\(\int (a+b x)^{-3+n} (c+d x)^{-n} \, dx\) [1860]

Optimal result

Integrand size = 19, antiderivative size = 86 \[ \int (a+b x)^{-3+n} (c+d x)^{-n} \, dx=-\frac {(a+b x)^{-2+n} (c+d x)^{1-n}}{(b c-a d) (2-n)}+\frac {d (a+b x)^{-1+n} (c+d x)^{1-n}}{(b c-a d)^2 (1-n) (2-n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 86, 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 = {47, 37} \[ \int (a+b x)^{-3+n} (c+d x)^{-n} \, dx=\frac {d (a+b x)^{n-1} (c+d x)^{1-n}}{(1-n) (2-n) (b c-a d)^2}-\frac {(a+b x)^{n-2} (c+d x)^{1-n}}{(2-n) (b c-a d)} \]

[In]

Int[(a + b*x)^(-3 + n)/(c + d*x)^n,x]

[Out]

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

Rule 47

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] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = -\frac {(a+b x)^{-2+n} (c+d x)^{1-n}}{(b c-a d) (2-n)}-\frac {d \int (a+b x)^{-2+n} (c+d x)^{-n} \, dx}{(b c-a d) (2-n)} \\ & = -\frac {(a+b x)^{-2+n} (c+d x)^{1-n}}{(b c-a d) (2-n)}+\frac {d (a+b x)^{-1+n} (c+d x)^{1-n}}{(b c-a d)^2 (1-n) (2-n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.69 \[ \int (a+b x)^{-3+n} (c+d x)^{-n} \, dx=\frac {(a+b x)^{-2+n} (c+d x)^{1-n} (-a d (-2+n)+b c (-1+n)+b d x)}{(b c-a d)^2 (-2+n) (-1+n)} \]

[In]

Integrate[(a + b*x)^(-3 + n)/(c + d*x)^n,x]

[Out]

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

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.48

method	result	size
gosper	\(-\frac {\left (b x +a \right )^{-2+n} \left (d x +c \right ) \left (d x +c \right )^{-n} \left (a d n -b c n -b d x -2 a d +b c \right )}{a^{2} d^{2} n^{2}-2 a b c d \,n^{2}+b^{2} c^{2} n^{2}-3 a^{2} d^{2} n +6 a b c d n -3 b^{2} c^{2} n +2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}}\)	\(127\)
parallelrisch	\(-\frac {\left (-x^{3} \left (b x +a \right )^{-3+n} b^{3} d^{3} n +x^{2} \left (b x +a \right )^{-3+n} a \,b^{2} d^{3} n^{2}-x^{2} \left (b x +a \right )^{-3+n} b^{3} c \,d^{2} n^{2}-3 x^{2} \left (b x +a \right )^{-3+n} a \,b^{2} d^{3} n +x \left (b x +a \right )^{-3+n} a^{2} b \,d^{3} n^{2}-x \left (b x +a \right )^{-3+n} b^{3} c^{2} d \,n^{2}-2 x \left (b x +a \right )^{-3+n} a^{2} b \,d^{3} n +x \left (b x +a \right )^{-3+n} b^{3} c^{2} d n +\left (b x +a \right )^{-3+n} a^{2} b c \,d^{2} n^{2}-\left (b x +a \right )^{-3+n} a \,b^{2} c^{2} d \,n^{2}-2 \left (b x +a \right )^{-3+n} a^{2} b c \,d^{2} n +\left (b x +a \right )^{-3+n} a \,b^{2} c^{2} d n -2 x \left (b x +a \right )^{-3+n} a \,b^{2} c \,d^{2} n \right ) \left (d x +c \right )^{-n}}{n b d \left (a^{2} d^{2} n^{2}-2 a b c d \,n^{2}+b^{2} c^{2} n^{2}-3 a^{2} d^{2} n +6 a b c d n -3 b^{2} c^{2} n +2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}\right )}\)	\(376\)

[In]

int((b*x+a)^(-3+n)/((d*x+c)^n),x,method=_RETURNVERBOSE)

[Out]

-(b*x+a)^(-2+n)*(d*x+c)/((d*x+c)^n)*(a*d*n-b*c*n-b*d*x-2*a*d+b*c)/(a^2*d^2*n^2-2*a*b*c*d*n^2+b^2*c^2*n^2-3*a^2
*d^2*n+6*a*b*c*d*n-3*b^2*c^2*n+2*a^2*d^2-4*a*b*c*d+2*b^2*c^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (79) = 158\).

