{a e^2 (n+1)} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 116.988, size = 282, normalized size = 0.96 \[ \frac{f \left (e + f x\right )^{n + 1}{{}_{2}F_{1}\left (\begin{matrix} 2, n + 1 \\ n + 2 \end{matrix}\middle |{1 + \frac{f x}{e}} \right )}}{a e^{2} \left (n + 1\right )} + \frac{b \left (e + f x\right )^{n + 1}{{}_{2}F_{1}\left (\begin{matrix} 1, n + 1 \\ n + 2 \end{matrix}\middle |{1 + \frac{f x}{e}} \right )}}{a^{2} e \left (n + 1\right )} - \frac{c \left (e + f x\right )^{n + 1} \left (- 2 a c + b^{2} - b \sqrt{- 4 a c + b^{2}}\right ){{}_{2}F_{1}\left (\begin{matrix} 1, n + 1 \\ n + 2 \end{matrix}\middle |{\frac{c \left (- 2 e - 2 f x\right )}{b f - 2 c e + f \sqrt{- 4 a c + b^{2}}}} \right )}}{a^{2} \left (n + 1\right ) \sqrt{- 4 a c + b^{2}} \left (b f - 2 c e + f \sqrt{- 4 a c + b^{2}}\right )} - \frac{c \left (e + f x\right )^{n + 1} \left (- 2 a c + b^{2} + b \sqrt{- 4 a c + b^{2}}\right ){{}_{2}F_{1}\left (\begin{matrix} 1, n + 1 \\ n + 2 \end{matrix}\middle |{\frac{c \left (- 2 e - 2 f x\right )}{b f - 2 c e - f \sqrt{- 4 a c + b^{2}}}} \right )}}{a^{2} \left (n + 1\right ) \sqrt{- 4 a c + b^{2}} \left (2 c e - f \left (b - \sqrt{- 4 a c + b^{2}}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

f*(e + f*x)**(n + 1)*hyper((2, n + 1), (n + 2,), 1 + f*x/e)/(a*e**2*(n + 1)) + b
*(e + f*x)**(n + 1)*hyper((1, n + 1), (n + 2,), 1 + f*x/e)/(a**2*e*(n + 1)) - c*
(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)))/(a**2*(n + 1)*sqr
t(-4*a*c + b**2)*(b*f - 2*c*e + f*sqrt(-4*a*c + b**2))) - c*(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)))/(a**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 = 3.73591, size = 431, normalized size = 1.46 \[ \frac{\left (\frac{e}{f x}+1\right )^{-n} (e+f x)^n \left (\frac{2^{-n} \left (b \sqrt{f^2 \left (b^2-4 a c\right )}-2 a c f+b^2 f\right ) \left (\frac{e}{f x}+1\right )^n \left (\frac{c (e+f x)}{-\sqrt{f^2 \left (b^2-4 a c\right )}+b f+2 c f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;\frac{2 c e-b f+\sqrt{\left (b^2-4 a c\right ) f^2}}{-b f-2 c x f+\sqrt{\left (b^2-4 a c\right ) f^2}}\right )}{n \sqrt{f^2 \left (b^2-4 a c\right )}}+\frac{2^{-n} \left (b \sqrt{f^2 \left (b^2-4 a c\right )}+2 a c f+b^2 (-f)\right ) \left (\frac{e}{f x}+1\right )^n \left (\frac{c (e+f x)}{\sqrt{f^2 \left (b^2-4 a c\right )}+b f+2 c f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;\frac{-2 c e+b f+\sqrt{\left (b^2-4 a c\right ) f^2}}{b f+2 c x f+\sqrt{\left (b^2-4 a c\right ) f^2}}\right )}{n \sqrt{f^2 \left (b^2-4 a c\right )}}+\frac{2 a \, _2F_1\left (1-n,-n;2-n;-\frac{e}{f x}\right )}{(n-1) x}-\frac{2 b \, _2F_1\left (-n,-n;1-n;-\frac{e}{f x}\right )}{n}\right )}{2 a^2} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [F]  time = 0.143, size = 0, normalized size = 0. \[ \int{\frac{ \left ( fx+e \right ) ^{n}}{{x}^{2} \left ( c{x}^{2}+bx+a \right ) }}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (f x + e\right )}^{n}}{c x^{4} + b x^{3} + a x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral((f*x + e)^n/(c*x^4 + b*x^3 + a*x^2), x)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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