3.73.26 \(\int \frac {-4 e^{2+\frac {4 (81 e^{25}-162 x)}{e^2 x}} x+(-324 e^{25+\frac {4 (81 e^{25}-162 x)}{e^2 x}}+12 e^{2+\frac {3 (81 e^{25}-162 x)}{e^2 x}} x^2) \log (x)+(-12 e^{2+\frac {2 (81 e^{25}-162 x)}{e^2 x}} x^3+e^{\frac {3 (81 e^{25}-162 x)}{e^2 x}} (972 e^{25} x-4 e^2 x^2)) \log ^2(x)+(4 e^{2+\frac {81 e^{25}-162 x}{e^2 x}} x^4+e^{\frac {2 (81 e^{25}-162 x)}{e^2 x}} (-972 e^{25} x^2+12 e^2 x^3)) \log ^3(x)+e^{\frac {81 e^{25}-162 x}{e^2 x}} (324 e^{25} x^3-12 e^2 x^4) \log ^4(x)+4 e^2 x^5 \log ^5(x)}{e^2 x^2 \log ^5(x)} \, dx\)

Optimal. Leaf size=27 \[ \left (-x+\frac {e^{\frac {81 \left (-2+\frac {e^{25}}{x}\right )}{e^2}}}{\log (x)}\right )^4 \]

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Rubi [B]  time = 2.54, antiderivative size = 98, normalized size of antiderivative = 3.63, number of steps used = 9, number of rules used = 4, integrand size = 276, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {12, 6688, 6742, 2288} \begin {gather*} x^4-\frac {4 e^{\frac {81 e^{23}}{x}-\frac {162}{e^2}} x^3}{\log (x)}+\frac {6 e^{\frac {162 e^{23}}{x}-\frac {324}{e^2}} x^2}{\log ^2(x)}+\frac {e^{\frac {324 e^{23}}{x}-\frac {648}{e^2}}}{\log ^4(x)}-\frac {4 e^{\frac {243 e^{23}}{x}-\frac {486}{e^2}} x}{\log ^3(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*E^(2 + (4*(81*E^25 - 162*x))/(E^2*x))*x + (-324*E^(25 + (4*(81*E^25 - 162*x))/(E^2*x)) + 12*E^(2 + (3*
(81*E^25 - 162*x))/(E^2*x))*x^2)*Log[x] + (-12*E^(2 + (2*(81*E^25 - 162*x))/(E^2*x))*x^3 + E^((3*(81*E^25 - 16
2*x))/(E^2*x))*(972*E^25*x - 4*E^2*x^2))*Log[x]^2 + (4*E^(2 + (81*E^25 - 162*x)/(E^2*x))*x^4 + E^((2*(81*E^25
- 162*x))/(E^2*x))*(-972*E^25*x^2 + 12*E^2*x^3))*Log[x]^3 + E^((81*E^25 - 162*x)/(E^2*x))*(324*E^25*x^3 - 12*E
^2*x^4)*Log[x]^4 + 4*E^2*x^5*Log[x]^5)/(E^2*x^2*Log[x]^5),x]

[Out]

