3.235 \(\int \frac{(a+b x^3)^2}{x^{13}} \, dx\)

Optimal. Leaf size=30 \[ -\frac{a^2}{12 x^{12}}-\frac{2 a b}{9 x^9}-\frac{b^2}{6 x^6} \]

[Out]

-a^2/(12*x^12) - (2*a*b)/(9*x^9) - b^2/(6*x^6)

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Rubi [A]  time = 0.0135996, antiderivative size = 30, 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} \[ -\frac{a^2}{12 x^{12}}-\frac{2 a b}{9 x^9}-\frac{b^2}{6 x^6} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^2/x^13,x]

[Out]

-a^2/(12*x^12) - (2*a*b)/(9*x^9) - b^2/(6*x^6)

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 )^2}{x^{13}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(a+b x)^2}{x^5} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{a^2}{x^5}+\frac{2 a b}{x^4}+\frac{b^2}{x^3}\right ) \, dx,x,x^3\right )\\ &=-\frac{a^2}{12 x^{12}}-\frac{2 a b}{9 x^9}-\frac{b^2}{6 x^6}\\ \end{align*}

Mathematica [A]  time = 0.000766, size = 30, normalized size = 1. \[ -\frac{a^2}{12 x^{12}}-\frac{2 a b}{9 x^9}-\frac{b^2}{6 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^2/x^13,x]

[Out]

-a^2/(12*x^12) - (2*a*b)/(9*x^9) - b^2/(6*x^6)

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Maple [A]  time = 0.004, size = 25, normalized size = 0.8 \begin{align*} -{\frac{{a}^{2}}{12\,{x}^{12}}}-{\frac{2\,ab}{9\,{x}^{9}}}-{\frac{{b}^{2}}{6\,{x}^{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^2/x^13,x)

[Out]

-1/12*a^2/x^12-2/9*a*b/x^9-1/6*b^2/x^6

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Maxima [A]  time = 0.951495, size = 35, normalized size = 1.17 \begin{align*} -\frac{6 \, b^{2} x^{6} + 8 \, a b x^{3} + 3 \, a^{2}}{36 \, x^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2/x^13,x, algorithm="maxima")

[Out]

-1/36*(6*b^2*x^6 + 8*a*b*x^3 + 3*a^2)/x^12

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Fricas [A]  time = 1.99672, size = 59, normalized size = 1.97 \begin{align*} -\frac{6 \, b^{2} x^{6} + 8 \, a b x^{3} + 3 \, a^{2}}{36 \, x^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2/x^13,x, algorithm="fricas")

[Out]

-1/36*(6*b^2*x^6 + 8*a*b*x^3 + 3*a^2)/x^12

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Sympy [A]  time = 0.439586, size = 27, normalized size = 0.9 \begin{align*} - \frac{3 a^{2} + 8 a b x^{3} + 6 b^{2} x^{6}}{36 x^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**2/x**13,x)

[Out]

-(3*a**2 + 8*a*b*x**3 + 6*b**2*x**6)/(36*x**12)

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Giac [A]  time = 1.10653, size = 35, normalized size = 1.17 \begin{align*} -\frac{6 \, b^{2} x^{6} + 8 \, a b x^{3} + 3 \, a^{2}}{36 \, x^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2/x^13,x, algorithm="giac")

[Out]

-1/36*(6*b^2*x^6 + 8*a*b*x^3 + 3*a^2)/x^12