Integrand size = 167, antiderivative size = 40 \[ \int \frac {-32+e^{\frac {-1+9 x}{x}}+14 x+64 x^2-16 x^3+\left (16-4 x-32 x^2+4 x^3\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+\left (-2+4 x^2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )}{32 x-8 x^2+\left (-16 x+2 x^2\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+2 x \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )} \, dx=x^2-\log \left (x-\frac {x^2}{4-\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )}\right ) \]
Leaf count is larger than twice the leaf count of optimal. \(145\) vs. \(2(40)=80\).
Time = 0.11 (sec) , antiderivative size = 145, normalized size of antiderivative = 3.62 \[ \int \frac {-32+e^{\frac {-1+9 x}{x}}+14 x+64 x^2-16 x^3+\left (16-4 x-32 x^2+4 x^3\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+\left (-2+4 x^2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )}{32 x-8 x^2+\left (-16 x+2 x^2\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+2 x \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )} \, dx=\frac {1}{2} \left (2 x^2-2 \log (x)+2 \log \left (8-e^{9-\frac {1}{x}}-2 \log \left (\frac {1}{x}\right )-2 \left (-\frac {1}{2} e^{9-\frac {1}{x}}-\log \left (\frac {1}{x}\right )+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )\right )-2 \log \left (8-e^{9-\frac {1}{x}}-2 x-2 \log \left (\frac {1}{x}\right )-2 \left (-\frac {1}{2} e^{9-\frac {1}{x}}-\log \left (\frac {1}{x}\right )+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right )\right )\right ) \]
Integrate[(-32 + E^((-1 + 9*x)/x) + 14*x + 64*x^2 - 16*x^3 + (16 - 4*x - 3 2*x^2 + 4*x^3)*Log[E^(E^((-1 + 9*x)/x)/2)/x] + (-2 + 4*x^2)*Log[E^(E^((-1 + 9*x)/x)/2)/x]^2)/(32*x - 8*x^2 + (-16*x + 2*x^2)*Log[E^(E^((-1 + 9*x)/x) /2)/x] + 2*x*Log[E^(E^((-1 + 9*x)/x)/2)/x]^2),x]
(2*x^2 - 2*Log[x] + 2*Log[8 - E^(9 - x^(-1)) - 2*Log[x^(-1)] - 2*(-1/2*E^( 9 - x^(-1)) - Log[x^(-1)] + Log[E^(E^(9 - x^(-1))/2)/x])] - 2*Log[8 - E^(9 - x^(-1)) - 2*x - 2*Log[x^(-1)] - 2*(-1/2*E^(9 - x^(-1)) - Log[x^(-1)] + Log[E^(E^(9 - x^(-1))/2)/x])])/2
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-16 x^3+64 x^2+\left (4 x^2-2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {9 x-1}{x}}}}{x}\right )+\left (4 x^3-32 x^2-4 x+16\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {9 x-1}{x}}}}{x}\right )+14 x+e^{\frac {9 x-1}{x}}-32}{-8 x^2+\left (2 x^2-16 x\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {9 x-1}{x}}}}{x}\right )+32 x+2 x \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {9 x-1}{x}}}}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-16 x^3+64 x^2+\left (4 x^2-2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {9 x-1}{x}}}}{x}\right )+\left (4 x^3-32 x^2-4 x+16\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {9 x-1}{x}}}}{x}\right )+14 x+e^{\frac {9 x-1}{x}}-32}{2 x \left (4-\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-x-\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )+4\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int -\frac {16 x^3-64 x^2-14 x-e^{-\frac {1-9 x}{x}}+2 \left (1-2 x^2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{-\frac {1-9 x}{x}}}}{x}\right )-4 \left (x^3-8 x^2-x+4\right ) \log \left (\frac {e^{\frac {1}{2} e^{-\frac {1-9 x}{x}}}}{x}\right )+32}{x \left (4-\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-x-\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )+4\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {16 x^3-64 x^2-14 x-e^{-\frac {1-9 x}{x}}+2 \left (1-2 x^2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{-\frac {1-9 x}{x}}}}{x}\right )-4 \left (x^3-8 x^2-x+4\right ) \log \left (\frac {e^{\frac {1}{2} e^{-\frac {1-9 x}{x}}}}{x}\right )+32}{x \left (4-\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )\right ) \left (-x-\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )+4\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {16 x^2}{\left (\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right ) \left (x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right )}-\frac {64 x}{\left (\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right ) \left (x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right )}-\frac {14}{\left (\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right ) \left (x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right )}-\frac {2 \left (2 x^2-1\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )}{\left (\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right ) \left (x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right ) x}-\frac {4 \left (x^3-8 x^2-x+4\right ) \log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )}{\left (\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right ) \left (x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right ) x}-\frac {e^{9-\frac {1}{x}}}{\left (\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right ) \left (x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right ) x}+\frac {32}{\left (\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right ) \left (x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right ) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\text {Subst}\left (\int \frac {e^{9-x}}{\log \left (e^{\frac {e^{9-x}}{2}} x\right )-4}dx,x,\frac {1}{x}\right )+2 \text {Subst}\left (\int \frac {1}{x \left (\log \left (e^{\frac {e^{9-x}}{2}} x\right )-4\right )}dx,x,\frac {1}{x}\right )-\int \frac {e^{9-\frac {1}{x}}}{x^2 \left (x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right )}dx+64 \int \frac {1}{-x-\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )+4}dx+62 \int \frac {1}{x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4}dx+2 \int \frac {1}{x \left (x+\log \left (\frac {e^{\frac {1}{2} e^{9-\frac {1}{x}}}}{x}\right )-4\right )}dx+2 x^2-2 \log (x)\right )\) |
Int[(-32 + E^((-1 + 9*x)/x) + 14*x + 64*x^2 - 16*x^3 + (16 - 4*x - 32*x^2 + 4*x^3)*Log[E^(E^((-1 + 9*x)/x)/2)/x] + (-2 + 4*x^2)*Log[E^(E^((-1 + 9*x) /x)/2)/x]^2)/(32*x - 8*x^2 + (-16*x + 2*x^2)*Log[E^(E^((-1 + 9*x)/x)/2)/x] + 2*x*Log[E^(E^((-1 + 9*x)/x)/2)/x]^2),x]
3.23.20.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.90 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.35
method | result | size |
parallelrisch | \(x^{2}-\ln \left (x \right )+\ln \left (\ln \left (\frac {{\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )-4\right )-\ln \left (\ln \left (\frac {{\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )+x -4\right )\) | \(54\) |
risch | \(x^{2}-\ln \left (x \right )+\ln \left (\ln \left ({\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}\right )+\frac {i \left (-\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )+\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )^{2}+\pi \,\operatorname {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )^{2}-\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )^{3}+2 i \ln \left (x \right )+8 i\right )}{2}\right )-\ln \left (\ln \left ({\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}\right )-\frac {i \left (\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{\frac {9 x -1}{x}}}{2}}}{x}\right )^{3}+2 i x -2 i \ln \left (x \right )-8 i\right )}{2}\right )\) | \(357\) |
int(((4*x^2-2)*ln(exp(1/2*exp((9*x-1)/x))/x)^2+(4*x^3-32*x^2-4*x+16)*ln(ex p(1/2*exp((9*x-1)/x))/x)+exp((9*x-1)/x)-16*x^3+64*x^2+14*x-32)/(2*x*ln(exp (1/2*exp((9*x-1)/x))/x)^2+(2*x^2-16*x)*ln(exp(1/2*exp((9*x-1)/x))/x)-8*x^2 +32*x),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (33) = 66\).
Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.98 \[ \int \frac {-32+e^{\frac {-1+9 x}{x}}+14 x+64 x^2-16 x^3+\left (16-4 x-32 x^2+4 x^3\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+\left (-2+4 x^2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )}{32 x-8 x^2+\left (-16 x+2 x^2\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+2 x \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )} \, dx=x^{2} - \frac {1}{2} \, e^{\left (\frac {9 \, x - 1}{x}\right )} - \log \left (x + \log \left (\frac {e^{\left (\frac {1}{2} \, e^{\left (\frac {9 \, x - 1}{x}\right )}\right )}}{x}\right ) - 4\right ) + \log \left (\frac {e^{\left (\frac {1}{2} \, e^{\left (\frac {9 \, x - 1}{x}\right )}\right )}}{x}\right ) + \log \left (\log \left (\frac {e^{\left (\frac {1}{2} \, e^{\left (\frac {9 \, x - 1}{x}\right )}\right )}}{x}\right ) - 4\right ) \]
integrate(((4*x^2-2)*log(exp(1/2*exp((9*x-1)/x))/x)^2+(4*x^3-32*x^2-4*x+16 )*log(exp(1/2*exp((9*x-1)/x))/x)+exp((9*x-1)/x)-16*x^3+64*x^2+14*x-32)/(2* x*log(exp(1/2*exp((9*x-1)/x))/x)^2+(2*x^2-16*x)*log(exp(1/2*exp((9*x-1)/x) )/x)-8*x^2+32*x),x, algorithm=\
x^2 - 1/2*e^((9*x - 1)/x) - log(x + log(e^(1/2*e^((9*x - 1)/x))/x) - 4) + log(e^(1/2*e^((9*x - 1)/x))/x) + log(log(e^(1/2*e^((9*x - 1)/x))/x) - 4)
Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.10 \[ \int \frac {-32+e^{\frac {-1+9 x}{x}}+14 x+64 x^2-16 x^3+\left (16-4 x-32 x^2+4 x^3\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+\left (-2+4 x^2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )}{32 x-8 x^2+\left (-16 x+2 x^2\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+2 x \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )} \, dx=x^{2} - \log {\left (x \right )} + \log {\left (\log {\left (\frac {e^{\frac {e^{\frac {9 x - 1}{x}}}{2}}}{x} \right )} - 4 \right )} - \log {\left (x + \log {\left (\frac {e^{\frac {e^{\frac {9 x - 1}{x}}}{2}}}{x} \right )} - 4 \right )} \]
integrate(((4*x**2-2)*ln(exp(1/2*exp((9*x-1)/x))/x)**2+(4*x**3-32*x**2-4*x +16)*ln(exp(1/2*exp((9*x-1)/x))/x)+exp((9*x-1)/x)-16*x**3+64*x**2+14*x-32) /(2*x*ln(exp(1/2*exp((9*x-1)/x))/x)**2+(2*x**2-16*x)*ln(exp(1/2*exp((9*x-1 )/x))/x)-8*x**2+32*x),x)
x**2 - log(x) + log(log(exp(exp((9*x - 1)/x)/2)/x) - 4) - log(x + log(exp( exp((9*x - 1)/x)/2)/x) - 4)
Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (33) = 66\).
