Integrand size = 69, antiderivative size = 27 \[ \int \frac {-10+e^{e^5+3 x} \left (e^4 \left (-2-6 x+3 x^2\right )+e^2 \left (-4-12 x+6 x^2\right ) \log (5)+\left (-2-6 x+3 x^2\right ) \log ^2(5)\right )}{4-4 x+x^2} \, dx=\frac {x \left (5+e^{e^5+3 x} \left (e^2+\log (5)\right )^2\right )}{-2+x} \] Output:
(exp(exp(5)+3*x)*(ln(5)+exp(2))^2+5)*x/(-2+x)
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-10+e^{e^5+3 x} \left (e^4 \left (-2-6 x+3 x^2\right )+e^2 \left (-4-12 x+6 x^2\right ) \log (5)+\left (-2-6 x+3 x^2\right ) \log ^2(5)\right )}{4-4 x+x^2} \, dx=\frac {10+e^{e^5+3 x} x \left (e^2+\log (5)\right )^2}{-2+x} \] Input:
Integrate[(-10 + E^(E^5 + 3*x)*(E^4*(-2 - 6*x + 3*x^2) + E^2*(-4 - 12*x + 6*x^2)*Log[5] + (-2 - 6*x + 3*x^2)*Log[5]^2))/(4 - 4*x + x^2),x]
Output:
(10 + E^(E^5 + 3*x)*x*(E^2 + Log[5])^2)/(-2 + x)
Time = 0.72 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {7239, 7293, 2009}
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.
\(\displaystyle \int \frac {e^{3 x+e^5} \left (e^4 \left (3 x^2-6 x-2\right )+\left (3 x^2-6 x-2\right ) \log ^2(5)+e^2 \left (6 x^2-12 x-4\right ) \log (5)\right )-10}{x^2-4 x+4} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{3 x+e^5} \left (3 x^2-6 x-2\right ) \left (e^2+\log (5)\right )^2-10}{(2-x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{3 x+e^5} \left (3 x^2-6 x-2\right ) \left (e^2+\log (5)\right )^2}{(x-2)^2}-\frac {10}{(x-2)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {10}{2-x}+e^{3 x+e^5} \left (e^2+\log (5)\right )^2-\frac {2 e^{3 x+e^5} \left (e^2+\log (5)\right )^2}{2-x}\) |
Input:
Int[(-10 + E^(E^5 + 3*x)*(E^4*(-2 - 6*x + 3*x^2) + E^2*(-4 - 12*x + 6*x^2) *Log[5] + (-2 - 6*x + 3*x^2)*Log[5]^2))/(4 - 4*x + x^2),x]
Output:
-10/(2 - x) + E^(E^5 + 3*x)*(E^2 + Log[5])^2 - (2*E^(E^5 + 3*x)*(E^2 + Log [5])^2)/(2 - x)
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.69 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22
method | result | size |
norman | \(\frac {\left ({\mathrm e}^{4}+2 \,{\mathrm e}^{2} \ln \left (5\right )+\ln \left (5\right )^{2}\right ) x \,{\mathrm e}^{{\mathrm e}^{5}+3 x}+10}{-2+x}\) | \(33\) |
risch | \(\frac {10}{-2+x}+\frac {\left ({\mathrm e}^{4}+2 \,{\mathrm e}^{2} \ln \left (5\right )+\ln \left (5\right )^{2}\right ) x \,{\mathrm e}^{{\mathrm e}^{5}+3 x}}{-2+x}\) | \(36\) |
parallelrisch | \(\frac {\ln \left (5\right )^{2} {\mathrm e}^{{\mathrm e}^{5}+3 x} x +2 \ln \left (5\right ) {\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{5}+3 x} x +{\mathrm e}^{{\mathrm e}^{5}+3 x} {\mathrm e}^{4} x +10}{-2+x}\) | \(49\) |
