\(\int x^3 (a+b x)^2 \, dx\) [52]

Optimal result

Integrand size = 11, antiderivative size = 30 \[ \int x^3 (a+b x)^2 \, dx=\frac {a^2 x^4}{4}+\frac {2}{5} a b x^5+\frac {b^2 x^6}{6} \]

[Out]

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

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

[In]

Int[x^3*(a + b*x)^2,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int x^3 (a+b x)^2 \, dx=\frac {a^2 x^4}{4}+\frac {2}{5} a b x^5+\frac {b^2 x^6}{6} \]

[In]

Integrate[x^3*(a + b*x)^2,x]

[Out]

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

Maple [A] (verified)

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

method	result	size
gosper	\(\frac {1}{4} a^{2} x^{4}+\frac {2}{5} a b \,x^{5}+\frac {1}{6} b^{2} x^{6}\)	\(25\)
default	\(\frac {1}{4} a^{2} x^{4}+\frac {2}{5} a b \,x^{5}+\frac {1}{6} b^{2} x^{6}\)	\(25\)
norman	\(\frac {1}{4} a^{2} x^{4}+\frac {2}{5} a b \,x^{5}+\frac {1}{6} b^{2} x^{6}\)	\(25\)
risch	\(\frac {1}{4} a^{2} x^{4}+\frac {2}{5} a b \,x^{5}+\frac {1}{6} b^{2} x^{6}\)	\(25\)
parallelrisch	\(\frac {1}{4} a^{2} x^{4}+\frac {2}{5} a b \,x^{5}+\frac {1}{6} b^{2} x^{6}\)	\(25\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int x^3 (a+b x)^2 \, dx=\frac {1}{6} \, b^{2} x^{6} + \frac {2}{5} \, a b x^{5} + \frac {1}{4} \, a^{2} x^{4} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int x^3 (a+b x)^2 \, dx=\frac {a^{2} x^{4}}{4} + \frac {2 a b x^{5}}{5} + \frac {b^{2} x^{6}}{6} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int x^3 (a+b x)^2 \, dx=\frac {1}{6} \, b^{2} x^{6} + \frac {2}{5} \, a b x^{5} + \frac {1}{4} \, a^{2} x^{4} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int x^3 (a+b x)^2 \, dx=\frac {1}{6} \, b^{2} x^{6} + \frac {2}{5} \, a b x^{5} + \frac {1}{4} \, a^{2} x^{4} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int x^3 (a+b x)^2 \, dx=\frac {a^2\,x^4}{4}+\frac {2\,a\,b\,x^5}{5}+\frac {b^2\,x^6}{6} \]

[In]

int(x^3*(a + b*x)^2,x)

[Out]

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