3.90.36 \(\int \frac {-3 e^2 x^2-\log (5)}{(2 e^2 x^3+2 x \log (5)) \log (\frac {1}{4} (e^2 x^3+x \log (5)))} \, dx\)

Optimal. Leaf size=23 \[ -\frac {1}{2} \log \left (2 \log \left (\frac {1}{4} x \left (e^2 x^2+\log (5)\right )\right )\right ) \]

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Rubi [A]  time = 0.19, antiderivative size = 22, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {1593, 6684} \begin {gather*} -\frac {1}{2} \log \left (\log \left (\frac {1}{4} \left (e^2 x^3+x \log (5)\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*E^2*x^2 - Log[5])/((2*E^2*x^3 + 2*x*Log[5])*Log[(E^2*x^3 + x*Log[5])/4]),x]

[Out]

-1/2*Log[Log[(E^2*x^3 + x*Log[5])/4]]

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 {-3 e^2 x^2-\log (5)}{x \left (2 e^2 x^2+2 \log (5)\right ) \log \left (\frac {1}{4} \left (e^2 x^3+x \log (5)\right )\right )} \, dx\\ &=-\frac {1}{2} \log \left (\log \left (\frac {1}{4} \left (e^2 x^3+x \log (5)\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 21, normalized size = 0.91 \begin {gather*} -\frac {1}{2} \log \left (\log \left (\frac {1}{4} x \left (e^2 x^2+\log (5)\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*E^2*x^2 - Log[5])/((2*E^2*x^3 + 2*x*Log[5])*Log[(E^2*x^3 + x*Log[5])/4]),x]

[Out]

-1/2*Log[Log[(x*(E^2*x^2 + Log[5]))/4]]

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fricas [A]  time = 0.66, size = 17, normalized size = 0.74 \begin {gather*} -\frac {1}{2} \, \log \left (\log \left (\frac {1}{4} \, x^{3} e^{2} + \frac {1}{4} \, x \log \relax (5)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(5)-3*x^2*exp(2))/(2*x*log(5)+2*x^3*exp(2))/log(1/4*x*log(5)+1/4*x^3*exp(2)),x, algorithm="fric
as")

[Out]

-1/2*log(log(1/4*x^3*e^2 + 1/4*x*log(5)))

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giac [A]  time = 0.13, size = 22, normalized size = 0.96 \begin {gather*} -\frac {1}{2} \, \log \left (2 \, \log \relax (2) - \log \left (x^{3} e^{2} + x \log \relax (5)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(5)-3*x^2*exp(2))/(2*x*log(5)+2*x^3*exp(2))/log(1/4*x*log(5)+1/4*x^3*exp(2)),x, algorithm="giac
")

[Out]

-1/2*log(2*log(2) - log(x^3*e^2 + x*log(5)))

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maple [A]  time = 0.43, size = 18, normalized size = 0.78

method	result	size
norman	\(-\frac {\ln \left (\ln \left (\frac {x \ln \relax (5)}{4}+\frac {x^{3} {\mathrm e}^{2}}{4}\right )\right )}{2}\)	\(18\)
risch	\(-\frac {\ln \left (\ln \left (\frac {x \ln \relax (5)}{4}+\frac {x^{3} {\mathrm e}^{2}}{4}\right )\right )}{2}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(5)-3*x^2*exp(2))/(2*x*ln(5)+2*x^3*exp(2))/ln(1/4*x*ln(5)+1/4*x^3*exp(2)),x,method=_RETURNVERBOSE)

[Out]

-1/2*ln(ln(1/4*x*ln(5)+1/4*x^3*exp(2)))

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maxima [A]  time = 0.57, size = 20, normalized size = 0.87 \begin {gather*} -\frac {1}{2} \, \log \left (-2 \, \log \relax (2) + \log \left (x^{2} e^{2} + \log \relax (5)\right ) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(5)-3*x^2*exp(2))/(2*x*log(5)+2*x^3*exp(2))/log(1/4*x*log(5)+1/4*x^3*exp(2)),x, algorithm="maxi
ma")

[Out]

-1/2*log(-2*log(2) + log(x^2*e^2 + log(5)) + log(x))

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mupad [B]  time = 9.11, size = 17, normalized size = 0.74 \begin {gather*} -\frac {\ln \left (\ln \left (\frac {{\mathrm {e}}^2\,x^3}{4}+\frac {\ln \relax (5)\,x}{4}\right )\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5) + 3*x^2*exp(2))/(log((x*log(5))/4 + (x^3*exp(2))/4)*(2*x*log(5) + 2*x^3*exp(2))),x)

[Out]

-log(log((x*log(5))/4 + (x^3*exp(2))/4))/2

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sympy [A]  time = 0.25, size = 20, normalized size = 0.87 \begin {gather*} - \frac {\log {\left (\log {\left (\frac {x^{3} e^{2}}{4} + \frac {x \log {\relax (5 )}}{4} \right )} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(5)-3*x**2*exp(2))/(2*x*ln(5)+2*x**3*exp(2))/ln(1/4*x*ln(5)+1/4*x**3*exp(2)),x)

[Out]

-log(log(x**3*exp(2)/4 + x*log(5)/4))/2

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