\(\int x^2 (a+b x) \, dx\) [44]

Optimal result

Integrand size = 9, antiderivative size = 17 \[ \int x^2 (a+b x) \, dx=\frac {a x^3}{3}+\frac {b x^4}{4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, 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} \[ \int x^2 (a+b x) \, dx=\frac {a x^3}{3}+\frac {b x^4}{4} \]

[In]

Int[x^2*(a + b*x),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int x^2 (a+b x) \, dx=\frac {a x^3}{3}+\frac {b x^4}{4} \]

[In]

Integrate[x^2*(a + b*x),x]

[Out]

(a*x^3)/3 + (b*x^4)/4

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
gosper	\(\frac {1}{3} a \,x^{3}+\frac {1}{4} b \,x^{4}\)	\(14\)
default	\(\frac {1}{3} a \,x^{3}+\frac {1}{4} b \,x^{4}\)	\(14\)
norman	\(\frac {1}{3} a \,x^{3}+\frac {1}{4} b \,x^{4}\)	\(14\)
risch	\(\frac {1}{3} a \,x^{3}+\frac {1}{4} b \,x^{4}\)	\(14\)
parallelrisch	\(\frac {1}{3} a \,x^{3}+\frac {1}{4} b \,x^{4}\)	\(14\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int x^2 (a+b x) \, dx=\frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} \]

[In]

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

[Out]

1/4*b*x^4 + 1/3*a*x^3

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int x^2 (a+b x) \, dx=\frac {a x^{3}}{3} + \frac {b x^{4}}{4} \]

[In]

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

[Out]

a*x**3/3 + b*x**4/4

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int x^2 (a+b x) \, dx=\frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} \]

[In]

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

[Out]

1/4*b*x^4 + 1/3*a*x^3

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int x^2 (a+b x) \, dx=\frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} \]

[In]

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

[Out]

1/4*b*x^4 + 1/3*a*x^3

Mupad [B] (verification not implemented)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int x^2 (a+b x) \, dx=\frac {x^3\,\left (4\,a+3\,b\,x\right )}{12} \]

[In]

int(x^2*(a + b*x),x)

[Out]

(x^3*(4*a + 3*b*x))/12