3.17.44 \(\int \frac {e^{3+x \log (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)})} (-e+320 x+80 x \log (x)+5 x \log ^2(x)+(8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)) \log (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}))}{8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ e^{3+x \log \left (5 (x+\log (2))+\frac {e}{8+\log (x)}\right )} \]

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Rubi [F]  time = 3.78, 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 (3+x \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )\right ) \left (-e+320 x+80 x \log (x)+5 x \log ^2(x)+\left (8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)\right ) \log \left (\frac {e+40 x+40 \log (2)+(5 x+5 \log (2)) \log (x)}{8+\log (x)}\right )\right )}{8 e+320 x+320 \log (2)+(e+80 x+80 \log (2)) \log (x)+(5 x+5 \log (2)) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(3 + x*Log[(E + 40*x + 40*Log[2] + (5*x + 5*Log[2])*Log[x])/(8 + Log[x])])*(-E + 320*x + 80*x*Log[x] +
5*x*Log[x]^2 + (8*E + 320*x + 320*Log[2] + (E + 80*x + 80*Log[2])*Log[x] + (5*x + 5*Log[2])*Log[x]^2)*Log[(E +
40*x + 40*Log[2] + (5*x + 5*Log[2])*Log[x])/(8 + Log[x])]))/(8*E + 320*x + 320*Log[2] + (E + 80*x + 80*Log[2]
)*Log[x] + (5*x + 5*Log[2])*Log[x]^2),x]

[Out]

5*E^3*Defer[Int][x*((E + 40*(x + Log[2]) + 5*(x + Log[2])*Log[x])/(8 + Log[x]))^(-1 + x), x] - E^4*Defer[Int][
((E + 40*(x + Log[2]) + 5*(x + Log[2])*Log[x])/(8 + Log[x]))^(-1 + x)/(8 + Log[x])^2, x] + E^3*Defer[Int][((E
+ 40*(x + Log[2]) + 5*(x + Log[2])*Log[x])/(8 + Log[x]))^x*Log[(E + 40*(x + Log[2]) + 5*(x + Log[2])*Log[x])/(
8 + Log[x])], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^3 \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x} \left (-e+320 x+80 x \log (x)+5 x \log ^2(x)+(8+\log (x)) (e+40 (x+\log (2))+5 (x+\log (2)) \log (x)) \log \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )\right )}{(8+\log (x))^2} \, dx\\ &=e^3 \int \frac {\left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x} \left (-e+320 x+80 x \log (x)+5 x \log ^2(x)+(8+\log (x)) (e+40 (x+\log (2))+5 (x+\log (2)) \log (x)) \log \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )\right )}{(8+\log (x))^2} \, dx\\ &=e^3 \int \left (-\frac {e \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{(8+\log (x))^2}+\frac {320 x \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{(8+\log (x))^2}+\frac {80 x \log (x) \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{(8+\log (x))^2}+\frac {5 x \log ^2(x) \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{(8+\log (x))^2}+\frac {\left (40 x+e \left (1+\frac {40 \log (2)}{e}\right )+5 x \log (x)+5 \log (2) \log (x)\right ) \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x} \log \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )}{8+\log (x)}\right ) \, dx\\ &=e^3 \int \frac {\left (40 x+e \left (1+\frac {40 \log (2)}{e}\right )+5 x \log (x)+5 \log (2) \log (x)\right ) \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x} \log \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )}{8+\log (x)} \, dx+\left (5 e^3\right ) \int \frac {x \log ^2(x) \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{(8+\log (x))^2} \, dx+\left (80 e^3\right ) \int \frac {x \log (x) \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{(8+\log (x))^2} \, dx+\left (320 e^3\right ) \int \frac {x \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{(8+\log (x))^2} \, dx-e^4 \int \frac {\left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{(8+\log (x))^2} \, dx\\ &=e^3 \int \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^x \log \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right ) \, dx+\left (5 e^3\right ) \int \left (x \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}+\frac {64 x \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{(8+\log (x))^2}-\frac {16 x \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{8+\log (x)}\right ) \, dx+\left (80 e^3\right ) \int \left (-\frac {8 x \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{(8+\log (x))^2}+\frac {x \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{8+\log (x)}\right ) \, dx+\left (320 e^3\right ) \int \frac {x \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{(8+\log (x))^2} \, dx-e^4 \int \frac {\left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{(8+\log (x))^2} \, dx\\ &=e^3 \int \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^x \log \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right ) \, dx+\left (5 e^3\right ) \int x \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x} \, dx+2 \left (\left (320 e^3\right ) \int \frac {x \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{(8+\log (x))^2} \, dx\right )-\left (640 e^3\right ) \int \frac {x \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{(8+\log (x))^2} \, dx-e^4 \int \frac {\left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^{-1+x}}{(8+\log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 29, normalized size = 1.32 \begin {gather*} e^3 \left (\frac {e+40 (x+\log (2))+5 (x+\log (2)) \log (x)}{8+\log (x)}\right )^x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3 + x*Log[(E + 40*x + 40*Log[2] + (5*x + 5*Log[2])*Log[x])/(8 + Log[x])])*(-E + 320*x + 80*x*Log
[x] + 5*x*Log[x]^2 + (8*E + 320*x + 320*Log[2] + (E + 80*x + 80*Log[2])*Log[x] + (5*x + 5*Log[2])*Log[x]^2)*Lo
g[(E + 40*x + 40*Log[2] + (5*x + 5*Log[2])*Log[x])/(8 + Log[x])]))/(8*E + 320*x + 320*Log[2] + (E + 80*x + 80*
Log[2])*Log[x] + (5*x + 5*Log[2])*Log[x]^2),x]

