\(\int (a+b x+c x^2)^4 \, dx\) [2150]

Optimal result

Integrand size = 12, antiderivative size = 133 \[ \int \left (a+b x+c x^2\right )^4 \, dx=a^4 x+2 a^3 b x^2+\frac {2}{3} a^2 \left (3 b^2+2 a c\right ) x^3+a b \left (b^2+3 a c\right ) x^4+\frac {1}{5} \left (b^4+12 a b^2 c+6 a^2 c^2\right ) x^5+\frac {2}{3} b c \left (b^2+3 a c\right ) x^6+\frac {2}{7} c^2 \left (3 b^2+2 a c\right ) x^7+\frac {1}{2} b c^3 x^8+\frac {c^4 x^9}{9} \]

[Out]

a^4*x+2*a^3*b*x^2+2/3*a^2*(2*a*c+3*b^2)*x^3+a*b*(3*a*c+b^2)*x^4+1/5*(6*a^2*c^2+12*a*b^2*c+b^4)*x^5+2/3*b*c*(3*
a*c+b^2)*x^6+2/7*c^2*(2*a*c+3*b^2)*x^7+1/2*b*c^3*x^8+1/9*c^4*x^9

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {625} \[ \int \left (a+b x+c x^2\right )^4 \, dx=a^4 x+2 a^3 b x^2+\frac {2}{3} a^2 x^3 \left (2 a c+3 b^2\right )+\frac {1}{5} x^5 \left (6 a^2 c^2+12 a b^2 c+b^4\right )+\frac {2}{7} c^2 x^7 \left (2 a c+3 b^2\right )+\frac {2}{3} b c x^6 \left (3 a c+b^2\right )+a b x^4 \left (3 a c+b^2\right )+\frac {1}{2} b c^3 x^8+\frac {c^4 x^9}{9} \]

[In]

Int[(a + b*x + c*x^2)^4,x]

[Out]

a^4*x + 2*a^3*b*x^2 + (2*a^2*(3*b^2 + 2*a*c)*x^3)/3 + a*b*(b^2 + 3*a*c)*x^4 + ((b^4 + 12*a*b^2*c + 6*a^2*c^2)*
x^5)/5 + (2*b*c*(b^2 + 3*a*c)*x^6)/3 + (2*c^2*(3*b^2 + 2*a*c)*x^7)/7 + (b*c^3*x^8)/2 + (c^4*x^9)/9

Rule 625

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && (EqQ[a, 0] ||  !PerfectSquareQ[b^2 - 4*a*c])

Rubi steps \begin{align*} \text {integral}& = \int \left (a^4+4 a^3 b x+6 a^2 b^2 \left (1+\frac {2 a c}{3 b^2}\right ) x^2+4 a b^3 \left (1+\frac {3 a c}{b^2}\right ) x^3+b^4 \left (1+\frac {6 a c \left (2 b^2+a c\right )}{b^4}\right ) x^4+4 b^3 c \left (1+\frac {3 a c}{b^2}\right ) x^5+6 b^2 c^2 \left (1+\frac {2 a c}{3 b^2}\right ) x^6+4 b c^3 x^7+c^4 x^8\right ) \, dx \\ & = a^4 x+2 a^3 b x^2+\frac {2}{3} a^2 \left (3 b^2+2 a c\right ) x^3+a b \left (b^2+3 a c\right ) x^4+\frac {1}{5} \left (b^4+12 a b^2 c+6 a^2 c^2\right ) x^5+\frac {2}{3} b c \left (b^2+3 a c\right ) x^6+\frac {2}{7} c^2 \left (3 b^2+2 a c\right ) x^7+\frac {1}{2} b c^3 x^8+\frac {c^4 x^9}{9} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00 \[ \int \left (a+b x+c x^2\right )^4 \, dx=a^4 x+2 a^3 b x^2+\frac {2}{3} a^2 \left (3 b^2+2 a c\right ) x^3+a b \left (b^2+3 a c\right ) x^4+\frac {1}{5} \left (b^4+12 a b^2 c+6 a^2 c^2\right ) x^5+\frac {2}{3} b c \left (b^2+3 a c\right ) x^6+\frac {2}{7} c^2 \left (3 b^2+2 a c\right ) x^7+\frac {1}{2} b c^3 x^8+\frac {c^4 x^9}{9} \]

