3.7.5 \(\int \frac {e^{\frac {-10+(2 x^4+\log (x)) \log (4 x \log (x))}{2 \log (4 x \log (x))}} (10+10 \log (x)+(1+8 x^4) \log (x) \log ^2(4 x \log (x)))}{2 x \log (x) \log ^2(4 x \log (x))} \, dx\)

Optimal. Leaf size=22 \[ e^{x^4-\frac {5}{\log (4 x \log (x))}} \sqrt {x} \]

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Rubi [F]  time = 3.71, 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 {\exp \left (\frac {-10+\left (2 x^4+\log (x)\right ) \log (4 x \log (x))}{2 \log (4 x \log (x))}\right ) \left (10+10 \log (x)+\left (1+8 x^4\right ) \log (x) \log ^2(4 x \log (x))\right )}{2 x \log (x) \log ^2(4 x \log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-10 + (2*x^4 + Log[x])*Log[4*x*Log[x]])/(2*Log[4*x*Log[x]]))*(10 + 10*Log[x] + (1 + 8*x^4)*Log[x]*Log
[4*x*Log[x]]^2))/(2*x*Log[x]*Log[4*x*Log[x]]^2),x]

[Out]

Defer[Subst][Defer[Int][E^(x^8 - 5/Log[4*x^2*Log[x^2]]), x], x, Sqrt[x]] + 8*Defer[Subst][Defer[Int][E^(x^8 -
5/Log[4*x^2*Log[x^2]])*x^8, x], x, Sqrt[x]] + 10*Defer[Subst][Defer[Int][E^(x^8 - 5/Log[4*x^2*Log[x^2]])/Log[4
*x^2*Log[x^2]]^2, x], x, Sqrt[x]] + 10*Defer[Subst][Defer[Int][E^(x^8 - 5/Log[4*x^2*Log[x^2]])/(Log[x^2]*Log[4
*x^2*Log[x^2]]^2), x], x, Sqrt[x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {\exp \left (\frac {-10+\left (2 x^4+\log (x)\right ) \log (4 x \log (x))}{2 \log (4 x \log (x))}\right ) \left (10+10 \log (x)+\left (1+8 x^4\right ) \log (x) \log ^2(4 x \log (x))\right )}{x \log (x) \log ^2(4 x \log (x))} \, dx\\ &=\frac {1}{2} \int \left (\frac {\exp \left (\frac {-10+\left (2 x^4+\log (x)\right ) \log (4 x \log (x))}{2 \log (4 x \log (x))}\right ) \left (1+8 x^4\right )}{x}+\frac {10 \exp \left (\frac {-10+\left (2 x^4+\log (x)\right ) \log (4 x \log (x))}{2 \log (4 x \log (x))}\right ) (1+\log (x))}{x \log (x) \log ^2(4 x \log (x))}\right ) \, dx\\ &=\frac {1}{2} \int \frac {\exp \left (\frac {-10+\left (2 x^4+\log (x)\right ) \log (4 x \log (x))}{2 \log (4 x \log (x))}\right ) \left (1+8 x^4\right )}{x} \, dx+5 \int \frac {\exp \left (\frac {-10+\left (2 x^4+\log (x)\right ) \log (4 x \log (x))}{2 \log (4 x \log (x))}\right ) (1+\log (x))}{x \log (x) \log ^2(4 x \log (x))} \, dx\\ &=\frac {1}{2} \int \frac {e^{x^4-\frac {5}{\log (4 x \log (x))}} \left (1+8 x^4\right )}{\sqrt {x}} \, dx+5 \int \frac {e^{x^4-\frac {5}{\log (4 x \log (x))}} (1+\log (x))}{\sqrt {x} \log (x) \log ^2(4 x \log (x))} \, dx\\ &=10 \operatorname {Subst}\left (\int \frac {e^{x^8-\frac {5}{\log \left (4 x^2 \log \left (x^2\right )\right )}} \left (1+\log \left (x^2\right )\right )}{\log \left (x^2\right ) \log ^2\left (4 x^2 \log \left (x^2\right )\right )} \, dx,x,\sqrt {x}\right )+\operatorname {Subst}\left (\int e^{x^8-\frac {5}{\log \left (4 x^2 \log \left (x^2\right )\right )}} \left (1+8 x^8\right ) \, dx,x,\sqrt {x}\right )\\ &=10 \operatorname {Subst}\left (\int \left (\frac {e^{x^8-\frac {5}{\log \left (4 x^2 \log \left (x^2\right )\right )}}}{\log ^2\left (4 x^2 \log \left (x^2\right )\right )}+\frac {e^{x^8-\frac {5}{\log \left (4 x^2 \log \left (x^2\right )\right )}}}{\log \left (x^2\right ) \log ^2\left (4 x^2 \log \left (x^2\right )\right )}\right ) \, dx,x,\sqrt {x}\right )+\operatorname {Subst}\left (\int \left (e^{x^8-\frac {5}{\log \left (4 x^2 \log \left (x^2\right )\right )}}+8 e^{x^8-\frac {5}{\log \left (4 x^2 \log \left (x^2\right )\right )}} x^8\right ) \, dx,x,\sqrt {x}\right )\\ &=8 \operatorname {Subst}\left (\int e^{x^8-\frac {5}{\log \left (4 x^2 \log \left (x^2\right )\right )}} x^8 \, dx,x,\sqrt {x}\right )+10 \operatorname {Subst}\left (\int \frac {e^{x^8-\frac {5}{\log \left (4 x^2 \log \left (x^2\right )\right )}}}{\log ^2\left (4 x^2 \log \left (x^2\right )\right )} \, dx,x,\sqrt {x}\right )+10 \operatorname {Subst}\left (\int \frac {e^{x^8-\frac {5}{\log \left (4 x^2 \log \left (x^2\right )\right )}}}{\log \left (x^2\right ) \log ^2\left (4 x^2 \log \left (x^2\right )\right )} \, dx,x,\sqrt {x}\right )+\operatorname {Subst}\left (\int e^{x^8-\frac {5}{\log \left (4 x^2 \log \left (x^2\right )\right )}} \, dx,x,\sqrt {x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.67, size = 22, normalized size = 1.00 \begin {gather*} e^{x^4-\frac {5}{\log (4 x \log (x))}} \sqrt {x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-10 + (2*x^4 + Log[x])*Log[4*x*Log[x]])/(2*Log[4*x*Log[x]]))*(10 + 10*Log[x] + (1 + 8*x^4)*Log[
x]*Log[4*x*Log[x]]^2))/(2*x*Log[x]*Log[4*x*Log[x]]^2),x]

