3.3055 \(\int (a+b x)^{-n} (c+d x) (e+f x)^{-2+n} \, dx\)

Optimal. Leaf size=118 \[ \frac{d (a+b x)^{-n} (e+f x)^n \left (-\frac{f (a+b x)}{b e-a f}\right )^n \, _2F_1\left (n,n;n+1;\frac{b (e+f x)}{b e-a f}\right )}{f^2 n}-\frac{(a+b x)^{1-n} (d e-c f) (e+f x)^{n-1}}{f (1-n) (b e-a f)} \]

[Out]

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

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Rubi [A]  time = 0.0539389, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {79, 70, 69} \[ \frac{d (a+b x)^{-n} (e+f x)^n \left (-\frac{f (a+b x)}{b e-a f}\right )^n \, _2F_1\left (n,n;n+1;\frac{b (e+f x)}{b e-a f}\right )}{f^2 n}-\frac{(a+b x)^{1-n} (d e-c f) (e+f x)^{n-1}}{f (1-n) (b e-a f)} \]

Antiderivative was successfully verified.

[In]

Int[((c + d*x)*(e + f*x)^(-2 + n))/(a + b*x)^n,x]

[Out]

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

Rule 79

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

Rule 70

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

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

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

Mathematica [A]  time = 0.157032, size = 105, normalized size = 0.89 \[ \frac{(a+b x)^{-n} (e+f x)^n \left (\frac{f (a+b x) (d e-c f)}{(n-1) (e+f x) (b e-a f)}+\frac{d \left (\frac{f (a+b x)}{a f-b e}\right )^n \, _2F_1\left (n,n;n+1;\frac{b (e+f x)}{b e-a f}\right )}{n}\right )}{f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.051, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) \left ( fx+e \right ) ^{-2+n}}{ \left ( bx+a \right ) ^{n}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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