x^4 + E^(-648/E^2 + (324*E^23)/x)/Log[x]^4 - (4*E^(-486/E^2 + (243*E^23)/x)*x)/Log[x]^3 + (6*E^(-324/E^2 + (16
2*E^23)/x)*x^2)/Log[x]^2 - (4*E^(-162/E^2 + (81*E^23)/x)*x^3)/Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-4 e^{2+\frac {4 \left (81 e^{25}-162 x\right )}{e^2 x}} x+\left (-324 e^{25+\frac {4 \left (81 e^{25}-162 x\right )}{e^2 x}}+12 e^{2+\frac {3 \left (81 e^{25}-162 x\right )}{e^2 x}} x^2\right ) \log (x)+\left (-12 e^{2+\frac {2 \left (81 e^{25}-162 x\right )}{e^2 x}} x^3+e^{\frac {3 \left (81 e^{25}-162 x\right )}{e^2 x}} \left (972 e^{25} x-4 e^2 x^2\right )\right ) \log ^2(x)+\left (4 e^{2+\frac {81 e^{25}-162 x}{e^2 x}} x^4+e^{\frac {2 \left (81 e^{25}-162 x\right )}{e^2 x}} \left (-972 e^{25} x^2+12 e^2 x^3\right )\right ) \log ^3(x)+e^{\frac {81 e^{25}-162 x}{e^2 x}} \left (324 e^{25} x^3-12 e^2 x^4\right ) \log ^4(x)+4 e^2 x^5 \log ^5(x)}{x^2 \log ^5(x)} \, dx}{e^2}\\ &=\frac {\int \frac {4 e^{2-\frac {648}{e^2}} \left (e^{\frac {81 e^{23}}{x}}-e^{\frac {162}{e^2}} x \log (x)\right )^3 \left (-e^{\frac {81 e^{23}}{x}} x-81 e^{23+\frac {81 e^{23}}{x}} \log (x)-e^{\frac {162}{e^2}} x^2 \log ^2(x)\right )}{x^2 \log ^5(x)} \, dx}{e^2}\\ &=\left (4 e^{-\frac {648}{e^2}}\right ) \int \frac {\left (e^{\frac {81 e^{23}}{x}}-e^{\frac {162}{e^2}} x \log (x)\right )^3 \left (-e^{\frac {81 e^{23}}{x}} x-81 e^{23+\frac {81 e^{23}}{x}} \log (x)-e^{\frac {162}{e^2}} x^2 \log ^2(x)\right )}{x^2 \log ^5(x)} \, dx\\ &=\left (4 e^{-\frac {648}{e^2}}\right ) \int \left (e^{\frac {648}{e^2}} x^3-\frac {e^{\frac {324 e^{23}}{x}} \left (x+81 e^{23} \log (x)\right )}{x^2 \log ^5(x)}-\frac {e^{\frac {162}{e^2}+\frac {243 e^{23}}{x}} \left (-3 x-243 e^{23} \log (x)+x \log (x)\right )}{x \log ^4(x)}+\frac {3 e^{\frac {324}{e^2}+\frac {162 e^{23}}{x}} \left (-x-81 e^{23} \log (x)+x \log (x)\right )}{\log ^3(x)}-\frac {e^{\frac {486}{e^2}+\frac {81 e^{23}}{x}} x \left (-x-81 e^{23} \log (x)+3 x \log (x)\right )}{\log ^2(x)}\right ) \, dx\\ &=x^4-\left (4 e^{-\frac {648}{e^2}}\right ) \int \frac {e^{\frac {324 e^{23}}{x}} \left (x+81 e^{23} \log (x)\right )}{x^2 \log ^5(x)} \, dx-\left (4 e^{-\frac {648}{e^2}}\right ) \int \frac {e^{\frac {162}{e^2}+\frac {243 e^{23}}{x}} \left (-3 x-243 e^{23} \log (x)+x \log (x)\right )}{x \log ^4(x)} \, dx-\left (4 e^{-\frac {648}{e^2}}\right ) \int \frac {e^{\frac {486}{e^2}+\frac {81 e^{23}}{x}} x \left (-x-81 e^{23} \log (x)+3 x \log (x)\right )}{\log ^2(x)} \, dx+\left (12 e^{-\frac {648}{e^2}}\right ) \int \frac {e^{\frac {324}{e^2}+\frac {162 e^{23}}{x}} \left (-x-81 e^{23} \log (x)+x \log (x)\right )}{\log ^3(x)} \, dx\\ &=x^4+\frac {e^{-\frac {648}{e^2}+\frac {324 e^{23}}{x}}}{\log ^4(x)}-\frac {4 e^{-\frac {486}{e^2}+\frac {243 e^{23}}{x}} x}{\log ^3(x)}+\frac {6 e^{-\frac {324}{e^2}+\frac {162 e^{23}}{x}} x^2}{\log ^2(x)}-\frac {4 e^{-\frac {162}{e^2}+\frac {81 e^{23}}{x}} x^3}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 37, normalized size = 1.37 \begin {gather*} \frac {e^{-\frac {648}{e^2}} \left (e^{\frac {81 e^{23}}{x}}-e^{\frac {162}{e^2}} x \log (x)\right )^4}{\log ^4(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*E^(2 + (4*(81*E^25 - 162*x))/(E^2*x))*x + (-324*E^(25 + (4*(81*E^25 - 162*x))/(E^2*x)) + 12*E^(2
+ (3*(81*E^25 - 162*x))/(E^2*x))*x^2)*Log[x] + (-12*E^(2 + (2*(81*E^25 - 162*x))/(E^2*x))*x^3 + E^((3*(81*E^2
5 - 162*x))/(E^2*x))*(972*E^25*x - 4*E^2*x^2))*Log[x]^2 + (4*E^(2 + (81*E^25 - 162*x)/(E^2*x))*x^4 + E^((2*(81
*E^25 - 162*x))/(E^2*x))*(-972*E^25*x^2 + 12*E^2*x^3))*Log[x]^3 + E^((81*E^25 - 162*x)/(E^2*x))*(324*E^25*x^3
- 12*E^2*x^4)*Log[x]^4 + 4*E^2*x^5*Log[x]^5)/(E^2*x^2*Log[x]^5),x]

