3.34.83 \(\int \frac {(4-4 e^4+e^8) \log ^2(x)+(-4+2 e^4) \log ^2(x) \log (4 x^2)+\log ^2(x) \log ^2(4 x^2)+e^{\frac {4}{-2+e^4+\log (4 x^2)}} (4-4 e^4+e^8+(4+4 e^4-e^8) \log (x)+(-4+2 e^4+(4-2 e^4) \log (x)) \log (4 x^2)+(1-\log (x)) \log ^2(4 x^2))}{(4-4 e^4+e^8) \log ^2(x)+(-4+2 e^4) \log ^2(x) \log (4 x^2)+\log ^2(x) \log ^2(4 x^2)} \, dx\)

Optimal. Leaf size=36 \[ x \left (2+\frac {1-x}{x}-\frac {e^{\frac {4}{-2+e^4+\log \left (4 x^2\right )}}}{\log (x)}\right ) \]

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Rubi [F]  time = 4.10, 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 {\left (4-4 e^4+e^8\right ) \log ^2(x)+\left (-4+2 e^4\right ) \log ^2(x) \log \left (4 x^2\right )+\log ^2(x) \log ^2\left (4 x^2\right )+e^{\frac {4}{-2+e^4+\log \left (4 x^2\right )}} \left (4-4 e^4+e^8+\left (4+4 e^4-e^8\right ) \log (x)+\left (-4+2 e^4+\left (4-2 e^4\right ) \log (x)\right ) \log \left (4 x^2\right )+(1-\log (x)) \log ^2\left (4 x^2\right )\right )}{\left (4-4 e^4+e^8\right ) \log ^2(x)+\left (-4+2 e^4\right ) \log ^2(x) \log \left (4 x^2\right )+\log ^2(x) \log ^2\left (4 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((4 - 4*E^4 + E^8)*Log[x]^2 + (-4 + 2*E^4)*Log[x]^2*Log[4*x^2] + Log[x]^2*Log[4*x^2]^2 + E^(4/(-2 + E^4 +
Log[4*x^2]))*(4 - 4*E^4 + E^8 + (4 + 4*E^4 - E^8)*Log[x] + (-4 + 2*E^4 + (4 - 2*E^4)*Log[x])*Log[4*x^2] + (1 -
Log[x])*Log[4*x^2]^2))/((4 - 4*E^4 + E^8)*Log[x]^2 + (-4 + 2*E^4)*Log[x]^2*Log[4*x^2] + Log[x]^2*Log[4*x^2]^2
),x]

[Out]