Time = 0.24 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.40 \[ \int (a+b x)^{-3+n} (c+d x)^{-n} \, dx=\frac {{\left (b^{2} d^{2} x^{3} - a b c^{2} + 2 \, a^{2} c d + {\left (3 \, a b d^{2} + {\left (b^{2} c d - a b d^{2}\right )} n\right )} x^{2} + {\left (a b c^{2} - a^{2} c d\right )} n - {\left (b^{2} c^{2} - 2 \, a b c d - 2 \, a^{2} d^{2} - {\left (b^{2} c^{2} - a^{2} d^{2}\right )} n\right )} x\right )} {\left (b x + a\right )}^{n - 3}}{{\left (2 \, b^{2} c^{2} - 4 \, a b c d + 2 \, a^{2} d^{2} + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} n^{2} - 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} n\right )} {\left (d x + c\right )}^{n}} \]

[In]

integrate((b*x+a)^(-3+n)/((d*x+c)^n),x, algorithm="fricas")

[Out]

(b^2*d^2*x^3 - a*b*c^2 + 2*a^2*c*d + (3*a*b*d^2 + (b^2*c*d - a*b*d^2)*n)*x^2 + (a*b*c^2 - a^2*c*d)*n - (b^2*c^
2 - 2*a*b*c*d - 2*a^2*d^2 - (b^2*c^2 - a^2*d^2)*n)*x)*(b*x + a)^(n - 3)/((2*b^2*c^2 - 4*a*b*c*d + 2*a^2*d^2 +
(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*n^2 - 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*n)*(d*x + c)^n)

Sympy [F(-2)]

Exception generated. \[ \int (a+b x)^{-3+n} (c+d x)^{-n} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

integrate((b*x+a)**(-3+n)/((d*x+c)**n),x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

\[ \int (a+b x)^{-3+n} (c+d x)^{-n} \, dx=\int { \frac {{\left (b x + a\right )}^{n - 3}}{{\left (d x + c\right )}^{n}} \,d x } \]

[In]

integrate((b*x+a)^(-3+n)/((d*x+c)^n),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(n - 3)/(d*x + c)^n, x)

Giac [F]

\[ \int (a+b x)^{-3+n} (c+d x)^{-n} \, dx=\int { \frac {{\left (b x + a\right )}^{n - 3}}{{\left (d x + c\right )}^{n}} \,d x } \]

[In]

integrate((b*x+a)^(-3+n)/((d*x+c)^n),x, algorithm="giac")

[Out]

integrate((b*x + a)^(n - 3)/(d*x + c)^n, x)

Mupad [B] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.56 \[ \int (a+b x)^{-3+n} (c+d x)^{-n} \, dx={\left (a+b\,x\right )}^{n-3}\,\left (\frac {x\,\left (2\,a^2\,d^2-b^2\,c^2-a^2\,d^2\,n+b^2\,c^2\,n+2\,a\,b\,c\,d\right )}{{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^n\,\left (n^2-3\,n+2\right )}+\frac {b^2\,d^2\,x^3}{{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^n\,\left (n^2-3\,n+2\right )}+\frac {a\,c\,\left (2\,a\,d-b\,c-a\,d\,n+b\,c\,n\right )}{{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^n\,\left (n^2-3\,n+2\right )}+\frac {b\,d\,x^2\,\left (3\,a\,d-a\,d\,n+b\,c\,n\right )}{{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^n\,\left (n^2-3\,n+2\right )}\right ) \]

[In]

int((a + b*x)^(n - 3)/(c + d*x)^n,x)

[Out]

(a + b*x)^(n - 3)*((x*(2*a^2*d^2 - b^2*c^2 - a^2*d^2*n + b^2*c^2*n + 2*a*b*c*d))/((a*d - b*c)^2*(c + d*x)^n*(n
^2 - 3*n + 2)) + (b^2*d^2*x^3)/((a*d - b*c)^2*(c + d*x)^n*(n^2 - 3*n + 2)) + (a*c*(2*a*d - b*c - a*d*n + b*c*n
))/((a*d - b*c)^2*(c + d*x)^n*(n^2 - 3*n + 2)) + (b*d*x^2*(3*a*d - a*d*n + b*c*n))/((a*d - b*c)^2*(c + d*x)^n*
(n^2 - 3*n + 2)))