Integrand size = 19, antiderivative size = 80 \[ \int (a+b x)^{-3-n} (c+d x)^n \, dx=-\frac {(a+b x)^{-2-n} (c+d x)^{1+n}}{(b c-a d) (2+n)}+\frac {d (a+b x)^{-1-n} (c+d x)^{1+n}}{(b c-a d)^2 (1+n) (2+n)} \] Output:
-(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)
Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.75 \[ \int (a+b x)^{-3-n} (c+d x)^n \, dx=\frac {(a+b x)^{-2-n} (c+d x)^{1+n} (a d (2+n)-b (c+c n-d x))}{(b c-a d)^2 (1+n) (2+n)} \] Input:
Integrate[(a + b*x)^(-3 - n)*(c + d*x)^n,x]
Output:
((a + b*x)^(-2 - n)*(c + d*x)^(1 + n)*(a*d*(2 + n) - b*(c + c*n - d*x)))/( (b*c - a*d)^2*(1 + n)*(2 + n))
Time = 0.18 (sec) , antiderivative size = 80, 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 = {55, 48}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (a+b x)^{-n-3} (c+d x)^n \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {d \int (a+b x)^{-n-2} (c+d x)^ndx}{(n+2) (b c-a d)}-\frac {(a+b x)^{-n-2} (c+d x)^{n+1}}{(n+2) (b c-a d)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {d (a+b x)^{-n-1} (c+d x)^{n+1}}{(n+1) (n+2) (b c-a d)^2}-\frac {(a+b x)^{-n-2} (c+d x)^{n+1}}{(n+2) (b c-a d)}\) |
Input:
Int[(a + b*x)^(-3 - n)*(c + d*x)^n,x]
Output:
-(((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))
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] && EqQ[m + n + 2, 0] && NeQ[m, -1]
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] - Simp[d*(S implify[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] && 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] || !SumSimp lerQ[n, 1])
Time = 0.34 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.54
method | result | size |
gosper | \(\frac {\left (b x +a \right )^{-2-n} \left (x d +c \right )^{1+n} \left (a d n -b c n +b d x +2 a d -b c \right )}{a^{2} d^{2} n^{2}-2 a b c d \,n^{2}+b^{2} c^{2} n^{2}+3 a^{2} d^{2} n -6 a b c d n +3 b^{2} c^{2} n +2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}}\) | \(123\) |
orering | \(\frac {\left (b x +a \right ) \left (x d +c \right ) \left (a d n -b c n +b d x +2 a d -b c \right ) \left (b x +a \right )^{-3-n} \left (x d +c \right )^{n}}{a^{2} d^{2} n^{2}-2 a b c d \,n^{2}+b^{2} c^{2} n^{2}+3 a^{2} d^{2} n -6 a b c d n +3 b^{2} c^{2} n +2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}}\) | \(131\) |
parallelrisch | \(\frac {x^{3} \left (x d +c \right )^{n} \left (b x +a \right )^{-3-n} b^{3} d^{3}+x^{2} \left (x d +c \right )^{n} \left (b x +a \right )^{-3-n} a \,b^{2} d^{3} n -x^{2} \left (x d +c \right )^{n} \left (b x +a \right )^{-3-n} b^{3} c \,d^{2} n +3 x^{2} \left (x d +c \right )^{n} \left (b x +a \right )^{-3-n} a \,b^{2} d^{3}+x \left (x d +c \right )^{n} \left (b x +a \right )^{-3-n} a^{2} b \,d^{3} n -x \left (x d +c \right )^{n} \left (b x +a \right )^{-3-n} b^{3} c^{2} d n +2 x \left (x d +c \right )^{n} \left (b x +a \right )^{-3-n} a^{2} b \,d^{3}+2 x \left (x d +c \right )^{n} \left (b x +a \right )^{-3-n} a \,b^{2} c \,d^{2}-x \left (x d +c \right )^{n} \left (b x +a \right )^{-3-n} b^{3} c^{2} d +\left (x d +c \right )^{n} \left (b x +a \right )^{-3-n} a^{2} b c \,d^{2} n -\left (x d +c \right )^{n} \left (b x +a \right )^{-3-n} a \,b^{2} c^{2} d n +2 \left (x d +c \right )^{n} \left (b x +a \right )^{-3-n} a^{2} b c \,d^{2}-\left (x d +c \right )^{n} \left (b x +a \right )^{-3-n} a \,b^{2} c^{2} d}{d b \left (a^{2} d^{2} n^{2}-2 a b c d \,n^{2}+b^{2} c^{2} n^{2}+3 a^{2} d^{2} n -6 a b c d n +3 b^{2} c^{2} n +2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}\right )}\) | \(462\) |
Input:
int((b*x+a)^(-3-n)*(d*x+c)^n,x,method=_RETURNVERBOSE)
Output:
(b*x+a)^(-2-n)*(d*x+c)^(1+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)
Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (80) = 160\).
