\(\int (a+b x)^{-2-n} (c+d x)^n \, dx\) [1868]

Optimal result

Integrand size = 19, antiderivative size = 37 \[ \int (a+b x)^{-2-n} (c+d x)^n \, dx=-\frac {(a+b x)^{-1-n} (c+d x)^{1+n}}{(b c-a d) (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {37} \[ \int (a+b x)^{-2-n} (c+d x)^n \, dx=-\frac {(a+b x)^{-n-1} (c+d x)^{n+1}}{(n+1) (b c-a d)} \]

[In]

Int[(a + b*x)^(-2 - n)*(c + d*x)^n,x]

[Out]

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

Rule 37

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(a+b x)^{-1-n} (c+d x)^{1+n}}{(b c-a d) (1+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.03 \[ \int (a+b x)^{-2-n} (c+d x)^n \, dx=\frac {(a+b x)^{-1-n} (c+d x)^{1+n}}{(b c-a d) (-1-n)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11

method	result	size
gosper	\(\frac {\left (b x +a \right )^{-1-n} \left (d x +c \right )^{1+n}}{a d n -b c n +a d -b c}\)	\(41\)
parallelrisch	\(\frac {x^{2} \left (d x +c \right )^{n} \left (b x +a \right )^{-2-n} b^{2} d^{2}+x \left (d x +c \right )^{n} \left (b x +a \right )^{-2-n} a b \,d^{2}+x \left (d x +c \right )^{n} \left (b x +a \right )^{-2-n} b^{2} c d +\left (d x +c \right )^{n} \left (b x +a \right )^{-2-n} a b c d}{\left (a d n -b c n +a d -b c \right ) b d}\)	\(129\)

[In]

int((b*x+a)^(-2-n)*(d*x+c)^n,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.59 \[ \int (a+b x)^{-2-n} (c+d x)^n \, dx=-\frac {{\left (b d x^{2} + a c + {\left (b c + a d\right )} x\right )} {\left (b x + a\right )}^{-n - 2} {\left (d x + c\right )}^{n}}{b c - a d + {\left (b c - a d\right )} n} \]

[In]

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

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int (a+b x)^{-2-n} (c+d x)^n \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

\[ \int (a+b x)^{-2-n} (c+d x)^n \, dx=\int { {\left (b x + a\right )}^{-n - 2} {\left (d x + c\right )}^{n} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (a+b x)^{-2-n} (c+d x)^n \, dx=\int { {\left (b x + a\right )}^{-n - 2} {\left (d x + c\right )}^{n} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.62 \[ \int (a+b x)^{-2-n} (c+d x)^n \, dx=\frac {\frac {a\,c\,{\left (c+d\,x\right )}^n}{\left (a\,d-b\,c\right )\,\left (n+1\right )}+\frac {x\,\left (a\,d+b\,c\right )\,{\left (c+d\,x\right )}^n}{\left (a\,d-b\,c\right )\,\left (n+1\right )}+\frac {b\,d\,x^2\,{\left (c+d\,x\right )}^n}{\left (a\,d-b\,c\right )\,\left (n+1\right )}}{{\left (a+b\,x\right )}^{n+2}} \]

[In]

int((c + d*x)^n/(a + b*x)^(n + 2),x)

[Out]

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