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

Optimal. Leaf size=191 \[ \frac{2 c (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{(n+1) \sqrt{b^2-4 a c} \left (2 c e-f \left (\sqrt{b^2-4 a c}+b\right )\right )}-\frac{2 c (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b-\sqrt{b^2-4 a c}\right ) f}\right )}{(n+1) \sqrt{b^2-4 a c} \left (2 c e-f \left (b-\sqrt{b^2-4 a c}\right )\right )} \]

[Out]

(-2*c*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(2*c*e - (b - 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 = 0.259567, antiderivative size = 191, 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, Rules used = {711, 68} \[ \frac{2 c (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{(n+1) \sqrt{b^2-4 a c} \left (2 c e-f \left (\sqrt{b^2-4 a c}+b\right )\right )}-\frac{2 c (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b-\sqrt{b^2-4 a c}\right ) f}\right )}{(n+1) \sqrt{b^2-4 a c} \left (2 c e-f \left (b-\sqrt{b^2-4 a c}\right )\right )} \]

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 - 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))

Rule 711

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^
m, 1/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a
*e^2, 0] && NeQ[2*c*d - b*e, 0] &&  !IntegerQ[m]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{(e+f x)^n}{a+b x+c x^2} \, dx &=\int \left (\frac{2 c (e+f x)^n}{\sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}+2 c x\right )}-\frac{2 c (e+f x)^n}{\sqrt{b^2-4 a c} \left (b+\sqrt{b^2-4 a c}+2 c x\right )}\right ) \, dx\\ &=\frac{(2 c) \int \frac{(e+f x)^n}{b-\sqrt{b^2-4 a c}+2 c x} \, dx}{\sqrt{b^2-4 a c}}-\frac{(2 c) \int \frac{(e+f x)^n}{b+\sqrt{b^2-4 a c}+2 c x} \, dx}{\sqrt{b^2-4 a c}}\\ &=-\frac{2 c (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{2 c (e+f x)}{2 c e-\left (b-\sqrt{b^2-4 a c}\right ) f}\right )}{\sqrt{b^2-4 a c} \left (2 c e-\left (b-\sqrt{b^2-4 a c}\right ) f\right ) (1+n)}+\frac{2 c (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{\sqrt{b^2-4 a c} \left (2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f\right ) (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.246821, size = 163, normalized size = 0.85 \[ \frac{2 c (e+f x)^{n+1} \left (\frac{\, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{2 c e-f \left (\sqrt{b^2-4 a c}+b\right )}-\frac{\, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e+\left (\sqrt{b^2-4 a c}-b\right ) f}\right )}{f \left (\sqrt{b^2-4 a c}-b\right )+2 c e}\right )}{(n+1) \sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

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

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Maple [F]  time = 1.303, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{n}}{c{x}^{2}+bx+a}}\, dx \end{align*}

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. \begin{align*} \int \frac{{\left (f x + e\right )}^{n}}{c x^{2} + b x + a}\,{d x} \end{align*}

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. \begin{align*}{\rm integral}\left (\frac{{\left (f x + e\right )}^{n}}{c x^{2} + b x + a}, x\right ) \end{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

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)