3.50.8 \(\int \frac {(625+1500 x+1350 x^2+540 x^3+81 x^4) \log (2)+(625+1500 x+1350 x^2+540 x^3+81 x^4) \log ^2(2)+e^{2 x} (4 x^2+4 x^3+(4 x^3+x^4) \log (2)+x^4 \log ^2(2))+e^x (-10 x^2+22 x^3+18 x^4+(-100 x-170 x^2-96 x^3-18 x^4) \log (2)+(-50 x^2-60 x^3-18 x^4) \log ^2(2))+((50 x^2+60 x^3+18 x^4) \log (2)+(50 x^2+60 x^3+18 x^4) \log ^2(2)+e^x (-4 x^3+2 x^4+(-4 x^3-2 x^4) \log (2)-2 x^4 \log ^2(2))) \log (x)+(x^4 \log (2)+x^4 \log ^2(2)) \log ^2(x)}{(625+1500 x+1350 x^2+540 x^3+81 x^4) \log ^2(2)+e^{2 x} (4 x^2+4 x^3 \log (2)+x^4 \log ^2(2))+e^x ((-100 x-120 x^2-36 x^3) \log (2)+(-50 x^2-60 x^3-18 x^4) \log ^2(2))+((50 x^2+60 x^3+18 x^4) \log ^2(2)+e^x (-4 x^3 \log (2)-2 x^4 \log ^2(2))) \log (x)+x^4 \log ^2(2) \log ^2(x)} \, dx\)

Optimal. Leaf size=36 \[ x+\frac {x}{\log (2)-\frac {2 e^x}{x \left (-e^x+\left (3+\frac {5}{x}\right )^2+\log (x)\right )}} \]

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Rubi [F]  time = 180.00, 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*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Int[((625 + 1500*x + 1350*x^2 + 540*x^3 + 81*x^4)*Log[2] + (625 + 1500*x + 1350*x^2 + 540*x^3 + 81*x^4)*Log[2]
^2 + E^(2*x)*(4*x^2 + 4*x^3 + (4*x^3 + x^4)*Log[2] + x^4*Log[2]^2) + E^x*(-10*x^2 + 22*x^3 + 18*x^4 + (-100*x
- 170*x^2 - 96*x^3 - 18*x^4)*Log[2] + (-50*x^2 - 60*x^3 - 18*x^4)*Log[2]^2) + ((50*x^2 + 60*x^3 + 18*x^4)*Log[
2] + (50*x^2 + 60*x^3 + 18*x^4)*Log[2]^2 + E^x*(-4*x^3 + 2*x^4 + (-4*x^3 - 2*x^4)*Log[2] - 2*x^4*Log[2]^2))*Lo
g[x] + (x^4*Log[2] + x^4*Log[2]^2)*Log[x]^2)/((625 + 1500*x + 1350*x^2 + 540*x^3 + 81*x^4)*Log[2]^2 + E^(2*x)*
(4*x^2 + 4*x^3*Log[2] + x^4*Log[2]^2) + E^x*((-100*x - 120*x^2 - 36*x^3)*Log[2] + (-50*x^2 - 60*x^3 - 18*x^4)*
Log[2]^2) + ((50*x^2 + 60*x^3 + 18*x^4)*Log[2]^2 + E^x*(-4*x^3*Log[2] - 2*x^4*Log[2]^2))*Log[x] + x^4*Log[2]^2
*Log[x]^2),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [B]  time = 0.55, size = 140, normalized size = 3.89 \begin {gather*} \frac {x \left (1+\log (2)-\frac {e^x x \left (100 \log (2)+60 x \log (2)-x^2 \left (36 \log ^2(2)+\log (4)-18 \log (2) \log (4)\right )+e^x x \left (-4+x^2 \log (4)+x \left (4-\log ^2(4)+\log (2) \log (16)\right )\right )\right )}{\left (\left (50+30 x-x^2\right ) \log (2)+e^x x \left (-2+2 x+x^2 \log (2)\right )\right ) \left (-(5+3 x)^2 \log (2)+e^x x (2+x \log (2))-x^2 \log (2) \log (x)\right )}\right )}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((625 + 1500*x + 1350*x^2 + 540*x^3 + 81*x^4)*Log[2] + (625 + 1500*x + 1350*x^2 + 540*x^3 + 81*x^4)*
Log[2]^2 + E^(2*x)*(4*x^2 + 4*x^3 + (4*x^3 + x^4)*Log[2] + x^4*Log[2]^2) + E^x*(-10*x^2 + 22*x^3 + 18*x^4 + (-
100*x - 170*x^2 - 96*x^3 - 18*x^4)*Log[2] + (-50*x^2 - 60*x^3 - 18*x^4)*Log[2]^2) + ((50*x^2 + 60*x^3 + 18*x^4
)*Log[2] + (50*x^2 + 60*x^3 + 18*x^4)*Log[2]^2 + E^x*(-4*x^3 + 2*x^4 + (-4*x^3 - 2*x^4)*Log[2] - 2*x^4*Log[2]^
2))*Log[x] + (x^4*Log[2] + x^4*Log[2]^2)*Log[x]^2)/((625 + 1500*x + 1350*x^2 + 540*x^3 + 81*x^4)*Log[2]^2 + E^
(2*x)*(4*x^2 + 4*x^3*Log[2] + x^4*Log[2]^2) + E^x*((-100*x - 120*x^2 - 36*x^3)*Log[2] + (-50*x^2 - 60*x^3 - 18
*x^4)*Log[2]^2) + ((50*x^2 + 60*x^3 + 18*x^4)*Log[2]^2 + E^x*(-4*x^3*Log[2] - 2*x^4*Log[2]^2))*Log[x] + x^4*Lo
g[2]^2*Log[x]^2),x]

