\(\int x^2 (b x+c x^2)^p \, dx\) [126]

Optimal result

Integrand size = 15, antiderivative size = 49 \[ \int x^2 \left (b x+c x^2\right )^p \, dx=\frac {x^3 \left (1+\frac {c x}{b}\right )^{-p} \left (b x+c x^2\right )^p \operatorname {Hypergeometric2F1}\left (-p,3+p,4+p,-\frac {c x}{b}\right )}{3+p} \]

[Out]

x^3*(c*x^2+b*x)^p*hypergeom([-p, 3+p],[4+p],-c*x/b)/(3+p)/((1+c*x/b)^p)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {688, 68, 66} \[ \int x^2 \left (b x+c x^2\right )^p \, dx=\frac {x^3 \left (\frac {c x}{b}+1\right )^{-p} \left (b x+c x^2\right )^p \operatorname {Hypergeometric2F1}\left (-p,p+3,p+4,-\frac {c x}{b}\right )}{p+3} \]

[In]

Int[x^2*(b*x + c*x^2)^p,x]

[Out]

(x^3*(b*x + c*x^2)^p*Hypergeometric2F1[-p, 3 + p, 4 + p, -((c*x)/b)])/((3 + p)*(1 + (c*x)/b)^p)

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))

Rule 68

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(
x/c))^FracPart[n]), Int[(b*x)^m*(1 + d*(x/c))^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0] && ((RationalQ[m] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0])) |
|  !RationalQ[n])

Rule 688

Int[((e_.)*(x_))^(m_)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*
(b + c*x)^p)), Int[x^(m + p)*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, m}, x] &&  !IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \left (x^{-p} (b+c x)^{-p} \left (b x+c x^2\right )^p\right ) \int x^{2+p} (b+c x)^p \, dx \\ & = \left (x^{-p} \left (1+\frac {c x}{b}\right )^{-p} \left (b x+c x^2\right )^p\right ) \int x^{2+p} \left (1+\frac {c x}{b}\right )^p \, dx \\ & = \frac {x^3 \left (1+\frac {c x}{b}\right )^{-p} \left (b x+c x^2\right )^p \, _2F_1\left (-p,3+p;4+p;-\frac {c x}{b}\right )}{3+p} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.96 \[ \int x^2 \left (b x+c x^2\right )^p \, dx=\frac {x^3 (x (b+c x))^p \left (1+\frac {c x}{b}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,3+p,4+p,-\frac {c x}{b}\right )}{3+p} \]

[In]

Integrate[x^2*(b*x + c*x^2)^p,x]

[Out]

(x^3*(x*(b + c*x))^p*Hypergeometric2F1[-p, 3 + p, 4 + p, -((c*x)/b)])/((3 + p)*(1 + (c*x)/b)^p)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((c*x^2 + b*x)^p*x^2, x)

Sympy [F]

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

[In]

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

[Out]

Integral(x**2*(x*(b + c*x))**p, x)

Maxima [F]

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

[In]

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

[Out]

integrate((c*x^2 + b*x)^p*x^2, x)

Giac [F]

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

[In]

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

[Out]

integrate((c*x^2 + b*x)^p*x^2, x)

Mupad [F(-1)]

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

[In]

int(x^2*(b*x + c*x^2)^p,x)

[Out]

int(x^2*(b*x + c*x^2)^p, x)