Integrand size = 16, antiderivative size = 104 \[ \int x^2 (a+b x)^n (c+d x) \, dx=\frac {a^2 (b c-a d) (a+b x)^{1+n}}{b^4 (1+n)}-\frac {a (2 b c-3 a d) (a+b x)^{2+n}}{b^4 (2+n)}+\frac {(b c-3 a d) (a+b x)^{3+n}}{b^4 (3+n)}+\frac {d (a+b x)^{4+n}}{b^4 (4+n)} \] Output:
a^2*(-a*d+b*c)*(b*x+a)^(1+n)/b^4/(1+n)-a*(-3*a*d+2*b*c)*(b*x+a)^(2+n)/b^4/ (2+n)+(-3*a*d+b*c)*(b*x+a)^(3+n)/b^4/(3+n)+d*(b*x+a)^(4+n)/b^4/(4+n)
Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.84 \[ \int x^2 (a+b x)^n (c+d x) \, dx=\frac {(a+b x)^{1+n} \left (\frac {a^2 (b c-a d)}{1+n}+\frac {a (-2 b c+3 a d) (a+b x)}{2+n}+\frac {(b c-3 a d) (a+b x)^2}{3+n}+\frac {d (a+b x)^3}{4+n}\right )}{b^4} \] Input:
Integrate[x^2*(a + b*x)^n*(c + d*x),x]
Output:
((a + b*x)^(1 + n)*((a^2*(b*c - a*d))/(1 + n) + (a*(-2*b*c + 3*a*d)*(a + b *x))/(2 + n) + ((b*c - 3*a*d)*(a + b*x)^2)/(3 + n) + (d*(a + b*x)^3)/(4 + n)))/b^4
Time = 0.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {86, 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.
| \(\displaystyle \int x^2 (c+d x) (a+b x)^n \, dx\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \int \left (-\frac {a^2 (a d-b c) (a+b x)^n}{b^3}+\frac {a (3 a d-2 b c) (a+b x)^{n+1}}{b^3}+\frac {(b c-3 a d) (a+b x)^{n+2}}{b^3}+\frac {d (a+b x)^{n+3}}{b^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^2 (b c-a d) (a+b x)^{n+1}}{b^4 (n+1)}-\frac {a (2 b c-3 a d) (a+b x)^{n+2}}{b^4 (n+2)}+\frac {(b c-3 a d) (a+b x)^{n+3}}{b^4 (n+3)}+\frac {d (a+b x)^{n+4}}{b^4 (n+4)}\) |
Input:
Int[x^2*(a + b*x)^n*(c + d*x),x]
Output:
(a^2*(b*c - a*d)*(a + b*x)^(1 + n))/(b^4*(1 + n)) - (a*(2*b*c - 3*a*d)*(a + b*x)^(2 + n))/(b^4*(2 + n)) + ((b*c - 3*a*d)*(a + b*x)^(3 + n))/(b^4*(3 + n)) + (d*(a + b*x)^(4 + n))/(b^4*(4 + n))
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Leaf count of result is larger than twice the leaf count of optimal. \(211\) vs. \(2(104)=208\).
