Integrand size = 19, antiderivative size = 83 \[ \int (a+b x)^2 (a c-b c x)^n \, dx=-\frac {4 a^2 (a c-b c x)^{1+n}}{b c (1+n)}+\frac {4 a (a c-b c x)^{2+n}}{b c^2 (2+n)}-\frac {(a c-b c x)^{3+n}}{b c^3 (3+n)} \] Output:
-4*a^2*(-b*c*x+a*c)^(1+n)/b/c/(1+n)+4*a*(-b*c*x+a*c)^(2+n)/b/c^2/(2+n)-(-b *c*x+a*c)^(3+n)/b/c^3/(3+n)
Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.93 \[ \int (a+b x)^2 (a c-b c x)^n \, dx=\frac {(c (a-b x))^n (-a+b x) \left (a^2 \left (14+7 n+n^2\right )+2 a b \left (4+5 n+n^2\right ) x+b^2 \left (2+3 n+n^2\right ) x^2\right )}{b (1+n) (2+n) (3+n)} \] Input:
Integrate[(a + b*x)^2*(a*c - b*c*x)^n,x]
Output:
((c*(a - b*x))^n*(-a + b*x)*(a^2*(14 + 7*n + n^2) + 2*a*b*(4 + 5*n + n^2)* x + b^2*(2 + 3*n + n^2)*x^2))/(b*(1 + n)*(2 + n)*(3 + n))
Time = 0.20 (sec) , antiderivative size = 83, 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 = {53, 2009}
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.
\(\displaystyle \int (a+b x)^2 (a c-b c x)^n \, dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \int \left (4 a^2 (a c-b c x)^n+\frac {(a c-b c x)^{n+2}}{c^2}-\frac {4 a (a c-b c x)^{n+1}}{c}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 a^2 (a c-b c x)^{n+1}}{b c (n+1)}-\frac {(a c-b c x)^{n+3}}{b c^3 (n+3)}+\frac {4 a (a c-b c x)^{n+2}}{b c^2 (n+2)}\) |
Input:
Int[(a + b*x)^2*(a*c - b*c*x)^n,x]
Output:
(-4*a^2*(a*c - b*c*x)^(1 + n))/(b*c*(1 + n)) + (4*a*(a*c - b*c*x)^(2 + n)) /(b*c^2*(2 + n)) - (a*c - b*c*x)^(3 + n)/(b*c^3*(3 + n))
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Time = 0.15 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.24
method | result | size |
gosper | \(-\frac {\left (-b x +a \right ) \left (b^{2} n^{2} x^{2}+2 a b \,n^{2} x +3 b^{2} n \,x^{2}+a^{2} n^{2}+10 a b n x +2 b^{2} x^{2}+7 a^{2} n +8 a b x +14 a^{2}\right ) \left (-b c x +a c \right )^{n}}{b \left (n^{3}+6 n^{2}+11 n +6\right )}\) | \(103\) |
orering | \(-\frac {\left (-b x +a \right ) \left (b^{2} n^{2} x^{2}+2 a b \,n^{2} x +3 b^{2} n \,x^{2}+a^{2} n^{2}+10 a b n x +2 b^{2} x^{2}+7 a^{2} n +8 a b x +14 a^{2}\right ) \left (-b c x +a c \right )^{n}}{b \left (n^{3}+6 n^{2}+11 n +6\right )}\) | \(103\) |
risch | \(-\frac {\left (-b^{3} n^{2} x^{3}-a \,b^{2} n^{2} x^{2}-3 n \,b^{3} x^{3}+a^{2} b \,n^{2} x -7 a \,b^{2} n \,x^{2}-2 b^{3} x^{3}+a^{3} n^{2}+3 a^{2} b n x -6 a \,b^{2} x^{2}+7 a^{3} n -6 a^{2} b x +14 a^{3}\right ) \left (c \left (-b x +a \right )\right )^{n}}{\left (2+n \right ) \left (3+n \right ) b \left (1+n \right )}\) | \(133\) |
