\(\int x (b x+c x^2)^p \, dx\) [127]

Optimal result

Integrand size = 13, antiderivative size = 49 \[ \int x \left (b x+c x^2\right )^p \, dx=\frac {x^2 \left (1+\frac {c x}{b}\right )^{-p} \left (b x+c x^2\right )^p \operatorname {Hypergeometric2F1}\left (-p,2+p,3+p,-\frac {c x}{b}\right )}{2+p} \]

[Out]

x^2*(c*x^2+b*x)^p*hypergeom([-p, 2+p],[3+p],-c*x/b)/(2+p)/((1+c*x/b)^p)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.69, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {654, 638} \[ \int x \left (b x+c x^2\right )^p \, dx=\frac {\left (b x+c x^2\right )^{p+1} \left (-\frac {c x}{b}\right )^{-p-1} \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {b+c x}{b}\right )}{2 c (p+1)}+\frac {\left (b x+c x^2\right )^{p+1}}{2 c (p+1)} \]

[In]

Int[x*(b*x + c*x^2)^p,x]

[Out]

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

Rule 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.96 \[ \int x \left (b x+c x^2\right )^p \, dx=\frac {x^2 (x (b+c x))^p \left (1+\frac {c x}{b}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,2+p,3+p,-\frac {c x}{b}\right )}{2+p} \]

[In]

Integrate[x*(b*x + c*x^2)^p,x]

[Out]

(x^2*(x*(b + c*x))^p*Hypergeometric2F1[-p, 2 + p, 3 + p, -((c*x)/b)])/((2 + p)*(1 + (c*x)/b)^p)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((c*x^2 + b*x)^p*x, x)

Sympy [F]

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

[In]

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

[Out]

Integral(x*(x*(b + c*x))**p, x)

Maxima [F]

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

[In]

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

[Out]

integrate((c*x^2 + b*x)^p*x, x)

Giac [F]

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

[In]

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

[Out]

integrate((c*x^2 + b*x)^p*x, x)

Mupad [F(-1)]

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

[In]

int(x*(b*x + c*x^2)^p,x)

[Out]

int(x*(b*x + c*x^2)^p, x)