Integrand size = 14, antiderivative size = 85 \[ \int \left (a+b \sqrt {x}+c x\right )^3 \, dx=a^3 x+2 a^2 b x^{3/2}+\frac {3}{2} a \left (b^2+a c\right ) x^2+\frac {2}{5} b \left (b^2+6 a c\right ) x^{5/2}+c \left (b^2+a c\right ) x^3+\frac {6}{7} b c^2 x^{7/2}+\frac {c^3 x^4}{4} \] Output:
a^3*x+2*a^2*b*x^(3/2)+3/2*a*(a*c+b^2)*x^2+2/5*b*(6*a*c+b^2)*x^(5/2)+c*(a*c +b^2)*x^3+6/7*b*c^2*x^(7/2)+1/4*c^3*x^4
Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.14 \[ \int \left (a+b \sqrt {x}+c x\right )^3 \, dx=\frac {1}{140} \left (140 a^3 x+280 a^2 b x^{3/2}+210 a b^2 x^2+210 a^2 c x^2+56 b^3 x^{5/2}+336 a b c x^{5/2}+140 b^2 c x^3+140 a c^2 x^3+120 b c^2 x^{7/2}+35 c^3 x^4\right ) \] Input:
Integrate[(a + b*Sqrt[x] + c*x)^3,x]
Output:
(140*a^3*x + 280*a^2*b*x^(3/2) + 210*a*b^2*x^2 + 210*a^2*c*x^2 + 56*b^3*x^ (5/2) + 336*a*b*c*x^(5/2) + 140*b^2*c*x^3 + 140*a*c^2*x^3 + 120*b*c^2*x^(7 /2) + 35*c^3*x^4)/140
Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1680, 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.
\(\displaystyle \int \left (a+b \sqrt {x}+c x\right )^3 \, dx\) |
\(\Big \downarrow \) 1680 |
\(\displaystyle 2 \int \sqrt {x} \left (a+c x+b \sqrt {x}\right )^3d\sqrt {x}\) |
\(\Big \downarrow \) 1140 |
\(\displaystyle 2 \int \left (c^3 x^{7/2}+3 b c^2 x^3+3 c \left (b^2+a c\right ) x^{5/2}+b \left (b^2+6 a c\right ) x^2+3 a \left (b^2+a c\right ) x^{3/2}+3 a^2 b x+a^3 \sqrt {x}\right )d\sqrt {x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {a^3 x}{2}+a^2 b x^{3/2}+\frac {1}{5} b x^{5/2} \left (6 a c+b^2\right )+\frac {1}{2} c x^3 \left (a c+b^2\right )+\frac {3}{4} a x^2 \left (a c+b^2\right )+\frac {3}{7} b c^2 x^{7/2}+\frac {c^3 x^4}{8}\right )\) |
Input:
Int[(a + b*Sqrt[x] + c*x)^3,x]
Output:
2*((a^3*x)/2 + a^2*b*x^(3/2) + (3*a*(b^2 + a*c)*x^2)/4 + (b*(b^2 + 6*a*c)* x^(5/2))/5 + (c*(b^2 + a*c)*x^3)/2 + (3*b*c^2*x^(7/2))/7 + (c^3*x^4)/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]
Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Simp[k Subst[Int[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k* n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] && Fr actionQ[n]
Time = 0.72 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.79
method | result | size |
default | \(\frac {2 b^{3} x^{\frac {5}{2}}}{5}+3 b^{2} \left (\frac {1}{3} c \,x^{3}+\frac {1}{2} a \,x^{2}\right )+3 b \left (\frac {2 c^{2} x^{\frac {7}{2}}}{7}+\frac {4 a c \,x^{\frac {5}{2}}}{5}+\frac {2 a^{2} x^{\frac {3}{2}}}{3}\right )+\frac {\left (c x +a \right )^{4}}{4 c}\) | \(67\) |
