3.14.20 \(\int \frac {240 x+120 e^4 x+15 e^8 x-5 x^2+(-240-120 e^4-15 e^8+5 x) \log (5)+(5 x^2+(-240-120 e^4-15 e^8) \log (5)) \log (x) \log (\log (x))}{(2304 x^2+144 e^{12} x^2+9 e^{16} x^2-96 x^3+x^4+e^4 (2304 x^2-48 x^3)+e^8 (864 x^2-6 x^3)+(-4608 x-288 e^{12} x-18 e^{16} x+192 x^2-2 x^3+e^8 (-1728 x+12 x^2)+e^4 (-4608 x+96 x^2)) \log (5)+(2304+144 e^{12}+9 e^{16}+e^4 (2304-48 x)+e^8 (864-6 x)-96 x+x^2) \log ^2(5)) \log (x)} \, dx\)

Optimal. Leaf size=29 \[ -\frac {5 x \log (\log (x))}{\left (3 \left (4+e^4\right )^2-x\right ) (-x+\log (5))} \]

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Rubi [F]  time = 2.43, 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 {240 x+120 e^4 x+15 e^8 x-5 x^2+\left (-240-120 e^4-15 e^8+5 x\right ) \log (5)+\left (5 x^2+\left (-240-120 e^4-15 e^8\right ) \log (5)\right ) \log (x) \log (\log (x))}{\left (2304 x^2+144 e^{12} x^2+9 e^{16} x^2-96 x^3+x^4+e^4 \left (2304 x^2-48 x^3\right )+e^8 \left (864 x^2-6 x^3\right )+\left (-4608 x-288 e^{12} x-18 e^{16} x+192 x^2-2 x^3+e^8 \left (-1728 x+12 x^2\right )+e^4 \left (-4608 x+96 x^2\right )\right ) \log (5)+\left (2304+144 e^{12}+9 e^{16}+e^4 (2304-48 x)+e^8 (864-6 x)-96 x+x^2\right ) \log ^2(5)\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(240*x + 120*E^4*x + 15*E^8*x - 5*x^2 + (-240 - 120*E^4 - 15*E^8 + 5*x)*Log[5] + (5*x^2 + (-240 - 120*E^4
- 15*E^8)*Log[5])*Log[x]*Log[Log[x]])/((2304*x^2 + 144*E^12*x^2 + 9*E^16*x^2 - 96*x^3 + x^4 + E^4*(2304*x^2 -
48*x^3) + E^8*(864*x^2 - 6*x^3) + (-4608*x - 288*E^12*x - 18*E^16*x + 192*x^2 - 2*x^3 + E^8*(-1728*x + 12*x^2)
+ E^4*(-4608*x + 96*x^2))*Log[5] + (2304 + 144*E^12 + 9*E^16 + E^4*(2304 - 48*x) + E^8*(864 - 6*x) - 96*x + x
^2)*Log[5]^2)*Log[x]),x]

[Out]

