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

Optimal. Leaf size=296 \[ -\frac{c \left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b-\sqrt{b^2-4 a c}\right ) f}\right )}{a^2 (n+1) \left (2 c e-f \left (b-\sqrt{b^2-4 a c}\right )\right )}-\frac{c \left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{a^2 (n+1) \left (2 c e-f \left (\sqrt{b^2-4 a c}+b\right )\right )}+\frac{b (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{f x}{e}+1\right )}{a^2 e (n+1)}+\frac{f (e+f x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{f x}{e}+1\right )}{a e^2 (n+1)} \]

[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 - 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 + Sqr
t[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)*Hypergeometric2F
1[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 = 0.477245, antiderivative size = 296, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {960, 65, 830, 68} \[ -\frac{c \left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b-\sqrt{b^2-4 a c}\right ) f}\right )}{a^2 (n+1) \left (2 c e-f \left (b-\sqrt{b^2-4 a c}\right )\right )}-\frac{c \left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{a^2 (n+1) \left (2 c e-f \left (\sqrt{b^2-4 a c}+b\right )\right )}+\frac{b (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{f x}{e}+1\right )}{a^2 e (n+1)}+\frac{f (e+f x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{f x}{e}+1\right )}{a e^2 (n+1)} \]

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 - 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 + Sqr
t[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)*Hypergeometric2F
1[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))

Rule 960

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] &&
ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rule 830

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m, (f + g*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !RationalQ[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}{x^2 \left (a+b x+c x^2\right )} \, dx &=\int \left (\frac{(e+f x)^n}{a x^2}-\frac{b (e+f x)^n}{a^2 x}+\frac{\left (b^2-a c+b c x\right ) (e+f x)^n}{a^2 \left (a+b x+c x^2\right )}\right ) \, dx\\ &=\frac{\int \frac{\left (b^2-a c+b c x\right ) (e+f x)^n}{a+b x+c x^2} \, dx}{a^2}+\frac{\int \frac{(e+f x)^n}{x^2} \, dx}{a}-\frac{b \int \frac{(e+f x)^n}{x} \, dx}{a^2}\\ &=\frac{b (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac{f x}{e}\right )}{a^2 e (1+n)}+\frac{f (e+f x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac{f x}{e}\right )}{a e^2 (1+n)}+\frac{\int \left (\frac{\left (b c+\frac{c \left (b^2-2 a c\right )}{\sqrt{b^2-4 a c}}\right ) (e+f x)^n}{b-\sqrt{b^2-4 a c}+2 c x}+\frac{\left (b c-\frac{c \left (b^2-2 a c\right )}{\sqrt{b^2-4 a c}}\right ) (e+f x)^n}{b+\sqrt{b^2-4 a c}+2 c x}\right ) \, dx}{a^2}\\ &=\frac{b (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac{f x}{e}\right )}{a^2 e (1+n)}+\frac{f (e+f x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac{f x}{e}\right )}{a e^2 (1+n)}+\frac{\left (c \left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{(e+f x)^n}{b+\sqrt{b^2-4 a c}+2 c x} \, dx}{a^2}+\frac{\left (c \left (b+\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{(e+f x)^n}{b-\sqrt{b^2-4 a c}+2 c x} \, dx}{a^2}\\ &=-\frac{c \left (b+\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) (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 )}{a^2 \left (2 c e-\left (b-\sqrt{b^2-4 a c}\right ) f\right ) (1+n)}-\frac{c \left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) (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 )}{a^2 \left (2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f\right ) (1+n)}+\frac{b (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac{f x}{e}\right )}{a^2 e (1+n)}+\frac{f (e+f x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac{f x}{e}\right )}{a e^2 (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.438454, size = 246, normalized size = 0.83 \[ \frac{(e+f x)^{n+1} \left (-\frac{c \left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e+\left (\sqrt{b^2-4 a c}-b\right ) f}\right )}{f \left (\sqrt{b^2-4 a c}-b\right )+2 c e}-\frac{c \left (\frac{2 a c-b^2}{\sqrt{b^2-4 a c}}+b\right ) \, _2F_1\left (1,n+1;n+2;\frac{2 c (e+f x)}{2 c e-\left (b+\sqrt{b^2-4 a c}\right ) f}\right )}{2 c e-f \left (\sqrt{b^2-4 a c}+b\right )}+\frac{a f \, _2F_1\left (2,n+1;n+2;\frac{f x}{e}+1\right )}{e^2}+\frac{b \, _2F_1\left (1,n+1;n+2;\frac{f x}{e}+1\right )}{e}\right )}{a^2 (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.307, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{n}}{{x}^{2} \left ( c{x}^{2}+bx+a \right ) }}\, dx \end{align*}

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

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/((c*x^2 + b*x + a)*x^2), 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^{4} + b x^{3} + a x^{2}}, x\right ) \end{align*}

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/(c*x^4 + b*x^3 + a*x^2), 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/x**2/(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}}{{\left (c x^{2} + b x + a\right )} x^{2}}\,{d x} \end{align*}

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/((c*x^2 + b*x + a)*x^2), x)