Time = 0.09 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.59 \[ \int (a+b x)^{-3-n} (c+d x)^n \, dx=\frac {{\left (b^{2} d^{2} x^{3} - a b c^{2} + 2 \, a^{2} c d + {\left (3 \, a b d^{2} - {\left (b^{2} c d - a b d^{2}\right )} n\right )} x^{2} - {\left (a b c^{2} - a^{2} c d\right )} n - {\left (b^{2} c^{2} - 2 \, a b c d - 2 \, a^{2} d^{2} + {\left (b^{2} c^{2} - a^{2} d^{2}\right )} n\right )} x\right )} {\left (b x + a\right )}^{-n - 3} {\left (d x + c\right )}^{n}}{2 \, b^{2} c^{2} - 4 \, a b c d + 2 \, a^{2} d^{2} + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} n^{2} + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} n} \] Input:
integrate((b*x+a)^(-3-n)*(d*x+c)^n,x, algorithm="fricas")
Output:
(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)*(d*x + c)^n/(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)
Exception generated. \[ \int (a+b x)^{-3-n} (c+d x)^n \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:
integrate((b*x+a)**(-3-n)*(d*x+c)**n,x)
Output:
Exception raised: HeuristicGCDFailed >> no luck
\[ \int (a+b x)^{-3-n} (c+d x)^n \, dx=\int { {\left (b x + a\right )}^{-n - 3} {\left (d x + c\right )}^{n} \,d x } \] Input:
integrate((b*x+a)^(-3-n)*(d*x+c)^n,x, algorithm="maxima")
Output:
integrate((b*x + a)^(-n - 3)*(d*x + c)^n, x)
\[ \int (a+b x)^{-3-n} (c+d x)^n \, dx=\int { {\left (b x + a\right )}^{-n - 3} {\left (d x + c\right )}^{n} \,d x } \] Input:
integrate((b*x+a)^(-3-n)*(d*x+c)^n,x, algorithm="giac")
Output:
integrate((b*x + a)^(-n - 3)*(d*x + c)^n, x)
Time = 0.42 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.68 \[ \int (a+b x)^{-3-n} (c+d x)^n \, dx=\frac {\frac {x\,{\left (c+d\,x\right )}^n\,\left (2\,a^2\,d^2-b^2\,c^2+a^2\,d^2\,n-b^2\,c^2\,n+2\,a\,b\,c\,d\right )}{{\left (a\,d-b\,c\right )}^2\,\left (n^2+3\,n+2\right )}+\frac {a\,c\,{\left (c+d\,x\right )}^n\,\left (2\,a\,d-b\,c+a\,d\,n-b\,c\,n\right )}{{\left (a\,d-b\,c\right )}^2\,\left (n^2+3\,n+2\right )}+\frac {b^2\,d^2\,x^3\,{\left (c+d\,x\right )}^n}{{\left (a\,d-b\,c\right )}^2\,\left (n^2+3\,n+2\right )}+\frac {b\,d\,x^2\,{\left (c+d\,x\right )}^n\,\left (3\,a\,d+a\,d\,n-b\,c\,n\right )}{{\left (a\,d-b\,c\right )}^2\,\left (n^2+3\,n+2\right )}}{{\left (a+b\,x\right )}^{n+3}} \] Input:
int((c + d*x)^n/(a + b*x)^(n + 3),x)
Output:
((x*(c + d*x)^n*(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*(3*n + n^2 + 2)) + (a*c*(c + d*x)^n*(2*a*d - b*c + a*d*n - b*c*n))/((a*d - b*c)^2*(3*n + n^2 + 2)) + (b^2*d^2*x^3*(c + d*x)^n)/((a*d - b*c)^2*(3*n + n^2 + 2)) + (b*d*x^2*(c + d*x)^n*(3*a*d + a*d*n - b*c*n)) /((a*d - b*c)^2*(3*n + n^2 + 2)))/(a + b*x)^(n + 3)
\[ \int (a+b x)^{-3-n} (c+d x)^n \, dx=\int \frac {\left (d x +c \right )^{n}}{\left (b x +a \right )^{n} a^{3}+3 \left (b x +a \right )^{n} a^{2} b x +3 \left (b x +a \right )^{n} a \,b^{2} x^{2}+\left (b x +a \right )^{n} b^{3} x^{3}}d x \] Input:
int((b*x+a)^(-3-n)*(d*x+c)^n,x)
Output:
int((c + d*x)**n/((a + b*x)**n*a**3 + 3*(a + b*x)**n*a**2*b*x + 3*(a + b*x )**n*a*b**2*x**2 + (a + b*x)**n*b**3*x**3),x)