5*Defer[Int][1/((48 + 24*E^4 + 3*E^8 - x)*(x - Log[5])*Log[x]), x] + (15*(4 + E^4)^2*Defer[Int][Log[Log[x]]/(4
8 + 24*E^4 + 3*E^8 - x)^2, x])/(48 + 24*E^4 + 3*E^8 - Log[5]) - (5*Log[5]*Defer[Int][Log[Log[x]]/(-x + Log[5])
^2, x])/(48 + 24*E^4 + 3*E^8 - Log[5])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {240 x+120 e^4 x+15 e^8 x-5 x^2+\left (-240-120 e^4-15 e^8+5 x\right ) \log (5)+\left (5 x^2+\left (-240-120 e^4-15 e^8\right ) \log (5)\right ) \log (x) \log (\log (x))}{\left (9 e^{16} x^2+\left (2304+144 e^{12}\right ) x^2-96 x^3+x^4+e^4 \left (2304 x^2-48 x^3\right )+e^8 \left (864 x^2-6 x^3\right )+\left (-4608 x-288 e^{12} x-18 e^{16} x+192 x^2-2 x^3+e^8 \left (-1728 x+12 x^2\right )+e^4 \left (-4608 x+96 x^2\right )\right ) \log (5)+\left (2304+144 e^{12}+9 e^{16}+e^4 (2304-48 x)+e^8 (864-6 x)-96 x+x^2\right ) \log ^2(5)\right ) \log (x)} \, dx\\ &=\int \frac {240 x+120 e^4 x+15 e^8 x-5 x^2+\left (-240-120 e^4-15 e^8+5 x\right ) \log (5)+\left (5 x^2+\left (-240-120 e^4-15 e^8\right ) \log (5)\right ) \log (x) \log (\log (x))}{\left (\left (2304+144 e^{12}+9 e^{16}\right ) x^2-96 x^3+x^4+e^4 \left (2304 x^2-48 x^3\right )+e^8 \left (864 x^2-6 x^3\right )+\left (-4608 x-288 e^{12} x-18 e^{16} x+192 x^2-2 x^3+e^8 \left (-1728 x+12 x^2\right )+e^4 \left (-4608 x+96 x^2\right )\right ) \log (5)+\left (2304+144 e^{12}+9 e^{16}+e^4 (2304-48 x)+e^8 (864-6 x)-96 x+x^2\right ) \log ^2(5)\right ) \log (x)} \, dx\\ &=\int \frac {15 e^8 x+\left (240+120 e^4\right ) x-5 x^2+\left (-240-120 e^4-15 e^8+5 x\right ) \log (5)+\left (5 x^2+\left (-240-120 e^4-15 e^8\right ) \log (5)\right ) \log (x) \log (\log (x))}{\left (\left (2304+144 e^{12}+9 e^{16}\right ) x^2-96 x^3+x^4+e^4 \left (2304 x^2-48 x^3\right )+e^8 \left (864 x^2-6 x^3\right )+\left (-4608 x-288 e^{12} x-18 e^{16} x+192 x^2-2 x^3+e^8 \left (-1728 x+12 x^2\right )+e^4 \left (-4608 x+96 x^2\right )\right ) \log (5)+\left (2304+144 e^{12}+9 e^{16}+e^4 (2304-48 x)+e^8 (864-6 x)-96 x+x^2\right ) \log ^2(5)\right ) \log (x)} \, dx\\ &=\int \frac {\left (240+120 e^4+15 e^8\right ) x-5 x^2+\left (-240-120 e^4-15 e^8+5 x\right ) \log (5)+\left (5 x^2+\left (-240-120 e^4-15 e^8\right ) \log (5)\right ) \log (x) \log (\log (x))}{\left (\left (2304+144 e^{12}+9 e^{16}\right ) x^2-96 x^3+x^4+e^4 \left (2304 x^2-48 x^3\right )+e^8 \left (864 x^2-6 x^3\right )+\left (-4608 x-288 e^{12} x-18 e^{16} x+192 x^2-2 x^3+e^8 \left (-1728 x+12 x^2\right )+e^4 \left (-4608 x+96 x^2\right )\right ) \log (5)+\left (2304+144 e^{12}+9 e^{16}+e^4 (2304-48 x)+e^8 (864-6 x)-96 x+x^2\right ) \log ^2(5)\right ) \log (x)} \, dx\\ &=\int \frac {5 \left (\left (48+24 e^4+3 e^8-x\right ) (x-\log (5))+\left (x^2-3 \left (4+e^4\right )^2 \log (5)\right ) \log (x) \log (\log (x))\right )}{\left (48+24 e^4+3 e^8-x\right )^2 (x-\log (5))^2 \log (x)} \, dx\\ &=5 \int \frac {\left (48+24 e^4+3 e^8-x\right ) (x-\log (5))+\left (x^2-3 \left (4+e^4\right )^2 \log (5)\right ) \log (x) \log (\log (x))}{\left (48+24 e^4+3 e^8-x\right )^2 (x-\log (5))^2 \log (x)} \, dx\\ &=5 \int \left (\frac {1}{\left (48+24 e^4+3 e^8-x\right ) (x-\log (5)) \log (x)}-\frac {\left (-x^2+48 \log (5)+24 e^4 \log (5)+3 e^8 \log (5)\right ) \log (\log (x))}{\left (48+24 e^4+3 e^8-x\right )^2 (x-\log (5))^2}\right ) \, dx\\ &=5 \int \frac {1}{\left (48+24 e^4+3 e^8-x\right ) (x-\log (5)) \log (x)} \, dx-5 \int \frac {\left (-x^2+48 \log (5)+24 e^4 \log (5)+3 e^8 \log (5)\right ) \log (\log (x))}{\left (48+24 e^4+3 e^8-x\right )^2 (x-\log (5))^2} \, dx\\ &=5 \int \frac {1}{\left (48+24 e^4+3 e^8-x\right ) (x-\log (5)) \log (x)} \, dx-5 \int \frac {\left (-x^2+3 \left (4+e^4\right )^2 \log (5)\right ) \log (\log (x))}{\left (48+24 e^4+3 e^8-x\right )^2 (x-\log (5))^2} \, dx\\ &=5 \int \frac {1}{\left (48+24 e^4+3 e^8-x\right ) (x-\log (5)) \log (x)} \, dx-5 \int \left (-\frac {3 \left (4+e^4\right )^2 \log (\log (x))}{\left (48+24 e^4+3 e^8-x\right )^2 \left (48+24 e^4+3 e^8-\log (5)\right )}+\frac {\log (5) \log (\log (x))}{\left (48+24 e^4+3 e^8-\log (5)\right ) (-x+\log (5))^2}\right ) \, dx\\ &=5 \int \frac {1}{\left (48+24 e^4+3 e^8-x\right ) (x-\log (5)) \log (x)} \, dx+\frac {\left (15 \left (4+e^4\right )^2\right ) \int \frac {\log (\log (x))}{\left (48+24 e^4+3 e^8-x\right )^2} \, dx}{48+24 e^4+3 e^8-\log (5)}-\frac {(5 \log (5)) \int \frac {\log (\log (x))}{(-x+\log (5))^2} \, dx}{48+24 e^4+3 e^8-\log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 31, normalized size = 1.07 \begin {gather*} \frac {5 x \log (\log (x))}{\left (48+24 e^4+3 e^8-x\right ) (x-\log (5))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(240*x + 120*E^4*x + 15*E^8*x - 5*x^2 + (-240 - 120*E^4 - 15*E^8 + 5*x)*Log[5] + (5*x^2 + (-240 - 12
0*E^4 - 15*E^8)*Log[5])*Log[x]*Log[Log[x]])/((2304*x^2 + 144*E^12*x^2 + 9*E^16*x^2 - 96*x^3 + x^4 + E^4*(2304*
x^2 - 48*x^3) + E^8*(864*x^2 - 6*x^3) + (-4608*x - 288*E^12*x - 18*E^16*x + 192*x^2 - 2*x^3 + E^8*(-1728*x + 1
2*x^2) + E^4*(-4608*x + 96*x^2))*Log[5] + (2304 + 144*E^12 + 9*E^16 + E^4*(2304 - 48*x) + E^8*(864 - 6*x) - 96
*x + x^2)*Log[5]^2)*Log[x]),x]

