3.1.37 \(\int \frac {(b x^2+c x^4)^2}{x^{10}} \, dx\)

Optimal. Leaf size=28 \[ -\frac {b^2}{5 x^5}-\frac {2 b c}{3 x^3}-\frac {c^2}{x} \]

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Rubi [A]  time = 0.02, antiderivative size = 28, 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}{5 x^5}-\frac {2 b c}{3 x^3}-\frac {c^2}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^2/x^10,x]

[Out]

-b^2/(5*x^5) - (2*b*c)/(3*x^3) - c^2/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 )^2}{x^{10}} \, dx &=\int \frac {\left (b+c x^2\right )^2}{x^6} \, dx\\ &=\int \left (\frac {b^2}{x^6}+\frac {2 b c}{x^4}+\frac {c^2}{x^2}\right ) \, dx\\ &=-\frac {b^2}{5 x^5}-\frac {2 b c}{3 x^3}-\frac {c^2}{x}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 28, normalized size = 1.00 \begin {gather*} -\frac {b^2}{5 x^5}-\frac {2 b c}{3 x^3}-\frac {c^2}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^2/x^10,x]

[Out]

-1/5*b^2/x^5 - (2*b*c)/(3*x^3) - c^2/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 )^2}{x^{10}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(b*x^2 + c*x^4)^2/x^10,x]

[Out]

IntegrateAlgebraic[(b*x^2 + c*x^4)^2/x^10, x]

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fricas [A]  time = 0.61, size = 26, normalized size = 0.93 \begin {gather*} -\frac {15 \, c^{2} x^{4} + 10 \, b c x^{2} + 3 \, b^{2}}{15 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^2/x^10,x, algorithm="fricas")

[Out]

-1/15*(15*c^2*x^4 + 10*b*c*x^2 + 3*b^2)/x^5

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giac [A]  time = 0.17, size = 26, normalized size = 0.93 \begin {gather*} -\frac {15 \, c^{2} x^{4} + 10 \, b c x^{2} + 3 \, b^{2}}{15 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^2/x^10,x, algorithm="giac")

[Out]

-1/15*(15*c^2*x^4 + 10*b*c*x^2 + 3*b^2)/x^5

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maple [A]  time = 0.00, size = 25, normalized size = 0.89 \begin {gather*} -\frac {c^{2}}{x}-\frac {2 b c}{3 x^{3}}-\frac {b^{2}}{5 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^2/x^10,x)

[Out]

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

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maxima [A]  time = 1.30, size = 26, normalized size = 0.93 \begin {gather*} -\frac {15 \, c^{2} x^{4} + 10 \, b c x^{2} + 3 \, b^{2}}{15 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^2/x^10,x, algorithm="maxima")

[Out]

-1/15*(15*c^2*x^4 + 10*b*c*x^2 + 3*b^2)/x^5

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mupad [B]  time = 0.04, size = 25, normalized size = 0.89 \begin {gather*} -\frac {\frac {b^2}{5}+\frac {2\,b\,c\,x^2}{3}+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)^2/x^10,x)

[Out]

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

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sympy [A]  time = 0.20, size = 27, normalized size = 0.96 \begin {gather*} \frac {- 3 b^{2} - 10 b c x^{2} - 15 c^{2} x^{4}}{15 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**2/x**10,x)

[Out]

(-3*b**2 - 10*b*c*x**2 - 15*c**2*x**4)/(15*x**5)

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