\(\int \frac {x^2 (e+f x)^n}{a+b x+c x^2} \, dx\) [543]

Optimal result

Integrand size = 23, antiderivative size = 237 \[ \int \frac {x^2 (e+f x)^n}{a+b x+c x^2} \, dx=\frac {(e+f x)^{1+n}}{c f (1+n)}+\frac {\left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{c \left (2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}+\frac {\left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{c \left (2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1642, 70} \[ \int \frac {x^2 (e+f x)^n}{a+b x+c x^2} \, dx=\frac {\left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (b-\sqrt {b^2-4 a c}\right )\right )}+\frac {\left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (\sqrt {b^2-4 a c}+b\right )\right )}+\frac {(e+f x)^{n+1}}{c f (n+1)} \]

[In]

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

[Out]

(e + f*x)^(1 + n)/(c*f*(1 + n)) + ((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)])/(c*(2*c*e - (b - Sqrt[b^2 - 4*a*c])*f)*(
1 + n)) + ((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)])/(c*(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 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.22 \[ \int \frac {x^2 (e+f x)^n}{a+b x+c x^2} \, dx=-\frac {2 (e+f x)^{1+n} \left (2 \sqrt {b^2-4 a c} \left (c e^2+f (-b e+a f)\right )+f \left (-b^2 e+2 a c e+b \sqrt {b^2-4 a c} e+a b f-a \sqrt {b^2-4 a c} f\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}\right )+f \left (b^2 e-2 a c e+b \sqrt {b^2-4 a c} e-a b f-a \sqrt {b^2-4 a c} f\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )\right )}{\sqrt {b^2-4 a c} f \left (2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f\right ) \left (-2 c e+\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (1+n)} \]

[In]

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

[Out]

(-2*(e + f*x)^(1 + n)*(2*Sqrt[b^2 - 4*a*c]*(c*e^2 + f*(-(b*e) + a*f)) + f*(-(b^2*e) + 2*a*c*e + b*Sqrt[b^2 - 4
*a*c]*e + a*b*f - a*Sqrt[b^2 - 4*a*c]*f)*Hypergeometric2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(2*c*e + (-b + Sqr
t[b^2 - 4*a*c])*f)] + f*(b^2*e - 2*a*c*e + b*Sqrt[b^2 - 4*a*c]*e - a*b*f - a*Sqrt[b^2 - 4*a*c]*f)*Hypergeometr
ic2F1[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]*f*(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]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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