3.56.49 \(\int \frac {-10+3 x-2 \log (3)-x \log (x^2)}{-25 x-5 x^2-5 x \log (3)+(5 x+x^2+x \log (3)) \log (x^2)} \, dx\)

Optimal. Leaf size=24 \[ \log \left (\frac {\frac {11}{2}+e^4}{(5+x+\log (3)) \left (-5+\log \left (x^2\right )\right )}\right ) \]

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Rubi [A]  time = 0.39, antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 5, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {6, 6741, 6742, 2302, 29} \begin {gather*} -\log \left (5-\log \left (x^2\right )\right )-\log (x+5+\log (3)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10 + 3*x - 2*Log[3] - x*Log[x^2])/(-25*x - 5*x^2 - 5*x*Log[3] + (5*x + x^2 + x*Log[3])*Log[x^2]),x]

[Out]

-Log[5 + x + Log[3]] - Log[5 - Log[x^2]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10+3 x-2 \log (3)-x \log \left (x^2\right )}{-5 x^2+x (-25-5 \log (3))+\left (5 x+x^2+x \log (3)\right ) \log \left (x^2\right )} \, dx\\ &=\int \frac {-3 x+10 \left (1+\frac {\log (3)}{5}\right )+x \log \left (x^2\right )}{x (5+x+\log (3)) \left (5-\log \left (x^2\right )\right )} \, dx\\ &=\int \left (\frac {1}{-5-x-\log (3)}-\frac {2}{x \left (-5+\log \left (x^2\right )\right )}\right ) \, dx\\ &=-\log (5+x+\log (3))-2 \int \frac {1}{x \left (-5+\log \left (x^2\right )\right )} \, dx\\ &=-\log (5+x+\log (3))-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,-5+\log \left (x^2\right )\right )\\ &=-\log (5+x+\log (3))-\log \left (5-\log \left (x^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 20, normalized size = 0.83 \begin {gather*} -\log (5+x+\log (3))-\log \left (5-\log \left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 + 3*x - 2*Log[3] - x*Log[x^2])/(-25*x - 5*x^2 - 5*x*Log[3] + (5*x + x^2 + x*Log[3])*Log[x^2]),x
]

[Out]

-Log[5 + x + Log[3]] - Log[5 - Log[x^2]]

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fricas [A]  time = 0.94, size = 18, normalized size = 0.75 \begin {gather*} -\log \left (x + \log \relax (3) + 5\right ) - \log \left (\log \left (x^{2}\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x^2)-2*log(3)+3*x-10)/((x*log(3)+x^2+5*x)*log(x^2)-5*x*log(3)-5*x^2-25*x),x, algorithm="fric
as")

[Out]

-log(x + log(3) + 5) - log(log(x^2) - 5)

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giac [A]  time = 0.16, size = 18, normalized size = 0.75 \begin {gather*} -\log \left (x + \log \relax (3) + 5\right ) - \log \left (\log \left (x^{2}\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x^2)-2*log(3)+3*x-10)/((x*log(3)+x^2+5*x)*log(x^2)-5*x*log(3)-5*x^2-25*x),x, algorithm="giac
")

[Out]

-log(x + log(3) + 5) - log(log(x^2) - 5)

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maple [A]  time = 0.07, size = 19, normalized size = 0.79

method	result	size
norman	\(-\ln \left (\ln \left (x^{2}\right )-5\right )-\ln \left (\ln \relax (3)+5+x \right )\)	\(19\)
risch	\(-\ln \left (\ln \left (x^{2}\right )-5\right )-\ln \left (\ln \relax (3)+5+x \right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*ln(x^2)-2*ln(3)+3*x-10)/((x*ln(3)+x^2+5*x)*ln(x^2)-5*x*ln(3)-5*x^2-25*x),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(x^2)-5)-ln(ln(3)+5+x)

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maxima [A]  time = 0.49, size = 16, normalized size = 0.67 \begin {gather*} -\log \left (x + \log \relax (3) + 5\right ) - \log \left (\log \relax (x) - \frac {5}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x^2)-2*log(3)+3*x-10)/((x*log(3)+x^2+5*x)*log(x^2)-5*x*log(3)-5*x^2-25*x),x, algorithm="maxi
ma")

[Out]

-log(x + log(3) + 5) - log(log(x) - 5/2)

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mupad [B]  time = 3.56, size = 18, normalized size = 0.75 \begin {gather*} -\ln \left (\ln \left (x^2\right )-5\right )-\ln \left (x+\ln \relax (3)+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*log(3) - 3*x + x*log(x^2) + 10)/(25*x + 5*x*log(3) + 5*x^2 - log(x^2)*(5*x + x*log(3) + x^2)),x)

[Out]

- log(log(x^2) - 5) - log(x + log(3) + 5)

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sympy [A]  time = 0.12, size = 17, normalized size = 0.71 \begin {gather*} - \log {\left (\log {\left (x^{2} \right )} - 5 \right )} - \log {\left (x + \log {\relax (3 )} + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*ln(x**2)-2*ln(3)+3*x-10)/((x*ln(3)+x**2+5*x)*ln(x**2)-5*x*ln(3)-5*x**2-25*x),x)

[Out]

-log(log(x**2) - 5) - log(x + log(3) + 5)

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