Integrand size = 166, antiderivative size = 36 \[ \int \frac {-16 x+\left (32 x^2+8 e^x x^2\right ) \log ^2(x)+e^{\frac {2 e^{2 x}-10 x}{x}} \left (8 x+e^{2 x} (-16+32 x) \log (x)+\left (-16 x^2-4 e^x x^2+e^{2 x} \left (-32 x+64 x^2+e^x (-8+16 x)\right )\right ) \log ^2(x)\right )}{4 x^2 \log ^2(x)-4 e^{\frac {2 e^{2 x}-10 x}{x}} x^2 \log ^2(x)+e^{\frac {2 \left (2 e^{2 x}-10 x\right )}{x}} x^2 \log ^2(x)} \, dx=\frac {4 \left (e^x+4 x+\frac {2}{\log (x)}\right )}{2-e^{2 \left (-5+\frac {e^{2 x}}{x}\right )}} \]
Time = 0.32 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.17 \[ \int \frac {-16 x+\left (32 x^2+8 e^x x^2\right ) \log ^2(x)+e^{\frac {2 e^{2 x}-10 x}{x}} \left (8 x+e^{2 x} (-16+32 x) \log (x)+\left (-16 x^2-4 e^x x^2+e^{2 x} \left (-32 x+64 x^2+e^x (-8+16 x)\right )\right ) \log ^2(x)\right )}{4 x^2 \log ^2(x)-4 e^{\frac {2 e^{2 x}-10 x}{x}} x^2 \log ^2(x)+e^{\frac {2 \left (2 e^{2 x}-10 x\right )}{x}} x^2 \log ^2(x)} \, dx=-\frac {4 e^{10} \left (2+e^x \log (x)+4 x \log (x)\right )}{\left (-2 e^{10}+e^{\frac {2 e^{2 x}}{x}}\right ) \log (x)} \]
Integrate[(-16*x + (32*x^2 + 8*E^x*x^2)*Log[x]^2 + E^((2*E^(2*x) - 10*x)/x )*(8*x + E^(2*x)*(-16 + 32*x)*Log[x] + (-16*x^2 - 4*E^x*x^2 + E^(2*x)*(-32 *x + 64*x^2 + E^x*(-8 + 16*x)))*Log[x]^2))/(4*x^2*Log[x]^2 - 4*E^((2*E^(2* x) - 10*x)/x)*x^2*Log[x]^2 + E^((2*(2*E^(2*x) - 10*x))/x)*x^2*Log[x]^2),x]
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.
| \(\displaystyle \int \frac {\left (8 e^x x^2+32 x^2\right ) \log ^2(x)+e^{\frac {2 e^{2 x}-10 x}{x}} \left (\left (-4 e^x x^2-16 x^2+e^{2 x} \left (64 x^2-32 x+e^x (16 x-8)\right )\right ) \log ^2(x)+8 x+e^{2 x} (32 x-16) \log (x)\right )-16 x}{-4 e^{\frac {2 e^{2 x}-10 x}{x}} x^2 \log ^2(x)+e^{\frac {2 \left (2 e^{2 x}-10 x\right )}{x}} x^2 \log ^2(x)+4 x^2 \log ^2(x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{20} \left (\left (8 e^x x^2+32 x^2\right ) \log ^2(x)+e^{\frac {2 e^{2 x}-10 x}{x}} \left (\left (-4 e^x x^2-16 x^2+e^{2 x} \left (64 x^2-32 x+e^x (16 x-8)\right )\right ) \log ^2(x)+8 x+e^{2 x} (32 x-16) \log (x)\right )-16 x\right )}{\left (2 e^{10}-e^{\frac {2 e^{2 x}}{x}}\right )^2 x^2 \log ^2(x)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle e^{20} \int -\frac {4 \left (-2 \left (e^x x^2+4 x^2\right ) \log ^2(x)+4 x-e^{\frac {2 \left (e^{2 x}-5 x\right )}{x}} \left (-\left (\left (e^x x^2+4 x^2+2 e^{2 x} \left (-8 x^2+4 x+e^x (1-2 x)\right )\right ) \log ^2(x)\right )-4 e^{2 x} (1-2 x) \log (x)+2 x\right )\right )}{\left (2 e^{10}-e^{\frac {2 e^{2 x}}{x}}\right )^2 x^2 \log ^2(x)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -4 e^{20} \int \frac {-2 \left (e^x x^2+4 x^2\right ) \log ^2(x)+4 x-e^{\frac {2 \left (e^{2 x}-5 x\right )}{x}} \left (-\left (\left (e^x x^2+4 x^2+2 e^{2 x} \left (-8 x^2+4 x+e^x (1-2 x)\right )\right ) \log ^2(x)\right )-4 e^{2 x} (1-2 x) \log (x)+2 x\right )}{\left (2 e^{10}-e^{\frac {2 e^{2 x}}{x}}\right )^2 x^2 \log ^2(x)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -4 e^{20} \int \left (-\frac {2 e^{3 x-10+\frac {2 e^{2 x}}{x}} (2 x-1)}{\left (-2 e^{10}+e^{\frac {2 e^{2 x}}{x}}\right )^2 x^2}+\frac {4 e^{\frac {2 \left (x^2-5 x+e^{2 x}\right )}{x}} (1-2 x) (2 x \log (x)+1)}{\left (2 e^{10}-e^{\frac {2 e^{2 x}}{x}}\right )^2 x^2 \log (x)}-\frac {e^{x-10}}{2 e^{10}-e^{\frac {2 e^{2 x}}{x}}}+\frac {4 e^{\frac {2 \left (e^{2 x}-5 x\right )}{x}}}{\left (2 e^{10}-e^{\frac {2 e^{2 x}}{x}}\right )^2}-\frac {8}{\left (-2 e^{10}+e^{\frac {2 e^{2 x}}{x}}\right )^2}-\frac {2 e^{\frac {2 \left (e^{2 x}-5 x\right )}{x}}}{\left (2 e^{10}-e^{\frac {2 e^{2 x}}{x}}\right )^2 x \log ^2(x)}+\frac {4}{\left (-2 e^{10}+e^{\frac {2 e^{2 x}}{x}}\right )^2 x \log ^2(x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -4 e^{20} \left (-16 \int \frac {e^{\frac {2 \left (x^2-5 x+e^{2 x}\right )}{x}}}{\left (-2 e^{10}+e^{\frac {2 e^{2 x}}{x}}\right )^2}dx+2 \int \frac {e^{3 x-10+\frac {2 e^{2 x}}{x}}}{\left (-2 e^{10}+e^{\frac {2 e^{2 x}}{x}}\right )^2 x^2}dx+8 \int \frac {e^{\frac {2 \left (x^2-5 x+e^{2 x}\right )}{x}}}{\left (-2 e^{10}+e^{\frac {2 e^{2 x}}{x}}\right )^2 x}dx+4 \int \frac {e^{\frac {2 \left (x^2-5 x+e^{2 x}\right )}{x}}}{\left (-2 e^{10}+e^{\frac {2 e^{2 x}}{x}}\right )^2 x^2 \log (x)}dx-8 \int \frac {e^{\frac {2 \left (x^2-5 x+e^{2 x}\right )}{x}}}{\left (-2 e^{10}+e^{\frac {2 e^{2 x}}{x}}\right )^2 x \log (x)}dx+4 \int \frac {e^{\frac {2 \left (e^{2 x}-5 x\right )}{x}}}{\left (2 e^{10}-e^{\frac {2 e^{2 x}}{x}}\right )^2}dx-\int \frac {e^{x-10}}{2 e^{10}-e^{\frac {2 e^{2 x}}{x}}}dx-8 \int \frac {1}{\left (-2 e^{10}+e^{\frac {2 e^{2 x}}{x}}\right )^2}dx-4 \int \frac {e^{3 x-10+\frac {2 e^{2 x}}{x}}}{\left (-2 e^{10}+e^{\frac {2 e^{2 x}}{x}}\right )^2 x}dx-2 \int \frac {e^{\frac {2 \left (e^{2 x}-5 x\right )}{x}}}{\left (2 e^{10}-e^{\frac {2 e^{2 x}}{x}}\right )^2 x \log ^2(x)}dx+4 \int \frac {1}{\left (-2 e^{10}+e^{\frac {2 e^{2 x}}{x}}\right )^2 x \log ^2(x)}dx\right )\) |