[In]

Integrate[(a + b*x + c*x^2)^4,x]

[Out]

a^4*x + 2*a^3*b*x^2 + (2*a^2*(3*b^2 + 2*a*c)*x^3)/3 + a*b*(b^2 + 3*a*c)*x^4 + ((b^4 + 12*a*b^2*c + 6*a^2*c^2)*
x^5)/5 + (2*b*c*(b^2 + 3*a*c)*x^6)/3 + (2*c^2*(3*b^2 + 2*a*c)*x^7)/7 + (b*c^3*x^8)/2 + (c^4*x^9)/9

Maple [A] (verified)

Time = 2.91 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.98

method	result	size
norman	\(\frac {c^{4} x^{9}}{9}+\frac {b \,c^{3} x^{8}}{2}+\left (\frac {4}{7} c^{3} a +\frac {6}{7} b^{2} c^{2}\right ) x^{7}+\left (2 b \,c^{2} a +\frac {2}{3} b^{3} c \right ) x^{6}+\left (\frac {6}{5} a^{2} c^{2}+\frac {12}{5} a \,b^{2} c +\frac {1}{5} b^{4}\right ) x^{5}+\left (3 a^{2} b c +a \,b^{3}\right ) x^{4}+\left (\frac {4}{3} c \,a^{3}+2 a^{2} b^{2}\right ) x^{3}+2 a^{3} b \,x^{2}+a^{4} x\)	\(131\)
gosper	\(\frac {1}{9} c^{4} x^{9}+\frac {1}{2} b \,c^{3} x^{8}+\frac {4}{7} a \,c^{3} x^{7}+\frac {6}{7} x^{7} b^{2} c^{2}+2 x^{6} b \,c^{2} a +\frac {2}{3} b^{3} c \,x^{6}+\frac {6}{5} a^{2} c^{2} x^{5}+\frac {12}{5} a \,b^{2} c \,x^{5}+\frac {1}{5} b^{4} x^{5}+3 a^{2} b c \,x^{4}+a \,b^{3} x^{4}+\frac {4}{3} c \,a^{3} x^{3}+2 a^{2} b^{2} x^{3}+2 a^{3} b \,x^{2}+a^{4} x\)	\(139\)
risch	\(\frac {1}{9} c^{4} x^{9}+\frac {1}{2} b \,c^{3} x^{8}+\frac {4}{7} a \,c^{3} x^{7}+\frac {6}{7} x^{7} b^{2} c^{2}+2 x^{6} b \,c^{2} a +\frac {2}{3} b^{3} c \,x^{6}+\frac {6}{5} a^{2} c^{2} x^{5}+\frac {12}{5} a \,b^{2} c \,x^{5}+\frac {1}{5} b^{4} x^{5}+3 a^{2} b c \,x^{4}+a \,b^{3} x^{4}+\frac {4}{3} c \,a^{3} x^{3}+2 a^{2} b^{2} x^{3}+2 a^{3} b \,x^{2}+a^{4} x\)	\(139\)
parallelrisch	\(\frac {1}{9} c^{4} x^{9}+\frac {1}{2} b \,c^{3} x^{8}+\frac {4}{7} a \,c^{3} x^{7}+\frac {6}{7} x^{7} b^{2} c^{2}+2 x^{6} b \,c^{2} a +\frac {2}{3} b^{3} c \,x^{6}+\frac {6}{5} a^{2} c^{2} x^{5}+\frac {12}{5} a \,b^{2} c \,x^{5}+\frac {1}{5} b^{4} x^{5}+3 a^{2} b c \,x^{4}+a \,b^{3} x^{4}+\frac {4}{3} c \,a^{3} x^{3}+2 a^{2} b^{2} x^{3}+2 a^{3} b \,x^{2}+a^{4} x\)	\(139\)
default	\(\frac {c^{4} x^{9}}{9}+\frac {b \,c^{3} x^{8}}{2}+\frac {\left (2 \left (2 a c +b^{2}\right ) c^{2}+4 b^{2} c^{2}\right ) x^{7}}{7}+\frac {\left (4 b \,c^{2} a +4 \left (2 a c +b^{2}\right ) b c \right ) x^{6}}{6}+\frac {\left (2 a^{2} c^{2}+8 a \,b^{2} c +\left (2 a c +b^{2}\right )^{2}\right ) x^{5}}{5}+\frac {\left (4 a^{2} b c +4 a b \left (2 a c +b^{2}\right )\right ) x^{4}}{4}+\frac {\left (2 a^{2} \left (2 a c +b^{2}\right )+4 a^{2} b^{2}\right ) x^{3}}{3}+2 a^{3} b \,x^{2}+a^{4} x\)	\(168\)

