\(\int (d+e x)^2 (a^2+2 a b x+b^2 x^2) \, dx\) [1453]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 24, antiderivative size = 65 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {(b d-a e)^2 (a+b x)^3}{3 b^3}+\frac {e (b d-a e) (a+b x)^4}{2 b^3}+\frac {e^2 (a+b x)^5}{5 b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {27, 45} \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {e (a+b x)^4 (b d-a e)}{2 b^3}+\frac {(a+b x)^3 (b d-a e)^2}{3 b^3}+\frac {e^2 (a+b x)^5}{5 b^3} \]

[In]

Int[(d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

((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)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int (a+b x)^2 (d+e x)^2 \, dx \\ & = \int \left (\frac {(b d-a e)^2 (a+b x)^2}{b^2}+\frac {2 e (b d-a e) (a+b x)^3}{b^2}+\frac {e^2 (a+b x)^4}{b^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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.22 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=a^2 d^2 x+a d (b d+a e) x^2+\frac {1}{3} \left (b^2 d^2+4 a b d e+a^2 e^2\right ) x^3+\frac {1}{2} b e (b d+a e) x^4+\frac {1}{5} b^2 e^2 x^5 \]

[In]

Integrate[(d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

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

Maple [A] (verified)

Time = 2.07 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.29

method result size
norman \(\frac {b^{2} e^{2} x^{5}}{5}+\left (\frac {1}{2} a b \,e^{2}+\frac {1}{2} b^{2} d e \right ) x^{4}+\left (\frac {1}{3} a^{2} e^{2}+\frac {4}{3} a b d e +\frac {1}{3} b^{2} d^{2}\right ) x^{3}+\left (a^{2} d e +a b \,d^{2}\right ) x^{2}+a^{2} d^{2} x\) \(84\)
default \(\frac {b^{2} e^{2} x^{5}}{5}+\frac {\left (2 a b \,e^{2}+2 b^{2} d e \right ) x^{4}}{4}+\frac {\left (a^{2} e^{2}+4 a b d e +b^{2} d^{2}\right ) x^{3}}{3}+\frac {\left (2 a^{2} d e +2 a b \,d^{2}\right ) x^{2}}{2}+a^{2} d^{2} x\) \(87\)
risch \(\frac {1}{5} b^{2} e^{2} x^{5}+\frac {1}{2} x^{4} a b \,e^{2}+\frac {1}{2} x^{4} b^{2} d e +\frac {1}{3} x^{3} a^{2} e^{2}+\frac {4}{3} x^{3} a b d e +\frac {1}{3} d^{2} x^{3} b^{2}+a^{2} d e \,x^{2}+x^{2} a b \,d^{2}+a^{2} d^{2} x\) \(90\)
parallelrisch \(\frac {1}{5} b^{2} e^{2} x^{5}+\frac {1}{2} x^{4} a b \,e^{2}+\frac {1}{2} x^{4} b^{2} d e +\frac {1}{3} x^{3} a^{2} e^{2}+\frac {4}{3} x^{3} a b d e +\frac {1}{3} d^{2} x^{3} b^{2}+a^{2} d e \,x^{2}+x^{2} a b \,d^{2}+a^{2} d^{2} x\) \(90\)
gosper \(\frac {x \left (6 b^{2} e^{2} x^{4}+15 x^{3} a b \,e^{2}+15 x^{3} b^{2} d e +10 x^{2} a^{2} e^{2}+40 x^{2} a b d e +10 d^{2} x^{2} b^{2}+30 a^{2} d e x +30 x a b \,d^{2}+30 a^{2} d^{2}\right )}{30}\) \(91\)

[In]

int((e*x+d)^2*(b^2*x^2+2*a*b*x+a^2),x,method=_RETURNVERBOSE)

[Out]

1/5*b^2*e^2*x^5+(1/2*a*b*e^2+1/2*b^2*d*e)*x^4+(1/3*a^2*e^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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.25 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{5} \, b^{2} e^{2} x^{5} + a^{2} d^{2} x + \frac {1}{2} \, {\left (b^{2} d e + a b e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (b^{2} d^{2} + 4 \, a b d e + a^{2} e^{2}\right )} x^{3} + {\left (a b d^{2} + a^{2} d e\right )} x^{2} \]

[In]

integrate((e*x+d)^2*(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.34 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=a^{2} d^{2} x + \frac {b^{2} e^{2} x^{5}}{5} + x^{4} \left (\frac {a b e^{2}}{2} + \frac {b^{2} d e}{2}\right ) + x^{3} \left (\frac {a^{2} e^{2}}{3} + \frac {4 a b d e}{3} + \frac {b^{2} d^{2}}{3}\right ) + x^{2} \left (a^{2} d e + a b d^{2}\right ) \]

[In]

integrate((e*x+d)**2*(b**2*x**2+2*a*b*x+a**2),x)

[Out]

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)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.25 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{5} \, b^{2} e^{2} x^{5} + a^{2} d^{2} x + \frac {1}{2} \, {\left (b^{2} d e + a b e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (b^{2} d^{2} + 4 \, a b d e + a^{2} e^{2}\right )} x^{3} + {\left (a b d^{2} + a^{2} d e\right )} x^{2} \]

[In]

integrate((e*x+d)^2*(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.37 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{5} \, b^{2} e^{2} x^{5} + \frac {1}{2} \, b^{2} d e x^{4} + \frac {1}{2} \, a b e^{2} x^{4} + \frac {1}{3} \, b^{2} d^{2} x^{3} + \frac {4}{3} \, a b d e x^{3} + \frac {1}{3} \, a^{2} e^{2} x^{3} + a b d^{2} x^{2} + a^{2} d e x^{2} + a^{2} d^{2} x \]

[In]

integrate((e*x+d)^2*(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.14 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx=x^3\,\left (\frac {a^2\,e^2}{3}+\frac {4\,a\,b\,d\,e}{3}+\frac {b^2\,d^2}{3}\right )+a^2\,d^2\,x+\frac {b^2\,e^2\,x^5}{5}+a\,d\,x^2\,\left (a\,e+b\,d\right )+\frac {b\,e\,x^4\,\left (a\,e+b\,d\right )}{2} \]

[In]

int((d + e*x)^2*(a^2 + b^2*x^2 + 2*a*b*x),x)

[Out]

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