Time = 0.21 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.04
| method | result | size |
| norman | \(\frac {d \,x^{4} {\mathrm e}^{n \ln \left (b x +a \right )}}{4+n}+\frac {\left (a d n +b c n +4 b c \right ) x^{3} {\mathrm e}^{n \ln \left (b x +a \right )}}{b \left (n^{2}+7 n +12\right )}-\frac {2 a^{3} \left (-b c n +3 a d -4 b c \right ) {\mathrm e}^{n \ln \left (b x +a \right )}}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}+\frac {2 n \,a^{2} \left (-b c n +3 a d -4 b c \right ) x \,{\mathrm e}^{n \ln \left (b x +a \right )}}{b^{3} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}-\frac {\left (-b c n +3 a d -4 b c \right ) a n \,x^{2} {\mathrm e}^{n \ln \left (b x +a \right )}}{b^{2} \left (n^{3}+9 n^{2}+26 n +24\right )}\) | \(212\) |
| gosper | \(-\frac {\left (b x +a \right )^{1+n} \left (-b^{3} d \,n^{3} x^{3}-b^{3} c \,n^{3} x^{2}-6 b^{3} d \,n^{2} x^{3}+3 a \,b^{2} d \,n^{2} x^{2}-7 b^{3} c \,n^{2} x^{2}-11 b^{3} d n \,x^{3}+2 a \,b^{2} c \,n^{2} x +9 a \,b^{2} d n \,x^{2}-14 b^{3} c n \,x^{2}-6 d \,x^{3} b^{3}-6 a^{2} b d n x +10 a \,b^{2} c n x +6 x^{2} a \,b^{2} d -8 x^{2} b^{3} c -2 a^{2} b c n -6 x \,a^{2} b d +8 x a \,b^{2} c +6 a^{3} d -8 a^{2} b c \right )}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}\) | \(222\) |
| orering | \(-\frac {\left (b x +a \right )^{n} \left (-b^{3} d \,n^{3} x^{3}-b^{3} c \,n^{3} x^{2}-6 b^{3} d \,n^{2} x^{3}+3 a \,b^{2} d \,n^{2} x^{2}-7 b^{3} c \,n^{2} x^{2}-11 b^{3} d n \,x^{3}+2 a \,b^{2} c \,n^{2} x +9 a \,b^{2} d n \,x^{2}-14 b^{3} c n \,x^{2}-6 d \,x^{3} b^{3}-6 a^{2} b d n x +10 a \,b^{2} c n x +6 x^{2} a \,b^{2} d -8 x^{2} b^{3} c -2 a^{2} b c n -6 x \,a^{2} b d +8 x a \,b^{2} c +6 a^{3} d -8 a^{2} b c \right ) \left (b x +a \right )}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )}\) | \(225\) |
| risch | \(-\frac {\left (-b^{4} d \,n^{3} x^{4}-a \,b^{3} d \,n^{3} x^{3}-b^{4} c \,n^{3} x^{3}-6 b^{4} d \,n^{2} x^{4}-a \,b^{3} c \,n^{3} x^{2}-3 a \,b^{3} d \,n^{2} x^{3}-7 b^{4} c \,n^{2} x^{3}-11 b^{4} d n \,x^{4}+3 a^{2} b^{2} d \,n^{2} x^{2}-5 a \,b^{3} c \,n^{2} x^{2}-2 a \,b^{3} d n \,x^{3}-14 b^{4} c n \,x^{3}-6 d \,x^{4} b^{4}+2 a^{2} b^{2} c \,n^{2} x +3 a^{2} b^{2} d n \,x^{2}-4 a \,b^{3} c n \,x^{2}-8 x^{3} b^{4} c -6 a^{3} b d n x +8 a^{2} b^{2} c n x -2 a^{3} b c n +6 a^{4} d -8 a^{3} b c \right ) \left (b x +a \right )^{n}}{\left (3+n \right ) \left (4+n \right ) \left (2+n \right ) \left (1+n \right ) b^{4}}\) | \(276\) |
