3.40.44 \(\int \frac {-16 x+3 x^2}{(10-8 x^2+x^3) \log (\frac {1}{5} (10-8 x^2+x^3))} \, dx\)

Optimal. Leaf size=22 \[ \log \left (\log \left (3-x \left (\frac {1}{x}+x+\frac {1}{5} (3-x) x\right )\right )\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 16, normalized size of antiderivative = 0.73, number of steps used = 2, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {1593, 6684} \begin {gather*} \log \left (\log \left (\frac {1}{5} \left (x^3-8 x^2+10\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16*x + 3*x^2)/((10 - 8*x^2 + x^3)*Log[(10 - 8*x^2 + x^3)/5]),x]

[Out]

Log[Log[(10 - 8*x^2 + x^3)/5]]

Rule 1593

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

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x (-16+3 x)}{\left (10-8 x^2+x^3\right ) \log \left (\frac {1}{5} \left (10-8 x^2+x^3\right )\right )} \, dx\\ &=\log \left (\log \left (\frac {1}{5} \left (10-8 x^2+x^3\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 16, normalized size = 0.73 \begin {gather*} \log \left (\log \left (\frac {1}{5} \left (10-8 x^2+x^3\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*x + 3*x^2)/((10 - 8*x^2 + x^3)*Log[(10 - 8*x^2 + x^3)/5]),x]

[Out]

Log[Log[(10 - 8*x^2 + x^3)/5]]

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fricas [A]  time = 0.72, size = 14, normalized size = 0.64 \begin {gather*} \log \left (\log \left (\frac {1}{5} \, x^{3} - \frac {8}{5} \, x^{2} + 2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2-16*x)/(x^3-8*x^2+10)/log(1/5*x^3-8/5*x^2+2),x, algorithm="fricas")

[Out]

log(log(1/5*x^3 - 8/5*x^2 + 2))

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giac [A]  time = 0.21, size = 14, normalized size = 0.64 \begin {gather*} \log \left (\log \left (\frac {1}{5} \, x^{3} - \frac {8}{5} \, x^{2} + 2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2-16*x)/(x^3-8*x^2+10)/log(1/5*x^3-8/5*x^2+2),x, algorithm="giac")

[Out]

log(log(1/5*x^3 - 8/5*x^2 + 2))

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maple [A]  time = 0.02, size = 15, normalized size = 0.68

method	result	size
norman	\(\ln \left (\ln \left (\frac {1}{5} x^{3}-\frac {8}{5} x^{2}+2\right )\right )\)	\(15\)
risch	\(\ln \left (\ln \left (\frac {1}{5} x^{3}-\frac {8}{5} x^{2}+2\right )\right )\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2-16*x)/(x^3-8*x^2+10)/ln(1/5*x^3-8/5*x^2+2),x,method=_RETURNVERBOSE)

[Out]

ln(ln(1/5*x^3-8/5*x^2+2))

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maxima [A]  time = 0.48, size = 17, normalized size = 0.77 \begin {gather*} \log \left (-\log \relax (5) + \log \left (x^{3} - 8 \, x^{2} + 10\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2-16*x)/(x^3-8*x^2+10)/log(1/5*x^3-8/5*x^2+2),x, algorithm="maxima")

[Out]

log(-log(5) + log(x^3 - 8*x^2 + 10))

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mupad [B]  time = 2.36, size = 14, normalized size = 0.64 \begin {gather*} \ln \left (\ln \left (\frac {x^3}{5}-\frac {8\,x^2}{5}+2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*x - 3*x^2)/(log(x^3/5 - (8*x^2)/5 + 2)*(x^3 - 8*x^2 + 10)),x)

[Out]

log(log(x^3/5 - (8*x^2)/5 + 2))

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sympy [A]  time = 0.19, size = 15, normalized size = 0.68 \begin {gather*} \log {\left (\log {\left (\frac {x^{3}}{5} - \frac {8 x^{2}}{5} + 2 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2-16*x)/(x**3-8*x**2+10)/ln(1/5*x**3-8/5*x**2+2),x)

[Out]

log(log(x**3/5 - 8*x**2/5 + 2))

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