3.50.99 \(\int \frac {e^{-\frac {-3 x+5 e^2 x \log (x)}{\log (x)}} (4 \log ^2(x)+(-3 x+3 x \log (x)-5 e^2 x \log ^2(x)) \log (16 x^4))}{x \log ^2(x)} \, dx\)

Optimal. Leaf size=23 \[ e^{-5 e^2 x+\frac {3 x}{\log (x)}} \log \left (16 x^4\right ) \]

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Rubi [B]  time = 0.19, antiderivative size = 93, normalized size of antiderivative = 4.04, number of steps used = 1, number of rules used = 1, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {2288} \begin {gather*} -\frac {e^{\frac {3 x}{\log (x)}-5 e^2 x} \left (3 x+5 e^2 x \log ^2(x)-3 x \log (x)\right ) \log \left (16 x^4\right )}{x \log ^2(x) \left (\frac {-5 e^2 \log (x)-5 e^2+3}{\log (x)}-\frac {3 x-5 e^2 x \log (x)}{x \log ^2(x)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*Log[x]^2 + (-3*x + 3*x*Log[x] - 5*E^2*x*Log[x]^2)*Log[16*x^4])/(E^((-3*x + 5*E^2*x*Log[x])/Log[x])*x*Lo
g[x]^2),x]

[Out]

-((E^(-5*E^2*x + (3*x)/Log[x])*(3*x - 3*x*Log[x] + 5*E^2*x*Log[x]^2)*Log[16*x^4])/(x*Log[x]^2*((3 - 5*E^2 - 5*
E^2*Log[x])/Log[x] - (3*x - 5*E^2*x*Log[x])/(x*Log[x]^2))))

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {e^{-5 e^2 x+\frac {3 x}{\log (x)}} \left (3 x-3 x \log (x)+5 e^2 x \log ^2(x)\right ) \log \left (16 x^4\right )}{x \log ^2(x) \left (\frac {3-5 e^2-5 e^2 \log (x)}{\log (x)}-\frac {3 x-5 e^2 x \log (x)}{x \log ^2(x)}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 23, normalized size = 1.00 \begin {gather*} e^{-5 e^2 x+\frac {3 x}{\log (x)}} \log \left (16 x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*Log[x]^2 + (-3*x + 3*x*Log[x] - 5*E^2*x*Log[x]^2)*Log[16*x^4])/(E^((-3*x + 5*E^2*x*Log[x])/Log[x]
)*x*Log[x]^2),x]

[Out]

E^(-5*E^2*x + (3*x)/Log[x])*Log[16*x^4]

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fricas [A]  time = 0.56, size = 25, normalized size = 1.09 \begin {gather*} 4 \, {\left (\log \relax (2) + \log \relax (x)\right )} e^{\left (-\frac {5 \, x e^{2} \log \relax (x) - 3 \, x}{\log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x*exp(2)*log(x)^2+3*x*log(x)-3*x)*log(16*x^4)+4*log(x)^2)/x/log(x)^2/exp((5*x*exp(2)*log(x)-3*x
)/log(x)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.28, size = 45, normalized size = 1.96 \begin {gather*} 4 \, e^{\left (-\frac {5 \, x e^{2} \log \relax (x) - 3 \, x}{\log \relax (x)}\right )} \log \relax (2) + 4 \, e^{\left (-\frac {5 \, x e^{2} \log \relax (x) - 3 \, x}{\log \relax (x)}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x*exp(2)*log(x)^2+3*x*log(x)-3*x)*log(16*x^4)+4*log(x)^2)/x/log(x)^2/exp((5*x*exp(2)*log(x)-3*x
)/log(x)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.22, size = 46, normalized size = 2.00

method	result	size
default	\(\frac {\left (\left (\ln \left (16 x^{4}\right )-4 \ln \relax (x )\right ) \ln \relax (x )+4 \ln \relax (x )^{2}\right ) {\mathrm e}^{-\frac {5 x \,{\mathrm e}^{2} \ln \relax (x )-3 x}{\ln \relax (x )}}}{\ln \relax (x )}\)	\(46\)
risch	\(\left (4 \ln \relax (2)+4 \ln \relax (x )-\frac {i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )}{2}-\frac {i \pi \mathrm {csgn}\left (i x^{3}\right )^{3}}{2}+\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}}{2}+\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}}{2}-\frac {i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )}{2}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\frac {i \pi \,\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )^{2}}{2}+\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{4}\right )^{2}}{2}-\frac {i \pi \mathrm {csgn}\left (i x^{4}\right )^{3}}{2}\right ) {\mathrm e}^{-\frac {x \left (5 \,{\mathrm e}^{2} \ln \relax (x )-3\right )}{\ln \relax (x )}}\)	\(224\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x*exp(2)*ln(x)^2+3*x*ln(x)-3*x)*ln(16*x^4)+4*ln(x)^2)/x/ln(x)^2/exp((5*x*exp(2)*ln(x)-3*x)/ln(x)),x,m
ethod=_RETURNVERBOSE)

[Out]

((ln(16*x^4)-4*ln(x))*ln(x)+4*ln(x)^2)/ln(x)/exp((5*x*exp(2)*ln(x)-3*x)/ln(x))

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maxima [A]  time = 0.49, size = 21, normalized size = 0.91 \begin {gather*} 4 \, {\left (\log \relax (2) + \log \relax (x)\right )} e^{\left (-5 \, x e^{2} + \frac {3 \, x}{\log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x*exp(2)*log(x)^2+3*x*log(x)-3*x)*log(16*x^4)+4*log(x)^2)/x/log(x)^2/exp((5*x*exp(2)*log(x)-3*x
)/log(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.53, size = 21, normalized size = 0.91 \begin {gather*} {\mathrm {e}}^{\frac {3\,x}{\ln \relax (x)}-5\,x\,{\mathrm {e}}^2}\,\ln \left (16\,x^4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((3*x - 5*x*exp(2)*log(x))/log(x))*(4*log(x)^2 - log(16*x^4)*(3*x - 3*x*log(x) + 5*x*exp(2)*log(x)^2))
)/(x*log(x)^2),x)

[Out]

exp((3*x)/log(x) - 5*x*exp(2))*log(16*x^4)

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sympy [A]  time = 7.48, size = 27, normalized size = 1.17 \begin {gather*} \left (4 \log {\relax (x )} + 4 \log {\relax (2 )}\right ) e^{- \frac {5 x e^{2} \log {\relax (x )} - 3 x}{\log {\relax (x )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x*exp(2)*ln(x)**2+3*x*ln(x)-3*x)*ln(16*x**4)+4*ln(x)**2)/x/ln(x)**2/exp((5*x*exp(2)*ln(x)-3*x)/
ln(x)),x)

[Out]

(4*log(x) + 4*log(2))*exp(-(5*x*exp(2)*log(x) - 3*x)/log(x))

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