x + Defer[Int][E^(4/(-2*(1 - E^4/2) + Log[4*x^2]))/Log[x]^2, x] - Defer[Int][E^(4/(-2*(1 - E^4/2) + Log[4*x^2]
))/Log[x], x] + 8*Defer[Int][E^(4/(-2*(1 - E^4/2) + Log[4*x^2]))/(Log[x]*(2*(1 - E^4/2) - Log[4*x^2])^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (4-4 e^4+e^8\right ) \log ^2(x)+\left (-4+2 e^4\right ) \log ^2(x) \log \left (4 x^2\right )+\log ^2(x) \log ^2\left (4 x^2\right )+e^{\frac {4}{-2+e^4+\log \left (4 x^2\right )}} \left (4-4 e^4+e^8+\left (4+4 e^4-e^8\right ) \log (x)+\left (-4+2 e^4+\left (4-2 e^4\right ) \log (x)\right ) \log \left (4 x^2\right )+(1-\log (x)) \log ^2\left (4 x^2\right )\right )}{\log ^2(x) \left (2 \left (1-\frac {e^4}{2}\right )-\log \left (4 x^2\right )\right )^2} \, dx\\ &=\int \left (1+\frac {e^{\frac {4}{-2 \left (1-\frac {e^4}{2}\right )+\log \left (4 x^2\right )}} \left (4 \left (1+\frac {1}{4} e^4 \left (-4+e^4\right )\right )+4 \left (1+e^4-\frac {e^8}{4}\right ) \log (x)-4 \left (1-\frac {e^4}{2}\right ) \log \left (4 x^2\right )+4 \left (1-\frac {e^4}{2}\right ) \log (x) \log \left (4 x^2\right )+\log ^2\left (4 x^2\right )-\log (x) \log ^2\left (4 x^2\right )\right )}{\log ^2(x) \left (2 \left (1-\frac {e^4}{2}\right )-\log \left (4 x^2\right )\right )^2}\right ) \, dx\\ &=x+\int \frac {e^{\frac {4}{-2 \left (1-\frac {e^4}{2}\right )+\log \left (4 x^2\right )}} \left (4 \left (1+\frac {1}{4} e^4 \left (-4+e^4\right )\right )+4 \left (1+e^4-\frac {e^8}{4}\right ) \log (x)-4 \left (1-\frac {e^4}{2}\right ) \log \left (4 x^2\right )+4 \left (1-\frac {e^4}{2}\right ) \log (x) \log \left (4 x^2\right )+\log ^2\left (4 x^2\right )-\log (x) \log ^2\left (4 x^2\right )\right )}{\log ^2(x) \left (2 \left (1-\frac {e^4}{2}\right )-\log \left (4 x^2\right )\right )^2} \, dx\\ &=x+\int \frac {e^{\frac {4}{-2 \left (1-\frac {e^4}{2}\right )+\log \left (4 x^2\right )}} \left (\left (-2+e^4+\log \left (4 x^2\right )\right )^2-\log (x) \left (-4-4 e^4+e^8+2 \left (-2+e^4\right ) \log \left (4 x^2\right )+\log ^2\left (4 x^2\right )\right )\right )}{\log ^2(x) \left (2 \left (1-\frac {e^4}{2}\right )-\log \left (4 x^2\right )\right )^2} \, dx\\ &=x+\int \left (\frac {e^{\frac {4}{-2 \left (1-\frac {e^4}{2}\right )+\log \left (4 x^2\right )}} (1-\log (x))}{\log ^2(x)}+\frac {8 e^{\frac {4}{-2 \left (1-\frac {e^4}{2}\right )+\log \left (4 x^2\right )}}}{\log (x) \left (2 \left (1-\frac {e^4}{2}\right )-\log \left (4 x^2\right )\right )^2}\right ) \, dx\\ &=x+8 \int \frac {e^{\frac {4}{-2 \left (1-\frac {e^4}{2}\right )+\log \left (4 x^2\right )}}}{\log (x) \left (2 \left (1-\frac {e^4}{2}\right )-\log \left (4 x^2\right )\right )^2} \, dx+\int \frac {e^{\frac {4}{-2 \left (1-\frac {e^4}{2}\right )+\log \left (4 x^2\right )}} (1-\log (x))}{\log ^2(x)} \, dx\\ &=x+8 \int \frac {e^{\frac {4}{-2 \left (1-\frac {e^4}{2}\right )+\log \left (4 x^2\right )}}}{\log (x) \left (2 \left (1-\frac {e^4}{2}\right )-\log \left (4 x^2\right )\right )^2} \, dx+\int \left (\frac {e^{\frac {4}{-2 \left (1-\frac {e^4}{2}\right )+\log \left (4 x^2\right )}}}{\log ^2(x)}-\frac {e^{\frac {4}{-2 \left (1-\frac {e^4}{2}\right )+\log \left (4 x^2\right )}}}{\log (x)}\right ) \, dx\\ &=x+8 \int \frac {e^{\frac {4}{-2 \left (1-\frac {e^4}{2}\right )+\log \left (4 x^2\right )}}}{\log (x) \left (2 \left (1-\frac {e^4}{2}\right )-\log \left (4 x^2\right )\right )^2} \, dx+\int \frac {e^{\frac {4}{-2 \left (1-\frac {e^4}{2}\right )+\log \left (4 x^2\right )}}}{\log ^2(x)} \, dx-\int \frac {e^{\frac {4}{-2 \left (1-\frac {e^4}{2}\right )+\log \left (4 x^2\right )}}}{\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 26, normalized size = 0.72 \begin {gather*} x-\frac {e^{\frac {4}{-2+e^4+\log \left (4 x^2\right )}} x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((4 - 4*E^4 + E^8)*Log[x]^2 + (-4 + 2*E^4)*Log[x]^2*Log[4*x^2] + Log[x]^2*Log[4*x^2]^2 + E^(4/(-2 +
E^4 + Log[4*x^2]))*(4 - 4*E^4 + E^8 + (4 + 4*E^4 - E^8)*Log[x] + (-4 + 2*E^4 + (4 - 2*E^4)*Log[x])*Log[4*x^2]
+ (1 - Log[x])*Log[4*x^2]^2))/((4 - 4*E^4 + E^8)*Log[x]^2 + (-4 + 2*E^4)*Log[x]^2*Log[4*x^2] + Log[x]^2*Log[4*
x^2]^2),x]

[Out]

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

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fricas [A]  time = 0.66, size = 31, normalized size = 0.86 \begin {gather*} -\frac {x e^{\left (\frac {4}{e^{4} + 2 \, \log \relax (2) + 2 \, \log \relax (x) - 2}\right )} - x \log \relax (x)}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Simplification assuming t_nostep near 0Simplification assuming t_nostep near 0Simplification assuming t_nos
tep near 0S

