Integrand size = 14, antiderivative size = 86 \[ \int x \left (a+b x+c x^2\right )^3 \, dx=\frac {a^3 x^2}{2}+a^2 b x^3+\frac {3}{4} a \left (b^2+a c\right ) x^4+\frac {1}{5} b \left (b^2+6 a c\right ) x^5+\frac {1}{2} c \left (b^2+a c\right ) x^6+\frac {3}{7} b c^2 x^7+\frac {c^3 x^8}{8} \] Output:
1/2*a^3*x^2+a^2*b*x^3+3/4*a*(a*c+b^2)*x^4+1/5*b*(6*a*c+b^2)*x^5+1/2*c*(a*c +b^2)*x^6+3/7*b*c^2*x^7+1/8*c^3*x^8
Time = 0.01 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int x \left (a+b x+c x^2\right )^3 \, dx=\frac {a^3 x^2}{2}+a^2 b x^3+\frac {3}{4} a \left (b^2+a c\right ) x^4+\frac {1}{5} b \left (b^2+6 a c\right ) x^5+\frac {1}{2} c \left (b^2+a c\right ) x^6+\frac {3}{7} b c^2 x^7+\frac {c^3 x^8}{8} \] Input:
Integrate[x*(a + b*x + c*x^2)^3,x]
Output:
(a^3*x^2)/2 + a^2*b*x^3 + (3*a*(b^2 + a*c)*x^4)/4 + (b*(b^2 + 6*a*c)*x^5)/ 5 + (c*(b^2 + a*c)*x^6)/2 + (3*b*c^2*x^7)/7 + (c^3*x^8)/8
Time = 0.38 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1140, 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.
\(\displaystyle \int x \left (a+b x+c x^2\right )^3 \, dx\) |
\(\Big \downarrow \) 1140 |
\(\displaystyle \int \left (a^3 x+3 a^2 b x^2+3 c x^5 \left (a c+b^2\right )+b x^4 \left (6 a c+b^2\right )+3 a x^3 \left (a c+b^2\right )+3 b c^2 x^6+c^3 x^7\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 x^2}{2}+a^2 b x^3+\frac {1}{2} c x^6 \left (a c+b^2\right )+\frac {1}{5} b x^5 \left (6 a c+b^2\right )+\frac {3}{4} a x^4 \left (a c+b^2\right )+\frac {3}{7} b c^2 x^7+\frac {c^3 x^8}{8}\) |
Input:
Int[x*(a + b*x + c*x^2)^3,x]
Output:
(a^3*x^2)/2 + a^2*b*x^3 + (3*a*(b^2 + a*c)*x^4)/4 + (b*(b^2 + 6*a*c)*x^5)/ 5 + (c*(b^2 + a*c)*x^6)/2 + (3*b*c^2*x^7)/7 + (c^3*x^8)/8
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x _Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0]
Time = 0.60 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.98
method | result | size |
norman | \(\frac {c^{3} x^{8}}{8}+\frac {3 b \,c^{2} x^{7}}{7}+\left (\frac {1}{2} a \,c^{2}+\frac {1}{2} b^{2} c \right ) x^{6}+\left (\frac {6}{5} a b c +\frac {1}{5} b^{3}\right ) x^{5}+\left (\frac {3}{4} a^{2} c +\frac {3}{4} a \,b^{2}\right ) x^{4}+a^{2} b \,x^{3}+\frac {a^{3} x^{2}}{2}\) | \(84\) |
gosper | \(\frac {1}{8} c^{3} x^{8}+\frac {3}{7} b \,c^{2} x^{7}+\frac {1}{2} x^{6} a \,c^{2}+\frac {1}{2} b^{2} c \,x^{6}+\frac {6}{5} a b c \,x^{5}+\frac {1}{5} b^{3} x^{5}+\frac {3}{4} a^{2} c \,x^{4}+\frac {3}{4} a \,b^{2} x^{4}+a^{2} b \,x^{3}+\frac {1}{2} a^{3} x^{2}\) | \(87\) |
