3.1.50 \(\int \frac {(b x^2+c x^4)^3}{x^{12}} \, dx\)

Optimal. Leaf size=34 \[ -\frac {b^3}{5 x^5}-\frac {b^2 c}{x^3}-\frac {3 b c^2}{x}+c^3 x \]

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Rubi [A]  time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1584, 270} \begin {gather*} -\frac {b^2 c}{x^3}-\frac {b^3}{5 x^5}-\frac {3 b c^2}{x}+c^3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^3/x^12,x]

[Out]

-b^3/(5*x^5) - (b^2*c)/x^3 - (3*b*c^2)/x + c^3*x

Rule 270

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]

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]

Rubi steps

\begin {align*} \int \frac {\left (b x^2+c x^4\right )^3}{x^{12}} \, dx &=\int \frac {\left (b+c x^2\right )^3}{x^6} \, dx\\ &=\int \left (c^3+\frac {b^3}{x^6}+\frac {3 b^2 c}{x^4}+\frac {3 b c^2}{x^2}\right ) \, dx\\ &=-\frac {b^3}{5 x^5}-\frac {b^2 c}{x^3}-\frac {3 b c^2}{x}+c^3 x\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 34, normalized size = 1.00 \begin {gather*} -\frac {b^3}{5 x^5}-\frac {b^2 c}{x^3}-\frac {3 b c^2}{x}+c^3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^3/x^12,x]

[Out]

-1/5*b^3/x^5 - (b^2*c)/x^3 - (3*b*c^2)/x + c^3*x

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b x^2+c x^4\right )^3}{x^{12}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(b*x^2 + c*x^4)^3/x^12,x]

[Out]

IntegrateAlgebraic[(b*x^2 + c*x^4)^3/x^12, x]

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fricas [A]  time = 0.74, size = 37, normalized size = 1.09 \begin {gather*} \frac {5 \, c^{3} x^{6} - 15 \, b c^{2} x^{4} - 5 \, b^{2} c x^{2} - b^{3}}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^12,x, algorithm="fricas")

[Out]

1/5*(5*c^3*x^6 - 15*b*c^2*x^4 - 5*b^2*c*x^2 - b^3)/x^5

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giac [A]  time = 0.17, size = 33, normalized size = 0.97 \begin {gather*} c^{3} x - \frac {15 \, b c^{2} x^{4} + 5 \, b^{2} c x^{2} + b^{3}}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^12,x, algorithm="giac")

[Out]

c^3*x - 1/5*(15*b*c^2*x^4 + 5*b^2*c*x^2 + b^3)/x^5

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maple [A]  time = 0.01, size = 33, normalized size = 0.97 \begin {gather*} c^{3} x -\frac {3 b \,c^{2}}{x}-\frac {b^{2} c}{x^{3}}-\frac {b^{3}}{5 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^3/x^12,x)

[Out]

-1/5*b^3/x^5-b^2*c/x^3-3*b*c^2/x+c^3*x

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maxima [A]  time = 1.30, size = 33, normalized size = 0.97 \begin {gather*} c^{3} x - \frac {15 \, b c^{2} x^{4} + 5 \, b^{2} c x^{2} + b^{3}}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^12,x, algorithm="maxima")

[Out]

c^3*x - 1/5*(15*b*c^2*x^4 + 5*b^2*c*x^2 + b^3)/x^5

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mupad [B]  time = 0.03, size = 34, normalized size = 1.00 \begin {gather*} c^3\,x-\frac {\frac {b^3}{5}+b^2\,c\,x^2+3\,b\,c^2\,x^4}{x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^3/x^12,x)

[Out]

c^3*x - (b^3/5 + b^2*c*x^2 + 3*b*c^2*x^4)/x^5

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sympy [A]  time = 0.23, size = 34, normalized size = 1.00 \begin {gather*} c^{3} x + \frac {- b^{3} - 5 b^{2} c x^{2} - 15 b c^{2} x^{4}}{5 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**3/x**12,x)

[Out]

c**3*x + (-b**3 - 5*b**2*c*x**2 - 15*b*c**2*x**4)/(5*x**5)

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