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

Optimal. Leaf size=201 \[ \frac{2 c x^{m+1} (e+f x)^n \left (\frac{f x}{e}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{f x}{e},-\frac{2 c x}{b-\sqrt{b^2-4 a c}}\right )}{(m+1) \sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}\right )}-\frac{2 c x^{m+1} (e+f x)^n \left (\frac{f x}{e}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{f x}{e},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{(m+1) \sqrt{b^2-4 a c} \left (\sqrt{b^2-4 a c}+b\right )} \]

[Out]

(2*c*x^(1 + m)*(e + f*x)^n*AppellF1[1 + m, -n, 1, 2 + m, -((f*x)/e), (-2*c*x)/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[
b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*(1 + m)*(1 + (f*x)/e)^n) - (2*c*x^(1 + m)*(e + f*x)^n*AppellF1[1 + m, -n,
1, 2 + m, -((f*x)/e), (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*(1 + m)*(
1 + (f*x)/e)^n)

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Rubi [A]  time = 0.363151, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {911, 135, 133} \[ \frac{2 c x^{m+1} (e+f x)^n \left (\frac{f x}{e}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{f x}{e},-\frac{2 c x}{b-\sqrt{b^2-4 a c}}\right )}{(m+1) \sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}\right )}-\frac{2 c x^{m+1} (e+f x)^n \left (\frac{f x}{e}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{f x}{e},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{(m+1) \sqrt{b^2-4 a c} \left (\sqrt{b^2-4 a c}+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*x^(1 + m)*(e + f*x)^n*AppellF1[1 + m, -n, 1, 2 + m, -((f*x)/e), (-2*c*x)/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[
b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*(1 + m)*(1 + (f*x)/e)^n) - (2*c*x^(1 + m)*(e + f*x)^n*AppellF1[1 + m, -n,
1, 2 + m, -((f*x)/e), (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*(1 + m)*(
1 + (f*x)/e)^n)

Rule 911

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

Rule 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int \frac{x^m (e+f x)^n}{a+b x+c x^2} \, dx &=\int \left (\frac{2 c x^m (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 x^m (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{x^m (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{x^m (e+f x)^n}{b+\sqrt{b^2-4 a c}+2 c x} \, dx}{\sqrt{b^2-4 a c}}\\ &=\frac{\left (2 c (e+f x)^n \left (1+\frac{f x}{e}\right )^{-n}\right ) \int \frac{x^m \left (1+\frac{f x}{e}\right )^n}{b-\sqrt{b^2-4 a c}+2 c x} \, dx}{\sqrt{b^2-4 a c}}-\frac{\left (2 c (e+f x)^n \left (1+\frac{f x}{e}\right )^{-n}\right ) \int \frac{x^m \left (1+\frac{f x}{e}\right )^n}{b+\sqrt{b^2-4 a c}+2 c x} \, dx}{\sqrt{b^2-4 a c}}\\ &=\frac{2 c x^{1+m} (e+f x)^n \left (1+\frac{f x}{e}\right )^{-n} F_1\left (1+m;-n,1;2+m;-\frac{f x}{e},-\frac{2 c x}{b-\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}\right ) (1+m)}-\frac{2 c x^{1+m} (e+f x)^n \left (1+\frac{f x}{e}\right )^{-n} F_1\left (1+m;-n,1;2+m;-\frac{f x}{e},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{\sqrt{b^2-4 a c} \left (b+\sqrt{b^2-4 a c}\right ) (1+m)}\\ \end{align*}

Mathematica [F]  time = 0.309209, size = 0, normalized size = 0. \[ \int \frac{x^m (e+f x)^n}{a+b x+c x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 1.344, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{n}{x}^{m}}{c{x}^{2}+bx+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^m*(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} x^{m}}{c x^{2} + b x + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((f*x + e)^n*x^m/(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} x^{m}}{c x^{2} + b x + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((f*x + e)^n*x^m/(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(x**m*(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} x^{m}}{c x^{2} + b x + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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