3.76.73 \(\int \frac {16 x^2+(160 x+32 x^2) \log (2)+e^x (-8 x^2+(-40 x-36 x^2) \log (2)+(40 x+(100+180 x) \log (2)) \log (5))+e^{2 x} (x^2+2 x^2 \log (2)+(-10 x-20 x \log (2)) \log (5)+(25+50 \log (2)) \log ^2(5))}{16 x^3+(80 x^2+32 x^3) \log (2)+e^x (-8 x^3+(-20 x^2-16 x^3) \log (2)+(40 x^2+(100 x+80 x^2) \log (2)) \log (5))+e^{2 x} (x^3+2 x^3 \log (2)+(-10 x^2-20 x^2 \log (2)) \log (5)+(25 x+50 x \log (2)) \log ^2(5))} \, dx\)

Optimal. Leaf size=32 \[ \log \left (\frac {x}{2+\frac {1}{\log (2)}+\frac {5}{x+\frac {1}{4} e^x (-x+5 \log (5))}}\right ) \]

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Rubi [F]  time = 6.87, 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 {16 x^2+\left (160 x+32 x^2\right ) \log (2)+e^x \left (-8 x^2+\left (-40 x-36 x^2\right ) \log (2)+(40 x+(100+180 x) \log (2)) \log (5)\right )+e^{2 x} \left (x^2+2 x^2 \log (2)+(-10 x-20 x \log (2)) \log (5)+(25+50 \log (2)) \log ^2(5)\right )}{16 x^3+\left (80 x^2+32 x^3\right ) \log (2)+e^x \left (-8 x^3+\left (-20 x^2-16 x^3\right ) \log (2)+\left (40 x^2+\left (100 x+80 x^2\right ) \log (2)\right ) \log (5)\right )+e^{2 x} \left (x^3+2 x^3 \log (2)+\left (-10 x^2-20 x^2 \log (2)\right ) \log (5)+(25 x+50 x \log (2)) \log ^2(5)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(16*x^2 + (160*x + 32*x^2)*Log[2] + E^x*(-8*x^2 + (-40*x - 36*x^2)*Log[2] + (40*x + (100 + 180*x)*Log[2])*
Log[5]) + E^(2*x)*(x^2 + 2*x^2*Log[2] + (-10*x - 20*x*Log[2])*Log[5] + (25 + 50*Log[2])*Log[5]^2))/(16*x^3 + (
80*x^2 + 32*x^3)*Log[2] + E^x*(-8*x^3 + (-20*x^2 - 16*x^3)*Log[2] + (40*x^2 + (100*x + 80*x^2)*Log[2])*Log[5])
+ E^(2*x)*(x^3 + 2*x^3*Log[2] + (-10*x^2 - 20*x^2*Log[2])*Log[5] + (25*x + 50*x*Log[2])*Log[5]^2)),x]

[Out]

