\(\int (a+b x)^{-1+n} (c+d x)^{-n} \, dx\) [1862]

Optimal result

Integrand size = 19, antiderivative size = 66 \[ \int (a+b x)^{-1+n} (c+d x)^{-n} \, dx=\frac {(a+b x)^n (c+d x)^{-n} \left (\frac {b (c+d x)}{b c-a d}\right )^n \operatorname {Hypergeometric2F1}\left (n,n,1+n,-\frac {d (a+b x)}{b c-a d}\right )}{b n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {72, 71} \[ \int (a+b x)^{-1+n} (c+d x)^{-n} \, dx=\frac {(a+b x)^n (c+d x)^{-n} \left (\frac {b (c+d x)}{b c-a d}\right )^n \operatorname {Hypergeometric2F1}\left (n,n,n+1,-\frac {d (a+b x)}{b c-a d}\right )}{b n} \]

[In]

Int[(a + b*x)^(-1 + n)/(c + d*x)^n,x]

[Out]

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

Rule 71

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

Rule 72

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])

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.98 \[ \int (a+b x)^{-1+n} (c+d x)^{-n} \, dx=\frac {(a+b x)^n (c+d x)^{-n} \left (\frac {b (c+d x)}{b c-a d}\right )^n \operatorname {Hypergeometric2F1}\left (n,n,1+n,\frac {d (a+b x)}{-b c+a d}\right )}{b n} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (b x +a \right )^{-1+n} \left (d x +c \right )^{-n}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (a+b x)^{-1+n} (c+d x)^{-n} \, dx=\int { \frac {{\left (b x + a\right )}^{n - 1}}{{\left (d x + c\right )}^{n}} \,d x } \]

[In]

integrate((b*x+a)^(-1+n)/((d*x+c)^n),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int (a+b x)^{-1+n} (c+d x)^{-n} \, dx=\int \left (a + b x\right )^{n - 1} \left (c + d x\right )^{- n}\, dx \]

[In]

integrate((b*x+a)**(-1+n)/((d*x+c)**n),x)

[Out]

Integral((a + b*x)**(n - 1)/(c + d*x)**n, x)

Maxima [F]

\[ \int (a+b x)^{-1+n} (c+d x)^{-n} \, dx=\int { \frac {{\left (b x + a\right )}^{n - 1}}{{\left (d x + c\right )}^{n}} \,d x } \]

[In]

integrate((b*x+a)^(-1+n)/((d*x+c)^n),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int (a+b x)^{-1+n} (c+d x)^{-n} \, dx=\int { \frac {{\left (b x + a\right )}^{n - 1}}{{\left (d x + c\right )}^{n}} \,d x } \]

[In]

integrate((b*x+a)^(-1+n)/((d*x+c)^n),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (a+b x)^{-1+n} (c+d x)^{-n} \, dx=\int \frac {{\left (a+b\,x\right )}^{n-1}}{{\left (c+d\,x\right )}^n} \,d x \]

[In]

int((a + b*x)^(n - 1)/(c + d*x)^n,x)

[Out]

int((a + b*x)^(n - 1)/(c + d*x)^n, x)