| parallelrisch | \(\frac {-6 \left (b x +a \right )^{n} a^{5} d +x^{2} \left (b x +a \right )^{n} a^{2} b^{3} c \,n^{3}+2 x^{3} \left (b x +a \right )^{n} a^{2} b^{3} d n +14 x^{3} \left (b x +a \right )^{n} a \,b^{4} c n -3 x^{2} \left (b x +a \right )^{n} a^{3} b^{2} d \,n^{2}+5 x^{2} \left (b x +a \right )^{n} a^{2} b^{3} c \,n^{2}+8 \left (b x +a \right )^{n} a^{4} b c +x^{4} \left (b x +a \right )^{n} a \,b^{4} d \,n^{3}+6 x^{4} \left (b x +a \right )^{n} a \,b^{4} d \,n^{2}+x^{3} \left (b x +a \right )^{n} a^{2} b^{3} d \,n^{3}+x^{3} \left (b x +a \right )^{n} a \,b^{4} c \,n^{3}+11 x^{4} \left (b x +a \right )^{n} a \,b^{4} d n +3 x^{3} \left (b x +a \right )^{n} a^{2} b^{3} d \,n^{2}+7 x^{3} \left (b x +a \right )^{n} a \,b^{4} c \,n^{2}+6 x^{4} \left (b x +a \right )^{n} a \,b^{4} d +8 x^{3} \left (b x +a \right )^{n} a \,b^{4} c +2 \left (b x +a \right )^{n} a^{4} b c n -3 x^{2} \left (b x +a \right )^{n} a^{3} b^{2} d n +4 x^{2} \left (b x +a \right )^{n} a^{2} b^{3} c n -2 x \left (b x +a \right )^{n} a^{3} b^{2} c \,n^{2}+6 x \left (b x +a \right )^{n} a^{4} b d n -8 x \left (b x +a \right )^{n} a^{3} b^{2} c n}{\left (4+n \right ) \left (3+n \right ) \left (2+n \right ) \left (1+n \right ) a \,b^{4}}\) | \(441\) |
Input:
int(x^2*(b*x+a)^n*(d*x+c),x,method=_RETURNVERBOSE)
Output:
d/(4+n)*x^4*exp(n*ln(b*x+a))+(a*d*n+b*c*n+4*b*c)/b/(n^2+7*n+12)*x^3*exp(n* ln(b*x+a))-2*a^3*(-b*c*n+3*a*d-4*b*c)/b^4/(n^4+10*n^3+35*n^2+50*n+24)*exp( n*ln(b*x+a))+2/b^3*n*a^2*(-b*c*n+3*a*d-4*b*c)/(n^4+10*n^3+35*n^2+50*n+24)* x*exp(n*ln(b*x+a))-(-b*c*n+3*a*d-4*b*c)*a/b^2*n/(n^3+9*n^2+26*n+24)*x^2*ex p(n*ln(b*x+a))
Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (104) = 208\).
Time = 0.08 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.41 \[ \int x^2 (a+b x)^n (c+d x) \, dx=\frac {{\left (2 \, a^{3} b c n + 8 \, a^{3} b c - 6 \, a^{4} d + {\left (b^{4} d n^{3} + 6 \, b^{4} d n^{2} + 11 \, b^{4} d n + 6 \, b^{4} d\right )} x^{4} + {\left (8 \, b^{4} c + {\left (b^{4} c + a b^{3} d\right )} n^{3} + {\left (7 \, b^{4} c + 3 \, a b^{3} d\right )} n^{2} + 2 \, {\left (7 \, b^{4} c + a b^{3} d\right )} n\right )} x^{3} + {\left (a b^{3} c n^{3} + {\left (5 \, a b^{3} c - 3 \, a^{2} b^{2} d\right )} n^{2} + {\left (4 \, a b^{3} c - 3 \, a^{2} b^{2} d\right )} n\right )} x^{2} - 2 \, {\left (a^{2} b^{2} c n^{2} + {\left (4 \, a^{2} b^{2} c - 3 \, a^{3} b d\right )} n\right )} x\right )} {\left (b x + a\right )}^{n}}{b^{4} n^{4} + 10 \, b^{4} n^{3} + 35 \, b^{4} n^{2} + 50 \, b^{4} n + 24 \, b^{4}} \] Input:
integrate(x^2*(b*x+a)^n*(d*x+c),x, algorithm="fricas")
Output:
(2*a^3*b*c*n + 8*a^3*b*c - 6*a^4*d + (b^4*d*n^3 + 6*b^4*d*n^2 + 11*b^4*d*n + 6*b^4*d)*x^4 + (8*b^4*c + (b^4*c + a*b^3*d)*n^3 + (7*b^4*c + 3*a*b^3*d) *n^2 + 2*(7*b^4*c + a*b^3*d)*n)*x^3 + (a*b^3*c*n^3 + (5*a*b^3*c - 3*a^2*b^ 2*d)*n^2 + (4*a*b^3*c - 3*a^2*b^2*d)*n)*x^2 - 2*(a^2*b^2*c*n^2 + (4*a^2*b^ 2*c - 3*a^3*b*d)*n)*x)*(b*x + a)^n/(b^4*n^4 + 10*b^4*n^3 + 35*b^4*n^2 + 50 *b^4*n + 24*b^4)
Leaf count of result is larger than twice the leaf count of optimal. 2462 vs. \(2 (92) = 184\).
