3.29.96 \(\int \frac {e^4 (65536 x^4+49152 x^3 \log (2)+13824 x^2 \log ^2(2)+1728 x \log ^3(2)+81 \log ^4(2)) (-16 x-240 x^2+(-3+15 x) \log (2)+64 x \log (x))}{\log ^4(2) (5 x-\log (x)) (-80 x^3-15 x^2 \log (2)+(16 x^2+3 x \log (2)) \log (x))} \, dx\)

Optimal. Leaf size=25 \[ \frac {e^4 \left (3+\frac {16 x}{\log (2)}\right )^4}{5 x-\log (x)} \]

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Rubi [F]  time = 1.52, 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 \left (65536 x^4+49152 x^3 \log (2)+13824 x^2 \log ^2(2)+1728 x \log ^3(2)+81 \log ^4(2)\right ) \left (-16 x-240 x^2+(-3+15 x) \log (2)+64 x \log (x)\right )}{\log ^4(2) (5 x-\log (x)) \left (-80 x^3-15 x^2 \log (2)+\left (16 x^2+3 x \log (2)\right ) \log (x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^4*(65536*x^4 + 49152*x^3*Log[2] + 13824*x^2*Log[2]^2 + 1728*x*Log[2]^3 + 81*Log[2]^4)*(-16*x - 240*x^2
+ (-3 + 15*x)*Log[2] + 64*x*Log[x]))/(Log[2]^4*(5*x - Log[x])*(-80*x^3 - 15*x^2*Log[2] + (16*x^2 + 3*x*Log[2])
*Log[x])),x]

[Out]