derivativedivides | \(\frac {c^{3} x^{4}}{4}+\frac {6 b \,c^{2} x^{\frac {7}{2}}}{7}+\frac {\left (a \,c^{2}+2 b^{2} c +c \left (2 a c +b^{2}\right )\right ) x^{3}}{3}+\frac {2 \left (4 a b c +b \left (2 a c +b^{2}\right )\right ) x^{\frac {5}{2}}}{5}+\frac {\left (a \left (2 a c +b^{2}\right )+2 b^{2} a +a^{2} c \right ) x^{2}}{2}+2 a^{2} b \,x^{\frac {3}{2}}+a^{3} x\) | \(108\) |
trager | \(\frac {\left (c^{3} x^{3}+4 a \,c^{2} x^{2}+4 b^{2} c \,x^{2}+c^{3} x^{2}+6 a^{2} c x +6 a \,b^{2} x +4 a \,c^{2} x +4 b^{2} c x +c^{3} x +4 a^{3}+6 a^{2} c +6 b^{2} a +4 a \,c^{2}+4 b^{2} c +c^{3}\right ) \left (x -1\right )}{4}+\frac {2 b \,x^{\frac {3}{2}} \left (15 c^{2} x^{2}+42 a c x +7 b^{2} x +35 a^{2}\right )}{35}\) | \(136\) |
orering | \(-\frac {\left (65 x^{6} c^{7}+242 x^{5} c^{6} a -143 b^{2} c^{5} x^{5}+219 x^{4} c^{5} a^{2}-222 a \,b^{2} c^{4} x^{4}+84 b^{4} c^{3} x^{4}-308 x^{3} c^{4} a^{3}+294 a^{2} b^{2} c^{3} x^{3}-98 a \,b^{4} c^{2} x^{3}+448 x \,c^{2} a^{5}-154 a^{4} b^{2} c x -42 a^{3} b^{4} x +98 c \,a^{6}-42 a^{5} b^{2}\right ) \left (a +b \sqrt {x}+c x \right )^{3}}{140 \left (-c^{3} x^{3}-a \,c^{2} x^{2}+b^{2} c \,x^{2}+a^{2} c x -a \,b^{2} x +a^{3}\right ) c^{2} \left (c^{2} x^{2}+2 a c x -b^{2} x +a^{2}\right )}+\frac {\left (15 x^{6} c^{7}+66 x^{5} c^{6} a -39 b^{2} c^{5} x^{5}+73 x^{4} c^{5} a^{2}-74 a \,b^{2} c^{4} x^{4}+28 b^{4} c^{3} x^{4}-132 x^{3} c^{4} a^{3}+126 a^{2} b^{2} c^{3} x^{3}-42 a \,b^{4} c^{2} x^{3}+448 x \,c^{2} a^{5}-154 a^{4} b^{2} c x -42 a^{3} b^{4} x +294 c \,a^{6}-126 a^{5} b^{2}\right ) x \left (a +b \sqrt {x}+c x \right )^{2} \left (\frac {b}{2 \sqrt {x}}+c \right )}{70 c^{2} \left (c^{2} x^{2}+2 a c x -b^{2} x +a^{2}\right )^{2} \left (-c x +a \right )}\) | \(429\) |
Input:
int((a+b*x^(1/2)+c*x)^3,x,method=_RETURNVERBOSE)
Output:
2/5*b^3*x^(5/2)+3*b^2*(1/3*c*x^3+1/2*a*x^2)+3*b*(2/7*c^2*x^(7/2)+4/5*a*c*x ^(5/2)+2/3*a^2*x^(3/2))+1/4*(c*x+a)^4/c
Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.95 \[ \int \left (a+b \sqrt {x}+c x\right )^3 \, dx=\frac {1}{4} \, c^{3} x^{4} + a^{3} x + {\left (b^{2} c + a c^{2}\right )} x^{3} + \frac {3}{2} \, {\left (a b^{2} + a^{2} c\right )} x^{2} + \frac {2}{35} \, {\left (15 \, b c^{2} x^{3} + 35 \, a^{2} b x + 7 \, {\left (b^{3} + 6 \, a b c\right )} x^{2}\right )} \sqrt {x} \] Input:
integrate((a+b*x^(1/2)+c*x)^3,x, algorithm="fricas")
Output:
1/4*c^3*x^4 + a^3*x + (b^2*c + a*c^2)*x^3 + 3/2*(a*b^2 + a^2*c)*x^2 + 2/35 *(15*b*c^2*x^3 + 35*a^2*b*x + 7*(b^3 + 6*a*b*c)*x^2)*sqrt(x)
Time = 0.12 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.22 \[ \int \left (a+b \sqrt {x}+c x\right )^3 \, dx=a^{3} x + 2 a^{2} b x^{\frac {3}{2}} + \frac {3 a^{2} c x^{2}}{2} + \frac {3 a b^{2} x^{2}}{2} + \frac {12 a b c x^{\frac {5}{2}}}{5} + a c^{2} x^{3} + \frac {2 b^{3} x^{\frac {5}{2}}}{5} + b^{2} c x^{3} + \frac {6 b c^{2} x^{\frac {7}{2}}}{7} + \frac {c^{3} x^{4}}{4} \] Input:
