3.34.49 \(\int \frac {e^3 (4 x^3+x^4)+e^3 (4 x^2+x^3) \log (4+x)+(4 x^3+x^4+(4 x^2+x^3) \log (4+x)) \log (x+\log (4+x))+e^{e^{\frac {\log ^2(e^3+\log (x+\log (4+x)))}{x}}+\frac {\log ^2(e^3+\log (x+\log (4+x)))}{x}} ((-10 x-2 x^2) \log (e^3+\log (x+\log (4+x)))+(e^3 (4 x+x^2)+e^3 (4+x) \log (4+x)+(4 x+x^2+(4+x) \log (4+x)) \log (x+\log (4+x))) \log ^2(e^3+\log (x+\log (4+x))))}{e^3 (4 x^3+x^4) \log (4)+e^3 (4 x^2+x^3) \log (4) \log (4+x)+((4 x^3+x^4) \log (4)+(4 x^2+x^3) \log (4) \log (4+x)) \log (x+\log (4+x))} \, dx\)

Optimal. Leaf size=31 \[ \frac {-e^{e^{\frac {\log ^2\left (e^3+\log (x+\log (4+x))\right )}{x}}}+x}{\log (4)} \]

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Rubi [F]  time = 180.00, 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*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^3*(4*x^3 + x^4) + E^3*(4*x^2 + x^3)*Log[4 + x] + (4*x^3 + x^4 + (4*x^2 + x^3)*Log[4 + x])*Log[x + Log[4
+ x]] + E^(E^(Log[E^3 + Log[x + Log[4 + x]]]^2/x) + Log[E^3 + Log[x + Log[4 + x]]]^2/x)*((-10*x - 2*x^2)*Log[
E^3 + Log[x + Log[4 + x]]] + (E^3*(4*x + x^2) + E^3*(4 + x)*Log[4 + x] + (4*x + x^2 + (4 + x)*Log[4 + x])*Log[
x + Log[4 + x]])*Log[E^3 + Log[x + Log[4 + x]]]^2))/(E^3*(4*x^3 + x^4)*Log[4] + E^3*(4*x^2 + x^3)*Log[4]*Log[4
+ x] + ((4*x^3 + x^4)*Log[4] + (4*x^2 + x^3)*Log[4]*Log[4 + x])*Log[x + Log[4 + x]]),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [A]  time = 0.24, size = 31, normalized size = 1.00 \begin {gather*} \frac {-e^{e^{\frac {\log ^2\left (e^3+\log (x+\log (4+x))\right )}{x}}}+x}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3*(4*x^3 + x^4) + E^3*(4*x^2 + x^3)*Log[4 + x] + (4*x^3 + x^4 + (4*x^2 + x^3)*Log[4 + x])*Log[x +
Log[4 + x]] + E^(E^(Log[E^3 + Log[x + Log[4 + x]]]^2/x) + Log[E^3 + Log[x + Log[4 + x]]]^2/x)*((-10*x - 2*x^2
)*Log[E^3 + Log[x + Log[4 + x]]] + (E^3*(4*x + x^2) + E^3*(4 + x)*Log[4 + x] + (4*x + x^2 + (4 + x)*Log[4 + x]
)*Log[x + Log[4 + x]])*Log[E^3 + Log[x + Log[4 + x]]]^2))/(E^3*(4*x^3 + x^4)*Log[4] + E^3*(4*x^2 + x^3)*Log[4]
*Log[4 + x] + ((4*x^3 + x^4)*Log[4] + (4*x^2 + x^3)*Log[4]*Log[4 + x])*Log[x + Log[4 + x]]),x]

[Out]

(-E^E^(Log[E^3 + Log[x + Log[4 + x]]]^2/x) + x)/Log[4]