Int[(-16*x + (32*x^2 + 8*E^x*x^2)*Log[x]^2 + E^((2*E^(2*x) - 10*x)/x)*(8*x + E^(2*x)*(-16 + 32*x)*Log[x] + (-16*x^2 - 4*E^x*x^2 + E^(2*x)*(-32*x + 6 4*x^2 + E^x*(-8 + 16*x)))*Log[x]^2))/(4*x^2*Log[x]^2 - 4*E^((2*E^(2*x) - 1 0*x)/x)*x^2*Log[x]^2 + E^((2*(2*E^(2*x) - 10*x))/x)*x^2*Log[x]^2),x]
3.6.53.3.1 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 = 12.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.06
| method | result | size |
| parallelrisch | \(-\frac {8+16 x \ln \left (x \right )+4 \,{\mathrm e}^{x} \ln \left (x \right )}{\left ({\mathrm e}^{\frac {2 \,{\mathrm e}^{2 x}-10 x}{x}}-2\right ) \ln \left (x \right )}\) | \(38\) |
| risch | \(-\frac {4 \left (4 x \ln \left (x \right )+{\mathrm e}^{x} \ln \left (x \right )+2\right )}{\ln \left (x \right ) \left ({\mathrm e}^{-\frac {2 \left (-{\mathrm e}^{2 x}+5 x \right )}{x}}-2\right )}\) | \(39\) |
int((((((16*x-8)*exp(x)+64*x^2-32*x)*exp(2*x)-4*exp(x)*x^2-16*x^2)*ln(x)^2 +(32*x-16)*exp(2*x)*ln(x)+8*x)*exp((2*exp(2*x)-10*x)/x)+(8*exp(x)*x^2+32*x ^2)*ln(x)^2-16*x)/(x^2*ln(x)^2*exp((2*exp(2*x)-10*x)/x)^2-4*x^2*ln(x)^2*ex p((2*exp(2*x)-10*x)/x)+4*x^2*ln(x)^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.08 \[ \int \frac {-16 x+\left (32 x^2+8 e^x x^2\right ) \log ^2(x)+e^{\frac {2 e^{2 x}-10 x}{x}} \left (8 x+e^{2 x} (-16+32 x) \log (x)+\left (-16 x^2-4 e^x x^2+e^{2 x} \left (-32 x+64 x^2+e^x (-8+16 x)\right )\right ) \log ^2(x)\right )}{4 x^2 \log ^2(x)-4 e^{\frac {2 e^{2 x}-10 x}{x}} x^2 \log ^2(x)+e^{\frac {2 \left (2 e^{2 x}-10 x\right )}{x}} x^2 \log ^2(x)} \, dx=-\frac {4 \, {\left ({\left (4 \, x + e^{x}\right )} \log \left (x\right ) + 2\right )}}{e^{\left (-\frac {2 \, {\left (5 \, x - e^{\left (2 \, x\right )}\right )}}{x}\right )} \log \left (x\right ) - 2 \, \log \left (x\right )} \]
integrate((((((16*x-8)*exp(x)+64*x^2-32*x)*exp(2*x)-4*exp(x)*x^2-16*x^2)*l og(x)^2+(32*x-16)*exp(2*x)*log(x)+8*x)*exp((2*exp(2*x)-10*x)/x)+(8*exp(x)* x^2+32*x^2)*log(x)^2-16*x)/(x^2*log(x)^2*exp((2*exp(2*x)-10*x)/x)^2-4*x^2* log(x)^2*exp((2*exp(2*x)-10*x)/x)+4*x^2*log(x)^2),x, algorithm=\
Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.08 \[ \int \frac {-16 x+\left (32 x^2+8 e^x x^2\right ) \log ^2(x)+e^{\frac {2 e^{2 x}-10 x}{x}} \left (8 x+e^{2 x} (-16+32 x) \log (x)+\left (-16 x^2-4 e^x x^2+e^{2 x} \left (-32 x+64 x^2+e^x (-8+16 x)\right )\right ) \log ^2(x)\right )}{4 x^2 \log ^2(x)-4 e^{\frac {2 e^{2 x}-10 x}{x}} x^2 \log ^2(x)+e^{\frac {2 \left (2 e^{2 x}-10 x\right )}{x}} x^2 \log ^2(x)} \, dx=\frac {- 16 x \log {\left (x \right )} - 4 e^{x} \log {\left (x \right )} - 8}{e^{\frac {- 10 x + 2 e^{2 x}}{x}} \log {\left (x \right )} - 2 \log {\left (x \right )}} \]
integrate((((((16*x-8)*exp(x)+64*x**2-32*x)*exp(2*x)-4*exp(x)*x**2-16*x**2 )*ln(x)**2+(32*x-16)*exp(2*x)*ln(x)+8*x)*exp((2*exp(2*x)-10*x)/x)+(8*exp(x )*x**2+32*x**2)*ln(x)**2-16*x)/(x**2*ln(x)**2*exp((2*exp(2*x)-10*x)/x)**2- 4*x**2*ln(x)**2*exp((2*exp(2*x)-10*x)/x)+4*x**2*ln(x)**2),x)
Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.22 \[ \int \frac {-16 x+\left (32 x^2+8 e^x x^2\right ) \log ^2(x)+e^{\frac {2 e^{2 x}-10 x}{x}} \left (8 x+e^{2 x} (-16+32 x) \log (x)+\left (-16 x^2-4 e^x x^2+e^{2 x} \left (-32 x+64 x^2+e^x (-8+16 x)\right )\right ) \log ^2(x)\right )}{4 x^2 \log ^2(x)-4 e^{\frac {2 e^{2 x}-10 x}{x}} x^2 \log ^2(x)+e^{\frac {2 \left (2 e^{2 x}-10 x\right )}{x}} x^2 \log ^2(x)} \, dx=\frac {4 \, {\left (4 \, x e^{10} \log \left (x\right ) + e^{\left (x + 10\right )} \log \left (x\right ) + 2 \, e^{10}\right )}}{2 \, e^{10} \log \left (x\right ) - e^{\left (\frac {2 \, e^{\left (2 \, x\right )}}{x}\right )} \log \left (x\right )} \]
integrate((((((16*x-8)*exp(x)+64*x^2-32*x)*exp(2*x)-4*exp(x)*x^2-16*x^2)*l og(x)^2+(32*x-16)*exp(2*x)*log(x)+8*x)*exp((2*exp(2*x)-10*x)/x)+(8*exp(x)* x^2+32*x^2)*log(x)^2-16*x)/(x^2*log(x)^2*exp((2*exp(2*x)-10*x)/x)^2-4*x^2* log(x)^2*exp((2*exp(2*x)-10*x)/x)+4*x^2*log(x)^2),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 8877 vs. \(2 (30) = 60\).
Time = 0.34 (sec) , antiderivative size = 8877, normalized size of antiderivative = 246.58 \[ \int \frac {-16 x+\left (32 x^2+8 e^x x^2\right ) \log ^2(x)+e^{\frac {2 e^{2 x}-10 x}{x}} \left (8 x+e^{2 x} (-16+32 x) \log (x)+\left (-16 x^2-4 e^x x^2+e^{2 x} \left (-32 x+64 x^2+e^x (-8+16 x)\right )\right ) \log ^2(x)\right )}{4 x^2 \log ^2(x)-4 e^{\frac {2 e^{2 x}-10 x}{x}} x^2 \log ^2(x)+e^{\frac {2 \left (2 e^{2 x}-10 x\right )}{x}} x^2 \log ^2(x)} \, dx=\text {Too large to display} \]
integrate((((((16*x-8)*exp(x)+64*x^2-32*x)*exp(2*x)-4*exp(x)*x^2-16*x^2)*l og(x)^2+(32*x-16)*exp(2*x)*log(x)+8*x)*exp((2*exp(2*x)-10*x)/x)+(8*exp(x)* x^2+32*x^2)*log(x)^2-16*x)/(x^2*log(x)^2*exp((2*exp(2*x)-10*x)/x)^2-4*x^2* log(x)^2*exp((2*exp(2*x)-10*x)/x)+4*x^2*log(x)^2),x, algorithm=\