[Out]

(x*(1 + Log[2] - (E^x*x*(100*Log[2] + 60*x*Log[2] - x^2*(36*Log[2]^2 + Log[4] - 18*Log[2]*Log[4]) + E^x*x*(-4
+ x^2*Log[4] + x*(4 - Log[4]^2 + Log[2]*Log[16]))))/(((50 + 30*x - x^2)*Log[2] + E^x*x*(-2 + 2*x + x^2*Log[2])
)*(-((5 + 3*x)^2*Log[2]) + E^x*x*(2 + x*Log[2]) - x^2*Log[2]*Log[x]))))/Log[2]

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fricas [B]  time = 0.87, size = 102, normalized size = 2.83 \begin {gather*} \frac {9 \, x^{3} + 30 \, x^{2} - {\left (x^{3} \log \relax (2) + x^{3} + 2 \, x^{2}\right )} e^{x} + {\left (9 \, x^{3} + 30 \, x^{2} + 25 \, x\right )} \log \relax (2) + {\left (x^{3} \log \relax (2) + x^{3}\right )} \log \relax (x) + 25 \, x}{x^{2} \log \relax (2) \log \relax (x) - {\left (x^{2} \log \relax (2) + 2 \, x\right )} e^{x} + {\left (9 \, x^{2} + 30 \, x + 25\right )} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4*log(2)^2+x^4*log(2))*log(x)^2+((-2*x^4*log(2)^2+(-2*x^4-4*x^3)*log(2)+2*x^4-4*x^3)*exp(x)+(18*
x^4+60*x^3+50*x^2)*log(2)^2+(18*x^4+60*x^3+50*x^2)*log(2))*log(x)+(x^4*log(2)^2+(x^4+4*x^3)*log(2)+4*x^3+4*x^2
)*exp(x)^2+((-18*x^4-60*x^3-50*x^2)*log(2)^2+(-18*x^4-96*x^3-170*x^2-100*x)*log(2)+18*x^4+22*x^3-10*x^2)*exp(x
)+(81*x^4+540*x^3+1350*x^2+1500*x+625)*log(2)^2+(81*x^4+540*x^3+1350*x^2+1500*x+625)*log(2))/(x^4*log(2)^2*log
(x)^2+((-2*x^4*log(2)^2-4*x^3*log(2))*exp(x)+(18*x^4+60*x^3+50*x^2)*log(2)^2)*log(x)+(x^4*log(2)^2+4*x^3*log(2
)+4*x^2)*exp(x)^2+((-18*x^4-60*x^3-50*x^2)*log(2)^2+(-36*x^3-120*x^2-100*x)*log(2))*exp(x)+(81*x^4+540*x^3+135
0*x^2+1500*x+625)*log(2)^2),x, algorithm="fricas")

