\(\int x^2 (a+b x)^3 \, dx\) [66]

Optimal result

Integrand size = 11, antiderivative size = 43 \[ \int x^2 (a+b x)^3 \, dx=\frac {a^3 x^3}{3}+\frac {3}{4} a^2 b x^4+\frac {3}{5} a b^2 x^5+\frac {b^3 x^6}{6} \]

[Out]

1/3*a^3*x^3+3/4*a^2*b*x^4+3/5*a*b^2*x^5+1/6*b^3*x^6

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, 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} \[ \int x^2 (a+b x)^3 \, dx=\frac {a^3 x^3}{3}+\frac {3}{4} a^2 b x^4+\frac {3}{5} a b^2 x^5+\frac {b^3 x^6}{6} \]

[In]

Int[x^2*(a + b*x)^3,x]

[Out]

(a^3*x^3)/3 + (3*a^2*b*x^4)/4 + (3*a*b^2*x^5)/5 + (b^3*x^6)/6

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^3 x^2+3 a^2 b x^3+3 a b^2 x^4+b^3 x^5\right ) \, dx \\ & = \frac {a^3 x^3}{3}+\frac {3}{4} a^2 b x^4+\frac {3}{5} a b^2 x^5+\frac {b^3 x^6}{6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int x^2 (a+b x)^3 \, dx=\frac {a^3 x^3}{3}+\frac {3}{4} a^2 b x^4+\frac {3}{5} a b^2 x^5+\frac {b^3 x^6}{6} \]

[In]

Integrate[x^2*(a + b*x)^3,x]

[Out]

(a^3*x^3)/3 + (3*a^2*b*x^4)/4 + (3*a*b^2*x^5)/5 + (b^3*x^6)/6

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84

method	result	size
gosper	\(\frac {1}{3} a^{3} x^{3}+\frac {3}{4} a^{2} b \,x^{4}+\frac {3}{5} a \,b^{2} x^{5}+\frac {1}{6} b^{3} x^{6}\)	\(36\)
default	\(\frac {1}{3} a^{3} x^{3}+\frac {3}{4} a^{2} b \,x^{4}+\frac {3}{5} a \,b^{2} x^{5}+\frac {1}{6} b^{3} x^{6}\)	\(36\)
norman	\(\frac {1}{3} a^{3} x^{3}+\frac {3}{4} a^{2} b \,x^{4}+\frac {3}{5} a \,b^{2} x^{5}+\frac {1}{6} b^{3} x^{6}\)	\(36\)
risch	\(\frac {1}{3} a^{3} x^{3}+\frac {3}{4} a^{2} b \,x^{4}+\frac {3}{5} a \,b^{2} x^{5}+\frac {1}{6} b^{3} x^{6}\)	\(36\)
parallelrisch	\(\frac {1}{3} a^{3} x^{3}+\frac {3}{4} a^{2} b \,x^{4}+\frac {3}{5} a \,b^{2} x^{5}+\frac {1}{6} b^{3} x^{6}\)	\(36\)

[In]

int(x^2*(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/3*a^3*x^3+3/4*a^2*b*x^4+3/5*a*b^2*x^5+1/6*b^3*x^6

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^2 (a+b x)^3 \, dx=\frac {1}{6} \, b^{3} x^{6} + \frac {3}{5} \, a b^{2} x^{5} + \frac {3}{4} \, a^{2} b x^{4} + \frac {1}{3} \, a^{3} x^{3} \]

[In]

integrate(x^2*(b*x+a)^3,x, algorithm="fricas")

[Out]

1/6*b^3*x^6 + 3/5*a*b^2*x^5 + 3/4*a^2*b*x^4 + 1/3*a^3*x^3

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int x^2 (a+b x)^3 \, dx=\frac {a^{3} x^{3}}{3} + \frac {3 a^{2} b x^{4}}{4} + \frac {3 a b^{2} x^{5}}{5} + \frac {b^{3} x^{6}}{6} \]

[In]

integrate(x**2*(b*x+a)**3,x)

[Out]

a**3*x**3/3 + 3*a**2*b*x**4/4 + 3*a*b**2*x**5/5 + b**3*x**6/6

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^2 (a+b x)^3 \, dx=\frac {1}{6} \, b^{3} x^{6} + \frac {3}{5} \, a b^{2} x^{5} + \frac {3}{4} \, a^{2} b x^{4} + \frac {1}{3} \, a^{3} x^{3} \]

[In]

integrate(x^2*(b*x+a)^3,x, algorithm="maxima")

[Out]

1/6*b^3*x^6 + 3/5*a*b^2*x^5 + 3/4*a^2*b*x^4 + 1/3*a^3*x^3

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^2 (a+b x)^3 \, dx=\frac {1}{6} \, b^{3} x^{6} + \frac {3}{5} \, a b^{2} x^{5} + \frac {3}{4} \, a^{2} b x^{4} + \frac {1}{3} \, a^{3} x^{3} \]

[In]

integrate(x^2*(b*x+a)^3,x, algorithm="giac")

[Out]

1/6*b^3*x^6 + 3/5*a*b^2*x^5 + 3/4*a^2*b*x^4 + 1/3*a^3*x^3

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^2 (a+b x)^3 \, dx=\frac {a^3\,x^3}{3}+\frac {3\,a^2\,b\,x^4}{4}+\frac {3\,a\,b^2\,x^5}{5}+\frac {b^3\,x^6}{6} \]

[In]

int(x^2*(a + b*x)^3,x)

[Out]

(a^3*x^3)/3 + (b^3*x^6)/6 + (3*a^2*b*x^4)/4 + (3*a*b^2*x^5)/5