\(\int x (a+b x)^2 \, dx\) [54]

Optimal result

Integrand size = 9, antiderivative size = 30 \[ \int x (a+b x)^2 \, dx=\frac {a^2 x^2}{2}+\frac {2}{3} a b x^3+\frac {b^2 x^4}{4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {45} \[ \int x (a+b x)^2 \, dx=\frac {a^2 x^2}{2}+\frac {2}{3} a b x^3+\frac {b^2 x^4}{4} \]

[In]

Int[x*(a + b*x)^2,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int x (a+b x)^2 \, dx=\frac {a^2 x^2}{2}+\frac {2}{3} a b x^3+\frac {b^2 x^4}{4} \]

[In]

Integrate[x*(a + b*x)^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
gosper	\(\frac {1}{2} a^{2} x^{2}+\frac {2}{3} a b \,x^{3}+\frac {1}{4} b^{2} x^{4}\)	\(25\)
default	\(\frac {1}{2} a^{2} x^{2}+\frac {2}{3} a b \,x^{3}+\frac {1}{4} b^{2} x^{4}\)	\(25\)
norman	\(\frac {1}{2} a^{2} x^{2}+\frac {2}{3} a b \,x^{3}+\frac {1}{4} b^{2} x^{4}\)	\(25\)
risch	\(\frac {1}{2} a^{2} x^{2}+\frac {2}{3} a b \,x^{3}+\frac {1}{4} b^{2} x^{4}\)	\(25\)
parallelrisch	\(\frac {1}{2} a^{2} x^{2}+\frac {2}{3} a b \,x^{3}+\frac {1}{4} b^{2} x^{4}\)	\(25\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int x (a+b x)^2 \, dx=\frac {1}{4} \, b^{2} x^{4} + \frac {2}{3} \, a b x^{3} + \frac {1}{2} \, a^{2} x^{2} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int x (a+b x)^2 \, dx=\frac {a^{2} x^{2}}{2} + \frac {2 a b x^{3}}{3} + \frac {b^{2} x^{4}}{4} \]

[In]

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

[Out]

a**2*x**2/2 + 2*a*b*x**3/3 + b**2*x**4/4

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int x (a+b x)^2 \, dx=\frac {1}{4} \, b^{2} x^{4} + \frac {2}{3} \, a b x^{3} + \frac {1}{2} \, a^{2} x^{2} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int x (a+b x)^2 \, dx=\frac {1}{4} \, b^{2} x^{4} + \frac {2}{3} \, a b x^{3} + \frac {1}{2} \, a^{2} x^{2} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int x (a+b x)^2 \, dx=\frac {a^2\,x^2}{2}+\frac {2\,a\,b\,x^3}{3}+\frac {b^2\,x^4}{4} \]

[In]

int(x*(a + b*x)^2,x)

[Out]

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