Log[x] + (16*Log[32]*Defer[Int][x/(-4*x + E^x*x - 5*E^x*Log[5]), x])/Log[1048576] + (100*Log[5]*Log[16]*Defer[
Int][1/((x - 5*Log[5])*(-4*x + E^x*x - 5*E^x*Log[5])), x])/Log[1048576] + (20*(16*Log[2]^2*Log[5] + 4*Log[4]*L
og[5] + Log[256] - 4*Log[5]*Log[512] + Log[16]*Log[512] - 4*Log[2]*(2 + Log[5]*Log[16] - Log[78125/16]))*Defer
[Int][(4*x*(1 + Log[4]) - E^x*x*(1 + Log[4]) + 5*E^x*(1 + Log[4])*Log[5] + 5*Log[16])^(-1), x])/Log[1048576] +
(16*Log[2]*(5 + Log[1024])*Defer[Int][x/(4*x*(1 + Log[4]) - E^x*x*(1 + Log[4]) + 5*E^x*(1 + Log[4])*Log[5] +
5*Log[16]), x])/Log[1048576] + (25*(32*Log[5/4]*Log[2]*Log[5] + 64*Log[2]^2*Log[5]^2 + 16*Log[4]*Log[5]^2 - 4*
Log[5]*Log[16] + Log[16]^2 + 4*Log[5]*Log[256] - 16*Log[5]^2*Log[512] + 8*Log[2]*Log[5]*(Log[25]*Log[1024] + L
og[4096] + 2*Log[5]*(5 - Log[16384])))*Defer[Int][1/((x - 5*Log[5])*(4*x*(1 + Log[4]) - E^x*x*(1 + Log[4]) + 5
*E^x*(1 + Log[4])*Log[5] + 5*Log[16])), x])/Log[1048576]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 e^x \log (5) \left (e^x (1+\log (4)) \log (5)+\log (16)\right )+x^2 \left (16+32 \log (2)+e^{2 x} (1+\log (4))-4 e^x (2+\log (512))\right )-10 x \left (-16 \log (2)+e^{2 x} (1+\log (4)) \log (5)+e^x (\log (16)-2 \log (5) (2+\log (512)))\right )}{x \left (\left (-4+e^x\right ) x-5 e^x \log (5)\right ) \left (x \left (-4-8 \log (2)+e^x (1+\log (4))\right )-5 \left (e^x (1+\log (4)) \log (5)+\log (16)\right )\right )} \, dx\\ &=\int \left (\frac {1}{x}+\frac {4 \left (25 \log (5) \log (16)+4 x^2 \log (32)-10 x \log (25) \log (32)\right )}{(x-5 \log (5)) \left (-4 x+e^x x-5 e^x \log (5)\right ) \log (1048576)}+\frac {20 x \left (16 \log ^2(2) \log (5)+4 \log (4) \log (5)+\log (256)-4 \log (5) \log (512)+\log (16) \log (512)-8 \log (2) \left (1-\log \left (\frac {5}{4}\right )-\log (4) \log (5)+\log (5) \log (512)\right )\right )+16 x^2 \log (2) (5+\log (1024))+25 (\log (16) (\log (16)-4 \log (5) (1+\log (512)))+8 \log (2) \log (5) (4+\log (4096)))}{(x-5 \log (5)) \left (4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)\right ) \log (1048576)}\right ) \, dx\\ &=\log (x)+\frac {\int \frac {20 x \left (16 \log ^2(2) \log (5)+4 \log (4) \log (5)+\log (256)-4 \log (5) \log (512)+\log (16) \log (512)-8 \log (2) \left (1-\log \left (\frac {5}{4}\right )-\log (4) \log (5)+\log (5) \log (512)\right )\right )+16 x^2 \log (2) (5+\log (1024))+25 (\log (16) (\log (16)-4 \log (5) (1+\log (512)))+8 \log (2) \log (5) (4+\log (4096)))}{(x-5 \log (5)) \left (4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)\right )} \, dx}{\log (1048576)}+\frac {4 \int \frac {25 \log (5) \log (16)+4 x^2 \log (32)-10 x \log (25) \log (32)}{(x-5 \log (5)) \left (-4 x+e^x x-5 e^x \log (5)\right )} \, dx}{\log (1048576)}\\ &=\log (x)+\frac {\int \left (\frac {16 x \log (2) (5+\log (1024))}{4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)}+\frac {20 \left (16 \log ^2(2) \log (5)+4 \log (4) \log (5)+\log (256)-4 \log (5) \log (512)+\log (16) \log (512)-4 \log (2) \left (2+\log (5) \log (16)-\log \left (\frac {78125}{16}\right )\right )\right )}{4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)}+\frac {25 \left (32 \log \left (\frac {5}{4}\right ) \log (2) \log (5)+64 \log ^2(2) \log ^2(5)+16 \log (4) \log ^2(5)-4 \log (5) \log (16)+\log ^2(16)+4 \log (5) \log (256)-16 \log ^2(5) \log (512)+8 \log (2) \log (5) (\log (25) \log (1024)+\log (4096)+2 \log (5) (5-\log (16384)))\right )}{(x-5 \log (5)) \left (4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)\right )}\right ) \, dx}{\log (1048576)}+\frac {4 \int \left (\frac {25 \log (5) \log (16)}{(x-5 \log (5)) \left (-4 x+e^x x-5 e^x \log (5)\right )}+\frac {4 x \log (32)}{-4 x+e^x x-5 e^x \log (5)}\right ) \, dx}{\log (1048576)}\\ &=\log (x)+\frac {(100 \log (5) \log (16)) \int \frac {1}{(x-5 \log (5)) \left (-4 x+e^x x-5 e^x \log (5)\right )} \, dx}{\log (1048576)}+\frac {(16 \log (32)) \int \frac {x}{-4 x+e^x x-5 e^x \log (5)} \, dx}{\log (1048576)}+\frac {(16 \log (2) (5+\log (1024))) \int \frac {x}{4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)} \, dx}{\log (1048576)}+\frac {\left (20 \left (16 \log ^2(2) \log (5)+4 \log (4) \log (5)+\log (256)-4 \log (5) \log (512)+\log (16) \log (512)-4 \log (2) \left (2+\log (5) \log (16)-\log \left (\frac {78125}{16}\right )\right )\right )\right ) \int \frac {1}{4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)} \, dx}{\log (1048576)}+\frac {\left (25 \left (32 \log \left (\frac {5}{4}\right ) \log (2) \log (5)+64 \log ^2(2) \log ^2(5)+16 \log (4) \log ^2(5)-4 \log (5) \log (16)+\log ^2(16)+4 \log (5) \log (256)-16 \log ^2(5) \log (512)+8 \log (2) \log (5) (\log (25) \log (1024)+\log (4096)+2 \log (5) (5-\log (16384)))\right )\right ) \int \frac {1}{(x-5 \log (5)) \left (4 x (1+\log (4))-e^x x (1+\log (4))+5 e^x (1+\log (4)) \log (5)+5 \log (16)\right )} \, dx}{\log (1048576)}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 2.19, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {16 x^2+\left (160 x+32 x^2\right ) \log (2)+e^x \left (-8 x^2+\left (-40 x-36 x^2\right ) \log (2)+(40 x+(100+180 x) \log (2)) \log (5)\right )+e^{2 x} \left (x^2+2 x^2 \log (2)+(-10 x-20 x \log (2)) \log (5)+(25+50 \log (2)) \log ^2(5)\right )}{16 x^3+\left (80 x^2+32 x^3\right ) \log (2)+e^x \left (-8 x^3+\left (-20 x^2-16 x^3\right ) \log (2)+\left (40 x^2+\left (100 x+80 x^2\right ) \log (2)\right ) \log (5)\right )+e^{2 x} \left (x^3+2 x^3 \log (2)+\left (-10 x^2-20 x^2 \log (2)\right ) \log (5)+(25 x+50 x \log (2)) \log ^2(5)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(16*x^2 + (160*x + 32*x^2)*Log[2] + E^x*(-8*x^2 + (-40*x - 36*x^2)*Log[2] + (40*x + (100 + 180*x)*Lo
g[2])*Log[5]) + E^(2*x)*(x^2 + 2*x^2*Log[2] + (-10*x - 20*x*Log[2])*Log[5] + (25 + 50*Log[2])*Log[5]^2))/(16*x
^3 + (80*x^2 + 32*x^3)*Log[2] + E^x*(-8*x^3 + (-20*x^2 - 16*x^3)*Log[2] + (40*x^2 + (100*x + 80*x^2)*Log[2])*L
og[5]) + E^(2*x)*(x^3 + 2*x^3*Log[2] + (-10*x^2 - 20*x^2*Log[2])*Log[5] + (25*x + 50*x*Log[2])*Log[5]^2)),x]