[In]

int((c*x^2+b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

1/9*c^4*x^9+1/2*b*c^3*x^8+(4/7*c^3*a+6/7*b^2*c^2)*x^7+(2*b*c^2*a+2/3*b^3*c)*x^6+(6/5*a^2*c^2+12/5*a*b^2*c+1/5*
b^4)*x^5+(3*a^2*b*c+a*b^3)*x^4+(4/3*c*a^3+2*a^2*b^2)*x^3+2*a^3*b*x^2+a^4*x

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.98 \[ \int \left (a+b x+c x^2\right )^4 \, dx=\frac {1}{9} \, c^{4} x^{9} + \frac {1}{2} \, b c^{3} x^{8} + \frac {2}{7} \, {\left (3 \, b^{2} c^{2} + 2 \, a c^{3}\right )} x^{7} + \frac {2}{3} \, {\left (b^{3} c + 3 \, a b c^{2}\right )} x^{6} + 2 \, a^{3} b x^{2} + \frac {1}{5} \, {\left (b^{4} + 12 \, a b^{2} c + 6 \, a^{2} c^{2}\right )} x^{5} + a^{4} x + {\left (a b^{3} + 3 \, a^{2} b c\right )} x^{4} + \frac {2}{3} \, {\left (3 \, a^{2} b^{2} + 2 \, a^{3} c\right )} x^{3} \]

[In]

integrate((c*x^2+b*x+a)^4,x, algorithm="fricas")

[Out]

1/9*c^4*x^9 + 1/2*b*c^3*x^8 + 2/7*(3*b^2*c^2 + 2*a*c^3)*x^7 + 2/3*(b^3*c + 3*a*b*c^2)*x^6 + 2*a^3*b*x^2 + 1/5*
(b^4 + 12*a*b^2*c + 6*a^2*c^2)*x^5 + a^4*x + (a*b^3 + 3*a^2*b*c)*x^4 + 2/3*(3*a^2*b^2 + 2*a^3*c)*x^3

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.06 \[ \int \left (a+b x+c x^2\right )^4 \, dx=a^{4} x + 2 a^{3} b x^{2} + \frac {b c^{3} x^{8}}{2} + \frac {c^{4} x^{9}}{9} + x^{7} \cdot \left (\frac {4 a c^{3}}{7} + \frac {6 b^{2} c^{2}}{7}\right ) + x^{6} \cdot \left (2 a b c^{2} + \frac {2 b^{3} c}{3}\right ) + x^{5} \cdot \left (\frac {6 a^{2} c^{2}}{5} + \frac {12 a b^{2} c}{5} + \frac {b^{4}}{5}\right ) + x^{4} \cdot \left (3 a^{2} b c + a b^{3}\right ) + x^{3} \cdot \left (\frac {4 a^{3} c}{3} + 2 a^{2} b^{2}\right ) \]

[In]

integrate((c*x**2+b*x+a)**4,x)

[Out]

a**4*x + 2*a**3*b*x**2 + b*c**3*x**8/2 + c**4*x**9/9 + x**7*(4*a*c**3/7 + 6*b**2*c**2/7) + x**6*(2*a*b*c**2 +
2*b**3*c/3) + x**5*(6*a**2*c**2/5 + 12*a*b**2*c/5 + b**4/5) + x**4*(3*a**2*b*c + a*b**3) + x**3*(4*a**3*c/3 +
2*a**2*b**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.02 \[ \int \left (a+b x+c x^2\right )^4 \, dx=\frac {1}{9} \, c^{4} x^{9} + \frac {1}{2} \, b c^{3} x^{8} + \frac {6}{7} \, b^{2} c^{2} x^{7} + \frac {2}{3} \, b^{3} c x^{6} + \frac {1}{5} \, b^{4} x^{5} + a^{4} x + \frac {2}{3} \, {\left (2 \, c x^{3} + 3 \, b x^{2}\right )} a^{3} + \frac {1}{5} \, {\left (6 \, c^{2} x^{5} + 15 \, b c x^{4} + 10 \, b^{2} x^{3}\right )} a^{2} + \frac {1}{35} \, {\left (20 \, c^{3} x^{7} + 70 \, b c^{2} x^{6} + 84 \, b^{2} c x^{5} + 35 \, b^{3} x^{4}\right )} a \]