[Out]

(5*x*Log[Log[x]])/((48 + 24*E^4 + 3*E^8 - x)*(x - Log[5]))

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fricas [A]  time = 0.77, size = 40, normalized size = 1.38 \begin {gather*} -\frac {5 \, x \log \left (\log \relax (x)\right )}{x^{2} - 3 \, x e^{8} - 24 \, x e^{4} - {\left (x - 3 \, e^{8} - 24 \, e^{4} - 48\right )} \log \relax (5) - 48 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-15*exp(4)^2-120*exp(4)-240)*log(5)+5*x^2)*log(x)*log(log(x))+(-15*exp(4)^2-120*exp(4)+5*x-240)*l
og(5)+15*x*exp(4)^2+120*x*exp(4)-5*x^2+240*x)/((9*exp(4)^4+144*exp(4)^3+(-6*x+864)*exp(4)^2+(-48*x+2304)*exp(4
)+x^2-96*x+2304)*log(5)^2+(-18*x*exp(4)^4-288*x*exp(4)^3+(12*x^2-1728*x)*exp(4)^2+(96*x^2-4608*x)*exp(4)-2*x^3
+192*x^2-4608*x)*log(5)+9*x^2*exp(4)^4+144*x^2*exp(4)^3+(-6*x^3+864*x^2)*exp(4)^2+(-48*x^3+2304*x^2)*exp(4)+x^
4-96*x^3+2304*x^2)/log(x),x, algorithm="fricas")

[Out]

-5*x*log(log(x))/(x^2 - 3*x*e^8 - 24*x*e^4 - (x - 3*e^8 - 24*e^4 - 48)*log(5) - 48*x)

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giac [A]  time = 10.48, size = 46, normalized size = 1.59 \begin {gather*} -\frac {5 \, x \log \left (\log \relax (x)\right )}{x^{2} - 3 \, x e^{8} - 24 \, x e^{4} - x \log \relax (5) + 3 \, e^{8} \log \relax (5) + 24 \, e^{4} \log \relax (5) - 48 \, x + 48 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-15*exp(4)^2-120*exp(4)-240)*log(5)+5*x^2)*log(x)*log(log(x))+(-15*exp(4)^2-120*exp(4)+5*x-240)*l
og(5)+15*x*exp(4)^2+120*x*exp(4)-5*x^2+240*x)/((9*exp(4)^4+144*exp(4)^3+(-6*x+864)*exp(4)^2+(-48*x+2304)*exp(4
)+x^2-96*x+2304)*log(5)^2+(-18*x*exp(4)^4-288*x*exp(4)^3+(12*x^2-1728*x)*exp(4)^2+(96*x^2-4608*x)*exp(4)-2*x^3
+192*x^2-4608*x)*log(5)+9*x^2*exp(4)^4+144*x^2*exp(4)^3+(-6*x^3+864*x^2)*exp(4)^2+(-48*x^3+2304*x^2)*exp(4)+x^
4-96*x^3+2304*x^2)/log(x),x, algorithm="giac")

