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

Optimal. Leaf size=122 \[ \frac{c^2 (a+b x)^{n+1}}{d^2 (c+d x) (b c-a d)}+\frac{c (a+b x)^{n+1} (2 a d-b c (n+2)) \, _2F_1\left (1,n+1;n+2;-\frac{d (a+b x)}{b c-a d}\right )}{d^2 (n+1) (b c-a d)^2}+\frac{(a+b x)^{n+1}}{b d^2 (n+1)} \]

[Out]

(a + b*x)^(1 + n)/(b*d^2*(1 + n)) + (c^2*(a + b*x)^(1 + n))/(d^2*(b*c - a*d)*(c + d*x)) + (c*(2*a*d - b*c*(2 +
n))*(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)^2*(1
+ n))

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Rubi [A]  time = 0.082037, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {89, 80, 68} \[ \frac{c^2 (a+b x)^{n+1}}{d^2 (c+d x) (b c-a d)}+\frac{c (a+b x)^{n+1} (2 a d-b c (n+2)) \, _2F_1\left (1,n+1;n+2;-\frac{d (a+b x)}{b c-a d}\right )}{d^2 (n+1) (b c-a d)^2}+\frac{(a+b x)^{n+1}}{b d^2 (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x)^n)/(c + d*x)^2,x]

[Out]

(a + b*x)^(1 + n)/(b*d^2*(1 + n)) + (c^2*(a + b*x)^(1 + n))/(d^2*(b*c - a*d)*(c + d*x)) + (c*(2*a*d - b*c*(2 +
n))*(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)^2*(1
+ n))

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 80

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

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

Mathematica [A]  time = 0.0623849, size = 115, normalized size = 0.94 \[ \frac{(a+b x)^{n+1} \left ((b c-a d) (b c (c (n+2)+d x)-a d (c+d x))-b c (c+d x) (b c (n+2)-2 a d) \, _2F_1\left (1,n+1;n+2;\frac{d (a+b x)}{a d-b c}\right )\right )}{b d^2 (n+1) (c+d x) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.049, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{n}{x}^{2}}{ \left ( dx+c \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*(b*x+a)^n/(d*x+c)^2,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}}{{\left (d x + c\right )}^{2}}\,{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)^2,x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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}}{{\left (d x + c\right )}^{2}}\,{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)^2,x, algorithm="giac")

[Out]

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