[Out]

Integrate[(16*x^2 + (160*x + 32*x^2)*Log[2] + E^x*(-8*x^2 + (-40*x - 36*x^2)*Log[2] + (40*x + (100 + 180*x)*Lo
g[2])*Log[5]) + E^(2*x)*(x^2 + 2*x^2*Log[2] + (-10*x - 20*x*Log[2])*Log[5] + (25 + 50*Log[2])*Log[5]^2))/(16*x
^3 + (80*x^2 + 32*x^3)*Log[2] + E^x*(-8*x^3 + (-20*x^2 - 16*x^3)*Log[2] + (40*x^2 + (100*x + 80*x^2)*Log[2])*L
og[5]) + E^(2*x)*(x^3 + 2*x^3*Log[2] + (-10*x^2 - 20*x^2*Log[2])*Log[5] + (25*x + 50*x*Log[2])*Log[5]^2)), x]

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fricas [B]  time = 0.73, size = 74, normalized size = 2.31 \begin {gather*} \log \relax (x) - \log \left (\frac {{\left (5 \, {\left (2 \, \log \relax (2) + 1\right )} \log \relax (5) - 2 \, x \log \relax (2) - x\right )} e^{x} + 4 \, {\left (2 \, x + 5\right )} \log \relax (2) + 4 \, x}{x - 5 \, \log \relax (5)}\right ) + \log \left (-\frac {{\left (x - 5 \, \log \relax (5)\right )} e^{x} - 4 \, x}{x - 5 \, \log \relax (5)}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*log(2)+25)*log(5)^2+(-20*x*log(2)-10*x)*log(5)+2*x^2*log(2)+x^2)*exp(x)^2+(((180*x+100)*log(2)
+40*x)*log(5)+(-36*x^2-40*x)*log(2)-8*x^2)*exp(x)+(32*x^2+160*x)*log(2)+16*x^2)/(((50*x*log(2)+25*x)*log(5)^2+
(-20*x^2*log(2)-10*x^2)*log(5)+2*x^3*log(2)+x^3)*exp(x)^2+(((80*x^2+100*x)*log(2)+40*x^2)*log(5)+(-16*x^3-20*x
^2)*log(2)-8*x^3)*exp(x)+(32*x^3+80*x^2)*log(2)+16*x^3),x, algorithm="fricas")

[Out]

log(x) - log(((5*(2*log(2) + 1)*log(5) - 2*x*log(2) - x)*e^x + 4*(2*x + 5)*log(2) + 4*x)/(x - 5*log(5))) + log
(-((x - 5*log(5))*e^x - 4*x)/(x - 5*log(5)))

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giac [A]  time = 0.90, size = 60, normalized size = 1.88 \begin {gather*} -\log \left (-2 \, x e^{x} \log \relax (2) + 10 \, e^{x} \log \relax (5) \log \relax (2) - x e^{x} + 5 \, e^{x} \log \relax (5) + 8 \, x \log \relax (2) + 4 \, x + 20 \, \log \relax (2)\right ) + \log \left (x e^{x} - 5 \, e^{x} \log \relax (5) - 4 \, x\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*log(2)+25)*log(5)^2+(-20*x*log(2)-10*x)*log(5)+2*x^2*log(2)+x^2)*exp(x)^2+(((180*x+100)*log(2)
+40*x)*log(5)+(-36*x^2-40*x)*log(2)-8*x^2)*exp(x)+(32*x^2+160*x)*log(2)+16*x^2)/(((50*x*log(2)+25*x)*log(5)^2+
(-20*x^2*log(2)-10*x^2)*log(5)+2*x^3*log(2)+x^3)*exp(x)^2+(((80*x^2+100*x)*log(2)+40*x^2)*log(5)+(-16*x^3-20*x
^2)*log(2)-8*x^3)*exp(x)+(32*x^3+80*x^2)*log(2)+16*x^3),x, algorithm="giac")

[Out]

-log(-2*x*e^x*log(2) + 10*e^x*log(5)*log(2) - x*e^x + 5*e^x*log(5) + 8*x*log(2) + 4*x + 20*log(2)) + log(x*e^x
- 5*e^x*log(5) - 4*x) + log(x)