Time = 0.79 (sec) , antiderivative size = 2462, normalized size of antiderivative = 23.67 \[ \int x^2 (a+b x)^n (c+d x) \, dx=\text {Too large to display} \] Input:
integrate(x**2*(b*x+a)**n*(d*x+c),x)
Output:
Piecewise((a**n*(c*x**3/3 + d*x**4/4), Eq(b, 0)), (6*a**3*d*log(a/b + x)/( 6*a**3*b**4 + 18*a**2*b**5*x + 18*a*b**6*x**2 + 6*b**7*x**3) + 11*a**3*d/( 6*a**3*b**4 + 18*a**2*b**5*x + 18*a*b**6*x**2 + 6*b**7*x**3) - 2*a**2*b*c/ (6*a**3*b**4 + 18*a**2*b**5*x + 18*a*b**6*x**2 + 6*b**7*x**3) + 18*a**2*b* d*x*log(a/b + x)/(6*a**3*b**4 + 18*a**2*b**5*x + 18*a*b**6*x**2 + 6*b**7*x **3) + 27*a**2*b*d*x/(6*a**3*b**4 + 18*a**2*b**5*x + 18*a*b**6*x**2 + 6*b* *7*x**3) - 6*a*b**2*c*x/(6*a**3*b**4 + 18*a**2*b**5*x + 18*a*b**6*x**2 + 6 *b**7*x**3) + 18*a*b**2*d*x**2*log(a/b + x)/(6*a**3*b**4 + 18*a**2*b**5*x + 18*a*b**6*x**2 + 6*b**7*x**3) + 18*a*b**2*d*x**2/(6*a**3*b**4 + 18*a**2* b**5*x + 18*a*b**6*x**2 + 6*b**7*x**3) - 6*b**3*c*x**2/(6*a**3*b**4 + 18*a **2*b**5*x + 18*a*b**6*x**2 + 6*b**7*x**3) + 6*b**3*d*x**3*log(a/b + x)/(6 *a**3*b**4 + 18*a**2*b**5*x + 18*a*b**6*x**2 + 6*b**7*x**3), Eq(n, -4)), ( -6*a**3*d*log(a/b + x)/(2*a**2*b**4 + 4*a*b**5*x + 2*b**6*x**2) - 9*a**3*d /(2*a**2*b**4 + 4*a*b**5*x + 2*b**6*x**2) + 2*a**2*b*c*log(a/b + x)/(2*a** 2*b**4 + 4*a*b**5*x + 2*b**6*x**2) + 3*a**2*b*c/(2*a**2*b**4 + 4*a*b**5*x + 2*b**6*x**2) - 12*a**2*b*d*x*log(a/b + x)/(2*a**2*b**4 + 4*a*b**5*x + 2* b**6*x**2) - 12*a**2*b*d*x/(2*a**2*b**4 + 4*a*b**5*x + 2*b**6*x**2) + 4*a* b**2*c*x*log(a/b + x)/(2*a**2*b**4 + 4*a*b**5*x + 2*b**6*x**2) + 4*a*b**2* c*x/(2*a**2*b**4 + 4*a*b**5*x + 2*b**6*x**2) - 6*a*b**2*d*x**2*log(a/b + x )/(2*a**2*b**4 + 4*a*b**5*x + 2*b**6*x**2) + 2*b**3*c*x**2*log(a/b + x)...
Time = 0.07 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.65 \[ \int x^2 (a+b x)^n (c+d x) \, dx=\frac {{\left ({\left (n^{2} + 3 \, n + 2\right )} b^{3} x^{3} + {\left (n^{2} + n\right )} a b^{2} x^{2} - 2 \, a^{2} b n x + 2 \, a^{3}\right )} {\left (b x + a\right )}^{n} c}{{\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{3}} + \frac {{\left ({\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} b^{4} x^{4} + {\left (n^{3} + 3 \, n^{2} + 2 \, n\right )} a b^{3} x^{3} - 3 \, {\left (n^{2} + n\right )} a^{2} b^{2} x^{2} + 6 \, a^{3} b n x - 6 \, a^{4}\right )} {\left (b x + a\right )}^{n} d}{{\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} b^{4}} \] Input:
integrate(x^2*(b*x+a)^n*(d*x+c),x, algorithm="maxima")
Output:
((n^2 + 3*n + 2)*b^3*x^3 + (n^2 + n)*a*b^2*x^2 - 2*a^2*b*n*x + 2*a^3)*(b*x + a)^n*c/((n^3 + 6*n^2 + 11*n + 6)*b^3) + ((n^3 + 6*n^2 + 11*n + 6)*b^4*x ^4 + (n^3 + 3*n^2 + 2*n)*a*b^3*x^3 - 3*(n^2 + n)*a^2*b^2*x^2 + 6*a^3*b*n*x - 6*a^4)*(b*x + a)^n*d/((n^4 + 10*n^3 + 35*n^2 + 50*n + 24)*b^4)
Leaf count of result is larger than twice the leaf count of optimal. 431 vs. \(2 (104) = 208\).
