3.167 \(\int \frac{(b x^2+c x^4)^3}{x^{11}} \, dx\)

Optimal. Leaf size=40 \[ -\frac{3 b^2 c}{2 x^2}-\frac{b^3}{4 x^4}+3 b c^2 \log (x)+\frac{c^3 x^2}{2} \]

[Out]

-b^3/(4*x^4) - (3*b^2*c)/(2*x^2) + (c^3*x^2)/2 + 3*b*c^2*Log[x]

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Rubi [A]  time = 0.0256056, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {1584, 266, 43} \[ -\frac{3 b^2 c}{2 x^2}-\frac{b^3}{4 x^4}+3 b c^2 \log (x)+\frac{c^3 x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^3/x^11,x]

[Out]

-b^3/(4*x^4) - (3*b^2*c)/(2*x^2) + (c^3*x^2)/2 + 3*b*c^2*Log[x]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 (b x^2+c x^4\right )^3}{x^{11}} \, dx &=\int \frac{\left (b+c x^2\right )^3}{x^5} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(b+c x)^3}{x^3} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (c^3+\frac{b^3}{x^3}+\frac{3 b^2 c}{x^2}+\frac{3 b c^2}{x}\right ) \, dx,x,x^2\right )\\ &=-\frac{b^3}{4 x^4}-\frac{3 b^2 c}{2 x^2}+\frac{c^3 x^2}{2}+3 b c^2 \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0042707, size = 40, normalized size = 1. \[ -\frac{3 b^2 c}{2 x^2}-\frac{b^3}{4 x^4}+3 b c^2 \log (x)+\frac{c^3 x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^3/x^11,x]

[Out]

-b^3/(4*x^4) - (3*b^2*c)/(2*x^2) + (c^3*x^2)/2 + 3*b*c^2*Log[x]

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Maple [A]  time = 0.05, size = 35, normalized size = 0.9 \begin{align*} -{\frac{{b}^{3}}{4\,{x}^{4}}}-{\frac{3\,{b}^{2}c}{2\,{x}^{2}}}+{\frac{{c}^{3}{x}^{2}}{2}}+3\,b{c}^{2}\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^3/x^11,x)

[Out]

-1/4*b^3/x^4-3/2*b^2*c/x^2+1/2*c^3*x^2+3*b*c^2*ln(x)

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Maxima [A]  time = 0.934611, size = 50, normalized size = 1.25 \begin{align*} \frac{1}{2} \, c^{3} x^{2} + \frac{3}{2} \, b c^{2} \log \left (x^{2}\right ) - \frac{6 \, b^{2} c x^{2} + b^{3}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^11,x, algorithm="maxima")

[Out]

1/2*c^3*x^2 + 3/2*b*c^2*log(x^2) - 1/4*(6*b^2*c*x^2 + b^3)/x^4

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Fricas [A]  time = 1.42323, size = 85, normalized size = 2.12 \begin{align*} \frac{2 \, c^{3} x^{6} + 12 \, b c^{2} x^{4} \log \left (x\right ) - 6 \, b^{2} c x^{2} - b^{3}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^11,x, algorithm="fricas")

[Out]

1/4*(2*c^3*x^6 + 12*b*c^2*x^4*log(x) - 6*b^2*c*x^2 - b^3)/x^4

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Sympy [A]  time = 0.369361, size = 36, normalized size = 0.9 \begin{align*} 3 b c^{2} \log{\left (x \right )} + \frac{c^{3} x^{2}}{2} - \frac{b^{3} + 6 b^{2} c x^{2}}{4 x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**3/x**11,x)

[Out]

3*b*c**2*log(x) + c**3*x**2/2 - (b**3 + 6*b**2*c*x**2)/(4*x**4)

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Giac [A]  time = 1.23102, size = 62, normalized size = 1.55 \begin{align*} \frac{1}{2} \, c^{3} x^{2} + \frac{3}{2} \, b c^{2} \log \left (x^{2}\right ) - \frac{9 \, b c^{2} x^{4} + 6 \, b^{2} c x^{2} + b^{3}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^11,x, algorithm="giac")

[Out]

1/2*c^3*x^2 + 3/2*b*c^2*log(x^2) - 1/4*(9*b*c^2*x^4 + 6*b^2*c*x^2 + b^3)/x^4