[Out]

(9*x^3 + 30*x^2 - (x^3*log(2) + x^3 + 2*x^2)*e^x + (9*x^3 + 30*x^2 + 25*x)*log(2) + (x^3*log(2) + x^3)*log(x)
+ 25*x)/(x^2*log(2)*log(x) - (x^2*log(2) + 2*x)*e^x + (9*x^2 + 30*x + 25)*log(2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4*log(2)^2+x^4*log(2))*log(x)^2+((-2*x^4*log(2)^2+(-2*x^4-4*x^3)*log(2)+2*x^4-4*x^3)*exp(x)+(18*
x^4+60*x^3+50*x^2)*log(2)^2+(18*x^4+60*x^3+50*x^2)*log(2))*log(x)+(x^4*log(2)^2+(x^4+4*x^3)*log(2)+4*x^3+4*x^2
)*exp(x)^2+((-18*x^4-60*x^3-50*x^2)*log(2)^2+(-18*x^4-96*x^3-170*x^2-100*x)*log(2)+18*x^4+22*x^3-10*x^2)*exp(x
)+(81*x^4+540*x^3+1350*x^2+1500*x+625)*log(2)^2+(81*x^4+540*x^3+1350*x^2+1500*x+625)*log(2))/(x^4*log(2)^2*log
(x)^2+((-2*x^4*log(2)^2-4*x^3*log(2))*exp(x)+(18*x^4+60*x^3+50*x^2)*log(2)^2)*log(x)+(x^4*log(2)^2+4*x^3*log(2
)+4*x^2)*exp(x)^2+((-18*x^4-60*x^3-50*x^2)*log(2)^2+(-36*x^3-120*x^2-100*x)*log(2))*exp(x)+(81*x^4+540*x^3+135
0*x^2+1500*x+625)*log(2)^2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.08, size = 61, normalized size = 1.69

method	result	size
risch	\(x +\frac {x}{\ln \relax (2)}-\frac {2 x^{2} {\mathrm e}^{x}}{\ln \relax (2) \left (x^{2} \ln \relax (2) {\mathrm e}^{x}-x^{2} \ln \relax (2) \ln \relax (x )-9 x^{2} \ln \relax (2)-30 x \ln \relax (2)+2 \,{\mathrm e}^{x} x -25 \ln \relax (2)\right )}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4*ln(2)^2+x^4*ln(2))*ln(x)^2+((-2*x^4*ln(2)^2+(-2*x^4-4*x^3)*ln(2)+2*x^4-4*x^3)*exp(x)+(18*x^4+60*x^3+
50*x^2)*ln(2)^2+(18*x^4+60*x^3+50*x^2)*ln(2))*ln(x)+(x^4*ln(2)^2+(x^4+4*x^3)*ln(2)+4*x^3+4*x^2)*exp(x)^2+((-18
*x^4-60*x^3-50*x^2)*ln(2)^2+(-18*x^4-96*x^3-170*x^2-100*x)*ln(2)+18*x^4+22*x^3-10*x^2)*exp(x)+(81*x^4+540*x^3+
1350*x^2+1500*x+625)*ln(2)^2+(81*x^4+540*x^3+1350*x^2+1500*x+625)*ln(2))/(x^4*ln(2)^2*ln(x)^2+((-2*x^4*ln(2)^2
-4*x^3*ln(2))*exp(x)+(18*x^4+60*x^3+50*x^2)*ln(2)^2)*ln(x)+(x^4*ln(2)^2+4*x^3*ln(2)+4*x^2)*exp(x)^2+((-18*x^4-
60*x^3-50*x^2)*ln(2)^2+(-36*x^3-120*x^2-100*x)*ln(2))*exp(x)+(81*x^4+540*x^3+1350*x^2+1500*x+625)*ln(2)^2),x,m
ethod=_RETURNVERBOSE)