[In]

integrate((c*x^2+b*x+a)^4,x, algorithm="maxima")

[Out]

1/9*c^4*x^9 + 1/2*b*c^3*x^8 + 6/7*b^2*c^2*x^7 + 2/3*b^3*c*x^6 + 1/5*b^4*x^5 + a^4*x + 2/3*(2*c*x^3 + 3*b*x^2)*
a^3 + 1/5*(6*c^2*x^5 + 15*b*c*x^4 + 10*b^2*x^3)*a^2 + 1/35*(20*c^3*x^7 + 70*b*c^2*x^6 + 84*b^2*c*x^5 + 35*b^3*
x^4)*a

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.04 \[ \int \left (a+b x+c x^2\right )^4 \, dx=\frac {1}{9} \, c^{4} x^{9} + \frac {1}{2} \, b c^{3} x^{8} + \frac {6}{7} \, b^{2} c^{2} x^{7} + \frac {4}{7} \, a c^{3} x^{7} + \frac {2}{3} \, b^{3} c x^{6} + 2 \, a b c^{2} x^{6} + \frac {1}{5} \, b^{4} x^{5} + \frac {12}{5} \, a b^{2} c x^{5} + \frac {6}{5} \, a^{2} c^{2} x^{5} + a b^{3} x^{4} + 3 \, a^{2} b c x^{4} + 2 \, a^{2} b^{2} x^{3} + \frac {4}{3} \, a^{3} c x^{3} + 2 \, a^{3} b x^{2} + a^{4} x \]

[In]

integrate((c*x^2+b*x+a)^4,x, algorithm="giac")

[Out]

1/9*c^4*x^9 + 1/2*b*c^3*x^8 + 6/7*b^2*c^2*x^7 + 4/7*a*c^3*x^7 + 2/3*b^3*c*x^6 + 2*a*b*c^2*x^6 + 1/5*b^4*x^5 +
12/5*a*b^2*c*x^5 + 6/5*a^2*c^2*x^5 + a*b^3*x^4 + 3*a^2*b*c*x^4 + 2*a^2*b^2*x^3 + 4/3*a^3*c*x^3 + 2*a^3*b*x^2 +
a^4*x

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.93 \[ \int \left (a+b x+c x^2\right )^4 \, dx=x^5\,\left (\frac {6\,a^2\,c^2}{5}+\frac {12\,a\,b^2\,c}{5}+\frac {b^4}{5}\right )+a^4\,x+\frac {c^4\,x^9}{9}+x^3\,\left (\frac {4\,c\,a^3}{3}+2\,a^2\,b^2\right )+x^7\,\left (\frac {6\,b^2\,c^2}{7}+\frac {4\,a\,c^3}{7}\right )+2\,a^3\,b\,x^2+\frac {b\,c^3\,x^8}{2}+a\,b\,x^4\,\left (b^2+3\,a\,c\right )+\frac {2\,b\,c\,x^6\,\left (b^2+3\,a\,c\right )}{3} \]

[In]

int((a + b*x + c*x^2)^4,x)

[Out]

x^5*(b^4/5 + (6*a^2*c^2)/5 + (12*a*b^2*c)/5) + a^4*x + (c^4*x^9)/9 + x^3*((4*a^3*c)/3 + 2*a^2*b^2) + x^7*((4*a
*c^3)/7 + (6*b^2*c^2)/7) + 2*a^3*b*x^2 + (b*c^3*x^8)/2 + a*b*x^4*(3*a*c + b^2) + (2*b*c*x^6*(3*a*c + b^2))/3