\(\int (d+e x) (a^2+2 a b x+b^2 x^2) \, dx\) [1454]

Optimal result

Integrand size = 22, antiderivative size = 38 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {(b d-a e) (a+b x)^3}{3 b^2}+\frac {e (a+b x)^4}{4 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {27, 45} \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {(a+b x)^3 (b d-a e)}{3 b^2}+\frac {e (a+b x)^4}{4 b^2} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.21 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{12} x \left (6 a^2 (2 d+e x)+4 a b x (3 d+2 e x)+b^2 x^2 (4 d+3 e x)\right ) \]

[In]

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

[Out]

(x*(6*a^2*(2*d + e*x) + 4*a*b*x*(3*d + 2*e*x) + b^2*x^2*(4*d + 3*e*x)))/12

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26

method	result	size
norman	\(\frac {b^{2} e \,x^{4}}{4}+\left (\frac {2}{3} a e b +\frac {1}{3} b^{2} d \right ) x^{3}+\left (\frac {1}{2} a^{2} e +a b d \right ) x^{2}+a^{2} d x\)	\(48\)
default	\(\frac {b^{2} e \,x^{4}}{4}+\frac {\left (2 a e b +b^{2} d \right ) x^{3}}{3}+\frac {\left (a^{2} e +2 a b d \right ) x^{2}}{2}+a^{2} d x\)	\(49\)
gosper	\(\frac {x \left (3 b^{2} e \,x^{3}+8 a b e \,x^{2}+4 x^{2} b^{2} d +6 a^{2} e x +12 a x b d +12 a^{2} d \right )}{12}\)	\(50\)
risch	\(\frac {1}{4} b^{2} e \,x^{4}+\frac {2}{3} a b e \,x^{3}+\frac {1}{3} d \,x^{3} b^{2}+\frac {1}{2} a^{2} e \,x^{2}+x^{2} a b d +a^{2} d x\)	\(50\)
parallelrisch	\(\frac {1}{4} b^{2} e \,x^{4}+\frac {2}{3} a b e \,x^{3}+\frac {1}{3} d \,x^{3} b^{2}+\frac {1}{2} a^{2} e \,x^{2}+x^{2} a b d +a^{2} d x\)	\(50\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{4} \, b^{2} e x^{4} + a^{2} d x + \frac {1}{3} \, {\left (b^{2} d + 2 \, a b e\right )} x^{3} + \frac {1}{2} \, {\left (2 \, a b d + a^{2} e\right )} x^{2} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.29 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=a^{2} d x + \frac {b^{2} e x^{4}}{4} + x^{3} \cdot \left (\frac {2 a b e}{3} + \frac {b^{2} d}{3}\right ) + x^{2} \left (\frac {a^{2} e}{2} + a b d\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{4} \, b^{2} e x^{4} + a^{2} d x + \frac {1}{3} \, {\left (b^{2} d + 2 \, a b e\right )} x^{3} + \frac {1}{2} \, {\left (2 \, a b d + a^{2} e\right )} x^{2} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.29 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {1}{4} \, b^{2} e x^{4} + \frac {1}{3} \, b^{2} d x^{3} + \frac {2}{3} \, a b e x^{3} + a b d x^{2} + \frac {1}{2} \, a^{2} e x^{2} + a^{2} d x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=x^2\,\left (\frac {e\,a^2}{2}+b\,d\,a\right )+x^3\,\left (\frac {d\,b^2}{3}+\frac {2\,a\,e\,b}{3}\right )+\frac {b^2\,e\,x^4}{4}+a^2\,d\,x \]

[In]

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

[Out]

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