integrate((a+b*x**(1/2)+c*x)**3,x)
Output:
a**3*x + 2*a**2*b*x**(3/2) + 3*a**2*c*x**2/2 + 3*a*b**2*x**2/2 + 12*a*b*c* x**(5/2)/5 + a*c**2*x**3 + 2*b**3*x**(5/2)/5 + b**2*c*x**3 + 6*b*c**2*x**( 7/2)/7 + c**3*x**4/4
Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.99 \[ \int \left (a+b \sqrt {x}+c x\right )^3 \, dx=\frac {1}{4} \, c^{3} x^{4} + \frac {6}{7} \, b c^{2} x^{\frac {7}{2}} + b^{2} c x^{3} + \frac {2}{5} \, b^{3} x^{\frac {5}{2}} + a^{3} x + \frac {1}{2} \, {\left (3 \, c x^{2} + 4 \, b x^{\frac {3}{2}}\right )} a^{2} + \frac {1}{10} \, {\left (10 \, c^{2} x^{3} + 24 \, b c x^{\frac {5}{2}} + 15 \, b^{2} x^{2}\right )} a \] Input:
integrate((a+b*x^(1/2)+c*x)^3,x, algorithm="maxima")
Output:
1/4*c^3*x^4 + 6/7*b*c^2*x^(7/2) + b^2*c*x^3 + 2/5*b^3*x^(5/2) + a^3*x + 1/ 2*(3*c*x^2 + 4*b*x^(3/2))*a^2 + 1/10*(10*c^2*x^3 + 24*b*c*x^(5/2) + 15*b^2 *x^2)*a
Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.96 \[ \int \left (a+b \sqrt {x}+c x\right )^3 \, dx=\frac {1}{4} \, c^{3} x^{4} + \frac {6}{7} \, b c^{2} x^{\frac {7}{2}} + b^{2} c x^{3} + a c^{2} x^{3} + \frac {2}{5} \, b^{3} x^{\frac {5}{2}} + \frac {12}{5} \, a b c x^{\frac {5}{2}} + \frac {3}{2} \, a b^{2} x^{2} + \frac {3}{2} \, a^{2} c x^{2} + 2 \, a^{2} b x^{\frac {3}{2}} + a^{3} x \] Input:
integrate((a+b*x^(1/2)+c*x)^3,x, algorithm="giac")
Output:
1/4*c^3*x^4 + 6/7*b*c^2*x^(7/2) + b^2*c*x^3 + a*c^2*x^3 + 2/5*b^3*x^(5/2) + 12/5*a*b*c*x^(5/2) + 3/2*a*b^2*x^2 + 3/2*a^2*c*x^2 + 2*a^2*b*x^(3/2) + a ^3*x
Time = 19.60 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.85 \[ \int \left (a+b \sqrt {x}+c x\right )^3 \, dx=x^{5/2}\,\left (\frac {2\,b^3}{5}+\frac {12\,a\,c\,b}{5}\right )+a^3\,x+\frac {c^3\,x^4}{4}+2\,a^2\,b\,x^{3/2}+\frac {6\,b\,c^2\,x^{7/2}}{7}+\frac {3\,a\,x^2\,\left (b^2+a\,c\right )}{2}+c\,x^3\,\left (b^2+a\,c\right ) \] Input:
int((a + c*x + b*x^(1/2))^3,x)
Output:
x^(5/2)*((2*b^3)/5 + (12*a*b*c)/5) + a^3*x + (c^3*x^4)/4 + 2*a^2*b*x^(3/2) + (6*b*c^2*x^(7/2))/7 + (3*a*x^2*(a*c + b^2))/2 + c*x^3*(a*c + b^2)
Time = 0.20 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.99 \[ \int \left (a+b \sqrt {x}+c x\right )^3 \, dx=\frac {x \left (280 \sqrt {x}\, a^{2} b +336 \sqrt {x}\, a b c x +56 \sqrt {x}\, b^{3} x +120 \sqrt {x}\, b \,c^{2} x^{2}+140 a^{3}+210 a^{2} c x +210 a \,b^{2} x +140 a \,c^{2} x^{2}+140 b^{2} c \,x^{2}+35 c^{3} x^{3}\right )}{140} \] Input:
int((a+b*x^(1/2)+c*x)^3,x)
Output:
(x*(280*sqrt(x)*a**2*b + 336*sqrt(x)*a*b*c*x + 56*sqrt(x)*b**3*x + 120*sqr t(x)*b*c**2*x**2 + 140*a**3 + 210*a**2*c*x + 210*a*b**2*x + 140*a*c**2*x** 2 + 140*b**2*c*x**2 + 35*c**3*x**3))/140