3.1232 \(\int \frac{(a c-b c x)^n}{(a+b x)^2} \, dx\)

Optimal. Leaf size=52 \[ -\frac{(a c-b c x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{a-b x}{2 a}\right )}{4 a^2 b c (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.0114906, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {68} \[ -\frac{(a c-b c x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{a-b x}{2 a}\right )}{4 a^2 b c (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a*c - b*c*x)^n/(a + b*x)^2,x]

[Out]

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

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{(a c-b c x)^n}{(a+b x)^2} \, dx &=-\frac{(a c-b c x)^{1+n} \, _2F_1\left (2,1+n;2+n;\frac{a-b x}{2 a}\right )}{4 a^2 b c (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.0109877, size = 52, normalized size = 1. \[ -\frac{(a-b x) (c (a-b x))^n \, _2F_1\left (2,n+1;n+2;\frac{a-b x}{2 a}\right )}{4 a^2 b (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.047, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -bcx+ac \right ) ^{n}}{ \left ( bx+a \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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