parts | \({\mathrm e}^{{\mathrm e}^{5}+3 x} {\mathrm e}^{4}+\ln \left (5\right )^{2} {\mathrm e}^{{\mathrm e}^{5}+3 x}+2 \ln \left (5\right ) {\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{5}+3 x}-\frac {6 \ln \left (5\right )^{2} {\mathrm e}^{{\mathrm e}^{5}+3 x}}{-3 x +6}-\frac {6 \,{\mathrm e}^{4} {\mathrm e}^{{\mathrm e}^{5}+3 x}}{-3 x +6}-\frac {12 \,{\mathrm e}^{2} \ln \left (5\right ) {\mathrm e}^{{\mathrm e}^{5}+3 x}}{-3 x +6}+\frac {10}{-2+x}\) | \(230\) |
orering | \(-\frac {x \left (3 x^{2}-6 x -4\right ) \left (\left (\left (3 x^{2}-6 x -2\right ) \ln \left (5\right )^{2}+\left (6 x^{2}-12 x -4\right ) {\mathrm e}^{2} \ln \left (5\right )+\left (3 x^{2}-6 x -2\right ) {\mathrm e}^{4}\right ) {\mathrm e}^{{\mathrm e}^{5}+3 x}-10\right )}{2 \left (2+3 x \right ) \left (x^{2}-4 x +4\right )}+\frac {x^{2} \left (-2+x \right ) \left (\frac {\left (\left (6 x -6\right ) \ln \left (5\right )^{2}+\left (12 x -12\right ) {\mathrm e}^{2} \ln \left (5\right )+\left (6 x -6\right ) {\mathrm e}^{4}\right ) {\mathrm e}^{{\mathrm e}^{5}+3 x}+3 \left (\left (3 x^{2}-6 x -2\right ) \ln \left (5\right )^{2}+\left (6 x^{2}-12 x -4\right ) {\mathrm e}^{2} \ln \left (5\right )+\left (3 x^{2}-6 x -2\right ) {\mathrm e}^{4}\right ) {\mathrm e}^{{\mathrm e}^{5}+3 x}}{x^{2}-4 x +4}-\frac {\left (\left (\left (3 x^{2}-6 x -2\right ) \ln \left (5\right )^{2}+\left (6 x^{2}-12 x -4\right ) {\mathrm e}^{2} \ln \left (5\right )+\left (3 x^{2}-6 x -2\right ) {\mathrm e}^{4}\right ) {\mathrm e}^{{\mathrm e}^{5}+3 x}-10\right ) \left (2 x -4\right )}{\left (x^{2}-4 x +4\right )^{2}}\right )}{6 x +4}\) | \(283\) |
derivativedivides | \(\text {Expression too large to display}\) | \(802\) |
default | \(\text {Expression too large to display}\) | \(802\) |
Input:
int((((3*x^2-6*x-2)*ln(5)^2+(6*x^2-12*x-4)*exp(2)*ln(5)+(3*x^2-6*x-2)*exp( 2)^2)*exp(exp(5)+3*x)-10)/(x^2-4*x+4),x,method=_RETURNVERBOSE)
Output:
((ln(5)^2+2*exp(2)*ln(5)+exp(2)^2)*x*exp(exp(5)+3*x)+10)/(-2+x)
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {-10+e^{e^5+3 x} \left (e^4 \left (-2-6 x+3 x^2\right )+e^2 \left (-4-12 x+6 x^2\right ) \log (5)+\left (-2-6 x+3 x^2\right ) \log ^2(5)\right )}{4-4 x+x^2} \, dx=\frac {{\left (2 \, x e^{2} \log \left (5\right ) + x \log \left (5\right )^{2} + x e^{4}\right )} e^{\left (3 \, x + e^{5}\right )} + 10}{x - 2} \] Input:
integrate((((3*x^2-6*x-2)*log(5)^2+(6*x^2-12*x-4)*exp(2)*log(5)+(3*x^2-6*x -2)*exp(2)^2)*exp(exp(5)+3*x)-10)/(x^2-4*x+4),x, algorithm="fricas")
Output:
((2*x*e^2*log(5) + x*log(5)^2 + x*e^4)*e^(3*x + e^5) + 10)/(x - 2)
Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {-10+e^{e^5+3 x} \left (e^4 \left (-2-6 x+3 x^2\right )+e^2 \left (-4-12 x+6 x^2\right ) \log (5)+\left (-2-6 x+3 x^2\right ) \log ^2(5)\right )}{4-4 x+x^2} \, dx=\frac {\left (x \log {\left (5 \right )}^{2} + 2 x e^{2} \log {\left (5 \right )} + x e^{4}\right ) e^{3 x + e^{5}}}{x - 2} + \frac {10}{x - 2} \] Input:
integrate((((3*x**2-6*x-2)*ln(5)**2+(6*x**2-12*x-4)*exp(2)*ln(5)+(3*x**2-6 *x-2)*exp(2)**2)*exp(exp(5)+3*x)-10)/(x**2-4*x+4),x)