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fricas [B]  time = 0.65, size = 87, normalized size = 2.81 \begin {gather*} \frac {{\left (x e^{\left (\frac {\log \left (e^{3} + \log \left (x + \log \left (x + 4\right )\right )\right )^{2}}{x}\right )} - e^{\left (\frac {x e^{\left (\frac {\log \left (e^{3} + \log \left (x + \log \left (x + 4\right )\right )\right )^{2}}{x}\right )} + \log \left (e^{3} + \log \left (x + \log \left (x + 4\right )\right )\right )^{2}}{x}\right )}\right )} e^{\left (-\frac {\log \left (e^{3} + \log \left (x + \log \left (x + 4\right )\right )\right )^{2}}{x}\right )}}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((((4+x)*log(4+x)+x^2+4*x)*log(log(4+x)+x)+(4+x)*exp(3)*log(4+x)+(x^2+4*x)*exp(3))*log(log(log(4+x)
+x)+exp(3))^2+(-2*x^2-10*x)*log(log(log(4+x)+x)+exp(3)))*exp(log(log(log(4+x)+x)+exp(3))^2/x)*exp(exp(log(log(
log(4+x)+x)+exp(3))^2/x))+((x^3+4*x^2)*log(4+x)+x^4+4*x^3)*log(log(4+x)+x)+(x^3+4*x^2)*exp(3)*log(4+x)+(x^4+4*
x^3)*exp(3))/((2*(x^3+4*x^2)*log(2)*log(4+x)+2*(x^4+4*x^3)*log(2))*log(log(4+x)+x)+2*(x^3+4*x^2)*exp(3)*log(2)
*log(4+x)+2*(x^4+4*x^3)*exp(3)*log(2)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{3} + 4 \, x^{2}\right )} e^{3} \log \left (x + 4\right ) + {\left (x^{4} + 4 \, x^{3}\right )} e^{3} + {\left ({\left ({\left (x + 4\right )} e^{3} \log \left (x + 4\right ) + {\left (x^{2} + 4 \, x\right )} e^{3} + {\left (x^{2} + {\left (x + 4\right )} \log \left (x + 4\right ) + 4 \, x\right )} \log \left (x + \log \left (x + 4\right )\right )\right )} \log \left (e^{3} + \log \left (x + \log \left (x + 4\right )\right )\right )^{2} - 2 \, {\left (x^{2} + 5 \, x\right )} \log \left (e^{3} + \log \left (x + \log \left (x + 4\right )\right )\right )\right )} e^{\left (\frac {\log \left (e^{3} + \log \left (x + \log \left (x + 4\right )\right )\right )^{2}}{x} + e^{\left (\frac {\log \left (e^{3} + \log \left (x + \log \left (x + 4\right )\right )\right )^{2}}{x}\right )}\right )} + {\left (x^{4} + 4 \, x^{3} + {\left (x^{3} + 4 \, x^{2}\right )} \log \left (x + 4\right )\right )} \log \left (x + \log \left (x + 4\right )\right )}{2 \, {\left ({\left (x^{3} + 4 \, x^{2}\right )} e^{3} \log \relax (2) \log \left (x + 4\right ) + {\left (x^{4} + 4 \, x^{3}\right )} e^{3} \log \relax (2) + {\left ({\left (x^{3} + 4 \, x^{2}\right )} \log \relax (2) \log \left (x + 4\right ) + {\left (x^{4} + 4 \, x^{3}\right )} \log \relax (2)\right )} \log \left (x + \log \left (x + 4\right )\right )\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((((4+x)*log(4+x)+x^2+4*x)*log(log(4+x)+x)+(4+x)*exp(3)*log(4+x)+(x^2+4*x)*exp(3))*log(log(log(4+x)
+x)+exp(3))^2+(-2*x^2-10*x)*log(log(log(4+x)+x)+exp(3)))*exp(log(log(log(4+x)+x)+exp(3))^2/x)*exp(exp(log(log(
log(4+x)+x)+exp(3))^2/x))+((x^3+4*x^2)*log(4+x)+x^4+4*x^3)*log(log(4+x)+x)+(x^3+4*x^2)*exp(3)*log(4+x)+(x^4+4*
x^3)*exp(3))/((2*(x^3+4*x^2)*log(2)*log(4+x)+2*(x^4+4*x^3)*log(2))*log(log(4+x)+x)+2*(x^3+4*x^2)*exp(3)*log(2)
*log(4+x)+2*(x^4+4*x^3)*exp(3)*log(2)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.14, size = 34, normalized size = 1.10

