3.3.21 \(\int x (a+b x^3)^2 \, dx\) [221]

Optimal. Leaf size=30 \[ \frac {a^2 x^2}{2}+\frac {2}{5} a b x^5+\frac {b^2 x^8}{8} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {276} \begin {gather*} \frac {a^2 x^2}{2}+\frac {2}{5} a b x^5+\frac {b^2 x^8}{8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x^3)^2,x]

[Out]

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

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*} \int x \left (a+b x^3\right )^2 \, dx &=\int \left (a^2 x+2 a b x^4+b^2 x^7\right ) \, dx\\ &=\frac {a^2 x^2}{2}+\frac {2}{5} a b x^5+\frac {b^2 x^8}{8}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 30, normalized size = 1.00 \begin {gather*} \frac {a^2 x^2}{2}+\frac {2}{5} a b x^5+\frac {b^2 x^8}{8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*x^3)^2,x]

[Out]

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

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Maple [A]
time = 0.14, size = 25, normalized size = 0.83

method	result	size
gosper	\(\frac {1}{2} a^{2} x^{2}+\frac {2}{5} a b \,x^{5}+\frac {1}{8} b^{2} x^{8}\)	\(25\)
default	\(\frac {1}{2} a^{2} x^{2}+\frac {2}{5} a b \,x^{5}+\frac {1}{8} b^{2} x^{8}\)	\(25\)
norman	\(\frac {1}{2} a^{2} x^{2}+\frac {2}{5} a b \,x^{5}+\frac {1}{8} b^{2} x^{8}\)	\(25\)
risch	\(\frac {1}{2} a^{2} x^{2}+\frac {2}{5} a b \,x^{5}+\frac {1}{8} b^{2} x^{8}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{8} \, b^{2} x^{8} + \frac {2}{5} \, a b x^{5} + \frac {1}{2} \, a^{2} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{8} \, b^{2} x^{8} + \frac {2}{5} \, a b x^{5} + \frac {1}{2} \, a^{2} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.01, size = 26, normalized size = 0.87 \begin {gather*} \frac {a^{2} x^{2}}{2} + \frac {2 a b x^{5}}{5} + \frac {b^{2} x^{8}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.40, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{8} \, b^{2} x^{8} + \frac {2}{5} \, a b x^{5} + \frac {1}{2} \, a^{2} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.03, size = 24, normalized size = 0.80 \begin {gather*} \frac {a^2\,x^2}{2}+\frac {2\,a\,b\,x^5}{5}+\frac {b^2\,x^8}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*x^3)^2,x)

[Out]

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

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