[Out]

x+x/ln(2)-2/ln(2)*x^2*exp(x)/(x^2*ln(2)*exp(x)-x^2*ln(2)*ln(x)-9*x^2*ln(2)-30*x*ln(2)+2*exp(x)*x-25*ln(2))

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maxima [B]  time = 0.61, size = 174, normalized size = 4.83 \begin {gather*} \frac {9 \, {\left (\log \relax (2)^{3} + \log \relax (2)^{2}\right )} x^{3} + 6 \, {\left (5 \, \log \relax (2)^{3} + 5 \, \log \relax (2)^{2} + 3 \, \log \relax (2)\right )} x^{2} + 5 \, {\left (5 \, \log \relax (2)^{3} + 5 \, \log \relax (2)^{2} + 12 \, \log \relax (2)\right )} x - {\left ({\left (\log \relax (2)^{3} + \log \relax (2)^{2}\right )} x^{3} + 2 \, {\left (\log \relax (2)^{2} + \log \relax (2)\right )} x^{2} + 4 \, x\right )} e^{x} + {\left ({\left (\log \relax (2)^{3} + \log \relax (2)^{2}\right )} x^{3} + 2 \, x^{2} \log \relax (2)\right )} \log \relax (x) + 50 \, \log \relax (2)}{x^{2} \log \relax (2)^{3} \log \relax (x) + 9 \, x^{2} \log \relax (2)^{3} + 30 \, x \log \relax (2)^{3} + 25 \, \log \relax (2)^{3} - {\left (x^{2} \log \relax (2)^{3} + 2 \, x \log \relax (2)^{2}\right )} e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4*log(2)^2+x^4*log(2))*log(x)^2+((-2*x^4*log(2)^2+(-2*x^4-4*x^3)*log(2)+2*x^4-4*x^3)*exp(x)+(18*
x^4+60*x^3+50*x^2)*log(2)^2+(18*x^4+60*x^3+50*x^2)*log(2))*log(x)+(x^4*log(2)^2+(x^4+4*x^3)*log(2)+4*x^3+4*x^2
)*exp(x)^2+((-18*x^4-60*x^3-50*x^2)*log(2)^2+(-18*x^4-96*x^3-170*x^2-100*x)*log(2)+18*x^4+22*x^3-10*x^2)*exp(x
)+(81*x^4+540*x^3+1350*x^2+1500*x+625)*log(2)^2+(81*x^4+540*x^3+1350*x^2+1500*x+625)*log(2))/(x^4*log(2)^2*log
(x)^2+((-2*x^4*log(2)^2-4*x^3*log(2))*exp(x)+(18*x^4+60*x^3+50*x^2)*log(2)^2)*log(x)+(x^4*log(2)^2+4*x^3*log(2
)+4*x^2)*exp(x)^2+((-18*x^4-60*x^3-50*x^2)*log(2)^2+(-36*x^3-120*x^2-100*x)*log(2))*exp(x)+(81*x^4+540*x^3+135
0*x^2+1500*x+625)*log(2)^2),x, algorithm="maxima")

[Out]

