3.241 \(\int \frac {(a+b x^3)^3}{x} \, dx\)

Optimal. Leaf size=36 \[ a^3 \log (x)+a^2 b x^3+\frac {1}{2} a b^2 x^6+\frac {b^3 x^9}{9} \]

[Out]

a^2*b*x^3+1/2*a*b^2*x^6+1/9*b^3*x^9+a^3*ln(x)

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Rubi [A]  time = 0.02, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {266, 43} \[ a^2 b x^3+a^3 \log (x)+\frac {1}{2} a b^2 x^6+\frac {b^3 x^9}{9} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^3/x,x]

[Out]

a^2*b*x^3 + (a*b^2*x^6)/2 + (b^3*x^9)/9 + a^3*Log[x]

Rule 43

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])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^3\right )^3}{x} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {(a+b x)^3}{x} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (3 a^2 b+\frac {a^3}{x}+3 a b^2 x+b^3 x^2\right ) \, dx,x,x^3\right )\\ &=a^2 b x^3+\frac {1}{2} a b^2 x^6+\frac {b^3 x^9}{9}+a^3 \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 36, normalized size = 1.00 \[ a^3 \log (x)+a^2 b x^3+\frac {1}{2} a b^2 x^6+\frac {b^3 x^9}{9} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^3/x,x]

[Out]

a^2*b*x^3 + (a*b^2*x^6)/2 + (b^3*x^9)/9 + a^3*Log[x]

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fricas [A]  time = 0.84, size = 32, normalized size = 0.89 \[ \frac {1}{9} \, b^{3} x^{9} + \frac {1}{2} \, a b^{2} x^{6} + a^{2} b x^{3} + a^{3} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*b^3*x^9 + 1/2*a*b^2*x^6 + a^2*b*x^3 + a^3*log(x)

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giac [A]  time = 0.15, size = 33, normalized size = 0.92 \[ \frac {1}{9} \, b^{3} x^{9} + \frac {1}{2} \, a b^{2} x^{6} + a^{2} b x^{3} + a^{3} \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*b^3*x^9 + 1/2*a*b^2*x^6 + a^2*b*x^3 + a^3*log(abs(x))

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maple [A]  time = 0.00, size = 33, normalized size = 0.92 \[ \frac {b^{3} x^{9}}{9}+\frac {a \,b^{2} x^{6}}{2}+a^{2} b \,x^{3}+a^{3} \ln \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^3/x,x)

[Out]

a^2*b*x^3+1/2*a*b^2*x^6+1/9*b^3*x^9+a^3*ln(x)

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maxima [A]  time = 1.37, size = 35, normalized size = 0.97 \[ \frac {1}{9} \, b^{3} x^{9} + \frac {1}{2} \, a b^{2} x^{6} + a^{2} b x^{3} + \frac {1}{3} \, a^{3} \log \left (x^{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*b^3*x^9 + 1/2*a*b^2*x^6 + a^2*b*x^3 + 1/3*a^3*log(x^3)

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mupad [B]  time = 0.03, size = 32, normalized size = 0.89 \[ a^3\,\ln \relax (x)+\frac {b^3\,x^9}{9}+a^2\,b\,x^3+\frac {a\,b^2\,x^6}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^3/x,x)

[Out]

a^3*log(x) + (b^3*x^9)/9 + a^2*b*x^3 + (a*b^2*x^6)/2

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sympy [A]  time = 0.13, size = 32, normalized size = 0.89 \[ a^{3} \log {\relax (x )} + a^{2} b x^{3} + \frac {a b^{2} x^{6}}{2} + \frac {b^{3} x^{9}}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*log(x) + a**2*b*x**3 + a*b**2*x**6/2 + b**3*x**9/9

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