[Out]

E^(x^4 - 5/Log[4*x*Log[x]])*Sqrt[x]

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fricas [A]  time = 0.58, size = 28, normalized size = 1.27 \begin {gather*} e^{\left (\frac {{\left (2 \, x^{4} + \log \relax (x)\right )} \log \left (4 \, x \log \relax (x)\right ) - 10}{2 \, \log \left (4 \, x \log \relax (x)\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((8*x^4+1)*log(x)*log(4*x*log(x))^2+10*log(x)+10)*exp(1/2*((log(x)+2*x^4)*log(4*x*log(x))-10)/lo
g(4*x*log(x)))/x/log(x)/log(4*x*log(x))^2,x, algorithm="fricas")

[Out]

e^(1/2*((2*x^4 + log(x))*log(4*x*log(x)) - 10)/log(4*x*log(x)))

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giac [A]  time = 8.91, size = 19, normalized size = 0.86 \begin {gather*} e^{\left (x^{4} - \frac {5}{\log \left (4 \, x \log \relax (x)\right )} + \frac {1}{2} \, \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((8*x^4+1)*log(x)*log(4*x*log(x))^2+10*log(x)+10)*exp(1/2*((log(x)+2*x^4)*log(4*x*log(x))-10)/lo
g(4*x*log(x)))/x/log(x)/log(4*x*log(x))^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 6.74, size = 299, normalized size = 13.59

method	result	size
risch	\({\mathrm e}^{\frac {-2 i \pi \mathrm {csgn}\left (i x \ln \relax (x )\right )^{3} x^{4}+2 i \pi \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i x \right ) x^{4}+2 i \pi \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )\right ) x^{4}-2 i \pi \,\mathrm {csgn}\left (i x \ln \relax (x )\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \ln \relax (x )\right ) x^{4}-i \pi \mathrm {csgn}\left (i x \ln \relax (x )\right )^{3} \ln \relax (x )+i \pi \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i x \right ) \ln \relax (x )+i \pi \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )\right ) \ln \relax (x )-i \pi \,\mathrm {csgn}\left (i x \ln \relax (x )\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \ln \relax (x )\right ) \ln \relax (x )+8 x^{4} \ln \relax (2)+4 \ln \left (\ln \relax (x )\right ) x^{4}+4 x^{4} \ln \relax (x )+4 \ln \relax (2) \ln \relax (x )+2 \ln \relax (x ) \ln \left (\ln \relax (x )\right )+2 \ln \relax (x )^{2}-20}{-2 i \pi \mathrm {csgn}\left (i x \ln \relax (x )\right )^{3}+2 i \pi \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i x \right )+2 i \pi \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )\right )-2 i \pi \,\mathrm {csgn}\left (i x \ln \relax (x )\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \ln \relax (x )\right )+8 \ln \relax (2)+4 \ln \left (\ln \relax (x )\right )+4 \ln \relax (x )}}\)	\(299\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((8*x^4+1)*ln(x)*ln(4*x*ln(x))^2+10*ln(x)+10)*exp(1/2*((ln(x)+2*x^4)*ln(4*x*ln(x))-10)/ln(4*x*ln(x)))/
x/ln(x)/ln(4*x*ln(x))^2,x,method=_RETURNVERBOSE)

