\(\int x^2 (a+b x)^2 \, dx\) [53]

Optimal result

Integrand size = 11, antiderivative size = 30 \[ \int x^2 (a+b x)^2 \, dx=\frac {a^2 x^3}{3}+\frac {1}{2} a b x^4+\frac {b^2 x^5}{5} \]

[Out]

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

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int x^2 (a+b x)^2 \, dx=\frac {a^2 x^3}{3}+\frac {1}{2} a b x^4+\frac {b^2 x^5}{5} \]

[In]

Int[x^2*(a + b*x)^2,x]

[Out]

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

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+2 a b x^3+b^2 x^4\right ) \, dx \\ & = \frac {a^2 x^3}{3}+\frac {1}{2} a b x^4+\frac {b^2 x^5}{5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int x^2 (a+b x)^2 \, dx=\frac {a^2 x^3}{3}+\frac {1}{2} a b x^4+\frac {b^2 x^5}{5} \]

[In]

Integrate[x^2*(a + b*x)^2,x]

[Out]

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

Maple [A] (verified)

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

method	result	size
gosper	\(\frac {1}{3} a^{2} x^{3}+\frac {1}{2} a b \,x^{4}+\frac {1}{5} b^{2} x^{5}\)	\(25\)
default	\(\frac {1}{3} a^{2} x^{3}+\frac {1}{2} a b \,x^{4}+\frac {1}{5} b^{2} x^{5}\)	\(25\)
norman	\(\frac {1}{3} a^{2} x^{3}+\frac {1}{2} a b \,x^{4}+\frac {1}{5} b^{2} x^{5}\)	\(25\)
risch	\(\frac {1}{3} a^{2} x^{3}+\frac {1}{2} a b \,x^{4}+\frac {1}{5} b^{2} x^{5}\)	\(25\)
parallelrisch	\(\frac {1}{3} a^{2} x^{3}+\frac {1}{2} a b \,x^{4}+\frac {1}{5} b^{2} x^{5}\)	\(25\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int x^2 (a+b x)^2 \, dx=\frac {1}{5} \, b^{2} x^{5} + \frac {1}{2} \, a b x^{4} + \frac {1}{3} \, a^{2} x^{3} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int x^2 (a+b x)^2 \, dx=\frac {a^{2} x^{3}}{3} + \frac {a b x^{4}}{2} + \frac {b^{2} x^{5}}{5} \]

[In]

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

[Out]

a**2*x**3/3 + a*b*x**4/2 + b**2*x**5/5

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int x^2 (a+b x)^2 \, dx=\frac {1}{5} \, b^{2} x^{5} + \frac {1}{2} \, a b x^{4} + \frac {1}{3} \, a^{2} x^{3} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int x^2 (a+b x)^2 \, dx=\frac {1}{5} \, b^{2} x^{5} + \frac {1}{2} \, a b x^{4} + \frac {1}{3} \, a^{2} x^{3} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int x^2 (a+b x)^2 \, dx=\frac {a^2\,x^3}{3}+\frac {a\,b\,x^4}{2}+\frac {b^2\,x^5}{5} \]

[In]

int(x^2*(a + b*x)^2,x)

[Out]

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