[Out]

-5*x*log(log(x))/(x^2 - 3*x*e^8 - 24*x*e^4 - x*log(5) + 3*e^8*log(5) + 24*e^4*log(5) - 48*x + 48*log(5))

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maple [A]  time = 0.10, size = 47, normalized size = 1.62

method	result	size
risch	\(-\frac {5 x \ln \left (\ln \relax (x )\right )}{3 \ln \relax (5) {\mathrm e}^{8}-3 x \,{\mathrm e}^{8}+24 \,{\mathrm e}^{4} \ln \relax (5)-x \ln \relax (5)-24 x \,{\mathrm e}^{4}+x^{2}+48 \ln \relax (5)-48 x}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-15*exp(4)^2-120*exp(4)-240)*ln(5)+5*x^2)*ln(x)*ln(ln(x))+(-15*exp(4)^2-120*exp(4)+5*x-240)*ln(5)+15*x*
exp(4)^2+120*x*exp(4)-5*x^2+240*x)/((9*exp(4)^4+144*exp(4)^3+(-6*x+864)*exp(4)^2+(-48*x+2304)*exp(4)+x^2-96*x+
2304)*ln(5)^2+(-18*x*exp(4)^4-288*x*exp(4)^3+(12*x^2-1728*x)*exp(4)^2+(96*x^2-4608*x)*exp(4)-2*x^3+192*x^2-460
8*x)*ln(5)+9*x^2*exp(4)^4+144*x^2*exp(4)^3+(-6*x^3+864*x^2)*exp(4)^2+(-48*x^3+2304*x^2)*exp(4)+x^4-96*x^3+2304
*x^2)/ln(x),x,method=_RETURNVERBOSE)

[Out]

-5*x/(3*ln(5)*exp(8)-3*x*exp(8)+24*exp(4)*ln(5)-x*ln(5)-24*x*exp(4)+x^2+48*ln(5)-48*x)*ln(ln(x))

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maxima [A]  time = 0.92, size = 43, normalized size = 1.48 \begin {gather*} -\frac {5 \, x \log \left (\log \relax (x)\right )}{x^{2} - x {\left (3 \, e^{8} + 24 \, e^{4} + \log \relax (5) + 48\right )} + 3 \, e^{8} \log \relax (5) + 24 \, e^{4} \log \relax (5) + 48 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-15*exp(4)^2-120*exp(4)-240)*log(5)+5*x^2)*log(x)*log(log(x))+(-15*exp(4)^2-120*exp(4)+5*x-240)*l
og(5)+15*x*exp(4)^2+120*x*exp(4)-5*x^2+240*x)/((9*exp(4)^4+144*exp(4)^3+(-6*x+864)*exp(4)^2+(-48*x+2304)*exp(4
)+x^2-96*x+2304)*log(5)^2+(-18*x*exp(4)^4-288*x*exp(4)^3+(12*x^2-1728*x)*exp(4)^2+(96*x^2-4608*x)*exp(4)-2*x^3
+192*x^2-4608*x)*log(5)+9*x^2*exp(4)^4+144*x^2*exp(4)^3+(-6*x^3+864*x^2)*exp(4)^2+(-48*x^3+2304*x^2)*exp(4)+x^
4-96*x^3+2304*x^2)/log(x),x, algorithm="maxima")

[Out]

