Integrand size = 24, antiderivative size = 65 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {(b d-a e)^2 (a+b x)^3}{3 b^3}+\frac {e (b d-a e) (a+b x)^4}{2 b^3}+\frac {e^2 (a+b x)^5}{5 b^3} \] Output:
1/3*(-a*e+b*d)^2*(b*x+a)^3/b^3+1/2*e*(-a*e+b*d)*(b*x+a)^4/b^3+1/5*e^2*(b*x +a)^5/b^3
Time = 0.01 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.22 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=a^2 d^2 x+a d (b d+a e) x^2+\frac {1}{3} \left (b^2 d^2+4 a b d e+a^2 e^2\right ) x^3+\frac {1}{2} b e (b d+a e) x^4+\frac {1}{5} b^2 e^2 x^5 \] Input:
Integrate[(d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2),x]
Output:
a^2*d^2*x + a*d*(b*d + a*e)*x^2 + ((b^2*d^2 + 4*a*b*d*e + a^2*e^2)*x^3)/3 + (b*e*(b*d + a*e)*x^4)/2 + (b^2*e^2*x^5)/5
Time = 0.39 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1098, 27, 49, 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.
\(\displaystyle \int \left (a^2+2 a b x+b^2 x^2\right ) (d+e x)^2 \, dx\) |
\(\Big \downarrow \) 1098 |
\(\displaystyle \frac {\int b^2 (a+b x)^2 (d+e x)^2dx}{b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int (a+b x)^2 (d+e x)^2dx\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \int \left (\frac {2 e (a+b x)^3 (b d-a e)}{b^2}+\frac {(a+b x)^2 (b d-a e)^2}{b^2}+\frac {e^2 (a+b x)^4}{b^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e (a+b x)^4 (b d-a e)}{2 b^3}+\frac {(a+b x)^3 (b d-a e)^2}{3 b^3}+\frac {e^2 (a+b x)^5}{5 b^3}\) |
Input:
Int[(d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2),x]
Output:
((b*d - a*e)^2*(a + b*x)^3)/(3*b^3) + (e*(b*d - a*e)*(a + b*x)^4)/(2*b^3) + (e^2*(a + b*x)^5)/(5*b^3)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[ {a, b, c, d, e, m}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.55 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.29
method | result | size |
norman | \(\frac {b^{2} e^{2} x^{5}}{5}+\left (\frac {1}{2} a b \,e^{2}+\frac {1}{2} d e \,b^{2}\right ) x^{4}+\left (\frac {1}{3} e^{2} a^{2}+\frac {4}{3} a b d e +\frac {1}{3} b^{2} d^{2}\right ) x^{3}+\left (a^{2} d e +a b \,d^{2}\right ) x^{2}+a^{2} d^{2} x\) | \(84\) |
default | \(\frac {b^{2} e^{2} x^{5}}{5}+\frac {\left (2 a b \,e^{2}+2 d e \,b^{2}\right ) x^{4}}{4}+\frac {\left (e^{2} a^{2}+4 a b d e +b^{2} d^{2}\right ) x^{3}}{3}+\frac {\left (2 a^{2} d e +2 a b \,d^{2}\right ) x^{2}}{2}+a^{2} d^{2} x\) | \(87\) |
risch | \(\frac {1}{5} b^{2} e^{2} x^{5}+\frac {1}{2} x^{4} a b \,e^{2}+\frac {1}{2} x^{4} d e \,b^{2}+\frac {1}{3} x^{3} e^{2} a^{2}+\frac {4}{3} x^{3} a b d e +\frac {1}{3} x^{3} b^{2} d^{2}+d e \,a^{2} x^{2}+a b \,d^{2} x^{2}+a^{2} d^{2} x\) | \(90\) |
parallelrisch | \(\frac {1}{5} b^{2} e^{2} x^{5}+\frac {1}{2} x^{4} a b \,e^{2}+\frac {1}{2} x^{4} d e \,b^{2}+\frac {1}{3} x^{3} e^{2} a^{2}+\frac {4}{3} x^{3} a b d e +\frac {1}{3} x^{3} b^{2} d^{2}+d e \,a^{2} x^{2}+a b \,d^{2} x^{2}+a^{2} d^{2} x\) | \(90\) |
gosper | \(\frac {x \left (6 b^{2} e^{2} x^{4}+15 x^{3} a b \,e^{2}+15 x^{3} d e \,b^{2}+10 x^{2} e^{2} a^{2}+40 x^{2} a b d e +10 b^{2} d^{2} x^{2}+30 a^{2} d e x +30 a b \,d^{2} x +30 a^{2} d^{2}\right )}{30}\) | \(91\) |
orering | \(\frac {x \left (6 b^{2} e^{2} x^{4}+15 x^{3} a b \,e^{2}+15 x^{3} d e \,b^{2}+10 x^{2} e^{2} a^{2}+40 x^{2} a b d e +10 b^{2} d^{2} x^{2}+30 a^{2} d e x +30 a b \,d^{2} x +30 a^{2} d^{2}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )}{30 \left (b x +a \right )^{2}}\) | \(114\) |
Input:
int((e*x+d)^2*(b^2*x^2+2*a*b*x+a^2),x,method=_RETURNVERBOSE)
Output:
