\(\int (a+b x+c x^2)^p \, dx\) [2566]

Optimal result

Integrand size = 12, antiderivative size = 122 \[ \int \left (a+b x+c x^2\right )^p \, dx=-\frac {2^{1+p} \left (-\frac {b-\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-1-p} \left (a+b x+c x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {b+\sqrt {b^2-4 a c}+2 c x}{2 \sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {638} \[ \int \left (a+b x+c x^2\right )^p \, dx=-\frac {2^{p+1} \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \left (a+b x+c x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{(p+1) \sqrt {b^2-4 a c}} \]

[In]

Int[(a + b*x + c*x^2)^p,x]

[Out]

-((2^(1 + p)*(-((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c]))^(-1 - p)*(a + b*x + c*x^2)^(1 + p)*Hyperge
ometric2F1[-p, 1 + p, 2 + p, (b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(2*Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(1 + p
)))

Rule 638

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(-(a + b*x + c*
x^2)^(p + 1)/(q*(p + 1)*((q - b - 2*c*x)/(2*q))^(p + 1)))*Hypergeometric2F1[-p, p + 1, p + 2, (b + q + 2*c*x)/
(2*q)], x]] /; FreeQ[{a, b, c, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[4*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {2^{1+p} \left (-\frac {b-\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-1-p} \left (a+b x+c x^2\right )^{1+p} \, _2F_1\left (-p,1+p;2+p;\frac {b+\sqrt {b^2-4 a c}+2 c x}{2 \sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} (1+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.03 \[ \int \left (a+b x+c x^2\right )^p \, dx=\frac {2^{-1+p} \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p} (a+x (b+c x))^p \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {-b+\sqrt {b^2-4 a c}-2 c x}{2 \sqrt {b^2-4 a c}}\right )}{c (1+p)} \]

[In]

Integrate[(a + b*x + c*x^2)^p,x]

[Out]

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

Maple [F]

\[\int \left (c \,x^{2}+b x +a \right )^{p}d x\]

[In]

int((c*x^2+b*x+a)^p,x)

[Out]

int((c*x^2+b*x+a)^p,x)

Fricas [F]

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

[In]

integrate((c*x^2+b*x+a)^p,x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)^p, x)

Sympy [F]

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

[In]

integrate((c*x**2+b*x+a)**p,x)

[Out]

Integral((a + b*x + c*x**2)**p, x)

Maxima [F]

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

[In]

integrate((c*x^2+b*x+a)^p,x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^p, x)

Giac [F]

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

[In]

integrate((c*x^2+b*x+a)^p,x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^p, x)

Mupad [F(-1)]

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

[In]

int((a + b*x + c*x^2)^p,x)

[Out]

int((a + b*x + c*x^2)^p, x)