3.92.48 \(\int \frac {-4 x-4 e^{4 x} x+(-2-2 e^{4 x}) \log (1+e^{4 x})+(2 x+6 e^{4 x} x) \log (x^2)}{(2 x^2+2 e^{4 x} x^2) \log (x^2)+(x+e^{4 x} x) \log (1+e^{4 x}) \log (x^2)} \, dx\) [9148]

Optimal. Leaf size=22 \[ \log \left (\frac {-2 x-\log \left (1+e^{4 x}\right )}{\log \left (x^2\right )}\right ) \]

[Out]

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

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Rubi [F]
time = 2.36, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*x - 4*E^(4*x)*x + (-2 - 2*E^(4*x))*Log[1 + E^(4*x)] + (2*x + 6*E^(4*x)*x)*Log[x^2])/((2*x^2 + 2*E^(4*x
)*x^2)*Log[x^2] + (x + E^(4*x)*x)*Log[1 + E^(4*x)]*Log[x^2]),x]

[Out]

-Log[Log[x^2]] + 3*Defer[Subst][Defer[Int][(x + Log[1 + E^(2*x)])^(-1), x], x, 2*x] - 2*Defer[Subst][Defer[Int
][1/((1 + E^(2*x))*(x + Log[1 + E^(2*x)])), x], x, 2*x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (1+e^{4 x}\right ) x \left (2 x+\log \left (1+e^{4 x}\right )\right ) \log \left (x^2\right )} \, dx\\ &=\int \left (-\frac {4}{\left (1+e^{4 x}\right ) \left (2 x+\log \left (1+e^{4 x}\right )\right )}+\frac {2 \left (-2 x-\log \left (1+e^{4 x}\right )+3 x \log \left (x^2\right )\right )}{x \left (2 x+\log \left (1+e^{4 x}\right )\right ) \log \left (x^2\right )}\right ) \, dx\\ &=2 \int \frac {-2 x-\log \left (1+e^{4 x}\right )+3 x \log \left (x^2\right )}{x \left (2 x+\log \left (1+e^{4 x}\right )\right ) \log \left (x^2\right )} \, dx-4 \int \frac {1}{\left (1+e^{4 x}\right ) \left (2 x+\log \left (1+e^{4 x}\right )\right )} \, dx\\ &=2 \int \left (\frac {3}{2 x+\log \left (1+e^{4 x}\right )}-\frac {1}{x \log \left (x^2\right )}\right ) \, dx-2 \text {Subst}\left (\int \frac {1}{\left (1+e^{2 x}\right ) \left (x+\log \left (1+e^{2 x}\right )\right )} \, dx,x,2 x\right )\\ &=-\left (2 \int \frac {1}{x \log \left (x^2\right )} \, dx\right )-2 \text {Subst}\left (\int \frac {1}{\left (1+e^{2 x}\right ) \left (x+\log \left (1+e^{2 x}\right )\right )} \, dx,x,2 x\right )+6 \int \frac {1}{2 x+\log \left (1+e^{4 x}\right )} \, dx\\ &=-\left (2 \text {Subst}\left (\int \frac {1}{\left (1+e^{2 x}\right ) \left (x+\log \left (1+e^{2 x}\right )\right )} \, dx,x,2 x\right )\right )+3 \text {Subst}\left (\int \frac {1}{x+\log \left (1+e^{2 x}\right )} \, dx,x,2 x\right )-\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (x^2\right )\right )\\ &=-\log \left (\log \left (x^2\right )\right )-2 \text {Subst}\left (\int \frac {1}{\left (1+e^{2 x}\right ) \left (x+\log \left (1+e^{2 x}\right )\right )} \, dx,x,2 x\right )+3 \text {Subst}\left (\int \frac {1}{x+\log \left (1+e^{2 x}\right )} \, dx,x,2 x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.32, size = 21, normalized size = 0.95 \begin {gather*} \log \left (2 x+\log \left (1+e^{4 x}\right )\right )-\log \left (\log \left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x - 4*E^(4*x)*x + (-2 - 2*E^(4*x))*Log[1 + E^(4*x)] + (2*x + 6*E^(4*x)*x)*Log[x^2])/((2*x^2 + 2*
E^(4*x)*x^2)*Log[x^2] + (x + E^(4*x)*x)*Log[1 + E^(4*x)]*Log[x^2]),x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.33, size = 62, normalized size = 2.82