Time = 0.11 (sec) , antiderivative size = 431, normalized size of antiderivative = 4.14 \[ \int x^2 (a+b x)^n (c+d x) \, dx=\frac {{\left (b x + a\right )}^{n} b^{4} d n^{3} x^{4} + {\left (b x + a\right )}^{n} b^{4} c n^{3} x^{3} + {\left (b x + a\right )}^{n} a b^{3} d n^{3} x^{3} + 6 \, {\left (b x + a\right )}^{n} b^{4} d n^{2} x^{4} + {\left (b x + a\right )}^{n} a b^{3} c n^{3} x^{2} + 7 \, {\left (b x + a\right )}^{n} b^{4} c n^{2} x^{3} + 3 \, {\left (b x + a\right )}^{n} a b^{3} d n^{2} x^{3} + 11 \, {\left (b x + a\right )}^{n} b^{4} d n x^{4} + 5 \, {\left (b x + a\right )}^{n} a b^{3} c n^{2} x^{2} - 3 \, {\left (b x + a\right )}^{n} a^{2} b^{2} d n^{2} x^{2} + 14 \, {\left (b x + a\right )}^{n} b^{4} c n x^{3} + 2 \, {\left (b x + a\right )}^{n} a b^{3} d n x^{3} + 6 \, {\left (b x + a\right )}^{n} b^{4} d x^{4} - 2 \, {\left (b x + a\right )}^{n} a^{2} b^{2} c n^{2} x + 4 \, {\left (b x + a\right )}^{n} a b^{3} c n x^{2} - 3 \, {\left (b x + a\right )}^{n} a^{2} b^{2} d n x^{2} + 8 \, {\left (b x + a\right )}^{n} b^{4} c x^{3} - 8 \, {\left (b x + a\right )}^{n} a^{2} b^{2} c n x + 6 \, {\left (b x + a\right )}^{n} a^{3} b d n x + 2 \, {\left (b x + a\right )}^{n} a^{3} b c n + 8 \, {\left (b x + a\right )}^{n} a^{3} b c - 6 \, {\left (b x + a\right )}^{n} a^{4} d}{b^{4} n^{4} + 10 \, b^{4} n^{3} + 35 \, b^{4} n^{2} + 50 \, b^{4} n + 24 \, b^{4}} \] Input:
integrate(x^2*(b*x+a)^n*(d*x+c),x, algorithm="giac")
Output:
((b*x + a)^n*b^4*d*n^3*x^4 + (b*x + a)^n*b^4*c*n^3*x^3 + (b*x + a)^n*a*b^3 *d*n^3*x^3 + 6*(b*x + a)^n*b^4*d*n^2*x^4 + (b*x + a)^n*a*b^3*c*n^3*x^2 + 7 *(b*x + a)^n*b^4*c*n^2*x^3 + 3*(b*x + a)^n*a*b^3*d*n^2*x^3 + 11*(b*x + a)^ n*b^4*d*n*x^4 + 5*(b*x + a)^n*a*b^3*c*n^2*x^2 - 3*(b*x + a)^n*a^2*b^2*d*n^ 2*x^2 + 14*(b*x + a)^n*b^4*c*n*x^3 + 2*(b*x + a)^n*a*b^3*d*n*x^3 + 6*(b*x + a)^n*b^4*d*x^4 - 2*(b*x + a)^n*a^2*b^2*c*n^2*x + 4*(b*x + a)^n*a*b^3*c*n *x^2 - 3*(b*x + a)^n*a^2*b^2*d*n*x^2 + 8*(b*x + a)^n*b^4*c*x^3 - 8*(b*x + a)^n*a^2*b^2*c*n*x + 6*(b*x + a)^n*a^3*b*d*n*x + 2*(b*x + a)^n*a^3*b*c*n + 8*(b*x + a)^n*a^3*b*c - 6*(b*x + a)^n*a^4*d)/(b^4*n^4 + 10*b^4*n^3 + 35*b ^4*n^2 + 50*b^4*n + 24*b^4)