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maple [B]  time = 0.41, size = 76, normalized size = 2.11

method	result	size
default	\(x +\frac {\left (2-{\mathrm e}^{4}-\ln \left (4 x^{2}\right )+2 \ln \relax (x )\right ) x \,{\mathrm e}^{\frac {4}{\ln \left (4 x^{2}\right )+{\mathrm e}^{4}-2}}-2 x \ln \relax (x ) {\mathrm e}^{\frac {4}{\ln \left (4 x^{2}\right )+{\mathrm e}^{4}-2}}}{\left (\ln \left (4 x^{2}\right )+{\mathrm e}^{4}-2\right ) \ln \relax (x )}\)	\(76\)
risch	\(x -\frac {x \,{\mathrm e}^{\frac {8}{-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+4 \ln \relax (x )+2 \,{\mathrm e}^{4}+4 \ln \relax (2)-4}}}{\ln \relax (x )}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((1-ln(x))*ln(4*x^2)^2+((-2*exp(4)+4)*ln(x)+2*exp(4)-4)*ln(4*x^2)+(-exp(4)^2+4*exp(4)+4)*ln(x)+exp(4)^2-4
*exp(4)+4)*exp(4/(ln(4*x^2)+exp(4)-2))+ln(x)^2*ln(4*x^2)^2+(2*exp(4)-4)*ln(x)^2*ln(4*x^2)+(exp(4)^2-4*exp(4)+4
)*ln(x)^2)/(ln(x)^2*ln(4*x^2)^2+(2*exp(4)-4)*ln(x)^2*ln(4*x^2)+(exp(4)^2-4*exp(4)+4)*ln(x)^2),x,method=_RETURN
VERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} x - \int \frac {{\left (4 \, {\left (e^{4} + 2 \, \log \relax (2) - 3\right )} \log \relax (x)^{2} + 4 \, \log \relax (x)^{3} - 4 \, {\left (\log \relax (2) - 1\right )} e^{4} - 4 \, \log \relax (2)^{2} + {\left (4 \, {\left (\log \relax (2) - 2\right )} e^{4} + 4 \, \log \relax (2)^{2} + e^{8} - 16 \, \log \relax (2) + 4\right )} \log \relax (x) - e^{8} + 8 \, \log \relax (2) - 4\right )} e^{\left (\frac {4}{e^{4} + 2 \, \log \relax (2) + 2 \, \log \relax (x) - 2}\right )}}{4 \, {\left (e^{4} + 2 \, \log \relax (2) - 2\right )} \log \relax (x)^{3} + 4 \, \log \relax (x)^{4} + {\left (4 \, {\left (\log \relax (2) - 1\right )} e^{4} + 4 \, \log \relax (2)^{2} + e^{8} - 8 \, \log \relax (2) + 4\right )} \log \relax (x)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - integrate((4*(e^4 + 2*log(2) - 3)*log(x)^2 + 4*log(x)^3 - 4*(log(2) - 1)*e^4 - 4*log(2)^2 + (4*(log(2) - 2
)*e^4 + 4*log(2)^2 + e^8 - 16*log(2) + 4)*log(x) - e^8 + 8*log(2) - 4)*e^(4/(e^4 + 2*log(2) + 2*log(x) - 2))/(
4*(e^4 + 2*log(2) - 2)*log(x)^3 + 4*log(x)^4 + (4*(log(2) - 1)*e^4 + 4*log(2)^2 + e^8 - 8*log(2) + 4)*log(x)^2
), x)

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mupad [B]  time = 2.92, size = 24, normalized size = 0.67 \begin {gather*} x-\frac {x\,{\mathrm {e}}^{\frac {4}{{\mathrm {e}}^4+\ln \left (4\,x^2\right )-2}}}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)^2*(exp(8) - 4*exp(4) + 4) - exp(4/(exp(4) + log(4*x^2) - 2))*(4*exp(4) - exp(8) - log(x)*(4*exp(4)
- exp(8) + 4) + log(4*x^2)*(log(x)*(2*exp(4) - 4) - 2*exp(4) + 4) + log(4*x^2)^2*(log(x) - 1) - 4) + log(4*x^
2)^2*log(x)^2 + log(4*x^2)*log(x)^2*(2*exp(4) - 4))/(log(x)^2*(exp(8) - 4*exp(4) + 4) + log(4*x^2)^2*log(x)^2
+ log(4*x^2)*log(x)^2*(2*exp(4) - 4)),x)

[Out]

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

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sympy [A]  time = 0.72, size = 22, normalized size = 0.61 \begin {gather*} - \frac {x e^{\frac {4}{2 \log {\relax (x )} - 2 + \log {\relax (4 )} + e^{4}}}}{\log {\relax (x )}} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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