\(\int x (a+b x^3)^3 \, dx\) [251]

Optimal result

Integrand size = 11, antiderivative size = 43 \[ \int x \left (a+b x^3\right )^3 \, dx=\frac {a^3 x^2}{2}+\frac {3}{5} a^2 b x^5+\frac {3}{8} a b^2 x^8+\frac {b^3 x^{11}}{11} \]

[Out]

1/2*a^3*x^2+3/5*a^2*b*x^5+3/8*a*b^2*x^8+1/11*b^3*x^11

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 = {276} \[ \int x \left (a+b x^3\right )^3 \, dx=\frac {a^3 x^2}{2}+\frac {3}{5} a^2 b x^5+\frac {3}{8} a b^2 x^8+\frac {b^3 x^{11}}{11} \]

[In]

Int[x*(a + b*x^3)^3,x]

[Out]

(a^3*x^2)/2 + (3*a^2*b*x^5)/5 + (3*a*b^2*x^8)/8 + (b^3*x^11)/11

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (a^3 x+3 a^2 b x^4+3 a b^2 x^7+b^3 x^{10}\right ) \, dx \\ & = \frac {a^3 x^2}{2}+\frac {3}{5} a^2 b x^5+\frac {3}{8} a b^2 x^8+\frac {b^3 x^{11}}{11} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int x \left (a+b x^3\right )^3 \, dx=\frac {a^3 x^2}{2}+\frac {3}{5} a^2 b x^5+\frac {3}{8} a b^2 x^8+\frac {b^3 x^{11}}{11} \]

[In]

Integrate[x*(a + b*x^3)^3,x]

[Out]

(a^3*x^2)/2 + (3*a^2*b*x^5)/5 + (3*a*b^2*x^8)/8 + (b^3*x^11)/11

Maple [A] (verified)

Time = 3.66 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84

method	result	size
gosper	\(\frac {1}{2} a^{3} x^{2}+\frac {3}{5} a^{2} b \,x^{5}+\frac {3}{8} a \,b^{2} x^{8}+\frac {1}{11} b^{3} x^{11}\)	\(36\)
default	\(\frac {1}{2} a^{3} x^{2}+\frac {3}{5} a^{2} b \,x^{5}+\frac {3}{8} a \,b^{2} x^{8}+\frac {1}{11} b^{3} x^{11}\)	\(36\)
norman	\(\frac {1}{2} a^{3} x^{2}+\frac {3}{5} a^{2} b \,x^{5}+\frac {3}{8} a \,b^{2} x^{8}+\frac {1}{11} b^{3} x^{11}\)	\(36\)
risch	\(\frac {1}{2} a^{3} x^{2}+\frac {3}{5} a^{2} b \,x^{5}+\frac {3}{8} a \,b^{2} x^{8}+\frac {1}{11} b^{3} x^{11}\)	\(36\)
parallelrisch	\(\frac {1}{2} a^{3} x^{2}+\frac {3}{5} a^{2} b \,x^{5}+\frac {3}{8} a \,b^{2} x^{8}+\frac {1}{11} b^{3} x^{11}\)	\(36\)

[In]

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

[Out]

1/2*a^3*x^2+3/5*a^2*b*x^5+3/8*a*b^2*x^8+1/11*b^3*x^11

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x \left (a+b x^3\right )^3 \, dx=\frac {1}{11} \, b^{3} x^{11} + \frac {3}{8} \, a b^{2} x^{8} + \frac {3}{5} \, a^{2} b x^{5} + \frac {1}{2} \, a^{3} x^{2} \]

[In]

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

[Out]

1/11*b^3*x^11 + 3/8*a*b^2*x^8 + 3/5*a^2*b*x^5 + 1/2*a^3*x^2

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int x \left (a+b x^3\right )^3 \, dx=\frac {a^{3} x^{2}}{2} + \frac {3 a^{2} b x^{5}}{5} + \frac {3 a b^{2} x^{8}}{8} + \frac {b^{3} x^{11}}{11} \]

[In]

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

[Out]

a**3*x**2/2 + 3*a**2*b*x**5/5 + 3*a*b**2*x**8/8 + b**3*x**11/11

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x \left (a+b x^3\right )^3 \, dx=\frac {1}{11} \, b^{3} x^{11} + \frac {3}{8} \, a b^{2} x^{8} + \frac {3}{5} \, a^{2} b x^{5} + \frac {1}{2} \, a^{3} x^{2} \]

[In]

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

[Out]

1/11*b^3*x^11 + 3/8*a*b^2*x^8 + 3/5*a^2*b*x^5 + 1/2*a^3*x^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x \left (a+b x^3\right )^3 \, dx=\frac {1}{11} \, b^{3} x^{11} + \frac {3}{8} \, a b^{2} x^{8} + \frac {3}{5} \, a^{2} b x^{5} + \frac {1}{2} \, a^{3} x^{2} \]

[In]

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

[Out]

1/11*b^3*x^11 + 3/8*a*b^2*x^8 + 3/5*a^2*b*x^5 + 1/2*a^3*x^2

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x \left (a+b x^3\right )^3 \, dx=\frac {a^3\,x^2}{2}+\frac {3\,a^2\,b\,x^5}{5}+\frac {3\,a\,b^2\,x^8}{8}+\frac {b^3\,x^{11}}{11} \]

[In]

int(x*(a + b*x^3)^3,x)

[Out]

(a^3*x^2)/2 + (b^3*x^11)/11 + (3*a^2*b*x^5)/5 + (3*a*b^2*x^8)/8