1/5*b^2*e^2*x^5+(1/2*a*b*e^2+1/2*d*e*b^2)*x^4+(1/3*e^2*a^2+4/3*a*b*d*e+1/3 *b^2*d^2)*x^3+(a^2*d*e+a*b*d^2)*x^2+a^2*d^2*x
Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.25 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{5} \, b^{2} e^{2} x^{5} + a^{2} d^{2} x + \frac {1}{2} \, {\left (b^{2} d e + a b e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (b^{2} d^{2} + 4 \, a b d e + a^{2} e^{2}\right )} x^{3} + {\left (a b d^{2} + a^{2} d e\right )} x^{2} \] Input:
integrate((e*x+d)^2*(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")
Output:
1/5*b^2*e^2*x^5 + a^2*d^2*x + 1/2*(b^2*d*e + a*b*e^2)*x^4 + 1/3*(b^2*d^2 + 4*a*b*d*e + a^2*e^2)*x^3 + (a*b*d^2 + a^2*d*e)*x^2
Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.34 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=a^{2} d^{2} x + \frac {b^{2} e^{2} x^{5}}{5} + x^{4} \left (\frac {a b e^{2}}{2} + \frac {b^{2} d e}{2}\right ) + x^{3} \left (\frac {a^{2} e^{2}}{3} + \frac {4 a b d e}{3} + \frac {b^{2} d^{2}}{3}\right ) + x^{2} \left (a^{2} d e + a b d^{2}\right ) \] Input:
integrate((e*x+d)**2*(b**2*x**2+2*a*b*x+a**2),x)
Output:
a**2*d**2*x + b**2*e**2*x**5/5 + x**4*(a*b*e**2/2 + b**2*d*e/2) + x**3*(a* *2*e**2/3 + 4*a*b*d*e/3 + b**2*d**2/3) + x**2*(a**2*d*e + a*b*d**2)
Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.25 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{5} \, b^{2} e^{2} x^{5} + a^{2} d^{2} x + \frac {1}{2} \, {\left (b^{2} d e + a b e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (b^{2} d^{2} + 4 \, a b d e + a^{2} e^{2}\right )} x^{3} + {\left (a b d^{2} + a^{2} d e\right )} x^{2} \] Input:
integrate((e*x+d)^2*(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")
Output:
1/5*b^2*e^2*x^5 + a^2*d^2*x + 1/2*(b^2*d*e + a*b*e^2)*x^4 + 1/3*(b^2*d^2 + 4*a*b*d*e + a^2*e^2)*x^3 + (a*b*d^2 + a^2*d*e)*x^2
Time = 0.21 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.37 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{5} \, b^{2} e^{2} x^{5} + \frac {1}{2} \, b^{2} d e x^{4} + \frac {1}{2} \, a b e^{2} x^{4} + \frac {1}{3} \, b^{2} d^{2} x^{3} + \frac {4}{3} \, a b d e x^{3} + \frac {1}{3} \, a^{2} e^{2} x^{3} + a b d^{2} x^{2} + a^{2} d e x^{2} + a^{2} d^{2} x \] Input:
integrate((e*x+d)^2*(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")
Output:
1/5*b^2*e^2*x^5 + 1/2*b^2*d*e*x^4 + 1/2*a*b*e^2*x^4 + 1/3*b^2*d^2*x^3 + 4/ 3*a*b*d*e*x^3 + 1/3*a^2*e^2*x^3 + a*b*d^2*x^2 + a^2*d*e*x^2 + a^2*d^2*x
Time = 9.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.14 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=x^3\,\left (\frac {a^2\,e^2}{3}+\frac {4\,a\,b\,d\,e}{3}+\frac {b^2\,d^2}{3}\right )+a^2\,d^2\,x+\frac {b^2\,e^2\,x^5}{5}+a\,d\,x^2\,\left (a\,e+b\,d\right )+\frac {b\,e\,x^4\,\left (a\,e+b\,d\right )}{2} \] Input:
int((d + e*x)^2*(a^2 + b^2*x^2 + 2*a*b*x),x)
Output:
x^3*((a^2*e^2)/3 + (b^2*d^2)/3 + (4*a*b*d*e)/3) + a^2*d^2*x + (b^2*e^2*x^5 )/5 + a*d*x^2*(a*e + b*d) + (b*e*x^4*(a*e + b*d))/2
Time = 0.26 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.38 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {x \left (6 b^{2} e^{2} x^{4}+15 a b \,e^{2} x^{3}+15 b^{2} d e \,x^{3}+10 a^{2} e^{2} x^{2}+40 a b d e \,x^{2}+10 b^{2} d^{2} x^{2}+30 a^{2} d e x +30 a b \,d^{2} x +30 a^{2} d^{2}\right )}{30} \] Input:
int((e*x+d)^2*(b^2*x^2+2*a*b*x+a^2),x)
Output:
(x*(30*a**2*d**2 + 30*a**2*d*e*x + 10*a**2*e**2*x**2 + 30*a*b*d**2*x + 40* a*b*d*e*x**2 + 15*a*b*e**2*x**3 + 10*b**2*d**2*x**2 + 15*b**2*d*e*x**3 + 6 *b**2*e**2*x**4))/30