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

Optimal. Leaf size=25 \[ \frac {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )}}{x^2} \]

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Rubi [F]  time = 2.97, 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 {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \left (2 e^{5+e^x+3 x}+(-2+x) \log (x)+e^{5+e^x+3 x} \left (3 x+e^x x\right ) \log (x) \log \left (\log ^2(x)\right )\right )}{x^3 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-2*Defer[Int][E^(4 + x + E^(5 + E^x + 3*x)*Log[Log[x]^2])/x^3, x] + Defer[Int][E^(4 + x + E^(5 + E^x + 3*x)*Lo
g[Log[x]^2])/x^2, x] + 2*Defer[Int][E^(9 + E^x + 4*x + E^(5 + E^x + 3*x)*Log[Log[x]^2])/(x^3*Log[x]), x] + 3*D
efer[Int][(E^(9 + E^x + 4*x + E^(5 + E^x + 3*x)*Log[Log[x]^2])*Log[Log[x]^2])/x^2, x] + Defer[Int][(E^(9 + E^x
+ 5*x + E^(5 + E^x + 3*x)*Log[Log[x]^2])*Log[Log[x]^2])/x^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} (-2+x)}{x^3}+\frac {e^{9+e^x+5 x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \log \left (\log ^2(x)\right )}{x^2}+\frac {e^{9+e^x+4 x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \left (2+3 x \log (x) \log \left (\log ^2(x)\right )\right )}{x^3 \log (x)}\right ) \, dx\\ &=\int \frac {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} (-2+x)}{x^3} \, dx+\int \frac {e^{9+e^x+5 x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \log \left (\log ^2(x)\right )}{x^2} \, dx+\int \frac {e^{9+e^x+4 x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \left (2+3 x \log (x) \log \left (\log ^2(x)\right )\right )}{x^3 \log (x)} \, dx\\ &=\int \left (-\frac {2 e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )}}{x^3}+\frac {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )}}{x^2}\right ) \, dx+\int \frac {e^{9+e^x+5 x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \log \left (\log ^2(x)\right )}{x^2} \, dx+\int \left (\frac {2 e^{9+e^x+4 x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )}}{x^3 \log (x)}+\frac {3 e^{9+e^x+4 x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \log \left (\log ^2(x)\right )}{x^2}\right ) \, dx\\ &=-\left (2 \int \frac {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )}}{x^3} \, dx\right )+2 \int \frac {e^{9+e^x+4 x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )}}{x^3 \log (x)} \, dx+3 \int \frac {e^{9+e^x+4 x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \log \left (\log ^2(x)\right )}{x^2} \, dx+\int \frac {e^{4+x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )}}{x^2} \, dx+\int \frac {e^{9+e^x+5 x+e^{5+e^x+3 x} \log \left (\log ^2(x)\right )} \log \left (\log ^2(x)\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.93, size = 24, normalized size = 0.96 \begin {gather*} \frac {e^{4+x} \log ^2(x)^{e^{5+e^x+3 x}}}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.81, size = 22, normalized size = 0.88 \begin {gather*} \frac {e^{\left (e^{\left (3 \, x + e^{x} + 5\right )} \log \left (\log \relax (x)^{2}\right ) + x + 4\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [C]  time = 0.42, size = 100, normalized size = 4.00

method	result	size
risch	\(\frac {\ln \relax (x )^{2 \,{\mathrm e}^{{\mathrm e}^{x}+3 x +5}} {\mathrm e}^{4-\frac {i {\mathrm e}^{{\mathrm e}^{x}+3 x +5} \pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3}}{2}+i {\mathrm e}^{{\mathrm e}^{x}+3 x +5} \pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )\right )-\frac {i {\mathrm e}^{{\mathrm e}^{x}+3 x +5} \pi \,\mathrm {csgn}\left (i \ln \relax (x )^{2}\right ) \mathrm {csgn}\left (i \ln \relax (x )\right )^{2}}{2}+x}}{x^{2}}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/x^2*ln(x)^(2*exp(exp(x)+3*x+5))*exp(4-1/2*I*exp(exp(x)+3*x+5)*Pi*csgn(I*ln(x)^2)^3+I*exp(exp(x)+3*x+5)*Pi*cs
gn(I*ln(x)^2)^2*csgn(I*ln(x))-1/2*I*exp(exp(x)+3*x+5)*Pi*csgn(I*ln(x)^2)*csgn(I*ln(x))^2+x)

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maxima [A]  time = 0.48, size = 21, normalized size = 0.84 \begin {gather*} \frac {e^{\left (2 \, e^{\left (3 \, x + e^{x} + 5\right )} \log \left (\log \relax (x)\right ) + x + 4\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.50, size = 23, normalized size = 0.92 \begin {gather*} \frac {{\mathrm {e}}^4\,{\mathrm {e}}^x\,{\left ({\ln \relax (x)}^2\right )}^{{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^5}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 7.26, size = 24, normalized size = 0.96 \begin {gather*} \frac {e^{x + e^{3 x + e^{x} + 5} \log {\left (\log {\relax (x )}^{2} \right )} + 4}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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