3.247 \(\int \frac{(a+b x^3)^3}{x^{19}} \, dx\)

Optimal. Leaf size=43 \[ -\frac{a^2 b}{5 x^{15}}-\frac{a^3}{18 x^{18}}-\frac{a b^2}{4 x^{12}}-\frac{b^3}{9 x^9} \]

[Out]

-a^3/(18*x^18) - (a^2*b)/(5*x^15) - (a*b^2)/(4*x^12) - b^3/(9*x^9)

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Rubi [A]  time = 0.0184389, antiderivative size = 43, normalized size of antiderivative = 1., 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} \[ -\frac{a^2 b}{5 x^{15}}-\frac{a^3}{18 x^{18}}-\frac{a b^2}{4 x^{12}}-\frac{b^3}{9 x^9} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^3/x^19,x]

[Out]

-a^3/(18*x^18) - (a^2*b)/(5*x^15) - (a*b^2)/(4*x^12) - b^3/(9*x^9)

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]]

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])

Rubi steps

\begin{align*} \int \frac{\left (a+b x^3\right )^3}{x^{19}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(a+b x)^3}{x^7} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{a^3}{x^7}+\frac{3 a^2 b}{x^6}+\frac{3 a b^2}{x^5}+\frac{b^3}{x^4}\right ) \, dx,x,x^3\right )\\ &=-\frac{a^3}{18 x^{18}}-\frac{a^2 b}{5 x^{15}}-\frac{a b^2}{4 x^{12}}-\frac{b^3}{9 x^9}\\ \end{align*}

Mathematica [A]  time = 0.0037339, size = 43, normalized size = 1. \[ -\frac{a^2 b}{5 x^{15}}-\frac{a^3}{18 x^{18}}-\frac{a b^2}{4 x^{12}}-\frac{b^3}{9 x^9} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^3/x^19,x]

[Out]

-a^3/(18*x^18) - (a^2*b)/(5*x^15) - (a*b^2)/(4*x^12) - b^3/(9*x^9)

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Maple [A]  time = 0.006, size = 36, normalized size = 0.8 \begin{align*} -{\frac{{a}^{3}}{18\,{x}^{18}}}-{\frac{{a}^{2}b}{5\,{x}^{15}}}-{\frac{a{b}^{2}}{4\,{x}^{12}}}-{\frac{{b}^{3}}{9\,{x}^{9}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^3/x^19,x)

[Out]

-1/18*a^3/x^18-1/5*a^2*b/x^15-1/4*a*b^2/x^12-1/9*b^3/x^9

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Maxima [A]  time = 0.97121, size = 50, normalized size = 1.16 \begin{align*} -\frac{20 \, b^{3} x^{9} + 45 \, a b^{2} x^{6} + 36 \, a^{2} b x^{3} + 10 \, a^{3}}{180 \, x^{18}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^3/x^19,x, algorithm="maxima")

[Out]

-1/180*(20*b^3*x^9 + 45*a*b^2*x^6 + 36*a^2*b*x^3 + 10*a^3)/x^18

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Fricas [A]  time = 1.66253, size = 88, normalized size = 2.05 \begin{align*} -\frac{20 \, b^{3} x^{9} + 45 \, a b^{2} x^{6} + 36 \, a^{2} b x^{3} + 10 \, a^{3}}{180 \, x^{18}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^3/x^19,x, algorithm="fricas")

[Out]

-1/180*(20*b^3*x^9 + 45*a*b^2*x^6 + 36*a^2*b*x^3 + 10*a^3)/x^18

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Sympy [A]  time = 0.713632, size = 39, normalized size = 0.91 \begin{align*} - \frac{10 a^{3} + 36 a^{2} b x^{3} + 45 a b^{2} x^{6} + 20 b^{3} x^{9}}{180 x^{18}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**3/x**19,x)

[Out]

-(10*a**3 + 36*a**2*b*x**3 + 45*a*b**2*x**6 + 20*b**3*x**9)/(180*x**18)

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Giac [A]  time = 1.10448, size = 50, normalized size = 1.16 \begin{align*} -\frac{20 \, b^{3} x^{9} + 45 \, a b^{2} x^{6} + 36 \, a^{2} b x^{3} + 10 \, a^{3}}{180 \, x^{18}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^3/x^19,x, algorithm="giac")

[Out]

-1/180*(20*b^3*x^9 + 45*a*b^2*x^6 + 36*a^2*b*x^3 + 10*a^3)/x^18