(E^4*(64 - 15*Log[2])*Log[8]^3*Defer[Int][(5*x - Log[x])^(-2), x])/Log[2]^4 + (E^4*Log[8]^4*Defer[Int][1/(x*(5
*x - Log[x])^2), x])/Log[2]^4 + (16*E^4*(96 - 45*Log[2] - 5*Log[8])*Log[8]^2*Defer[Int][x/(5*x - Log[x])^2, x]
)/Log[2]^4 + (256*E^4*(64 - 45*Log[2] - 15*Log[8])*Log[8]*Defer[Int][x^2/(5*x - Log[x])^2, x])/Log[2]^4 + (409
6*E^4*(16 - 15*Log[2] - 15*Log[8])*Defer[Int][x^3/(5*x - Log[x])^2, x])/Log[2]^4 - (327680*E^4*Defer[Int][x^4/
(5*x - Log[x])^2, x])/Log[2]^4 + (64*E^4*Log[8]^3*Defer[Int][(5*x - Log[x])^(-1), x])/Log[2]^4 + (3072*E^4*Log
[8]^2*Defer[Int][x/(5*x - Log[x]), x])/Log[2]^4 + (49152*E^4*Log[8]*Defer[Int][x^2/(5*x - Log[x]), x])/Log[2]^
4 + (262144*E^4*Defer[Int][x^3/(5*x - Log[x]), x])/Log[2]^4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {e^4 \int \frac {\left (65536 x^4+49152 x^3 \log (2)+13824 x^2 \log ^2(2)+1728 x \log ^3(2)+81 \log ^4(2)\right ) \left (-16 x-240 x^2+(-3+15 x) \log (2)+64 x \log (x)\right )}{(5 x-\log (x)) \left (-80 x^3-15 x^2 \log (2)+\left (16 x^2+3 x \log (2)\right ) \log (x)\right )} \, dx}{\log ^4(2)}\\ &=\frac {e^4 \int \frac {(16 x+\log (8))^3 \left (240 x^2+x (16-15 \log (2))+\log (8)-64 x \log (x)\right )}{x (5 x-\log (x))^2} \, dx}{\log ^4(2)}\\ &=\frac {e^4 \int \left (\frac {(16 x+\log (8))^3 \left (-80 x^2+x (16-15 \log (2))+\log (8)\right )}{x (5 x-\log (x))^2}+\frac {64 (16 x+\log (8))^3}{5 x-\log (x)}\right ) \, dx}{\log ^4(2)}\\ &=\frac {e^4 \int \frac {(16 x+\log (8))^3 \left (-80 x^2+x (16-15 \log (2))+\log (8)\right )}{x (5 x-\log (x))^2} \, dx}{\log ^4(2)}+\frac {\left (64 e^4\right ) \int \frac {(16 x+\log (8))^3}{5 x-\log (x)} \, dx}{\log ^4(2)}\\ &=\frac {e^4 \int \left (-\frac {327680 x^4}{(5 x-\log (x))^2}-\frac {(-64+15 \log (2)) \log ^3(8)}{(5 x-\log (x))^2}+\frac {\log ^4(8)}{x (5 x-\log (x))^2}-\frac {16 x \log ^2(8) (-96+45 \log (2)+5 \log (8))}{(5 x-\log (x))^2}-\frac {4096 x^3 (-16+15 \log (2)+15 \log (8))}{(5 x-\log (x))^2}-\frac {256 x^2 \log (8) (-64+45 \log (2)+15 \log (8))}{(5 x-\log (x))^2}\right ) \, dx}{\log ^4(2)}+\frac {\left (64 e^4\right ) \int \left (\frac {4096 x^3}{5 x-\log (x)}+\frac {768 x^2 \log (8)}{5 x-\log (x)}+\frac {48 x \log ^2(8)}{5 x-\log (x)}+\frac {\log ^3(8)}{5 x-\log (x)}\right ) \, dx}{\log ^4(2)}\\ &=\frac {\left (262144 e^4\right ) \int \frac {x^3}{5 x-\log (x)} \, dx}{\log ^4(2)}-\frac {\left (327680 e^4\right ) \int \frac {x^4}{(5 x-\log (x))^2} \, dx}{\log ^4(2)}+\frac {\left (4096 e^4 (16-15 \log (2)-15 \log (8))\right ) \int \frac {x^3}{(5 x-\log (x))^2} \, dx}{\log ^4(2)}+\frac {\left (49152 e^4 \log (8)\right ) \int \frac {x^2}{5 x-\log (x)} \, dx}{\log ^4(2)}+\frac {\left (256 e^4 (64-45 \log (2)-15 \log (8)) \log (8)\right ) \int \frac {x^2}{(5 x-\log (x))^2} \, dx}{\log ^4(2)}+\frac {\left (3072 e^4 \log ^2(8)\right ) \int \frac {x}{5 x-\log (x)} \, dx}{\log ^4(2)}+\frac {\left (16 e^4 (96-45 \log (2)-5 \log (8)) \log ^2(8)\right ) \int \frac {x}{(5 x-\log (x))^2} \, dx}{\log ^4(2)}+\frac {\left (64 e^4 \log ^3(8)\right ) \int \frac {1}{5 x-\log (x)} \, dx}{\log ^4(2)}+\frac {\left (e^4 (64-15 \log (2)) \log ^3(8)\right ) \int \frac {1}{(5 x-\log (x))^2} \, dx}{\log ^4(2)}+\frac {\left (e^4 \log ^4(8)\right ) \int \frac {1}{x (5 x-\log (x))^2} \, dx}{\log ^4(2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.57, size = 50, normalized size = 2.00 \begin {gather*} -\frac {e^4 \left (-16 x+80 x^2+15 x \log (2)-\log (8)\right ) (16 x+\log (8))^3}{(-1+5 x) \log ^4(2) (-5 x+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^4*(65536*x^4 + 49152*x^3*Log[2] + 13824*x^2*Log[2]^2 + 1728*x*Log[2]^3 + 81*Log[2]^4)*(-16*x - 24
0*x^2 + (-3 + 15*x)*Log[2] + 64*x*Log[x]))/(Log[2]^4*(5*x - Log[x])*(-80*x^3 - 15*x^2*Log[2] + (16*x^2 + 3*x*L
og[2])*Log[x])),x]