risch | \(\frac {1}{8} c^{3} x^{8}+\frac {3}{7} b \,c^{2} x^{7}+\frac {1}{2} x^{6} a \,c^{2}+\frac {1}{2} b^{2} c \,x^{6}+\frac {6}{5} a b c \,x^{5}+\frac {1}{5} b^{3} x^{5}+\frac {3}{4} a^{2} c \,x^{4}+\frac {3}{4} a \,b^{2} x^{4}+a^{2} b \,x^{3}+\frac {1}{2} a^{3} x^{2}\) | \(87\) |
parallelrisch | \(\frac {1}{8} c^{3} x^{8}+\frac {3}{7} b \,c^{2} x^{7}+\frac {1}{2} x^{6} a \,c^{2}+\frac {1}{2} b^{2} c \,x^{6}+\frac {6}{5} a b c \,x^{5}+\frac {1}{5} b^{3} x^{5}+\frac {3}{4} a^{2} c \,x^{4}+\frac {3}{4} a \,b^{2} x^{4}+a^{2} b \,x^{3}+\frac {1}{2} a^{3} x^{2}\) | \(87\) |
orering | \(\frac {x^{2} \left (35 c^{3} x^{6}+120 b \,c^{2} x^{5}+140 a \,c^{2} x^{4}+140 b^{2} c \,x^{4}+336 a b c \,x^{3}+56 b^{3} x^{3}+210 a^{2} c \,x^{2}+210 a \,b^{2} x^{2}+280 a^{2} b x +140 a^{3}\right )}{280}\) | \(88\) |
default | \(\frac {c^{3} x^{8}}{8}+\frac {3 b \,c^{2} x^{7}}{7}+\frac {\left (a \,c^{2}+2 b^{2} c +c \left (2 a c +b^{2}\right )\right ) x^{6}}{6}+\frac {\left (4 a b c +b \left (2 a c +b^{2}\right )\right ) x^{5}}{5}+\frac {\left (a \left (2 a c +b^{2}\right )+2 a \,b^{2}+a^{2} c \right ) x^{4}}{4}+a^{2} b \,x^{3}+\frac {a^{3} x^{2}}{2}\) | \(110\) |
Input:
int(x*(c*x^2+b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/8*c^3*x^8+3/7*b*c^2*x^7+(1/2*a*c^2+1/2*b^2*c)*x^6+(6/5*a*b*c+1/5*b^3)*x^ 5+(3/4*a^2*c+3/4*a*b^2)*x^4+a^2*b*x^3+1/2*a^3*x^2
Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int x \left (a+b x+c x^2\right )^3 \, dx=\frac {1}{8} \, c^{3} x^{8} + \frac {3}{7} \, b c^{2} x^{7} + \frac {1}{2} \, {\left (b^{2} c + a c^{2}\right )} x^{6} + a^{2} b x^{3} + \frac {1}{5} \, {\left (b^{3} + 6 \, a b c\right )} x^{5} + \frac {1}{2} \, a^{3} x^{2} + \frac {3}{4} \, {\left (a b^{2} + a^{2} c\right )} x^{4} \] Input:
integrate(x*(c*x^2+b*x+a)^3,x, algorithm="fricas")
Output:
1/8*c^3*x^8 + 3/7*b*c^2*x^7 + 1/2*(b^2*c + a*c^2)*x^6 + a^2*b*x^3 + 1/5*(b ^3 + 6*a*b*c)*x^5 + 1/2*a^3*x^2 + 3/4*(a*b^2 + a^2*c)*x^4
Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.05 \[ \int x \left (a+b x+c x^2\right )^3 \, dx=\frac {a^{3} x^{2}}{2} + a^{2} b x^{3} + \frac {3 b c^{2} x^{7}}{7} + \frac {c^{3} x^{8}}{8} + x^{6} \left (\frac {a c^{2}}{2} + \frac {b^{2} c}{2}\right ) + x^{5} \cdot \left (\frac {6 a b c}{5} + \frac {b^{3}}{5}\right ) + x^{4} \cdot \left (\frac {3 a^{2} c}{4} + \frac {3 a b^{2}}{4}\right ) \] Input:
