3.13.32 \(\int \frac {(a c-b c x)^n}{(a+b x)^2} \, dx\) [1232]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 52, normalized size of antiderivative = 1.00, 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 = {70} \begin {gather*} -\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*c - b*c*x)^n/(a + b*x)^2,x]

[Out]

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

Rule 70

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

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Mathematica [A]
time = 0.04, size = 52, normalized size = 1.00 \begin {gather*} -\frac {(a-b x) (c (a-b x))^n \, _2F_1\left (2,1+n;2+n;\frac {a-b x}{2 a}\right )}{4 a^2 b (1+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (-b c x +a c \right )^{n}}{\left (b x +a \right )^{2}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- c \left (- a + b x\right )\right )^{n}}{\left (a + b x\right )^{2}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (a\,c-b\,c\,x\right )}^n}{{\left (a+b\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*c - b*c*x)^n/(a + b*x)^2,x)

[Out]

int((a*c - b*c*x)^n/(a + b*x)^2, x)

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