[Out]

exp(1/2*(-2*I*Pi*csgn(I*x*ln(x))^3*x^4+2*I*Pi*csgn(I*x*ln(x))^2*csgn(I*x)*x^4+2*I*Pi*csgn(I*x*ln(x))^2*csgn(I*
ln(x))*x^4-2*I*Pi*csgn(I*x*ln(x))*csgn(I*x)*csgn(I*ln(x))*x^4-I*Pi*csgn(I*x*ln(x))^3*ln(x)+I*Pi*csgn(I*x*ln(x)
)^2*csgn(I*x)*ln(x)+I*Pi*csgn(I*x*ln(x))^2*csgn(I*ln(x))*ln(x)-I*Pi*csgn(I*x*ln(x))*csgn(I*x)*csgn(I*ln(x))*ln
(x)+8*x^4*ln(2)+4*ln(ln(x))*x^4+4*x^4*ln(x)+4*ln(2)*ln(x)+2*ln(x)*ln(ln(x))+2*ln(x)^2-20)/(-I*Pi*csgn(I*x*ln(x
))^3+I*Pi*csgn(I*x*ln(x))^2*csgn(I*x)+I*Pi*csgn(I*x*ln(x))^2*csgn(I*ln(x))-I*Pi*csgn(I*x*ln(x))*csgn(I*x)*csgn
(I*ln(x))+4*ln(2)+2*ln(ln(x))+2*ln(x)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{2} \, \int \frac {{\left ({\left (8 \, x^{4} + 1\right )} \log \left (4 \, x \log \relax (x)\right )^{2} \log \relax (x) + 10 \, \log \relax (x) + 10\right )} e^{\left (\frac {{\left (2 \, x^{4} + \log \relax (x)\right )} \log \left (4 \, x \log \relax (x)\right ) - 10}{2 \, \log \left (4 \, x \log \relax (x)\right )}\right )}}{x \log \left (4 \, x \log \relax (x)\right )^{2} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((8*x^4+1)*log(x)*log(4*x*log(x))^2+10*log(x)+10)*exp(1/2*((log(x)+2*x^4)*log(4*x*log(x))-10)/lo
g(4*x*log(x)))/x/log(x)/log(4*x*log(x))^2,x, algorithm="maxima")

[Out]

1/2*integrate(((8*x^4 + 1)*log(4*x*log(x))^2*log(x) + 10*log(x) + 10)*e^(1/2*((2*x^4 + log(x))*log(4*x*log(x))
- 10)/log(4*x*log(x)))/(x*log(4*x*log(x))^2*log(x)), x)

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mupad [B]  time = 0.84, size = 62, normalized size = 2.82 \begin {gather*} x^{\frac {\ln \relax (2)}{\ln \left (4\,x\,\ln \relax (x)\right )}}\,{\mathrm {e}}^{\frac {\ln \left (x\,\ln \relax (x)\right )\,\ln \left (\sqrt {x}\right )}{\ln \left (4\,x\,\ln \relax (x)\right )}-\frac {5}{\ln \left (4\,x\,\ln \relax (x)\right )}}\,{\left (4\,x\,\ln \relax (x)\right )}^{\frac {x^4}{\ln \left (4\,x\,\ln \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(((log(4*x*log(x))*(log(x) + 2*x^4))/2 - 5)/log(4*x*log(x)))*(10*log(x) + log(4*x*log(x))^2*log(x)*(8*
x^4 + 1) + 10))/(2*x*log(4*x*log(x))^2*log(x)),x)

[Out]

x^(log(2)/log(4*x*log(x)))*exp((log(x*log(x))*log(x^(1/2)))/log(4*x*log(x)) - 5/log(4*x*log(x)))*(4*x*log(x))^
(x^4/log(4*x*log(x)))

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sympy [A]  time = 0.67, size = 29, normalized size = 1.32 \begin {gather*} e^{\frac {\frac {\left (2 x^{4} + \log {\relax (x )}\right ) \log {\left (4 x \log {\relax (x )} \right )}}{2} - 5}{\log {\left (4 x \log {\relax (x )} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((8*x**4+1)*ln(x)*ln(4*x*ln(x))**2+10*ln(x)+10)*exp(1/2*((ln(x)+2*x**4)*ln(4*x*ln(x))-10)/ln(4*x
*ln(x)))/x/ln(x)/ln(4*x*ln(x))**2,x)

[Out]

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

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