\(\int \frac {(e+f x)^n}{a+b x+c x^2} \, dx\) [545]

Optimal result

Integrand size = 20, antiderivative size = 191 \[ \int \frac {(e+f x)^n}{a+b x+c x^2} \, dx=-\frac {2 c (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{\sqrt {b^2-4 a c} \left (2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}+\frac {2 c (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{\sqrt {b^2-4 a c} \left (2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {725, 70} \[ \int \frac {(e+f x)^n}{a+b x+c x^2} \, dx=\frac {2 c (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{(n+1) \sqrt {b^2-4 a c} \left (2 c e-f \left (\sqrt {b^2-4 a c}+b\right )\right )}-\frac {2 c (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{(n+1) \sqrt {b^2-4 a c} \left (2 c e-f \left (b-\sqrt {b^2-4 a c}\right )\right )} \]

[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 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 725

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.18 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.85 \[ \int \frac {(e+f x)^n}{a+b x+c x^2} \, dx=\frac {2 c (e+f x)^{1+n} \left (-\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}\right )}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}+\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{\sqrt {b^2-4 a c} (1+n)} \]

[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))

Maple [F]

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

[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)

Fricas [F]

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

[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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[In]

int((e + f*x)^n/(a + b*x + c*x^2),x)

[Out]

int((e + f*x)^n/(a + b*x + c*x^2), x)