[Out]

E^3*((E + 40*(x + Log[2]) + 5*(x + Log[2])*Log[x])/(8 + Log[x]))^x

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fricas [A]  time = 0.88, size = 31, normalized size = 1.41 \begin {gather*} e^{\left (x \log \left (\frac {5 \, {\left (x + \log \relax (2)\right )} \log \relax (x) + 40 \, x + e + 40 \, \log \relax (2)}{\log \relax (x) + 8}\right ) + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log(2)+8*exp(1)+320*x)*log(((5*log(2)+5
*x)*log(x)+40*log(2)+exp(1)+40*x)/(log(x)+8))+5*x*log(x)^2+80*x*log(x)-exp(1)+320*x)*exp(x*log(((5*log(2)+5*x)
*log(x)+40*log(2)+exp(1)+40*x)/(log(x)+8))+3)/((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log(
2)+8*exp(1)+320*x),x, algorithm="fricas")

[Out]

e^(x*log((5*(x + log(2))*log(x) + 40*x + e + 40*log(2))/(log(x) + 8)) + 3)

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giac [B]  time = 0.88, size = 58, normalized size = 2.64 \begin {gather*} e^{\left (x \log \left (\frac {5 \, x \log \relax (x)}{\log \relax (x) + 8} + \frac {5 \, \log \relax (2) \log \relax (x)}{\log \relax (x) + 8} + \frac {40 \, x}{\log \relax (x) + 8} + \frac {e}{\log \relax (x) + 8} + \frac {40 \, \log \relax (2)}{\log \relax (x) + 8}\right ) + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log(2)+8*exp(1)+320*x)*log(((5*log(2)+5
*x)*log(x)+40*log(2)+exp(1)+40*x)/(log(x)+8))+5*x*log(x)^2+80*x*log(x)-exp(1)+320*x)*exp(x*log(((5*log(2)+5*x)
*log(x)+40*log(2)+exp(1)+40*x)/(log(x)+8))+3)/((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log(
2)+8*exp(1)+320*x),x, algorithm="giac")

[Out]

e^(x*log(5*x*log(x)/(log(x) + 8) + 5*log(2)*log(x)/(log(x) + 8) + 40*x/(log(x) + 8) + e/(log(x) + 8) + 40*log(
2)/(log(x) + 8)) + 3)

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maple [C]  time = 0.65, size = 249, normalized size = 11.32

method	result	size
risch	\(\left (\ln \relax (x )+8\right )^{-x} \left ({\mathrm e}+\left (5 \ln \relax (x )+40\right ) \ln \relax (2)+\left (5 \ln \relax (x )+40\right ) x \right )^{x} {\mathrm e}^{3-\frac {i x \pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}+\left (5 \ln \relax (x )+40\right ) \ln \relax (2)+\left (5 \ln \relax (x )+40\right ) x \right )}{\ln \relax (x )+8}\right )^{3}}{2}+\frac {i x \pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}+\left (5 \ln \relax (x )+40\right ) \ln \relax (2)+\left (5 \ln \relax (x )+40\right ) x \right )}{\ln \relax (x )+8}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )+8}\right )}{2}+\frac {i x \pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}+\left (5 \ln \relax (x )+40\right ) \ln \relax (2)+\left (5 \ln \relax (x )+40\right ) x \right )}{\ln \relax (x )+8}\right )^{2} \mathrm {csgn}\left (i \left ({\mathrm e}+\left (5 \ln \relax (x )+40\right ) \ln \relax (2)+\left (5 \ln \relax (x )+40\right ) x \right )\right )}{2}-\frac {i x \pi \,\mathrm {csgn}\left (\frac {i \left ({\mathrm e}+\left (5 \ln \relax (x )+40\right ) \ln \relax (2)+\left (5 \ln \relax (x )+40\right ) x \right )}{\ln \relax (x )+8}\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )+8}\right ) \mathrm {csgn}\left (i \left ({\mathrm e}+\left (5 \ln \relax (x )+40\right ) \ln \relax (2)+\left (5 \ln \relax (x )+40\right ) x \right )\right )}{2}}\)	\(249\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((5*ln(2)+5*x)*ln(x)^2+(80*ln(2)+exp(1)+80*x)*ln(x)+320*ln(2)+8*exp(1)+320*x)*ln(((5*ln(2)+5*x)*ln(x)+40*
ln(2)+exp(1)+40*x)/(ln(x)+8))+5*x*ln(x)^2+80*x*ln(x)-exp(1)+320*x)*exp(x*ln(((5*ln(2)+5*x)*ln(x)+40*ln(2)+exp(
1)+40*x)/(ln(x)+8))+3)/((5*ln(2)+5*x)*ln(x)^2+(80*ln(2)+exp(1)+80*x)*ln(x)+320*ln(2)+8*exp(1)+320*x),x,method=
_RETURNVERBOSE)

