3.937 \(\int \frac{x^2 (a+b x)^n}{c+d x} \, dx\)

Optimal. Leaf size=108 \[ -\frac{(a d+b c) (a+b x)^{n+1}}{b^2 d^2 (n+1)}+\frac{(a+b x)^{n+2}}{b^2 d (n+2)}+\frac{c^2 (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;-\frac{d (a+b x)}{b c-a d}\right )}{d^2 (n+1) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.0608744, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {88, 68} \[ -\frac{(a d+b c) (a+b x)^{n+1}}{b^2 d^2 (n+1)}+\frac{(a+b x)^{n+2}}{b^2 d (n+2)}+\frac{c^2 (a+b x)^{n+1} \, _2F_1\left (1,n+1;n+2;-\frac{d (a+b x)}{b c-a d}\right )}{d^2 (n+1) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x)^n)/(c + d*x),x]

[Out]

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{x^2 (a+b x)^n}{c+d x} \, dx &=\int \left (\frac{(-b c-a d) (a+b x)^n}{b d^2}+\frac{(a+b x)^{1+n}}{b d}+\frac{c^2 (a+b x)^n}{d^2 (c+d x)}\right ) \, dx\\ &=-\frac{(b c+a d) (a+b x)^{1+n}}{b^2 d^2 (1+n)}+\frac{(a+b x)^{2+n}}{b^2 d (2+n)}+\frac{c^2 \int \frac{(a+b x)^n}{c+d x} \, dx}{d^2}\\ &=-\frac{(b c+a d) (a+b x)^{1+n}}{b^2 d^2 (1+n)}+\frac{(a+b x)^{2+n}}{b^2 d (2+n)}+\frac{c^2 (a+b x)^{1+n} \, _2F_1\left (1,1+n;2+n;-\frac{d (a+b x)}{b c-a d}\right )}{d^2 (b c-a d) (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.0918826, size = 100, normalized size = 0.93 \[ \frac{(a+b x)^{n+1} \left (b^2 c^2 (n+2) \, _2F_1\left (1,n+1;n+2;\frac{d (a+b x)}{a d-b c}\right )-(b c-a d) (a d+b c (n+2)-b d (n+1) x)\right )}{b^2 d^2 (n+1) (n+2) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*x)^n)/(c + d*x),x]

[Out]

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

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Maple [F]  time = 0.045, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{n}{x}^{2}}{dx+c}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x+a)^n/(d*x+c),x)

[Out]

int(x^2*(b*x+a)^n/(d*x+c),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x + a)^n*x^2/(d*x + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x + a)^n*x^2/(d*x + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x + a)^n*x^2/(d*x + c), x)