[Out]

(E^((81*E^23)/x) - E^(162/E^2)*x*Log[x])^4/(E^(648/E^2)*Log[x]^4)

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fricas [B]  time = 0.87, size = 130, normalized size = 4.81 \begin {gather*} \frac {{\left (x^{4} e^{8} \log \relax (x)^{4} - 4 \, x^{3} e^{\left (\frac {{\left (2 \, x e^{2} - 162 \, x + 81 \, e^{25}\right )} e^{\left (-2\right )}}{x} + 6\right )} \log \relax (x)^{3} + 6 \, x^{2} e^{\left (\frac {2 \, {\left (2 \, x e^{2} - 162 \, x + 81 \, e^{25}\right )} e^{\left (-2\right )}}{x} + 4\right )} \log \relax (x)^{2} - 4 \, x e^{\left (\frac {3 \, {\left (2 \, x e^{2} - 162 \, x + 81 \, e^{25}\right )} e^{\left (-2\right )}}{x} + 2\right )} \log \relax (x) + e^{\left (\frac {4 \, {\left (2 \, x e^{2} - 162 \, x + 81 \, e^{25}\right )} e^{\left (-2\right )}}{x}\right )}\right )} e^{\left (-8\right )}}{\log \relax (x)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^5*exp(2)*log(x)^5+(324*x^3*exp(25)-12*x^4*exp(2))*exp((81*exp(25)-162*x)/exp(2)/x)*log(x)^4+((-
972*x^2*exp(25)+12*x^3*exp(2))*exp((81*exp(25)-162*x)/exp(2)/x)^2+4*x^4*exp(2)*exp((81*exp(25)-162*x)/exp(2)/x
))*log(x)^3+((972*x*exp(25)-4*x^2*exp(2))*exp((81*exp(25)-162*x)/exp(2)/x)^3-12*x^3*exp(2)*exp((81*exp(25)-162
*x)/exp(2)/x)^2)*log(x)^2+(-324*exp(25)*exp((81*exp(25)-162*x)/exp(2)/x)^4+12*x^2*exp(2)*exp((81*exp(25)-162*x
)/exp(2)/x)^3)*log(x)-4*x*exp(2)*exp((81*exp(25)-162*x)/exp(2)/x)^4)/x^2/exp(2)/log(x)^5,x, algorithm="fricas"
)

[Out]

(x^4*e^8*log(x)^4 - 4*x^3*e^((2*x*e^2 - 162*x + 81*e^25)*e^(-2)/x + 6)*log(x)^3 + 6*x^2*e^(2*(2*x*e^2 - 162*x
+ 81*e^25)*e^(-2)/x + 4)*log(x)^2 - 4*x*e^(3*(2*x*e^2 - 162*x + 81*e^25)*e^(-2)/x + 2)*log(x) + e^(4*(2*x*e^2
- 162*x + 81*e^25)*e^(-2)/x))*e^(-8)/log(x)^4