method	result	size
risch	\(-\ln \left (\ln \left (x \right )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (\mathrm {csgn}\left (i x \right )^{2}-2 \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )+\mathrm {csgn}\left (i x^{2}\right )^{2}\right )}{4}\right )+\ln \left (\ln \left ({\mathrm e}^{4 x}+1\right )+2 x \right )\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*exp(4*x)-2)*ln(exp(4*x)+1)+(6*x*exp(4*x)+2*x)*ln(x^2)-4*x*exp(4*x)-4*x)/((x*exp(4*x)+x)*ln(x^2)*ln(ex
p(4*x)+1)+(2*x^2*exp(4*x)+2*x^2)*ln(x^2)),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(x)-1/4*I*Pi*csgn(I*x^2)*(csgn(I*x)^2-2*csgn(I*x^2)*csgn(I*x)+csgn(I*x^2)^2))+ln(ln(exp(4*x)+1)+2*x)

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Maxima [A]
time = 0.52, size = 18, normalized size = 0.82 \begin {gather*} \log \left (2 \, x + \log \left (e^{\left (4 \, x\right )} + 1\right )\right ) - \log \left (\log \left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(4*x)-2)*log(exp(4*x)+1)+(6*x*exp(4*x)+2*x)*log(x^2)-4*x*exp(4*x)-4*x)/((x*exp(4*x)+x)*log(x
^2)*log(exp(4*x)+1)+(2*x^2*exp(4*x)+2*x^2)*log(x^2)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.44, size = 20, normalized size = 0.91 \begin {gather*} \log \left (2 \, x + \log \left (e^{\left (4 \, x\right )} + 1\right )\right ) - \log \left (\log \left (x^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(4*x)-2)*log(exp(4*x)+1)+(6*x*exp(4*x)+2*x)*log(x^2)-4*x*exp(4*x)-4*x)/((x*exp(4*x)+x)*log(x
^2)*log(exp(4*x)+1)+(2*x^2*exp(4*x)+2*x^2)*log(x^2)),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.16, size = 19, normalized size = 0.86 \begin {gather*} \log {\left (2 x + \log {\left (e^{4 x} + 1 \right )} \right )} - \log {\left (\log {\left (x^{2} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(4*x)-2)*ln(exp(4*x)+1)+(6*x*exp(4*x)+2*x)*ln(x**2)-4*x*exp(4*x)-4*x)/((x*exp(4*x)+x)*ln(x**
2)*ln(exp(4*x)+1)+(2*x**2*exp(4*x)+2*x**2)*ln(x**2)),x)

[Out]

log(2*x + log(exp(4*x) + 1)) - log(log(x**2))

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Giac [A]
time = 0.46, size = 20, normalized size = 0.91 \begin {gather*} \log \left (2 \, x + \log \left (e^{\left (4 \, x\right )} + 1\right )\right ) - \log \left (\log \left (x^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(4*x)-2)*log(exp(4*x)+1)+(6*x*exp(4*x)+2*x)*log(x^2)-4*x*exp(4*x)-4*x)/((x*exp(4*x)+x)*log(x
^2)*log(exp(4*x)+1)+(2*x^2*exp(4*x)+2*x^2)*log(x^2)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 7.90, size = 20, normalized size = 0.91 \begin {gather*} \ln \left (2\,x+\ln \left ({\mathrm {e}}^{4\,x}+1\right )\right )-\ln \left (\ln \left (x^2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x + 4*x*exp(4*x) + log(exp(4*x) + 1)*(2*exp(4*x) + 2) - log(x^2)*(2*x + 6*x*exp(4*x)))/(log(x^2)*(2*x^
2*exp(4*x) + 2*x^2) + log(exp(4*x) + 1)*log(x^2)*(x + x*exp(4*x))),x)

[Out]

log(2*x + log(exp(4*x) + 1)) - log(log(x^2))

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