\(\int \frac {x (e+f x)^n}{a+b x+c x^2} \, dx\) [544]

Optimal result

Integrand size = 21, antiderivative size = 198 \[ \int \frac {x (e+f x)^n}{a+b x+c x^2} \, dx=-\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{\left (2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}-\frac {\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{\left (2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {844, 70} \[ \int \frac {x (e+f x)^n}{a+b x+c x^2} \, dx=-\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{(n+1) \left (2 c e-f \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{(n+1) \left (2 c e-f \left (\sqrt {b^2-4 a c}+b\right )\right )} \]

[In]

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

[Out]

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

Rule 70

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

Rule 844

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]

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.92 \[ \int \frac {x (e+f x)^n}{a+b x+c x^2} \, dx=\frac {(e+f x)^{1+n} \left (-\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}\right )}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}-\frac {\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{1+n} \]

[In]

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

[Out]

((e + f*x)^(1 + n)*(-(((1 - b/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)) - ((1 + b/Sqrt[b^2 - 4*a*c])*Hypergeometric
2F1[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)
))/(1 + n)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x (e+f x)^n}{a+b x+c x^2} \, dx=\int \frac {x\,{\left (e+f\,x\right )}^n}{c\,x^2+b\,x+a} \,d x \]

[In]

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

[Out]

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