3.1453 \(\int (d+e x)^2 (a^2+2 a b x+b^2 x^2) \, dx\)

Optimal. Leaf size=65 \[ \frac {e (a+b x)^4 (b d-a e)}{2 b^3}+\frac {(a+b x)^3 (b d-a e)^2}{3 b^3}+\frac {e^2 (a+b x)^5}{5 b^3} \]

[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

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Rubi [A]  time = 0.08, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {27, 43} \[ \frac {e (a+b x)^4 (b d-a e)}{2 b^3}+\frac {(a+b x)^3 (b d-a e)^2}{3 b^3}+\frac {e^2 (a+b x)^5}{5 b^3} \]

Antiderivative was successfully verified.

[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 43

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*} \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right ) \, dx &=\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*}

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Mathematica [A]  time = 0.01, size = 79, normalized size = 1.22 \[ \frac {1}{3} x^3 \left (a^2 e^2+4 a b d e+b^2 d^2\right )+a^2 d^2 x+\frac {1}{2} b e x^4 (a e+b d)+a d x^2 (a e+b d)+\frac {1}{5} b^2 e^2 x^5 \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.93, size = 89, normalized size = 1.37 \[ \frac {1}{5} x^{5} e^{2} b^{2} + \frac {1}{2} x^{4} e d b^{2} + \frac {1}{2} x^{4} e^{2} b a + \frac {1}{3} x^{3} d^{2} b^{2} + \frac {4}{3} x^{3} e d b a + \frac {1}{3} x^{3} e^{2} a^{2} + x^{2} d^{2} b a + x^{2} e d a^{2} + x d^{2} a^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 89, normalized size = 1.37 \[ \frac {1}{5} \, b^{2} x^{5} e^{2} + \frac {1}{2} \, b^{2} d x^{4} e + \frac {1}{3} \, b^{2} d^{2} x^{3} + \frac {1}{2} \, a b x^{4} e^{2} + \frac {4}{3} \, a b d x^{3} e + a b d^{2} x^{2} + \frac {1}{3} \, a^{2} x^{3} e^{2} + a^{2} d x^{2} e + a^{2} d^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 87, normalized size = 1.34 \[ \frac {b^{2} e^{2} x^{5}}{5}+a^{2} d^{2} x +\frac {\left (2 e^{2} a b +2 b^{2} d e \right ) x^{4}}{4}+\frac {\left (a^{2} e^{2}+4 a b d e +b^{2} d^{2}\right ) x^{3}}{3}+\frac {\left (2 d e \,a^{2}+2 d^{2} a b \right ) x^{2}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.38, size = 81, normalized size = 1.25 \[ \frac {1}{5} \, b^{2} e^{2} x^{5} + a^{2} d^{2} x + \frac {1}{2} \, {\left (b^{2} d e + a b e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (b^{2} d^{2} + 4 \, a b d e + a^{2} e^{2}\right )} x^{3} + {\left (a b d^{2} + a^{2} d e\right )} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.04, size = 74, normalized size = 1.14 \[ x^3\,\left (\frac {a^2\,e^2}{3}+\frac {4\,a\,b\,d\,e}{3}+\frac {b^2\,d^2}{3}\right )+a^2\,d^2\,x+\frac {b^2\,e^2\,x^5}{5}+a\,d\,x^2\,\left (a\,e+b\,d\right )+\frac {b\,e\,x^4\,\left (a\,e+b\,d\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.08, size = 87, normalized size = 1.34 \[ a^{2} d^{2} x + \frac {b^{2} e^{2} x^{5}}{5} + x^{4} \left (\frac {a b e^{2}}{2} + \frac {b^{2} d e}{2}\right ) + x^{3} \left (\frac {a^{2} e^{2}}{3} + \frac {4 a b d e}{3} + \frac {b^{2} d^{2}}{3}\right ) + x^{2} \left (a^{2} d e + a b d^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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