(9*(log(2)^3 + log(2)^2)*x^3 + 6*(5*log(2)^3 + 5*log(2)^2 + 3*log(2))*x^2 + 5*(5*log(2)^3 + 5*log(2)^2 + 12*lo
g(2))*x - ((log(2)^3 + log(2)^2)*x^3 + 2*(log(2)^2 + log(2))*x^2 + 4*x)*e^x + ((log(2)^3 + log(2)^2)*x^3 + 2*x
^2*log(2))*log(x) + 50*log(2))/(x^2*log(2)^3*log(x) + 9*x^2*log(2)^3 + 30*x*log(2)^3 + 25*log(2)^3 - (x^2*log(
2)^3 + 2*x*log(2)^2)*e^x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\left (x^4\,{\ln \relax (2)}^2+x^4\,\ln \relax (2)\right )\,{\ln \relax (x)}^2+\left ({\ln \relax (2)}^2\,\left (18\,x^4+60\,x^3+50\,x^2\right )-{\mathrm {e}}^x\,\left (2\,x^4\,{\ln \relax (2)}^2+\ln \relax (2)\,\left (2\,x^4+4\,x^3\right )+4\,x^3-2\,x^4\right )+\ln \relax (2)\,\left (18\,x^4+60\,x^3+50\,x^2\right )\right )\,\ln \relax (x)+{\ln \relax (2)}^2\,\left (81\,x^4+540\,x^3+1350\,x^2+1500\,x+625\right )-{\mathrm {e}}^x\,\left ({\ln \relax (2)}^2\,\left (18\,x^4+60\,x^3+50\,x^2\right )+\ln \relax (2)\,\left (18\,x^4+96\,x^3+170\,x^2+100\,x\right )+10\,x^2-22\,x^3-18\,x^4\right )+{\mathrm {e}}^{2\,x}\,\left (x^4\,{\ln \relax (2)}^2+\ln \relax (2)\,\left (x^4+4\,x^3\right )+4\,x^2+4\,x^3\right )+\ln \relax (2)\,\left (81\,x^4+540\,x^3+1350\,x^2+1500\,x+625\right )}{{\ln \relax (2)}^2\,\left (81\,x^4+540\,x^3+1350\,x^2+1500\,x+625\right )+\ln \relax (x)\,\left ({\ln \relax (2)}^2\,\left (18\,x^4+60\,x^3+50\,x^2\right )-{\mathrm {e}}^x\,\left (2\,{\ln \relax (2)}^2\,x^4+4\,\ln \relax (2)\,x^3\right )\right )-{\mathrm {e}}^x\,\left ({\ln \relax (2)}^2\,\left (18\,x^4+60\,x^3+50\,x^2\right )+\ln \relax (2)\,\left (36\,x^3+120\,x^2+100\,x\right )\right )+{\mathrm {e}}^{2\,x}\,\left ({\ln \relax (2)}^2\,x^4+4\,\ln \relax (2)\,x^3+4\,x^2\right )+x^4\,{\ln \relax (2)}^2\,{\ln \relax (x)}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(2)^2*(1500*x + 1350*x^2 + 540*x^3 + 81*x^4 + 625) - exp(x)*(log(2)^2*(50*x^2 + 60*x^3 + 18*x^4) + log
(2)*(100*x + 170*x^2 + 96*x^3 + 18*x^4) + 10*x^2 - 22*x^3 - 18*x^4) + log(x)*(log(2)^2*(50*x^2 + 60*x^3 + 18*x
^4) - exp(x)*(2*x^4*log(2)^2 + log(2)*(4*x^3 + 2*x^4) + 4*x^3 - 2*x^4) + log(2)*(50*x^2 + 60*x^3 + 18*x^4)) +
exp(2*x)*(x^4*log(2)^2 + log(2)*(4*x^3 + x^4) + 4*x^2 + 4*x^3) + log(2)*(1500*x + 1350*x^2 + 540*x^3 + 81*x^4
+ 625) + log(x)^2*(x^4*log(2)^2 + x^4*log(2)))/(log(2)^2*(1500*x + 1350*x^2 + 540*x^3 + 81*x^4 + 625) + log(x)
*(log(2)^2*(50*x^2 + 60*x^3 + 18*x^4) - exp(x)*(2*x^4*log(2)^2 + 4*x^3*log(2))) - exp(x)*(log(2)^2*(50*x^2 + 6
0*x^3 + 18*x^4) + log(2)*(100*x + 120*x^2 + 36*x^3)) + exp(2*x)*(x^4*log(2)^2 + 4*x^3*log(2) + 4*x^2) + x^4*lo
g(2)^2*log(x)^2),x)