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maple [B]  time = 0.43, size = 61, normalized size = 1.91

method	result	size
risch	\(\ln \relax (x )+\ln \left ({\mathrm e}^{x}+\frac {4 x}{5 \ln \relax (5)-x}\right )-\ln \left ({\mathrm e}^{x}+\frac {8 x \ln \relax (2)+20 \ln \relax (2)+4 x}{10 \ln \relax (2) \ln \relax (5)-2 x \ln \relax (2)+5 \ln \relax (5)-x}\right )\)	\(61\)
norman	\(-\ln \left (10 \ln \relax (5) \ln \relax (2) {\mathrm e}^{x}-2 x \ln \relax (2) {\mathrm e}^{x}+8 x \ln \relax (2)+5 \,{\mathrm e}^{x} \ln \relax (5)-{\mathrm e}^{x} x +20 \ln \relax (2)+4 x \right )+\ln \relax (x )+\ln \left (5 \,{\mathrm e}^{x} \ln \relax (5)-{\mathrm e}^{x} x +4 x \right )\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((50*ln(2)+25)*ln(5)^2+(-20*x*ln(2)-10*x)*ln(5)+2*x^2*ln(2)+x^2)*exp(x)^2+(((180*x+100)*ln(2)+40*x)*ln(5)
+(-36*x^2-40*x)*ln(2)-8*x^2)*exp(x)+(32*x^2+160*x)*ln(2)+16*x^2)/(((50*x*ln(2)+25*x)*ln(5)^2+(-20*x^2*ln(2)-10
*x^2)*ln(5)+2*x^3*ln(2)+x^3)*exp(x)^2+(((80*x^2+100*x)*ln(2)+40*x^2)*ln(5)+(-16*x^3-20*x^2)*ln(2)-8*x^3)*exp(x
)+(32*x^3+80*x^2)*ln(2)+16*x^3),x,method=_RETURNVERBOSE)

[Out]

ln(x)+ln(exp(x)+4*x/(5*ln(5)-x))-ln(exp(x)+4*(2*x*ln(2)+5*ln(2)+x)/(10*ln(2)*ln(5)-2*x*ln(2)+5*ln(5)-x))

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maxima [B]  time = 0.57, size = 89, normalized size = 2.78 \begin {gather*} \log \relax (x) - \log \left (-\frac {4 \, x {\left (2 \, \log \relax (2) + 1\right )} - {\left (x {\left (2 \, \log \relax (2) + 1\right )} - 10 \, \log \relax (5) \log \relax (2) - 5 \, \log \relax (5)\right )} e^{x} + 20 \, \log \relax (2)}{x {\left (2 \, \log \relax (2) + 1\right )} - 10 \, \log \relax (5) \log \relax (2) - 5 \, \log \relax (5)}\right ) + \log \left (\frac {{\left (x - 5 \, \log \relax (5)\right )} e^{x} - 4 \, x}{x - 5 \, \log \relax (5)}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*log(2)+25)*log(5)^2+(-20*x*log(2)-10*x)*log(5)+2*x^2*log(2)+x^2)*exp(x)^2+(((180*x+100)*log(2)
+40*x)*log(5)+(-36*x^2-40*x)*log(2)-8*x^2)*exp(x)+(32*x^2+160*x)*log(2)+16*x^2)/(((50*x*log(2)+25*x)*log(5)^2+
(-20*x^2*log(2)-10*x^2)*log(5)+2*x^3*log(2)+x^3)*exp(x)^2+(((80*x^2+100*x)*log(2)+40*x^2)*log(5)+(-16*x^3-20*x
^2)*log(2)-8*x^3)*exp(x)+(32*x^3+80*x^2)*log(2)+16*x^3),x, algorithm="maxima")