-5*x*log(log(x))/(x^2 - x*(3*e^8 + 24*e^4 + log(5) + 48) + 3*e^8*log(5) + 24*e^4*log(5) + 48*log(5))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {240\,x+120\,x\,{\mathrm {e}}^4+15\,x\,{\mathrm {e}}^8-\ln \relax (5)\,\left (120\,{\mathrm {e}}^4-5\,x+15\,{\mathrm {e}}^8+240\right )-5\,x^2-\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\,\left (\ln \relax (5)\,\left (120\,{\mathrm {e}}^4+15\,{\mathrm {e}}^8+240\right )-5\,x^2\right )}{\ln \relax (x)\,\left ({\mathrm {e}}^8\,\left (864\,x^2-6\,x^3\right )-\ln \relax (5)\,\left (4608\,x+{\mathrm {e}}^8\,\left (1728\,x-12\,x^2\right )+{\mathrm {e}}^4\,\left (4608\,x-96\,x^2\right )+288\,x\,{\mathrm {e}}^{12}+18\,x\,{\mathrm {e}}^{16}-192\,x^2+2\,x^3\right )+{\mathrm {e}}^4\,\left (2304\,x^2-48\,x^3\right )+144\,x^2\,{\mathrm {e}}^{12}+9\,x^2\,{\mathrm {e}}^{16}+{\ln \relax (5)}^2\,\left (144\,{\mathrm {e}}^{12}-96\,x+9\,{\mathrm {e}}^{16}+x^2-{\mathrm {e}}^8\,\left (6\,x-864\right )-{\mathrm {e}}^4\,\left (48\,x-2304\right )+2304\right )+2304\,x^2-96\,x^3+x^4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((240*x + 120*x*exp(4) + 15*x*exp(8) - log(5)*(120*exp(4) - 5*x + 15*exp(8) + 240) - 5*x^2 - log(log(x))*lo
g(x)*(log(5)*(120*exp(4) + 15*exp(8) + 240) - 5*x^2))/(log(x)*(exp(8)*(864*x^2 - 6*x^3) - log(5)*(4608*x + exp
(8)*(1728*x - 12*x^2) + exp(4)*(4608*x - 96*x^2) + 288*x*exp(12) + 18*x*exp(16) - 192*x^2 + 2*x^3) + exp(4)*(2
304*x^2 - 48*x^3) + 144*x^2*exp(12) + 9*x^2*exp(16) + log(5)^2*(144*exp(12) - 96*x + 9*exp(16) + x^2 - exp(8)*
(6*x - 864) - exp(4)*(48*x - 2304) + 2304) + 2304*x^2 - 96*x^3 + x^4)),x)

[Out]

int((240*x + 120*x*exp(4) + 15*x*exp(8) - log(5)*(120*exp(4) - 5*x + 15*exp(8) + 240) - 5*x^2 - log(log(x))*lo
g(x)*(log(5)*(120*exp(4) + 15*exp(8) + 240) - 5*x^2))/(log(x)*(exp(8)*(864*x^2 - 6*x^3) - log(5)*(4608*x + exp
(8)*(1728*x - 12*x^2) + exp(4)*(4608*x - 96*x^2) + 288*x*exp(12) + 18*x*exp(16) - 192*x^2 + 2*x^3) + exp(4)*(2
304*x^2 - 48*x^3) + 144*x^2*exp(12) + 9*x^2*exp(16) + log(5)^2*(144*exp(12) - 96*x + 9*exp(16) + x^2 - exp(8)*
(6*x - 864) - exp(4)*(48*x - 2304) + 2304) + 2304*x^2 - 96*x^3 + x^4)), x)

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sympy [B]  time = 0.88, size = 56, normalized size = 1.93 \begin {gather*} - \frac {5 x \log {\left (\log {\relax (x )} \right )}}{x^{2} - 3 x e^{8} - 24 x e^{4} - 48 x - x \log {\relax (5 )} + 48 \log {\relax (5 )} + 24 e^{4} \log {\relax (5 )} + 3 e^{8} \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-15*exp(4)**2-120*exp(4)-240)*ln(5)+5*x**2)*ln(x)*ln(ln(x))+(-15*exp(4)**2-120*exp(4)+5*x-240)*ln
(5)+15*x*exp(4)**2+120*x*exp(4)-5*x**2+240*x)/((9*exp(4)**4+144*exp(4)**3+(-6*x+864)*exp(4)**2+(-48*x+2304)*ex
p(4)+x**2-96*x+2304)*ln(5)**2+(-18*x*exp(4)**4-288*x*exp(4)**3+(12*x**2-1728*x)*exp(4)**2+(96*x**2-4608*x)*exp
(4)-2*x**3+192*x**2-4608*x)*ln(5)+9*x**2*exp(4)**4+144*x**2*exp(4)**3+(-6*x**3+864*x**2)*exp(4)**2+(-48*x**3+2
304*x**2)*exp(4)+x**4-96*x**3+2304*x**2)/ln(x),x)

[Out]

-5*x*log(log(x))/(x**2 - 3*x*exp(8) - 24*x*exp(4) - 48*x - x*log(5) + 48*log(5) + 24*exp(4)*log(5) + 3*exp(8)*
log(5))

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