[Out]

int((log(2)^2*(1500*x + 1350*x^2 + 540*x^3 + 81*x^4 + 625) - exp(x)*(log(2)^2*(50*x^2 + 60*x^3 + 18*x^4) + log
(2)*(100*x + 170*x^2 + 96*x^3 + 18*x^4) + 10*x^2 - 22*x^3 - 18*x^4) + log(x)*(log(2)^2*(50*x^2 + 60*x^3 + 18*x
^4) - exp(x)*(2*x^4*log(2)^2 + log(2)*(4*x^3 + 2*x^4) + 4*x^3 - 2*x^4) + log(2)*(50*x^2 + 60*x^3 + 18*x^4)) +
exp(2*x)*(x^4*log(2)^2 + log(2)*(4*x^3 + x^4) + 4*x^2 + 4*x^3) + log(2)*(1500*x + 1350*x^2 + 540*x^3 + 81*x^4
+ 625) + log(x)^2*(x^4*log(2)^2 + x^4*log(2)))/(log(2)^2*(1500*x + 1350*x^2 + 540*x^3 + 81*x^4 + 625) + log(x)
*(log(2)^2*(50*x^2 + 60*x^3 + 18*x^4) - exp(x)*(2*x^4*log(2)^2 + 4*x^3*log(2))) - exp(x)*(log(2)^2*(50*x^2 + 6
0*x^3 + 18*x^4) + log(2)*(100*x + 120*x^2 + 36*x^3)) + exp(2*x)*(x^4*log(2)^2 + 4*x^3*log(2) + 4*x^2) + x^4*lo
g(2)^2*log(x)^2), x)

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sympy [B]  time = 1.10, size = 143, normalized size = 3.97 \begin {gather*} x \left (1 + \frac {1}{\log {\relax (2 )}}\right ) + \frac {- 2 x^{3} \log {\relax (x )} - 18 x^{3} - 60 x^{2} - 50 x}{- x^{3} \log {\relax (2 )}^{2} \log {\relax (x )} - 9 x^{3} \log {\relax (2 )}^{2} - 2 x^{2} \log {\relax (2 )} \log {\relax (x )} - 30 x^{2} \log {\relax (2 )}^{2} - 18 x^{2} \log {\relax (2 )} - 60 x \log {\relax (2 )} - 25 x \log {\relax (2 )}^{2} + \left (x^{3} \log {\relax (2 )}^{2} + 4 x^{2} \log {\relax (2 )} + 4 x\right ) e^{x} - 50 \log {\relax (2 )}} + \frac {4}{x \log {\relax (2 )}^{3} + 2 \log {\relax (2 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**4*ln(2)**2+x**4*ln(2))*ln(x)**2+((-2*x**4*ln(2)**2+(-2*x**4-4*x**3)*ln(2)+2*x**4-4*x**3)*exp(x)
+(18*x**4+60*x**3+50*x**2)*ln(2)**2+(18*x**4+60*x**3+50*x**2)*ln(2))*ln(x)+(x**4*ln(2)**2+(x**4+4*x**3)*ln(2)+
4*x**3+4*x**2)*exp(x)**2+((-18*x**4-60*x**3-50*x**2)*ln(2)**2+(-18*x**4-96*x**3-170*x**2-100*x)*ln(2)+18*x**4+
22*x**3-10*x**2)*exp(x)+(81*x**4+540*x**3+1350*x**2+1500*x+625)*ln(2)**2+(81*x**4+540*x**3+1350*x**2+1500*x+62
5)*ln(2))/(x**4*ln(2)**2*ln(x)**2+((-2*x**4*ln(2)**2-4*x**3*ln(2))*exp(x)+(18*x**4+60*x**3+50*x**2)*ln(2)**2)*
ln(x)+(x**4*ln(2)**2+4*x**3*ln(2)+4*x**2)*exp(x)**2+((-18*x**4-60*x**3-50*x**2)*ln(2)**2+(-36*x**3-120*x**2-10
0*x)*ln(2))*exp(x)+(81*x**4+540*x**3+1350*x**2+1500*x+625)*ln(2)**2),x)

[Out]

x*(1 + 1/log(2)) + (-2*x**3*log(x) - 18*x**3 - 60*x**2 - 50*x)/(-x**3*log(2)**2*log(x) - 9*x**3*log(2)**2 - 2*
x**2*log(2)*log(x) - 30*x**2*log(2)**2 - 18*x**2*log(2) - 60*x*log(2) - 25*x*log(2)**2 + (x**3*log(2)**2 + 4*x
**2*log(2) + 4*x)*exp(x) - 50*log(2)) + 4/(x*log(2)**3 + 2*log(2)**2)

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