Time = 0.24 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.92 \[ \int \frac {-32+e^{\frac {-1+9 x}{x}}+14 x+64 x^2-16 x^3+\left (16-4 x-32 x^2+4 x^3\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+\left (-2+4 x^2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )}{32 x-8 x^2+\left (-16 x+2 x^2\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+2 x \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )} \, dx=x^{2} - \log \left (x\right ) - \log \left (-x + \log \left (x\right ) + 4\right ) - \log \left (\frac {2 \, {\left (x - \log \left (x\right ) - 4\right )} e^{\frac {1}{x}} + e^{9}}{2 \, {\left (x - \log \left (x\right ) - 4\right )}}\right ) + \log \left (\frac {2 \, {\left (\log \left (x\right ) + 4\right )} e^{\frac {1}{x}} - e^{9}}{2 \, {\left (\log \left (x\right ) + 4\right )}}\right ) + \log \left (\log \left (x\right ) + 4\right ) \]
integrate(((4*x^2-2)*log(exp(1/2*exp((9*x-1)/x))/x)^2+(4*x^3-32*x^2-4*x+16 )*log(exp(1/2*exp((9*x-1)/x))/x)+exp((9*x-1)/x)-16*x^3+64*x^2+14*x-32)/(2* x*log(exp(1/2*exp((9*x-1)/x))/x)^2+(2*x^2-16*x)*log(exp(1/2*exp((9*x-1)/x) )/x)-8*x^2+32*x),x, algorithm=\
x^2 - log(x) - log(-x + log(x) + 4) - log(1/2*(2*(x - log(x) - 4)*e^(1/x) + e^9)/(x - log(x) - 4)) + log(1/2*(2*(log(x) + 4)*e^(1/x) - e^9)/(log(x) + 4)) + log(log(x) + 4)
Time = 0.36 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.32 \[ \int \frac {-32+e^{\frac {-1+9 x}{x}}+14 x+64 x^2-16 x^3+\left (16-4 x-32 x^2+4 x^3\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+\left (-2+4 x^2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )}{32 x-8 x^2+\left (-16 x+2 x^2\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+2 x \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )} \, dx=x^{2} - \log \left (2 \, x e^{\frac {1}{x}} - 2 \, e^{\frac {1}{x}} \log \left (x\right ) + e^{9} - 8 \, e^{\frac {1}{x}}\right ) + \log \left (-2 \, e^{\frac {1}{x}} \log \left (x\right ) + e^{9} - 8 \, e^{\frac {1}{x}}\right ) - \log \left (x\right ) \]
integrate(((4*x^2-2)*log(exp(1/2*exp((9*x-1)/x))/x)^2+(4*x^3-32*x^2-4*x+16 )*log(exp(1/2*exp((9*x-1)/x))/x)+exp((9*x-1)/x)-16*x^3+64*x^2+14*x-32)/(2* x*log(exp(1/2*exp((9*x-1)/x))/x)^2+(2*x^2-16*x)*log(exp(1/2*exp((9*x-1)/x) )/x)-8*x^2+32*x),x, algorithm=\
x^2 - log(2*x*e^(1/x) - 2*e^(1/x)*log(x) + e^9 - 8*e^(1/x)) + log(-2*e^(1/ x)*log(x) + e^9 - 8*e^(1/x)) - log(x)
Timed out. \[ \int \frac {-32+e^{\frac {-1+9 x}{x}}+14 x+64 x^2-16 x^3+\left (16-4 x-32 x^2+4 x^3\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+\left (-2+4 x^2\right ) \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )}{32 x-8 x^2+\left (-16 x+2 x^2\right ) \log \left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )+2 x \log ^2\left (\frac {e^{\frac {1}{2} e^{\frac {-1+9 x}{x}}}}{x}\right )} \, dx=\int \frac {14\,x+{\mathrm {e}}^{\frac {9\,x-1}{x}}-\ln \left (\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {9\,x-1}{x}}}{2}}}{x}\right )\,\left (-4\,x^3+32\,x^2+4\,x-16\right )+{\ln \left (\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {9\,x-1}{x}}}{2}}}{x}\right )}^2\,\left (4\,x^2-2\right )+64\,x^2-16\,x^3-32}{32\,x-\ln \left (\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {9\,x-1}{x}}}{2}}}{x}\right )\,\left (16\,x-2\,x^2\right )+2\,x\,{\ln \left (\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {9\,x-1}{x}}}{2}}}{x}\right )}^2-8\,x^2} \,d x \]
int((14*x + exp((9*x - 1)/x) - log(exp(exp((9*x - 1)/x)/2)/x)*(4*x + 32*x^ 2 - 4*x^3 - 16) + log(exp(exp((9*x - 1)/x)/2)/x)^2*(4*x^2 - 2) + 64*x^2 - 16*x^3 - 32)/(32*x - log(exp(exp((9*x - 1)/x)/2)/x)*(16*x - 2*x^2) + 2*x*l og(exp(exp((9*x - 1)/x)/2)/x)^2 - 8*x^2),x)