norman | \(\frac {b^{2} x^{3} {\mathrm e}^{n \ln \left (-b c x +a c \right )}}{3+n}+\frac {a b \left (n +6\right ) x^{2} {\mathrm e}^{n \ln \left (-b c x +a c \right )}}{n^{2}+5 n +6}-\frac {a^{2} \left (n^{2}+3 n -6\right ) x \,{\mathrm e}^{n \ln \left (-b c x +a c \right )}}{n^{3}+6 n^{2}+11 n +6}-\frac {a^{3} \left (n^{2}+7 n +14\right ) {\mathrm e}^{n \ln \left (-b c x +a c \right )}}{b \left (n^{3}+6 n^{2}+11 n +6\right )}\) | \(145\) |
parallelrisch | \(\frac {x^{3} \left (c \left (-b x +a \right )\right )^{n} b^{3} n^{2}+3 x^{3} \left (c \left (-b x +a \right )\right )^{n} b^{3} n +x^{2} \left (c \left (-b x +a \right )\right )^{n} a \,b^{2} n^{2}+2 x^{3} \left (c \left (-b x +a \right )\right )^{n} b^{3}+7 x^{2} \left (c \left (-b x +a \right )\right )^{n} a \,b^{2} n -x \left (c \left (-b x +a \right )\right )^{n} a^{2} b \,n^{2}+6 x^{2} \left (c \left (-b x +a \right )\right )^{n} a \,b^{2}-3 x \left (c \left (-b x +a \right )\right )^{n} a^{2} b n -\left (c \left (-b x +a \right )\right )^{n} a^{3} n^{2}+6 x \left (c \left (-b x +a \right )\right )^{n} a^{2} b -7 \left (c \left (-b x +a \right )\right )^{n} a^{3} n -14 \left (c \left (-b x +a \right )\right )^{n} a^{3}}{b \left (n^{3}+6 n^{2}+11 n +6\right )}\) | \(242\) |
Input:
int((b*x+a)^2*(-b*c*x+a*c)^n,x,method=_RETURNVERBOSE)
Output:
-(-b*x+a)*(b^2*n^2*x^2+2*a*b*n^2*x+3*b^2*n*x^2+a^2*n^2+10*a*b*n*x+2*b^2*x^ 2+7*a^2*n+8*a*b*x+14*a^2)*(-b*c*x+a*c)^n/b/(n^3+6*n^2+11*n+6)
Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.54 \[ \int (a+b x)^2 (a c-b c x)^n \, dx=-\frac {{\left (a^{3} n^{2} + 7 \, a^{3} n - {\left (b^{3} n^{2} + 3 \, b^{3} n + 2 \, b^{3}\right )} x^{3} + 14 \, a^{3} - {\left (a b^{2} n^{2} + 7 \, a b^{2} n + 6 \, a b^{2}\right )} x^{2} + {\left (a^{2} b n^{2} + 3 \, a^{2} b n - 6 \, a^{2} b\right )} x\right )} {\left (-b c x + a c\right )}^{n}}{b n^{3} + 6 \, b n^{2} + 11 \, b n + 6 \, b} \] Input:
integrate((b*x+a)^2*(-b*c*x+a*c)^n,x, algorithm="fricas")
Output:
-(a^3*n^2 + 7*a^3*n - (b^3*n^2 + 3*b^3*n + 2*b^3)*x^3 + 14*a^3 - (a*b^2*n^ 2 + 7*a*b^2*n + 6*a*b^2)*x^2 + (a^2*b*n^2 + 3*a^2*b*n - 6*a^2*b)*x)*(-b*c* x + a*c)^n/(b*n^3 + 6*b*n^2 + 11*b*n + 6*b)
Leaf count of result is larger than twice the leaf count of optimal. 819 vs. \(2 (66) = 132\).
