\(\int (a+b x)^n (c+d x)^{-3-n} \, dx\) [1875]

Optimal result

Integrand size = 19, antiderivative size = 79 \[ \int (a+b x)^n (c+d x)^{-3-n} \, dx=\frac {(a+b x)^{1+n} (c+d x)^{-2-n}}{(b c-a d) (2+n)}+\frac {b (a+b x)^{1+n} (c+d x)^{-1-n}}{(b c-a d)^2 (1+n) (2+n)} \]

[Out]

(b*x+a)^(1+n)*(d*x+c)^(-2-n)/(-a*d+b*c)/(2+n)+b*(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 = 79, 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} \[ \int (a+b x)^n (c+d x)^{-3-n} \, dx=\frac {(a+b x)^{n+1} (c+d x)^{-n-2}}{(n+2) (b c-a d)}+\frac {b (a+b x)^{n+1} (c+d x)^{-n-1}}{(n+1) (n+2) (b c-a d)^2} \]

[In]

Int[(a + b*x)^n*(c + d*x)^(-3 - n),x]

[Out]

((a + b*x)^(1 + n)*(c + d*x)^(-2 - n))/((b*c - a*d)*(2 + n)) + (b*(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)^{1+n} (c+d x)^{-2-n}}{(b c-a d) (2+n)}+\frac {b \int (a+b x)^n (c+d x)^{-2-n} \, dx}{(b c-a d) (2+n)} \\ & = \frac {(a+b x)^{1+n} (c+d x)^{-2-n}}{(b c-a d) (2+n)}+\frac {b (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 = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.75 \[ \int (a+b x)^n (c+d x)^{-3-n} \, dx=\frac {(a+b x)^{1+n} (c+d x)^{-2-n} (-a d (1+n)+b c (2+n)+b d x)}{(b c-a d)^2 (1+n) (2+n)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.57

method	result	size
gosper	\(-\frac {\left (b x +a \right )^{1+n} \left (d x +c \right )^{-2-n} \left (a d n -b c n -b d x +a d -2 b c \right )}{a^{2} d^{2} n^{2}-2 a b c d \,n^{2}+b^{2} c^{2} n^{2}+3 a^{2} d^{2} n -6 a b c d n +3 b^{2} c^{2} n +2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}}\)	\(124\)
parallelrisch	\(\frac {x^{3} \left (b x +a \right )^{n} \left (d x +c \right )^{-3-n} b^{3} d^{3}-x^{2} \left (b x +a \right )^{n} \left (d x +c \right )^{-3-n} a \,b^{2} d^{3} n +x^{2} \left (b x +a \right )^{n} \left (d x +c \right )^{-3-n} b^{3} c \,d^{2} n +3 x^{2} \left (b x +a \right )^{n} \left (d x +c \right )^{-3-n} b^{3} c \,d^{2}-x \left (b x +a \right )^{n} \left (d x +c \right )^{-3-n} a^{2} b \,d^{3} n +x \left (b x +a \right )^{n} \left (d x +c \right )^{-3-n} b^{3} c^{2} d n -x \left (b x +a \right )^{n} \left (d x +c \right )^{-3-n} a^{2} b \,d^{3}+2 x \left (b x +a \right )^{n} \left (d x +c \right )^{-3-n} a \,b^{2} c \,d^{2}+2 x \left (b x +a \right )^{n} \left (d x +c \right )^{-3-n} b^{3} c^{2} d -\left (b x +a \right )^{n} \left (d x +c \right )^{-3-n} a^{2} b c \,d^{2} n +\left (b x +a \right )^{n} \left (d x +c \right )^{-3-n} a \,b^{2} c^{2} d n -\left (b x +a \right )^{n} \left (d x +c \right )^{-3-n} a^{2} b c \,d^{2}+2 \left (b x +a \right )^{n} \left (d x +c \right )^{-3-n} a \,b^{2} c^{2} d}{b d \left (a^{2} d^{2} n^{2}-2 a b c d \,n^{2}+b^{2} c^{2} n^{2}+3 a^{2} d^{2} n -6 a b c d n +3 b^{2} c^{2} n +2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}\right )}\)	\(462\)

[In]

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

[Out]

-(b*x+a)^(1+n)*(d*x+c)^(-2-n)*(a*d*n-b*c*n-b*d*x+a*d-2*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. 205 vs. \(2 (79) = 158\).

Time = 0.24 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.59 \[ \int (a+b x)^n (c+d x)^{-3-n} \, dx=\frac {{\left (b^{2} d^{2} x^{3} + 2 \, a b c^{2} - a^{2} c d + {\left (3 \, b^{2} c d + {\left (b^{2} c d - a b d^{2}\right )} n\right )} x^{2} + {\left (a b c^{2} - a^{2} c d\right )} n + {\left (2 \, b^{2} c^{2} + 2 \, a b c d - a^{2} d^{2} + {\left (b^{2} c^{2} - a^{2} d^{2}\right )} n\right )} x\right )} {\left (b x + a\right )}^{n} {\left (d x + c\right )}^{-n - 3}}{2 \, b^{2} c^{2} - 4 \, a b c d + 2 \, a^{2} d^{2} + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} n^{2} + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} n} \]

[In]

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

[Out]

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

Sympy [F(-2)]

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

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.90 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.71 \[ \int (a+b x)^n (c+d x)^{-3-n} \, dx=\frac {\frac {x\,{\left (a+b\,x\right )}^n\,\left (2\,b^2\,c^2-a^2\,d^2-a^2\,d^2\,n+b^2\,c^2\,n+2\,a\,b\,c\,d\right )}{{\left (a\,d-b\,c\right )}^2\,\left (n^2+3\,n+2\right )}-\frac {a\,c\,{\left (a+b\,x\right )}^n\,\left (a\,d-2\,b\,c+a\,d\,n-b\,c\,n\right )}{{\left (a\,d-b\,c\right )}^2\,\left (n^2+3\,n+2\right )}+\frac {b^2\,d^2\,x^3\,{\left (a+b\,x\right )}^n}{{\left (a\,d-b\,c\right )}^2\,\left (n^2+3\,n+2\right )}+\frac {b\,d\,x^2\,{\left (a+b\,x\right )}^n\,\left (3\,b\,c-a\,d\,n+b\,c\,n\right )}{{\left (a\,d-b\,c\right )}^2\,\left (n^2+3\,n+2\right )}}{{\left (c+d\,x\right )}^{n+3}} \]

[In]

int((a + b*x)^n/(c + d*x)^(n + 3),x)

[Out]

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