[Out]

-((E^4*(-16*x + 80*x^2 + 15*x*Log[2] - Log[8])*(16*x + Log[8])^3)/((-1 + 5*x)*Log[2]^4*(-5*x + Log[x])))

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fricas [B]  time = 0.75, size = 64, normalized size = 2.56 \begin {gather*} \frac {65536 \, x^{4} e^{4} + 49152 \, x^{3} e^{4} \log \relax (2) + 13824 \, x^{2} e^{4} \log \relax (2)^{2} + 1728 \, x e^{4} \log \relax (2)^{3} + 81 \, e^{4} \log \relax (2)^{4}}{5 \, x \log \relax (2)^{4} - \log \relax (2)^{4} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((64*x*log(x)+(15*x-3)*log(2)-240*x^2-16*x)*exp(-log(-log(x)+5*x)+log((81*log(2)^4+1728*x*log(2)^3+13
824*x^2*log(2)^2+49152*x^3*log(2)+65536*x^4)/log(2)^4)+4)/((3*x*log(2)+16*x^2)*log(x)-15*x^2*log(2)-80*x^3),x,
algorithm="fricas")

[Out]

(65536*x^4*e^4 + 49152*x^3*e^4*log(2) + 13824*x^2*e^4*log(2)^2 + 1728*x*e^4*log(2)^3 + 81*e^4*log(2)^4)/(5*x*l
og(2)^4 - log(2)^4*log(x))

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giac [B]  time = 0.22, size = 64, normalized size = 2.56 \begin {gather*} \frac {65536 \, x^{4} e^{4} + 49152 \, x^{3} e^{4} \log \relax (2) + 13824 \, x^{2} e^{4} \log \relax (2)^{2} + 1728 \, x e^{4} \log \relax (2)^{3} + 81 \, e^{4} \log \relax (2)^{4}}{5 \, x \log \relax (2)^{4} - \log \relax (2)^{4} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((64*x*log(x)+(15*x-3)*log(2)-240*x^2-16*x)*exp(-log(-log(x)+5*x)+log((81*log(2)^4+1728*x*log(2)^3+13
824*x^2*log(2)^2+49152*x^3*log(2)+65536*x^4)/log(2)^4)+4)/((3*x*log(2)+16*x^2)*log(x)-15*x^2*log(2)-80*x^3),x,
algorithm="giac")

[Out]

(65536*x^4*e^4 + 49152*x^3*e^4*log(2) + 13824*x^2*e^4*log(2)^2 + 1728*x*e^4*log(2)^3 + 81*e^4*log(2)^4)/(5*x*l
og(2)^4 - log(2)^4*log(x))

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maple [B]  time = 0.33, size = 61, normalized size = 2.44

method	result	size
norman	\(\frac {1728 \,{\mathrm e}^{4} \ln \relax (2)^{2} x +49152 x^{3} {\mathrm e}^{4}+81 \,{\mathrm e}^{4} \ln \relax (2)^{3}+13824 x^{2} {\mathrm e}^{4} \ln \relax (2)+\frac {65536 \,{\mathrm e}^{4} x^{4}}{\ln \relax (2)}}{\left (-\ln \relax (x )+5 x \right ) \ln \relax (2)^{3}}\)	\(61\)
default	\(\frac {{\mathrm e}^{4} \left (\frac {65536 x^{4}}{-\ln \relax (x )+5 x}+\frac {1728 \ln \relax (2)^{3} \ln \relax (x )}{5 \left (-\ln \relax (x )+5 x \right )}+\frac {13824 \ln \relax (2)^{2} x^{2}}{-\ln \relax (x )+5 x}+\frac {49152 x^{3} \ln \relax (2)}{-\ln \relax (x )+5 x}+\frac {81 \ln \relax (2)^{4}}{-\ln \relax (x )+5 x}\right )}{\ln \relax (2)^{4}}\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((64*x*ln(x)+(15*x-3)*ln(2)-240*x^2-16*x)*exp(-ln(-ln(x)+5*x)+ln((81*ln(2)^4+1728*x*ln(2)^3+13824*x^2*ln(2)
^2+49152*x^3*ln(2)+65536*x^4)/ln(2)^4)+4)/((3*x*ln(2)+16*x^2)*ln(x)-15*x^2*ln(2)-80*x^3),x,method=_RETURNVERBO
SE)