Time = 0.41 (sec) , antiderivative size = 819, normalized size of antiderivative = 9.87 \[ \int (a+b x)^2 (a c-b c x)^n \, dx =\text {Too large to display} \] Input:
integrate((b*x+a)**2*(-b*c*x+a*c)**n,x)
Output:
Piecewise((a**2*x*(a*c)**n, Eq(b, 0)), (-a**2*log(-a/b + x)/(a**2*b*c**3 - 2*a*b**2*c**3*x + b**3*c**3*x**2) - 2*a**2/(a**2*b*c**3 - 2*a*b**2*c**3*x + b**3*c**3*x**2) + 2*a*b*x*log(-a/b + x)/(a**2*b*c**3 - 2*a*b**2*c**3*x + b**3*c**3*x**2) + 4*a*b*x/(a**2*b*c**3 - 2*a*b**2*c**3*x + b**3*c**3*x** 2) - b**2*x**2*log(-a/b + x)/(a**2*b*c**3 - 2*a*b**2*c**3*x + b**3*c**3*x* *2), Eq(n, -3)), (-4*a**2*log(-a/b + x)/(-a*b*c**2 + b**2*c**2*x) - 5*a**2 /(-a*b*c**2 + b**2*c**2*x) + 4*a*b*x*log(-a/b + x)/(-a*b*c**2 + b**2*c**2* x) + b**2*x**2/(-a*b*c**2 + b**2*c**2*x), Eq(n, -2)), (-4*a**2*log(-a/b + x)/(b*c) - 3*a*x/c - b*x**2/(2*c), Eq(n, -1)), (-a**3*n**2*(a*c - b*c*x)** n/(b*n**3 + 6*b*n**2 + 11*b*n + 6*b) - 7*a**3*n*(a*c - b*c*x)**n/(b*n**3 + 6*b*n**2 + 11*b*n + 6*b) - 14*a**3*(a*c - b*c*x)**n/(b*n**3 + 6*b*n**2 + 11*b*n + 6*b) - a**2*b*n**2*x*(a*c - b*c*x)**n/(b*n**3 + 6*b*n**2 + 11*b*n + 6*b) - 3*a**2*b*n*x*(a*c - b*c*x)**n/(b*n**3 + 6*b*n**2 + 11*b*n + 6*b) + 6*a**2*b*x*(a*c - b*c*x)**n/(b*n**3 + 6*b*n**2 + 11*b*n + 6*b) + a*b**2 *n**2*x**2*(a*c - b*c*x)**n/(b*n**3 + 6*b*n**2 + 11*b*n + 6*b) + 7*a*b**2* n*x**2*(a*c - b*c*x)**n/(b*n**3 + 6*b*n**2 + 11*b*n + 6*b) + 6*a*b**2*x**2 *(a*c - b*c*x)**n/(b*n**3 + 6*b*n**2 + 11*b*n + 6*b) + b**3*n**2*x**3*(a*c - b*c*x)**n/(b*n**3 + 6*b*n**2 + 11*b*n + 6*b) + 3*b**3*n*x**3*(a*c - b*c *x)**n/(b*n**3 + 6*b*n**2 + 11*b*n + 6*b) + 2*b**3*x**3*(a*c - b*c*x)**n/( b*n**3 + 6*b*n**2 + 11*b*n + 6*b), True))
Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (83) = 166\).
Time = 0.05 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.01 \[ \int (a+b x)^2 (a c-b c x)^n \, dx=\frac {2 \, {\left (b^{2} c^{n} {\left (n + 1\right )} x^{2} - a b c^{n} n x - a^{2} c^{n}\right )} {\left (-b x + a\right )}^{n} a}{{\left (n^{2} + 3 \, n + 2\right )} b} + \frac {{\left ({\left (n^{2} + 3 \, n + 2\right )} b^{3} c^{n} x^{3} - {\left (n^{2} + n\right )} a b^{2} c^{n} x^{2} - 2 \, a^{2} b c^{n} n x - 2 \, a^{3} c^{n}\right )} {\left (-b x + a\right )}^{n}}{{\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b} - \frac {{\left (-b c x + a c\right )}^{n + 1} a^{2}}{b c {\left (n + 1\right )}} \] Input:
integrate((b*x+a)^2*(-b*c*x+a*c)^n,x, algorithm="maxima")
Output:
2*(b^2*c^n*(n + 1)*x^2 - a*b*c^n*n*x - a^2*c^n)*(-b*x + a)^n*a/((n^2 + 3*n + 2)*b) + ((n^2 + 3*n + 2)*b^3*c^n*x^3 - (n^2 + n)*a*b^2*c^n*x^2 - 2*a^2* b*c^n*n*x - 2*a^3*c^n)*(-b*x + a)^n/((n^3 + 6*n^2 + 11*n + 6)*b) - (-b*c*x + a*c)^(n + 1)*a^2/(b*c*(n + 1))
Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (83) = 166\).
