### 3.126 $$\int x^2 (b x+c x^2)^p \, dx$$

Optimal. Leaf size=49 $\frac{x^3 \left (\frac{c x}{b}+1\right )^{-p} \left (b x+c x^2\right )^p \, _2F_1\left (-p,p+3;p+4;-\frac{c x}{b}\right )}{p+3}$

[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)

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Rubi [A]  time = 0.0203876, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {674, 66, 64} $\frac{x^3 \left (\frac{c x}{b}+1\right )^{-p} \left (b x+c x^2\right )^p \, _2F_1\left (-p,p+3;p+4;-\frac{c x}{b}\right )}{p+3}$

Antiderivative was successfully veriﬁed.

[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 674

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]

Rule 66

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 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), 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])))

Rubi steps

\begin{align*} \int x^2 \left (b x+c x^2\right )^p \, dx &=\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]  time = 0.0089366, size = 47, normalized size = 0.96 $\frac{x^3 (x (b+c x))^p \left (\frac{c x}{b}+1\right )^{-p} \, _2F_1\left (-p,p+3;p+4;-\frac{c x}{b}\right )}{p+3}$

Antiderivative was successfully veriﬁed.

[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)

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Maple [F]  time = 0.371, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( c{x}^{2}+bx \right ) ^{p}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x\right )}^{p} x^{2}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x)^p*x^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (c x^{2} + b x\right )}^{p} x^{2}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + b*x)^p*x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \left (x \left (b + c x\right )\right )^{p}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(x*(b + c*x))**p, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x\right )}^{p} x^{2}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x)^p*x^2, x)