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giac [B]  time = 0.29, size = 132, normalized size = 4.89 \begin {gather*} \frac {2 \, {\left (3 \, x^{2} e^{\left (\frac {2 \, {\left (x e^{2} - 162 \, x + 81 \, e^{25}\right )} e^{\left (-2\right )}}{x} + 1\right )} \log \relax (x) - 2 \, x e^{\left (\frac {3 \, {\left (x e^{2} - 162 \, x + 81 \, e^{25}\right )} e^{\left (-2\right )}}{x}\right )}\right )} e^{\left (-3\right )}}{\log \relax (x)^{3}} + \frac {{\left (x^{4} e^{8} \log \relax (x)^{4} - 4 \, x^{3} e^{\left (\frac {{\left (2 \, x e^{2} - 162 \, x + 81 \, e^{25}\right )} e^{\left (-2\right )}}{x} + 6\right )} \log \relax (x)^{3} + e^{\left (\frac {4 \, {\left (2 \, x e^{2} - 162 \, x + 81 \, e^{25}\right )} e^{\left (-2\right )}}{x}\right )}\right )} e^{\left (-8\right )}}{\log \relax (x)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^5*exp(2)*log(x)^5+(324*x^3*exp(25)-12*x^4*exp(2))*exp((81*exp(25)-162*x)/exp(2)/x)*log(x)^4+((-
972*x^2*exp(25)+12*x^3*exp(2))*exp((81*exp(25)-162*x)/exp(2)/x)^2+4*x^4*exp(2)*exp((81*exp(25)-162*x)/exp(2)/x
))*log(x)^3+((972*x*exp(25)-4*x^2*exp(2))*exp((81*exp(25)-162*x)/exp(2)/x)^3-12*x^3*exp(2)*exp((81*exp(25)-162
*x)/exp(2)/x)^2)*log(x)^2+(-324*exp(25)*exp((81*exp(25)-162*x)/exp(2)/x)^4+12*x^2*exp(2)*exp((81*exp(25)-162*x
)/exp(2)/x)^3)*log(x)-4*x*exp(2)*exp((81*exp(25)-162*x)/exp(2)/x)^4)/x^2/exp(2)/log(x)^5,x, algorithm="giac")

[Out]

2*(3*x^2*e^(2*(x*e^2 - 162*x + 81*e^25)*e^(-2)/x + 1)*log(x) - 2*x*e^(3*(x*e^2 - 162*x + 81*e^25)*e^(-2)/x))*e
^(-3)/log(x)^3 + (x^4*e^8*log(x)^4 - 4*x^3*e^((2*x*e^2 - 162*x + 81*e^25)*e^(-2)/x + 6)*log(x)^3 + e^(4*(2*x*e
^2 - 162*x + 81*e^25)*e^(-2)/x))*e^(-8)/log(x)^4

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maple [B]  time = 0.16, size = 93, normalized size = 3.44

method	result	size
risch	\(x^{4}-\frac {{\mathrm e}^{\frac {81 \left ({\mathrm e}^{25}-2 x \right ) {\mathrm e}^{-2}}{x}} \left (4 x^{3} \ln \relax (x )^{3}-6 x^{2} {\mathrm e}^{\frac {81 \left ({\mathrm e}^{25}-2 x \right ) {\mathrm e}^{-2}}{x}} \ln \relax (x )^{2}+4 x \,{\mathrm e}^{\frac {162 \left ({\mathrm e}^{25}-2 x \right ) {\mathrm e}^{-2}}{x}} \ln \relax (x )-{\mathrm e}^{\frac {243 \left ({\mathrm e}^{25}-2 x \right ) {\mathrm e}^{-2}}{x}}\right )}{\ln \relax (x )^{4}}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^5*exp(2)*ln(x)^5+(324*x^3*exp(25)-12*x^4*exp(2))*exp((81*exp(25)-162*x)/exp(2)/x)*ln(x)^4+((-972*x^2*
exp(25)+12*x^3*exp(2))*exp((81*exp(25)-162*x)/exp(2)/x)^2+4*x^4*exp(2)*exp((81*exp(25)-162*x)/exp(2)/x))*ln(x)
^3+((972*x*exp(25)-4*x^2*exp(2))*exp((81*exp(25)-162*x)/exp(2)/x)^3-12*x^3*exp(2)*exp((81*exp(25)-162*x)/exp(2
)/x)^2)*ln(x)^2+(-324*exp(25)*exp((81*exp(25)-162*x)/exp(2)/x)^4+12*x^2*exp(2)*exp((81*exp(25)-162*x)/exp(2)/x
)^3)*ln(x)-4*x*exp(2)*exp((81*exp(25)-162*x)/exp(2)/x)^4)/x^2/exp(2)/ln(x)^5,x,method=_RETURNVERBOSE)