[Out]

(ln(x)+8)^(-x)*(exp(1)+(5*ln(x)+40)*ln(2)+(5*ln(x)+40)*x)^x*exp(3-1/2*I*x*Pi*csgn(I/(ln(x)+8)*(exp(1)+(5*ln(x)
+40)*ln(2)+(5*ln(x)+40)*x))^3+1/2*I*x*Pi*csgn(I/(ln(x)+8)*(exp(1)+(5*ln(x)+40)*ln(2)+(5*ln(x)+40)*x))^2*csgn(I
/(ln(x)+8))+1/2*I*x*Pi*csgn(I/(ln(x)+8)*(exp(1)+(5*ln(x)+40)*ln(2)+(5*ln(x)+40)*x))^2*csgn(I*(exp(1)+(5*ln(x)+
40)*ln(2)+(5*ln(x)+40)*x))-1/2*I*x*Pi*csgn(I/(ln(x)+8)*(exp(1)+(5*ln(x)+40)*ln(2)+(5*ln(x)+40)*x))*csgn(I/(ln(
x)+8))*csgn(I*(exp(1)+(5*ln(x)+40)*ln(2)+(5*ln(x)+40)*x)))

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maxima [A]  time = 0.74, size = 32, normalized size = 1.45 \begin {gather*} e^{\left (x \log \left (5 \, {\left (x + \log \relax (2)\right )} \log \relax (x) + 40 \, x + e + 40 \, \log \relax (2)\right ) - x \log \left (\log \relax (x) + 8\right ) + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log(2)+8*exp(1)+320*x)*log(((5*log(2)+5
*x)*log(x)+40*log(2)+exp(1)+40*x)/(log(x)+8))+5*x*log(x)^2+80*x*log(x)-exp(1)+320*x)*exp(x*log(((5*log(2)+5*x)
*log(x)+40*log(2)+exp(1)+40*x)/(log(x)+8))+3)/((5*log(2)+5*x)*log(x)^2+(80*log(2)+exp(1)+80*x)*log(x)+320*log(
2)+8*exp(1)+320*x),x, algorithm="maxima")

[Out]

e^(x*log(5*(x + log(2))*log(x) + 40*x + e + 40*log(2)) - x*log(log(x) + 8) + 3)

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mupad [B]  time = 2.15, size = 33, normalized size = 1.50 \begin {gather*} {\mathrm {e}}^3\,{\left (\frac {40\,x+\mathrm {e}+40\,\ln \relax (2)+5\,\ln \relax (2)\,\ln \relax (x)+5\,x\,\ln \relax (x)}{\ln \relax (x)+8}\right )}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x*log((40*x + exp(1) + 40*log(2) + log(x)*(5*x + 5*log(2)))/(log(x) + 8)) + 3)*(320*x - exp(1) + 5*x*
log(x)^2 + log((40*x + exp(1) + 40*log(2) + log(x)*(5*x + 5*log(2)))/(log(x) + 8))*(320*x + 8*exp(1) + 320*log
(2) + log(x)^2*(5*x + 5*log(2)) + log(x)*(80*x + exp(1) + 80*log(2))) + 80*x*log(x)))/(320*x + 8*exp(1) + 320*
log(2) + log(x)^2*(5*x + 5*log(2)) + log(x)*(80*x + exp(1) + 80*log(2))),x)

[Out]

exp(3)*((40*x + exp(1) + 40*log(2) + 5*log(2)*log(x) + 5*x*log(x))/(log(x) + 8))^x

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sympy [A]  time = 2.69, size = 34, normalized size = 1.55 \begin {gather*} e^{x \log {\left (\frac {40 x + \left (5 x + 5 \log {\relax (2 )}\right ) \log {\relax (x )} + e + 40 \log {\relax (2 )}}{\log {\relax (x )} + 8} \right )} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*ln(2)+5*x)*ln(x)**2+(80*ln(2)+exp(1)+80*x)*ln(x)+320*ln(2)+8*exp(1)+320*x)*ln(((5*ln(2)+5*x)*ln
(x)+40*ln(2)+exp(1)+40*x)/(ln(x)+8))+5*x*ln(x)**2+80*x*ln(x)-exp(1)+320*x)*exp(x*ln(((5*ln(2)+5*x)*ln(x)+40*ln
(2)+exp(1)+40*x)/(ln(x)+8))+3)/((5*ln(2)+5*x)*ln(x)**2+(80*ln(2)+exp(1)+80*x)*ln(x)+320*ln(2)+8*exp(1)+320*x),
x)

[Out]

exp(x*log((40*x + (5*x + 5*log(2))*log(x) + E + 40*log(2))/(log(x) + 8)) + 3)

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