Time = 0.13 (sec) , antiderivative size = 256, normalized size of antiderivative = 3.08 \[ \int (a+b x)^2 (a c-b c x)^n \, dx=\frac {{\left (-b c x + a c\right )}^{n} b^{3} n^{2} x^{3} + {\left (-b c x + a c\right )}^{n} a b^{2} n^{2} x^{2} + 3 \, {\left (-b c x + a c\right )}^{n} b^{3} n x^{3} - {\left (-b c x + a c\right )}^{n} a^{2} b n^{2} x + 7 \, {\left (-b c x + a c\right )}^{n} a b^{2} n x^{2} + 2 \, {\left (-b c x + a c\right )}^{n} b^{3} x^{3} - {\left (-b c x + a c\right )}^{n} a^{3} n^{2} - 3 \, {\left (-b c x + a c\right )}^{n} a^{2} b n x + 6 \, {\left (-b c x + a c\right )}^{n} a b^{2} x^{2} - 7 \, {\left (-b c x + a c\right )}^{n} a^{3} n + 6 \, {\left (-b c x + a c\right )}^{n} a^{2} b x - 14 \, {\left (-b c x + a c\right )}^{n} a^{3}}{b n^{3} + 6 \, b n^{2} + 11 \, b n + 6 \, b} \] Input:
integrate((b*x+a)^2*(-b*c*x+a*c)^n,x, algorithm="giac")
Output:
((-b*c*x + a*c)^n*b^3*n^2*x^3 + (-b*c*x + a*c)^n*a*b^2*n^2*x^2 + 3*(-b*c*x + a*c)^n*b^3*n*x^3 - (-b*c*x + a*c)^n*a^2*b*n^2*x + 7*(-b*c*x + a*c)^n*a* b^2*n*x^2 + 2*(-b*c*x + a*c)^n*b^3*x^3 - (-b*c*x + a*c)^n*a^3*n^2 - 3*(-b* c*x + a*c)^n*a^2*b*n*x + 6*(-b*c*x + a*c)^n*a*b^2*x^2 - 7*(-b*c*x + a*c)^n *a^3*n + 6*(-b*c*x + a*c)^n*a^2*b*x - 14*(-b*c*x + a*c)^n*a^3)/(b*n^3 + 6* b*n^2 + 11*b*n + 6*b)
Time = 0.30 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.60 \[ \int (a+b x)^2 (a c-b c x)^n \, dx=-{\left (a\,c-b\,c\,x\right )}^n\,\left (\frac {a^2\,x\,\left (n^2+3\,n-6\right )}{n^3+6\,n^2+11\,n+6}+\frac {a^3\,\left (n^2+7\,n+14\right )}{b\,\left (n^3+6\,n^2+11\,n+6\right )}-\frac {b^2\,x^3\,\left (n^2+3\,n+2\right )}{n^3+6\,n^2+11\,n+6}-\frac {a\,b\,x^2\,\left (n^2+7\,n+6\right )}{n^3+6\,n^2+11\,n+6}\right ) \] Input:
int((a*c - b*c*x)^n*(a + b*x)^2,x)
Output:
-(a*c - b*c*x)^n*((a^2*x*(3*n + n^2 - 6))/(11*n + 6*n^2 + n^3 + 6) + (a^3* (7*n + n^2 + 14))/(b*(11*n + 6*n^2 + n^3 + 6)) - (b^2*x^3*(3*n + n^2 + 2)) /(11*n + 6*n^2 + n^3 + 6) - (a*b*x^2*(7*n + n^2 + 6))/(11*n + 6*n^2 + n^3 + 6))
Time = 0.16 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.59 \[ \int (a+b x)^2 (a c-b c x)^n \, dx=\frac {\left (-b c x +a c \right )^{n} \left (b^{3} n^{2} x^{3}+a \,b^{2} n^{2} x^{2}+3 b^{3} n \,x^{3}-a^{2} b \,n^{2} x +7 a \,b^{2} n \,x^{2}+2 b^{3} x^{3}-a^{3} n^{2}-3 a^{2} b n x +6 a \,b^{2} x^{2}-7 a^{3} n +6 a^{2} b x -14 a^{3}\right )}{b \left (n^{3}+6 n^{2}+11 n +6\right )} \] Input:
int((b*x+a)^2*(-b*c*x+a*c)^n,x)
Output:
((a*c - b*c*x)**n*( - a**3*n**2 - 7*a**3*n - 14*a**3 - a**2*b*n**2*x - 3*a **2*b*n*x + 6*a**2*b*x + a*b**2*n**2*x**2 + 7*a*b**2*n*x**2 + 6*a*b**2*x** 2 + b**3*n**2*x**3 + 3*b**3*n*x**3 + 2*b**3*x**3))/(b*(n**3 + 6*n**2 + 11* n + 6))