[Out]

x^4-exp(81*(exp(25)-2*x)*exp(-2)/x)*(4*x^3*ln(x)^3-6*x^2*exp(81*(exp(25)-2*x)*exp(-2)/x)*ln(x)^2+4*x*exp(162*(
exp(25)-2*x)*exp(-2)/x)*ln(x)-exp(243*(exp(25)-2*x)*exp(-2)/x))/ln(x)^4

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maxima [B]  time = 0.48, size = 99, normalized size = 3.67 \begin {gather*} {\left (x^{4} e^{2} - \frac {{\left (4 \, x^{3} e^{\left (\frac {81 \, e^{23}}{x} + 486 \, e^{\left (-2\right )} + 2\right )} \log \relax (x)^{3} - 6 \, x^{2} e^{\left (\frac {162 \, e^{23}}{x} + 324 \, e^{\left (-2\right )} + 2\right )} \log \relax (x)^{2} + 4 \, x e^{\left (\frac {243 \, e^{23}}{x} + 162 \, e^{\left (-2\right )} + 2\right )} \log \relax (x) - e^{\left (\frac {324 \, e^{23}}{x} + 2\right )}\right )} e^{\left (-648 \, e^{\left (-2\right )}\right )}}{\log \relax (x)^{4}}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^5*exp(2)*log(x)^5+(324*x^3*exp(25)-12*x^4*exp(2))*exp((81*exp(25)-162*x)/exp(2)/x)*log(x)^4+((-
972*x^2*exp(25)+12*x^3*exp(2))*exp((81*exp(25)-162*x)/exp(2)/x)^2+4*x^4*exp(2)*exp((81*exp(25)-162*x)/exp(2)/x
))*log(x)^3+((972*x*exp(25)-4*x^2*exp(2))*exp((81*exp(25)-162*x)/exp(2)/x)^3-12*x^3*exp(2)*exp((81*exp(25)-162
*x)/exp(2)/x)^2)*log(x)^2+(-324*exp(25)*exp((81*exp(25)-162*x)/exp(2)/x)^4+12*x^2*exp(2)*exp((81*exp(25)-162*x
)/exp(2)/x)^3)*log(x)-4*x*exp(2)*exp((81*exp(25)-162*x)/exp(2)/x)^4)/x^2/exp(2)/log(x)^5,x, algorithm="maxima"
)

[Out]

(x^4*e^2 - (4*x^3*e^(81*e^23/x + 486*e^(-2) + 2)*log(x)^3 - 6*x^2*e^(162*e^23/x + 324*e^(-2) + 2)*log(x)^2 + 4
*x*e^(243*e^23/x + 162*e^(-2) + 2)*log(x) - e^(324*e^23/x + 2))*e^(-648*e^(-2))/log(x)^4)*e^(-2)

