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

Optimal. Leaf size=26 \[ \log \left (x-\log \left (\frac {1}{2} \left (-x+x^3+\log \left (e^{2 x}+\log (x)\right )\right )\right )\right ) \]

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Rubi [A]  time = 1.11, antiderivative size = 24, normalized size of antiderivative = 0.92, number of steps used = 2, number of rules used = 2, integrand size = 198, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {6688, 6684} \begin {gather*} \log \left (-\log \left (x^3-x+\log \left (e^{2 x}+\log (x)\right )\right )+x+\log (2)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + E^(2*x)*(x + x^2 + 3*x^3 - x^4) + (-x + x^2 + 3*x^3 - x^4)*Log[x] + (-(E^(2*x)*x) - x*Log[x])*Log[E^(
2*x) + Log[x]])/(E^(2*x)*(x^3 - x^5) + (x^3 - x^5)*Log[x] + (-(E^(2*x)*x^2) - x^2*Log[x])*Log[E^(2*x) + Log[x]
] + (E^(2*x)*(-x^2 + x^4) + (-x^2 + x^4)*Log[x] + (E^(2*x)*x + x*Log[x])*Log[E^(2*x) + Log[x]])*Log[(-x + x^3
+ Log[E^(2*x) + Log[x]])/2]),x]

[Out]

Log[x + Log[2] - Log[-x + x^3 + Log[E^(2*x) + Log[x]]]]

Rule 6684

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

Rule 6688

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

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 24, normalized size = 0.92 \begin {gather*} \log \left (x+\log (2)-\log \left (-x+x^3+\log \left (e^{2 x}+\log (x)\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^(2*x)*(x + x^2 + 3*x^3 - x^4) + (-x + x^2 + 3*x^3 - x^4)*Log[x] + (-(E^(2*x)*x) - x*Log[x])*L
og[E^(2*x) + Log[x]])/(E^(2*x)*(x^3 - x^5) + (x^3 - x^5)*Log[x] + (-(E^(2*x)*x^2) - x^2*Log[x])*Log[E^(2*x) +
Log[x]] + (E^(2*x)*(-x^2 + x^4) + (-x^2 + x^4)*Log[x] + (E^(2*x)*x + x*Log[x])*Log[E^(2*x) + Log[x]])*Log[(-x
+ x^3 + Log[E^(2*x) + Log[x]])/2]),x]

[Out]

Log[x + Log[2] - Log[-x + x^3 + Log[E^(2*x) + Log[x]]]]

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fricas [A]  time = 0.50, size = 25, normalized size = 0.96 \begin {gather*} \log \left (-x + \log \left (\frac {1}{2} \, x^{3} - \frac {1}{2} \, x + \frac {1}{2} \, \log \left (e^{\left (2 \, x\right )} + \log \relax (x)\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(x)-x*exp(x)^2)*log(log(x)+exp(x)^2)+(-x^4+3*x^3+x^2-x)*log(x)+(-x^4+3*x^3+x^2+x)*exp(x)^2+1
)/(((x*log(x)+x*exp(x)^2)*log(log(x)+exp(x)^2)+(x^4-x^2)*log(x)+(x^4-x^2)*exp(x)^2)*log(1/2*log(log(x)+exp(x)^
2)+1/2*x^3-1/2*x)+(-x^2*log(x)-exp(x)^2*x^2)*log(log(x)+exp(x)^2)+(-x^5+x^3)*log(x)+(-x^5+x^3)*exp(x)^2),x, al
gorithm="fricas")

[Out]

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

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giac [A]  time = 1.16, size = 23, normalized size = 0.88 \begin {gather*} \log \left (x + \log \relax (2) - \log \left (x^{3} - x + \log \left (e^{\left (2 \, x\right )} + \log \relax (x)\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + log(2) - log(x^3 - x + log(e^(2*x) + log(x))))

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maple [A]  time = 0.07, size = 26, normalized size = 1.00

method	result	size
risch	\(\ln \left (\ln \left (\frac {\ln \left (\ln \relax (x )+{\mathrm e}^{2 x}\right )}{2}+\frac {x^{3}}{2}-\frac {x}{2}\right )-x \right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.72, size = 25, normalized size = 0.96 \begin {gather*} \log \left (-x - \log \relax (2) + \log \left (x^{3} - x + \log \left (e^{\left (2 \, x\right )} + \log \relax (x)\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(x)-x*exp(x)^2)*log(log(x)+exp(x)^2)+(-x^4+3*x^3+x^2-x)*log(x)+(-x^4+3*x^3+x^2+x)*exp(x)^2+1
)/(((x*log(x)+x*exp(x)^2)*log(log(x)+exp(x)^2)+(x^4-x^2)*log(x)+(x^4-x^2)*exp(x)^2)*log(1/2*log(log(x)+exp(x)^
2)+1/2*x^3-1/2*x)+(-x^2*log(x)-exp(x)^2*x^2)*log(log(x)+exp(x)^2)+(-x^5+x^3)*log(x)+(-x^5+x^3)*exp(x)^2),x, al
gorithm="maxima")

[Out]

log(-x - log(2) + log(x^3 - x + log(e^(2*x) + log(x))))

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mupad [B]  time = 2.90, size = 25, normalized size = 0.96 \begin {gather*} \ln \left (\ln \left (\frac {\ln \left ({\mathrm {e}}^{2\,x}+\ln \relax (x)\right )}{2}-\frac {x}{2}+\frac {x^3}{2}\right )-x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(x - x^2 - 3*x^3 + x^4) + log(exp(2*x) + log(x))*(x*exp(2*x) + x*log(x)) - exp(2*x)*(x + x^2 + 3*
x^3 - x^4) - 1)/(exp(2*x)*(x^3 - x^5) - log(log(exp(2*x) + log(x))/2 - x/2 + x^3/2)*(exp(2*x)*(x^2 - x^4) - lo
g(exp(2*x) + log(x))*(x*exp(2*x) + x*log(x)) + log(x)*(x^2 - x^4)) - log(exp(2*x) + log(x))*(x^2*log(x) + x^2*
exp(2*x)) + log(x)*(x^3 - x^5)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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