Integrand size = 20, antiderivative size = 75 \[ \int x^2 (A+B x) \left (b x+c x^2\right )^3 \, dx=\frac {1}{6} A b^3 x^6+\frac {1}{7} b^2 (b B+3 A c) x^7+\frac {3}{8} b c (b B+A c) x^8+\frac {1}{9} c^2 (3 b B+A c) x^9+\frac {1}{10} B c^3 x^{10} \] Output:
1/6*A*b^3*x^6+1/7*b^2*(3*A*c+B*b)*x^7+3/8*b*c*(A*c+B*b)*x^8+1/9*c^2*(A*c+3 *B*b)*x^9+1/10*B*c^3*x^10
Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int x^2 (A+B x) \left (b x+c x^2\right )^3 \, dx=\frac {1}{6} A b^3 x^6+\frac {1}{7} b^2 (b B+3 A c) x^7+\frac {3}{8} b c (b B+A c) x^8+\frac {1}{9} c^2 (3 b B+A c) x^9+\frac {1}{10} B c^3 x^{10} \] Input:
Integrate[x^2*(A + B*x)*(b*x + c*x^2)^3,x]
Output:
(A*b^3*x^6)/6 + (b^2*(b*B + 3*A*c)*x^7)/7 + (3*b*c*(b*B + A*c)*x^8)/8 + (c ^2*(3*b*B + A*c)*x^9)/9 + (B*c^3*x^10)/10
Time = 0.38 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {9, 85, 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.
\(\displaystyle \int x^2 (A+B x) \left (b x+c x^2\right )^3 \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int x^5 (A+B x) (b+c x)^3dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (A b^3 x^5+b^2 x^6 (3 A c+b B)+c^2 x^8 (A c+3 b B)+3 b c x^7 (A c+b B)+B c^3 x^9\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{6} A b^3 x^6+\frac {1}{7} b^2 x^7 (3 A c+b B)+\frac {1}{9} c^2 x^9 (A c+3 b B)+\frac {3}{8} b c x^8 (A c+b B)+\frac {1}{10} B c^3 x^{10}\) |
Input:
Int[x^2*(A + B*x)*(b*x + c*x^2)^3,x]
Output:
(A*b^3*x^6)/6 + (b^2*(b*B + 3*A*c)*x^7)/7 + (3*b*c*(b*B + A*c)*x^8)/8 + (c ^2*(3*b*B + A*c)*x^9)/9 + (B*c^3*x^10)/10
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
Time = 0.60 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\frac {B \,c^{3} x^{10}}{10}+\left (\frac {1}{9} A \,c^{3}+\frac {1}{3} B b \,c^{2}\right ) x^{9}+\left (\frac {3}{8} A b \,c^{2}+\frac {3}{8} B \,b^{2} c \right ) x^{8}+\left (\frac {3}{7} A \,b^{2} c +\frac {1}{7} B \,b^{3}\right ) x^{7}+\frac {A \,b^{3} x^{6}}{6}\) | \(75\) |
gosper | \(\frac {x^{6} \left (252 B \,c^{3} x^{4}+280 A \,c^{3} x^{3}+840 x^{3} B b \,c^{2}+945 A b \,c^{2} x^{2}+945 x^{2} B \,b^{2} c +1080 A \,b^{2} c x +360 x B \,b^{3}+420 A \,b^{3}\right )}{2520}\) | \(76\) |
default | \(\frac {B \,c^{3} x^{10}}{10}+\frac {\left (A \,c^{3}+3 B b \,c^{2}\right ) x^{9}}{9}+\frac {\left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{8}}{8}+\frac {\left (3 A \,b^{2} c +B \,b^{3}\right ) x^{7}}{7}+\frac {A \,b^{3} x^{6}}{6}\) | \(76\) |
risch | \(\frac {1}{10} B \,c^{3} x^{10}+\frac {1}{9} x^{9} A \,c^{3}+\frac {1}{3} x^{9} B b \,c^{2}+\frac {3}{8} x^{8} A b \,c^{2}+\frac {3}{8} x^{8} B \,b^{2} c +\frac {3}{7} x^{7} A \,b^{2} c +\frac {1}{7} B \,b^{3} x^{7}+\frac {1}{6} A \,b^{3} x^{6}\) | \(78\) |
parallelrisch | \(\frac {1}{10} B \,c^{3} x^{10}+\frac {1}{9} x^{9} A \,c^{3}+\frac {1}{3} x^{9} B b \,c^{2}+\frac {3}{8} x^{8} A b \,c^{2}+\frac {3}{8} x^{8} B \,b^{2} c +\frac {3}{7} x^{7} A \,b^{2} c +\frac {1}{7} B \,b^{3} x^{7}+\frac {1}{6} A \,b^{3} x^{6}\) | \(78\) |
orering | \(\frac {x^{3} \left (252 B \,c^{3} x^{4}+280 A \,c^{3} x^{3}+840 x^{3} B b \,c^{2}+945 A b \,c^{2} x^{2}+945 x^{2} B \,b^{2} c +1080 A \,b^{2} c x +360 x B \,b^{3}+420 A \,b^{3}\right ) \left (c \,x^{2}+b x \right )^{3}}{2520 \left (c x +b \right )^{3}}\) | \(94\) |
Input:
int(x^2*(B*x+A)*(c*x^2+b*x)^3,x,method=_RETURNVERBOSE)
Output:
1/10*B*c^3*x^10+(1/9*A*c^3+1/3*B*b*c^2)*x^9+(3/8*A*b*c^2+3/8*B*b^2*c)*x^8+ (3/7*A*b^2*c+1/7*B*b^3)*x^7+1/6*A*b^3*x^6
Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int x^2 (A+B x) \left (b x+c x^2\right )^3 \, dx=\frac {1}{10} \, B c^{3} x^{10} + \frac {1}{6} \, A b^{3} x^{6} + \frac {1}{9} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{9} + \frac {3}{8} \, {\left (B b^{2} c + A b c^{2}\right )} x^{8} + \frac {1}{7} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{7} \] Input:
integrate(x^2*(B*x+A)*(c*x^2+b*x)^3,x, algorithm="fricas")
Output:
1/10*B*c^3*x^10 + 1/6*A*b^3*x^6 + 1/9*(3*B*b*c^2 + A*c^3)*x^9 + 3/8*(B*b^2 *c + A*b*c^2)*x^8 + 1/7*(B*b^3 + 3*A*b^2*c)*x^7
Time = 0.02 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09 \[ \int x^2 (A+B x) \left (b x+c x^2\right )^3 \, dx=\frac {A b^{3} x^{6}}{6} + \frac {B c^{3} x^{10}}{10} + x^{9} \left (\frac {A c^{3}}{9} + \frac {B b c^{2}}{3}\right ) + x^{8} \cdot \left (\frac {3 A b c^{2}}{8} + \frac {3 B b^{2} c}{8}\right ) + x^{7} \cdot \left (\frac {3 A b^{2} c}{7} + \frac {B b^{3}}{7}\right ) \] Input:
integrate(x**2*(B*x+A)*(c*x**2+b*x)**3,x)
Output:
A*b**3*x**6/6 + B*c**3*x**10/10 + x**9*(A*c**3/9 + B*b*c**2/3) + x**8*(3*A *b*c**2/8 + 3*B*b**2*c/8) + x**7*(3*A*b**2*c/7 + B*b**3/7)
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int x^2 (A+B x) \left (b x+c x^2\right )^3 \, dx=\frac {1}{10} \, B c^{3} x^{10} + \frac {1}{6} \, A b^{3} x^{6} + \frac {1}{9} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{9} + \frac {3}{8} \, {\left (B b^{2} c + A b c^{2}\right )} x^{8} + \frac {1}{7} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{7} \] Input:
integrate(x^2*(B*x+A)*(c*x^2+b*x)^3,x, algorithm="maxima")
Output:
1/10*B*c^3*x^10 + 1/6*A*b^3*x^6 + 1/9*(3*B*b*c^2 + A*c^3)*x^9 + 3/8*(B*b^2 *c + A*b*c^2)*x^8 + 1/7*(B*b^3 + 3*A*b^2*c)*x^7
Time = 0.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03 \[ \int x^2 (A+B x) \left (b x+c x^2\right )^3 \, dx=\frac {1}{10} \, B c^{3} x^{10} + \frac {1}{3} \, B b c^{2} x^{9} + \frac {1}{9} \, A c^{3} x^{9} + \frac {3}{8} \, B b^{2} c x^{8} + \frac {3}{8} \, A b c^{2} x^{8} + \frac {1}{7} \, B b^{3} x^{7} + \frac {3}{7} \, A b^{2} c x^{7} + \frac {1}{6} \, A b^{3} x^{6} \] Input:
integrate(x^2*(B*x+A)*(c*x^2+b*x)^3,x, algorithm="giac")
Output:
1/10*B*c^3*x^10 + 1/3*B*b*c^2*x^9 + 1/9*A*c^3*x^9 + 3/8*B*b^2*c*x^8 + 3/8* A*b*c^2*x^8 + 1/7*B*b^3*x^7 + 3/7*A*b^2*c*x^7 + 1/6*A*b^3*x^6
Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92 \[ \int x^2 (A+B x) \left (b x+c x^2\right )^3 \, dx=x^7\,\left (\frac {B\,b^3}{7}+\frac {3\,A\,c\,b^2}{7}\right )+x^9\,\left (\frac {A\,c^3}{9}+\frac {B\,b\,c^2}{3}\right )+\frac {A\,b^3\,x^6}{6}+\frac {B\,c^3\,x^{10}}{10}+\frac {3\,b\,c\,x^8\,\left (A\,c+B\,b\right )}{8} \] Input:
int(x^2*(b*x + c*x^2)^3*(A + B*x),x)
Output:
x^7*((B*b^3)/7 + (3*A*b^2*c)/7) + x^9*((A*c^3)/9 + (B*b*c^2)/3) + (A*b^3*x ^6)/6 + (B*c^3*x^10)/10 + (3*b*c*x^8*(A*c + B*b))/8
Time = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99 \[ \int x^2 (A+B x) \left (b x+c x^2\right )^3 \, dx=\frac {x^{6} \left (252 b \,c^{3} x^{4}+280 a \,c^{3} x^{3}+840 b^{2} c^{2} x^{3}+945 a b \,c^{2} x^{2}+945 b^{3} c \,x^{2}+1080 a \,b^{2} c x +360 b^{4} x +420 a \,b^{3}\right )}{2520} \] Input:
int(x^2*(B*x+A)*(c*x^2+b*x)^3,x)
Output:
(x**6*(420*a*b**3 + 1080*a*b**2*c*x + 945*a*b*c**2*x**2 + 280*a*c**3*x**3 + 360*b**4*x + 945*b**3*c*x**2 + 840*b**2*c**2*x**3 + 252*b*c**3*x**4))/25 20