method	result	size
risch	\(\frac {x}{2 \ln \relax (2)}-\frac {{\mathrm e}^{{\mathrm e}^{\frac {\ln \left (\ln \left (\ln \left (4+x \right )+x \right )+{\mathrm e}^{3}\right )^{2}}{x}}}}{2 \ln \relax (2)}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((((4+x)*ln(4+x)+x^2+4*x)*ln(ln(4+x)+x)+(4+x)*exp(3)*ln(4+x)+(x^2+4*x)*exp(3))*ln(ln(ln(4+x)+x)+exp(3))^2
+(-2*x^2-10*x)*ln(ln(ln(4+x)+x)+exp(3)))*exp(ln(ln(ln(4+x)+x)+exp(3))^2/x)*exp(exp(ln(ln(ln(4+x)+x)+exp(3))^2/
x))+((x^3+4*x^2)*ln(4+x)+x^4+4*x^3)*ln(ln(4+x)+x)+(x^3+4*x^2)*exp(3)*ln(4+x)+(x^4+4*x^3)*exp(3))/((2*(x^3+4*x^
2)*ln(2)*ln(4+x)+2*(x^4+4*x^3)*ln(2))*ln(ln(4+x)+x)+2*(x^3+4*x^2)*exp(3)*ln(2)*ln(4+x)+2*(x^4+4*x^3)*exp(3)*ln
(2)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.83, size = 29, normalized size = 0.94 \begin {gather*} \frac {x - e^{\left (e^{\left (\frac {\log \left (e^{3} + \log \left (x + \log \left (x + 4\right )\right )\right )^{2}}{x}\right )}\right )}}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((((4+x)*log(4+x)+x^2+4*x)*log(log(4+x)+x)+(4+x)*exp(3)*log(4+x)+(x^2+4*x)*exp(3))*log(log(log(4+x)
+x)+exp(3))^2+(-2*x^2-10*x)*log(log(log(4+x)+x)+exp(3)))*exp(log(log(log(4+x)+x)+exp(3))^2/x)*exp(exp(log(log(
log(4+x)+x)+exp(3))^2/x))+((x^3+4*x^2)*log(4+x)+x^4+4*x^3)*log(log(4+x)+x)+(x^3+4*x^2)*exp(3)*log(4+x)+(x^4+4*
x^3)*exp(3))/((2*(x^3+4*x^2)*log(2)*log(4+x)+2*(x^4+4*x^3)*log(2))*log(log(4+x)+x)+2*(x^3+4*x^2)*exp(3)*log(2)
*log(4+x)+2*(x^4+4*x^3)*exp(3)*log(2)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.20, size = 29, normalized size = 0.94 \begin {gather*} \frac {x-{\mathrm {e}}^{{\mathrm {e}}^{\frac {{\ln \left ({\mathrm {e}}^3+\ln \left (x+\ln \left (x+4\right )\right )\right )}^2}{x}}}}{2\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3)*(4*x^3 + x^4) + log(x + log(x + 4))*(log(x + 4)*(4*x^2 + x^3) + 4*x^3 + x^4) - exp(exp(log(exp(3)
+ log(x + log(x + 4)))^2/x))*exp(log(exp(3) + log(x + log(x + 4)))^2/x)*(log(exp(3) + log(x + log(x + 4)))*(10
*x + 2*x^2) - log(exp(3) + log(x + log(x + 4)))^2*(log(x + log(x + 4))*(4*x + log(x + 4)*(x + 4) + x^2) + exp(
3)*(4*x + x^2) + log(x + 4)*exp(3)*(x + 4))) + log(x + 4)*exp(3)*(4*x^2 + x^3))/(log(x + log(x + 4))*(2*log(2)
*(4*x^3 + x^4) + 2*log(x + 4)*log(2)*(4*x^2 + x^3)) + 2*exp(3)*log(2)*(4*x^3 + x^4) + 2*log(x + 4)*exp(3)*log(
2)*(4*x^2 + x^3)),x)

[Out]

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

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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((((((4+x)*ln(4+x)+x**2+4*x)*ln(ln(4+x)+x)+(4+x)*exp(3)*ln(4+x)+(x**2+4*x)*exp(3))*ln(ln(ln(4+x)+x)+e
xp(3))**2+(-2*x**2-10*x)*ln(ln(ln(4+x)+x)+exp(3)))*exp(ln(ln(ln(4+x)+x)+exp(3))**2/x)*exp(exp(ln(ln(ln(4+x)+x)
+exp(3))**2/x))+((x**3+4*x**2)*ln(4+x)+x**4+4*x**3)*ln(ln(4+x)+x)+(x**3+4*x**2)*exp(3)*ln(4+x)+(x**4+4*x**3)*e
xp(3))/((2*(x**3+4*x**2)*ln(2)*ln(4+x)+2*(x**4+4*x**3)*ln(2))*ln(ln(4+x)+x)+2*(x**3+4*x**2)*exp(3)*ln(2)*ln(4+
x)+2*(x**4+4*x**3)*exp(3)*ln(2)),x)

[Out]

Timed out

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