integrate(x*(c*x**2+b*x+a)**3,x)
Output:
a**3*x**2/2 + a**2*b*x**3 + 3*b*c**2*x**7/7 + c**3*x**8/8 + x**6*(a*c**2/2 + b**2*c/2) + x**5*(6*a*b*c/5 + b**3/5) + x**4*(3*a**2*c/4 + 3*a*b**2/4)
Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int x \left (a+b x+c x^2\right )^3 \, dx=\frac {1}{8} \, c^{3} x^{8} + \frac {3}{7} \, b c^{2} x^{7} + \frac {1}{2} \, {\left (b^{2} c + a c^{2}\right )} x^{6} + a^{2} b x^{3} + \frac {1}{5} \, {\left (b^{3} + 6 \, a b c\right )} x^{5} + \frac {1}{2} \, a^{3} x^{2} + \frac {3}{4} \, {\left (a b^{2} + a^{2} c\right )} x^{4} \] Input:
integrate(x*(c*x^2+b*x+a)^3,x, algorithm="maxima")
Output:
1/8*c^3*x^8 + 3/7*b*c^2*x^7 + 1/2*(b^2*c + a*c^2)*x^6 + a^2*b*x^3 + 1/5*(b ^3 + 6*a*b*c)*x^5 + 1/2*a^3*x^2 + 3/4*(a*b^2 + a^2*c)*x^4
Time = 0.21 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int x \left (a+b x+c x^2\right )^3 \, dx=\frac {1}{8} \, c^{3} x^{8} + \frac {3}{7} \, b c^{2} x^{7} + \frac {1}{2} \, b^{2} c x^{6} + \frac {1}{2} \, a c^{2} x^{6} + \frac {1}{5} \, b^{3} x^{5} + \frac {6}{5} \, a b c x^{5} + \frac {3}{4} \, a b^{2} x^{4} + \frac {3}{4} \, a^{2} c x^{4} + a^{2} b x^{3} + \frac {1}{2} \, a^{3} x^{2} \] Input:
integrate(x*(c*x^2+b*x+a)^3,x, algorithm="giac")
Output:
1/8*c^3*x^8 + 3/7*b*c^2*x^7 + 1/2*b^2*c*x^6 + 1/2*a*c^2*x^6 + 1/5*b^3*x^5 + 6/5*a*b*c*x^5 + 3/4*a*b^2*x^4 + 3/4*a^2*c*x^4 + a^2*b*x^3 + 1/2*a^3*x^2
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.87 \[ \int x \left (a+b x+c x^2\right )^3 \, dx=x^5\,\left (\frac {b^3}{5}+\frac {6\,a\,c\,b}{5}\right )+\frac {a^3\,x^2}{2}+\frac {c^3\,x^8}{8}+a^2\,b\,x^3+\frac {3\,b\,c^2\,x^7}{7}+\frac {3\,a\,x^4\,\left (b^2+a\,c\right )}{4}+\frac {c\,x^6\,\left (b^2+a\,c\right )}{2} \] Input:
int(x*(a + b*x + c*x^2)^3,x)
Output:
x^5*(b^3/5 + (6*a*b*c)/5) + (a^3*x^2)/2 + (c^3*x^8)/8 + a^2*b*x^3 + (3*b*c ^2*x^7)/7 + (3*a*x^4*(a*c + b^2))/4 + (c*x^6*(a*c + b^2))/2
Time = 0.35 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.01 \[ \int x \left (a+b x+c x^2\right )^3 \, dx=\frac {x^{2} \left (35 c^{3} x^{6}+120 b \,c^{2} x^{5}+140 a \,c^{2} x^{4}+140 b^{2} c \,x^{4}+336 a b c \,x^{3}+56 b^{3} x^{3}+210 a^{2} c \,x^{2}+210 a \,b^{2} x^{2}+280 a^{2} b x +140 a^{3}\right )}{280} \] Input:
int(x*(c*x^2+b*x+a)^3,x)
Output:
(x**2*(140*a**3 + 280*a**2*b*x + 210*a**2*c*x**2 + 210*a*b**2*x**2 + 336*a *b*c*x**3 + 140*a*c**2*x**4 + 56*b**3*x**3 + 140*b**2*c*x**4 + 120*b*c**2* x**5 + 35*c**3*x**6))/280