Time = 1.05 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.15 \[ \int x^2 (a+b x)^n (c+d x) \, dx={\left (a+b\,x\right )}^n\,\left (\frac {d\,x^4\,\left (n^3+6\,n^2+11\,n+6\right )}{n^4+10\,n^3+35\,n^2+50\,n+24}+\frac {2\,a^3\,\left (4\,b\,c-3\,a\,d+b\,c\,n\right )}{b^4\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {x^3\,\left (4\,b\,c+a\,d\,n+b\,c\,n\right )\,\left (n^2+3\,n+2\right )}{b\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}-\frac {2\,a^2\,n\,x\,\left (4\,b\,c-3\,a\,d+b\,c\,n\right )}{b^3\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}+\frac {a\,n\,x^2\,\left (n+1\right )\,\left (4\,b\,c-3\,a\,d+b\,c\,n\right )}{b^2\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )}\right ) \] Input:
int(x^2*(a + b*x)^n*(c + d*x),x)
Output:
(a + b*x)^n*((d*x^4*(11*n + 6*n^2 + n^3 + 6))/(50*n + 35*n^2 + 10*n^3 + n^ 4 + 24) + (2*a^3*(4*b*c - 3*a*d + b*c*n))/(b^4*(50*n + 35*n^2 + 10*n^3 + n ^4 + 24)) + (x^3*(4*b*c + a*d*n + b*c*n)*(3*n + n^2 + 2))/(b*(50*n + 35*n^ 2 + 10*n^3 + n^4 + 24)) - (2*a^2*n*x*(4*b*c - 3*a*d + b*c*n))/(b^3*(50*n + 35*n^2 + 10*n^3 + n^4 + 24)) + (a*n*x^2*(n + 1)*(4*b*c - 3*a*d + b*c*n))/ (b^2*(50*n + 35*n^2 + 10*n^3 + n^4 + 24)))
Time = 0.15 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.60 \[ \int x^2 (a+b x)^n (c+d x) \, dx=\frac {\left (b x +a \right )^{n} \left (b^{4} d \,n^{3} x^{4}+a \,b^{3} d \,n^{3} x^{3}+b^{4} c \,n^{3} x^{3}+6 b^{4} d \,n^{2} x^{4}+a \,b^{3} c \,n^{3} x^{2}+3 a \,b^{3} d \,n^{2} x^{3}+7 b^{4} c \,n^{2} x^{3}+11 b^{4} d n \,x^{4}-3 a^{2} b^{2} d \,n^{2} x^{2}+5 a \,b^{3} c \,n^{2} x^{2}+2 a \,b^{3} d n \,x^{3}+14 b^{4} c n \,x^{3}+6 b^{4} d \,x^{4}-2 a^{2} b^{2} c \,n^{2} x -3 a^{2} b^{2} d n \,x^{2}+4 a \,b^{3} c n \,x^{2}+8 b^{4} c \,x^{3}+6 a^{3} b d n x -8 a^{2} b^{2} c n x +2 a^{3} b c n -6 a^{4} d +8 a^{3} b c \right )}{b^{4} \left (n^{4}+10 n^{3}+35 n^{2}+50 n +24\right )} \] Input:
int(x^2*(b*x+a)^n*(d*x+c),x)
Output:
((a + b*x)**n*( - 6*a**4*d + 2*a**3*b*c*n + 8*a**3*b*c + 6*a**3*b*d*n*x - 2*a**2*b**2*c*n**2*x - 8*a**2*b**2*c*n*x - 3*a**2*b**2*d*n**2*x**2 - 3*a** 2*b**2*d*n*x**2 + a*b**3*c*n**3*x**2 + 5*a*b**3*c*n**2*x**2 + 4*a*b**3*c*n *x**2 + a*b**3*d*n**3*x**3 + 3*a*b**3*d*n**2*x**3 + 2*a*b**3*d*n*x**3 + b* *4*c*n**3*x**3 + 7*b**4*c*n**2*x**3 + 14*b**4*c*n*x**3 + 8*b**4*c*x**3 + b **4*d*n**3*x**4 + 6*b**4*d*n**2*x**4 + 11*b**4*d*n*x**4 + 6*b**4*d*x**4))/ (b**4*(n**4 + 10*n**3 + 35*n**2 + 50*n + 24))