Output:
(x*log(5)**2 + 2*x*exp(2)*log(5) + x*exp(4))*exp(3*x + exp(5))/(x - 2) + 1 0/(x - 2)
\[ \int \frac {-10+e^{e^5+3 x} \left (e^4 \left (-2-6 x+3 x^2\right )+e^2 \left (-4-12 x+6 x^2\right ) \log (5)+\left (-2-6 x+3 x^2\right ) \log ^2(5)\right )}{4-4 x+x^2} \, dx=\int { \frac {{\left (2 \, {\left (3 \, x^{2} - 6 \, x - 2\right )} e^{2} \log \left (5\right ) + {\left (3 \, x^{2} - 6 \, x - 2\right )} \log \left (5\right )^{2} + {\left (3 \, x^{2} - 6 \, x - 2\right )} e^{4}\right )} e^{\left (3 \, x + e^{5}\right )} - 10}{x^{2} - 4 \, x + 4} \,d x } \] Input:
integrate((((3*x^2-6*x-2)*log(5)^2+(6*x^2-12*x-4)*exp(2)*log(5)+(3*x^2-6*x -2)*exp(2)^2)*exp(exp(5)+3*x)-10)/(x^2-4*x+4),x, algorithm="maxima")
Output:
2*e^(e^5 + 6)*exp_integral_e(2, -3*x + 6)*log(5)^2/(x - 2) + 2*integrate(e ^(3*x + e^5)/(x^2 - 4*x + 4), x)*log(5)^2 + (e^(e^5)*log(5)^2 + 2*e^(e^5 + 2)*log(5) + e^(e^5 + 4))*x*e^(3*x)/(x - 2) + 10/(x - 2)
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \frac {-10+e^{e^5+3 x} \left (e^4 \left (-2-6 x+3 x^2\right )+e^2 \left (-4-12 x+6 x^2\right ) \log (5)+\left (-2-6 x+3 x^2\right ) \log ^2(5)\right )}{4-4 x+x^2} \, dx=\frac {x e^{\left (3 \, x + e^{5}\right )} \log \left (5\right )^{2} + 2 \, x e^{\left (3 \, x + e^{5} + 2\right )} \log \left (5\right ) + x e^{\left (3 \, x + e^{5} + 4\right )} + 10}{x - 2} \] Input:
integrate((((3*x^2-6*x-2)*log(5)^2+(6*x^2-12*x-4)*exp(2)*log(5)+(3*x^2-6*x -2)*exp(2)^2)*exp(exp(5)+3*x)-10)/(x^2-4*x+4),x, algorithm="giac")
Output:
(x*e^(3*x + e^5)*log(5)^2 + 2*x*e^(3*x + e^5 + 2)*log(5) + x*e^(3*x + e^5 + 4) + 10)/(x - 2)
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-10+e^{e^5+3 x} \left (e^4 \left (-2-6 x+3 x^2\right )+e^2 \left (-4-12 x+6 x^2\right ) \log (5)+\left (-2-6 x+3 x^2\right ) \log ^2(5)\right )}{4-4 x+x^2} \, dx=\frac {5\,x+x\,{\mathrm {e}}^{3\,x+{\mathrm {e}}^5}\,{\left ({\mathrm {e}}^2+\ln \left (5\right )\right )}^2}{x-2} \] Input:
int(-(exp(3*x + exp(5))*(exp(4)*(6*x - 3*x^2 + 2) + log(5)^2*(6*x - 3*x^2 + 2) + exp(2)*log(5)*(12*x - 6*x^2 + 4)) + 10)/(x^2 - 4*x + 4),x)
Output:
(5*x + x*exp(3*x + exp(5))*(exp(2) + log(5))^2)/(x - 2)
Time = 0.23 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.93 \[ \int \frac {-10+e^{e^5+3 x} \left (e^4 \left (-2-6 x+3 x^2\right )+e^2 \left (-4-12 x+6 x^2\right ) \log (5)+\left (-2-6 x+3 x^2\right ) \log ^2(5)\right )}{4-4 x+x^2} \, dx=\frac {x \left (e^{e^{5}+3 x} \mathrm {log}\left (5\right )^{2}+2 e^{e^{5}+3 x} \mathrm {log}\left (5\right ) e^{2}+e^{e^{5}+3 x} e^{4}+5\right )}{x -2} \] Input:
int((((3*x^2-6*x-2)*log(5)^2+(6*x^2-12*x-4)*exp(2)*log(5)+(3*x^2-6*x-2)*ex p(2)^2)*exp(exp(5)+3*x)-10)/(x^2-4*x+4),x)
Output:
(x*(e**(e**5 + 3*x)*log(5)**2 + 2*e**(e**5 + 3*x)*log(5)*e**2 + e**(e**5 + 3*x)*e**4 + 5))/(x - 2)