[Out]

(1728*exp(4)*ln(2)^2*x+49152*x^3*exp(4)+81*exp(4)*ln(2)^3+13824*x^2*exp(4)*ln(2)+65536*exp(4)/ln(2)*x^4)/(-ln(
x)+5*x)/ln(2)^3

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maxima [A]  time = 0.98, size = 52, normalized size = 2.08 \begin {gather*} \frac {{\left (65536 \, x^{4} + 49152 \, x^{3} \log \relax (2) + 13824 \, x^{2} \log \relax (2)^{2} + 1728 \, x \log \relax (2)^{3} + 81 \, \log \relax (2)^{4}\right )} e^{4}}{{\left (5 \, x - \log \relax (x)\right )} \log \relax (2)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((64*x*log(x)+(15*x-3)*log(2)-240*x^2-16*x)*exp(-log(-log(x)+5*x)+log((81*log(2)^4+1728*x*log(2)^3+13
824*x^2*log(2)^2+49152*x^3*log(2)+65536*x^4)/log(2)^4)+4)/((3*x*log(2)+16*x^2)*log(x)-15*x^2*log(2)-80*x^3),x,
algorithm="maxima")

[Out]

(65536*x^4 + 49152*x^3*log(2) + 13824*x^2*log(2)^2 + 1728*x*log(2)^3 + 81*log(2)^4)*e^4/((5*x - log(x))*log(2)
^4)

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mupad [B]  time = 2.14, size = 25, normalized size = 1.00 \begin {gather*} \frac {{\mathrm {e}}^4\,{\left (16\,x+\ln \relax (8)\right )}^4}{{\ln \relax (2)}^4\,\left (5\,x-\ln \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log((13824*x^2*log(2)^2 + 1728*x*log(2)^3 + 49152*x^3*log(2) + 81*log(2)^4 + 65536*x^4)/log(2)^4) - l
og(5*x - log(x)) + 4)*(16*x - log(2)*(15*x - 3) - 64*x*log(x) + 240*x^2))/(15*x^2*log(2) - log(x)*(3*x*log(2)
+ 16*x^2) + 80*x^3),x)

[Out]

(exp(4)*(16*x + log(8))^4)/(log(2)^4*(5*x - log(x)))

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sympy [B]  time = 0.19, size = 73, normalized size = 2.92 \begin {gather*} \frac {- 65536 x^{4} e^{4} - 49152 x^{3} e^{4} \log {\relax (2 )} - 13824 x^{2} e^{4} \log {\relax (2 )}^{2} - 1728 x e^{4} \log {\relax (2 )}^{3} - 81 e^{4} \log {\relax (2 )}^{4}}{- 5 x \log {\relax (2 )}^{4} + \log {\relax (2 )}^{4} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((64*x*ln(x)+(15*x-3)*ln(2)-240*x**2-16*x)*exp(-ln(-ln(x)+5*x)+ln((81*ln(2)**4+1728*x*ln(2)**3+13824*
x**2*ln(2)**2+49152*x**3*ln(2)+65536*x**4)/ln(2)**4)+4)/((3*x*ln(2)+16*x**2)*ln(x)-15*x**2*ln(2)-80*x**3),x)

[Out]

(-65536*x**4*exp(4) - 49152*x**3*exp(4)*log(2) - 13824*x**2*exp(4)*log(2)**2 - 1728*x*exp(4)*log(2)**3 - 81*ex
p(4)*log(2)**4)/(-5*x*log(2)**4 + log(2)**4*log(x))

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