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mupad [B]  time = 5.26, size = 86, normalized size = 3.19 \begin {gather*} \frac {{\mathrm {e}}^{\frac {324\,{\mathrm {e}}^{23}}{x}-648\,{\mathrm {e}}^{-2}}}{{\ln \relax (x)}^4}+x^4-\frac {4\,x\,{\mathrm {e}}^{\frac {243\,{\mathrm {e}}^{23}}{x}-486\,{\mathrm {e}}^{-2}}}{{\ln \relax (x)}^3}-\frac {4\,x^3\,{\mathrm {e}}^{\frac {81\,{\mathrm {e}}^{23}}{x}-162\,{\mathrm {e}}^{-2}}}{\ln \relax (x)}+\frac {6\,x^2\,{\mathrm {e}}^{\frac {162\,{\mathrm {e}}^{23}}{x}-324\,{\mathrm {e}}^{-2}}}{{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2)*(log(x)^2*(exp(-(3*exp(-2)*(162*x - 81*exp(25)))/x)*(972*x*exp(25) - 4*x^2*exp(2)) - 12*x^3*exp(2
)*exp(-(2*exp(-2)*(162*x - 81*exp(25)))/x)) + log(x)^3*(exp(-(2*exp(-2)*(162*x - 81*exp(25)))/x)*(12*x^3*exp(2
) - 972*x^2*exp(25)) + 4*x^4*exp(2)*exp(-(exp(-2)*(162*x - 81*exp(25)))/x)) - log(x)*(324*exp(25)*exp(-(4*exp(
-2)*(162*x - 81*exp(25)))/x) - 12*x^2*exp(2)*exp(-(3*exp(-2)*(162*x - 81*exp(25)))/x)) - exp(-(exp(-2)*(162*x
- 81*exp(25)))/x)*log(x)^4*(12*x^4*exp(2) - 324*x^3*exp(25)) - 4*x*exp(2)*exp(-(4*exp(-2)*(162*x - 81*exp(25))
)/x) + 4*x^5*exp(2)*log(x)^5))/(x^2*log(x)^5),x)

[Out]

exp((324*exp(23))/x - 648*exp(-2))/log(x)^4 + x^4 - (4*x*exp((243*exp(23))/x - 486*exp(-2)))/log(x)^3 - (4*x^3
*exp((81*exp(23))/x - 162*exp(-2)))/log(x) + (6*x^2*exp((162*exp(23))/x - 324*exp(-2)))/log(x)^2

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sympy [B]  time = 0.66, size = 102, normalized size = 3.78 \begin {gather*} x^{4} + \frac {- 4 x^{3} e^{\frac {- 162 x + 81 e^{25}}{x e^{2}}} \log {\relax (x )}^{9} + 6 x^{2} e^{\frac {2 \left (- 162 x + 81 e^{25}\right )}{x e^{2}}} \log {\relax (x )}^{8} - 4 x e^{\frac {3 \left (- 162 x + 81 e^{25}\right )}{x e^{2}}} \log {\relax (x )}^{7} + e^{\frac {4 \left (- 162 x + 81 e^{25}\right )}{x e^{2}}} \log {\relax (x )}^{6}}{\log {\relax (x )}^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**5*exp(2)*ln(x)**5+(324*x**3*exp(25)-12*x**4*exp(2))*exp((81*exp(25)-162*x)/exp(2)/x)*ln(x)**4+
((-972*x**2*exp(25)+12*x**3*exp(2))*exp((81*exp(25)-162*x)/exp(2)/x)**2+4*x**4*exp(2)*exp((81*exp(25)-162*x)/e
xp(2)/x))*ln(x)**3+((972*x*exp(25)-4*x**2*exp(2))*exp((81*exp(25)-162*x)/exp(2)/x)**3-12*x**3*exp(2)*exp((81*e
xp(25)-162*x)/exp(2)/x)**2)*ln(x)**2+(-324*exp(25)*exp((81*exp(25)-162*x)/exp(2)/x)**4+12*x**2*exp(2)*exp((81*
exp(25)-162*x)/exp(2)/x)**3)*ln(x)-4*x*exp(2)*exp((81*exp(25)-162*x)/exp(2)/x)**4)/x**2/exp(2)/ln(x)**5,x)

[Out]

x**4 + (-4*x**3*exp((-162*x + 81*exp(25))*exp(-2)/x)*log(x)**9 + 6*x**2*exp(2*(-162*x + 81*exp(25))*exp(-2)/x)
*log(x)**8 - 4*x*exp(3*(-162*x + 81*exp(25))*exp(-2)/x)*log(x)**7 + exp(4*(-162*x + 81*exp(25))*exp(-2)/x)*log
(x)**6)/log(x)**10

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