2*(64*x^2*e^(8*x - 2*(5*x - e^(2*x))/x + 2*e^(2*x)/x + 30)*log(x)^2 - 128* x^2*e^(8*x - 2*(5*x - e^(2*x))/x + 40)*log(x)^2 - 128*x^2*e^(8*x + 2*e^(2* x)/x + 30)*log(x)^2 + 256*x^2*e^(8*x + 40)*log(x)^2 - 32*x^2*e^(7*x + (x^2 + 2*e^(2*x))/x - 2*(5*x - e^(2*x))/x + 2*e^(2*x)/x + 20)*log(x)^2 + 64*x^ 2*e^(7*x + (x^2 + 2*e^(2*x))/x - 2*(5*x - e^(2*x))/x + 30)*log(x)^2 + 64*x ^2*e^(7*x + (x^2 + 2*e^(2*x))/x + 2*e^(2*x)/x + 20)*log(x)^2 - 128*x^2*e^( 7*x + (x^2 + 2*e^(2*x))/x + 30)*log(x)^2 - 4*x^2*e^(7*x - 2*(5*x - e^(2*x) )/x + 2*e^(2*x)/x + 30)*log(x)^2 + 8*x^2*e^(7*x - 2*(5*x - e^(2*x))/x + 40 )*log(x)^2 + 4*x^2*e^(6*x + (x^2 + 2*e^(2*x))/x + 2*e^(2*x)/x + 20)*log(x) ^2 - 8*x^2*e^(6*x + (x^2 + 2*e^(2*x))/x + 30)*log(x)^2 + 32*x^2*e^(6*x - 2 *(5*x - e^(2*x))/x + 40)*log(x)^2 - 32*x^2*e^(6*x + 2*e^(2*x)/x + 30)*log( x)^2 - 32*x^2*e^(5*x + (3*x^2 + 2*e^(2*x))/x - 2*(5*x - e^(2*x))/x + 2*e^( 2*x)/x + 20)*log(x)^2 + 64*x^2*e^(5*x + (3*x^2 + 2*e^(2*x))/x - 2*(5*x - e ^(2*x))/x + 30)*log(x)^2 + 64*x^2*e^(5*x + (3*x^2 + 2*e^(2*x))/x + 2*e^(2* x)/x + 20)*log(x)^2 - 128*x^2*e^(5*x + (3*x^2 + 2*e^(2*x))/x + 30)*log(x)^ 2 - 16*x^2*e^(5*x + (x^2 + 2*e^(2*x))/x - 2*(5*x - e^(2*x))/x + 30)*log(x) ^2 + 16*x^2*e^(5*x + (x^2 + 2*e^(2*x))/x + 2*e^(2*x)/x + 20)*log(x)^2 + 2* x^2*e^(5*x + 2*(x^2 + e^(2*x))/x - 2*(5*x - e^(2*x))/x + 2*e^(2*x)/x + 20) *log(x)^2 - 4*x^2*e^(5*x + 2*(x^2 + e^(2*x))/x - 2*(5*x - e^(2*x))/x + 30) *log(x)^2 + 16*x^2*e^(4*x + (3*x^2 + 2*e^(2*x))/x + (x^2 + 2*e^(2*x))/x...
Time = 12.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {-16 x+\left (32 x^2+8 e^x x^2\right ) \log ^2(x)+e^{\frac {2 e^{2 x}-10 x}{x}} \left (8 x+e^{2 x} (-16+32 x) \log (x)+\left (-16 x^2-4 e^x x^2+e^{2 x} \left (-32 x+64 x^2+e^x (-8+16 x)\right )\right ) \log ^2(x)\right )}{4 x^2 \log ^2(x)-4 e^{\frac {2 e^{2 x}-10 x}{x}} x^2 \log ^2(x)+e^{\frac {2 \left (2 e^{2 x}-10 x\right )}{x}} x^2 \log ^2(x)} \, dx=-\frac {4\,\left ({\mathrm {e}}^x\,\ln \left (x\right )+4\,x\,\ln \left (x\right )+2\right )}{\ln \left (x\right )\,\left ({\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{2\,x}}{x}-10}-2\right )} \]
int((log(x)^2*(8*x^2*exp(x) + 32*x^2) - 16*x + exp(-(10*x - 2*exp(2*x))/x) *(8*x - log(x)^2*(4*x^2*exp(x) - exp(2*x)*(exp(x)*(16*x - 8) - 32*x + 64*x ^2) + 16*x^2) + exp(2*x)*log(x)*(32*x - 16)))/(4*x^2*log(x)^2 - 4*x^2*exp( -(10*x - 2*exp(2*x))/x)*log(x)^2 + x^2*exp(-(2*(10*x - 2*exp(2*x)))/x)*log (x)^2),x)