[Out]

log(x) - log(-(4*x*(2*log(2) + 1) - (x*(2*log(2) + 1) - 10*log(5)*log(2) - 5*log(5))*e^x + 20*log(2))/(x*(2*lo
g(2) + 1) - 10*log(5)*log(2) - 5*log(5))) + log(((x - 5*log(5))*e^x - 4*x)/(x - 5*log(5)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{2\,x}\,\left ({\ln \relax (5)}^2\,\left (50\,\ln \relax (2)+25\right )-\ln \relax (5)\,\left (10\,x+20\,x\,\ln \relax (2)\right )+2\,x^2\,\ln \relax (2)+x^2\right )-{\mathrm {e}}^x\,\left (\ln \relax (2)\,\left (36\,x^2+40\,x\right )-\ln \relax (5)\,\left (40\,x+\ln \relax (2)\,\left (180\,x+100\right )\right )+8\,x^2\right )+\ln \relax (2)\,\left (32\,x^2+160\,x\right )+16\,x^2}{{\mathrm {e}}^{2\,x}\,\left (2\,x^3\,\ln \relax (2)-\ln \relax (5)\,\left (20\,x^2\,\ln \relax (2)+10\,x^2\right )+{\ln \relax (5)}^2\,\left (25\,x+50\,x\,\ln \relax (2)\right )+x^3\right )-{\mathrm {e}}^x\,\left (\ln \relax (2)\,\left (16\,x^3+20\,x^2\right )-\ln \relax (5)\,\left (\ln \relax (2)\,\left (80\,x^2+100\,x\right )+40\,x^2\right )+8\,x^3\right )+\ln \relax (2)\,\left (32\,x^3+80\,x^2\right )+16\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)*(log(5)^2*(50*log(2) + 25) - log(5)*(10*x + 20*x*log(2)) + 2*x^2*log(2) + x^2) - exp(x)*(log(2)*
(40*x + 36*x^2) - log(5)*(40*x + log(2)*(180*x + 100)) + 8*x^2) + log(2)*(160*x + 32*x^2) + 16*x^2)/(exp(2*x)*
(2*x^3*log(2) - log(5)*(20*x^2*log(2) + 10*x^2) + log(5)^2*(25*x + 50*x*log(2)) + x^3) - exp(x)*(log(2)*(20*x^
2 + 16*x^3) - log(5)*(log(2)*(100*x + 80*x^2) + 40*x^2) + 8*x^3) + log(2)*(80*x^2 + 32*x^3) + 16*x^3),x)

[Out]

int((exp(2*x)*(log(5)^2*(50*log(2) + 25) - log(5)*(10*x + 20*x*log(2)) + 2*x^2*log(2) + x^2) - exp(x)*(log(2)*
(40*x + 36*x^2) - log(5)*(40*x + log(2)*(180*x + 100)) + 8*x^2) + log(2)*(160*x + 32*x^2) + 16*x^2)/(exp(2*x)*
(2*x^3*log(2) - log(5)*(20*x^2*log(2) + 10*x^2) + log(5)^2*(25*x + 50*x*log(2)) + x^3) - exp(x)*(log(2)*(20*x^
2 + 16*x^3) - log(5)*(log(2)*(100*x + 80*x^2) + 40*x^2) + 8*x^3) + log(2)*(80*x^2 + 32*x^3) + 16*x^3), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*ln(2)+25)*ln(5)**2+(-20*x*ln(2)-10*x)*ln(5)+2*x**2*ln(2)+x**2)*exp(x)**2+(((180*x+100)*ln(2)+4
0*x)*ln(5)+(-36*x**2-40*x)*ln(2)-8*x**2)*exp(x)+(32*x**2+160*x)*ln(2)+16*x**2)/(((50*x*ln(2)+25*x)*ln(5)**2+(-
20*x**2*ln(2)-10*x**2)*ln(5)+2*x**3*ln(2)+x**3)*exp(x)**2+(((80*x**2+100*x)*ln(2)+40*x**2)*ln(5)+(-16*x**3-20*
x**2)*ln(2)-8*x**3)*exp(x)+(32*x**3+80*x**2)*ln(2)+16*x**3),x)

[Out]

Exception raised: PolynomialError

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