Integrand size = 91, antiderivative size = 25 \[ \int \frac {e^{1+8 x+x^2+x \log \left (\frac {7-3 e^4+6 x}{3 x}\right )} \left (-49-62 x-12 x^2+e^4 (21+6 x)+\left (-7+3 e^4-6 x\right ) \log \left (\frac {7-3 e^4+6 x}{3 x}\right )\right )}{-7+3 e^4-6 x} \, dx=e^{1+x \left (8+x+\log \left (2+\frac {\frac {7}{3}-e^4}{x}\right )\right )} \] Output:
exp(x*(8+ln((7/3-exp(4))/x+2)+x)+1)
Time = 0.05 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {e^{1+8 x+x^2+x \log \left (\frac {7-3 e^4+6 x}{3 x}\right )} \left (-49-62 x-12 x^2+e^4 (21+6 x)+\left (-7+3 e^4-6 x\right ) \log \left (\frac {7-3 e^4+6 x}{3 x}\right )\right )}{-7+3 e^4-6 x} \, dx=3^{-x} e^{1+8 x+x^2} \left (\frac {7-3 e^4+6 x}{x}\right )^x \] Input:
Integrate[(E^(1 + 8*x + x^2 + x*Log[(7 - 3*E^4 + 6*x)/(3*x)])*(-49 - 62*x - 12*x^2 + E^4*(21 + 6*x) + (-7 + 3*E^4 - 6*x)*Log[(7 - 3*E^4 + 6*x)/(3*x) ]))/(-7 + 3*E^4 - 6*x),x]
Output:
(E^(1 + 8*x + x^2)*((7 - 3*E^4 + 6*x)/x)^x)/3^x
Time = 0.65 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {7257}
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.
\(\displaystyle \int \frac {e^{x^2+8 x+x \log \left (\frac {6 x-3 e^4+7}{3 x}\right )+1} \left (-12 x^2-62 x+e^4 (6 x+21)+\left (-6 x+3 e^4-7\right ) \log \left (\frac {6 x-3 e^4+7}{3 x}\right )-49\right )}{-6 x+3 e^4-7} \, dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle 3^{-x} e^{x^2+8 x+1} \left (\frac {6 x-3 e^4+7}{x}\right )^x\) |
Input:
Int[(E^(1 + 8*x + x^2 + x*Log[(7 - 3*E^4 + 6*x)/(3*x)])*(-49 - 62*x - 12*x ^2 + E^4*(21 + 6*x) + (-7 + 3*E^4 - 6*x)*Log[(7 - 3*E^4 + 6*x)/(3*x)]))/(- 7 + 3*E^4 - 6*x),x]
Output:
(E^(1 + 8*x + x^2)*((7 - 3*E^4 + 6*x)/x)^x)/3^x
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 0.84 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08
method | result | size |
norman | \({\mathrm e}^{x \ln \left (\frac {-3 \,{\mathrm e}^{4}+6 x +7}{3 x}\right )+x^{2}+8 x +1}\) | \(27\) |
risch | \(\left (\frac {-3 \,{\mathrm e}^{4}+6 x +7}{3 x}\right )^{x} {\mathrm e}^{x^{2}+8 x +1}\) | \(27\) |
parallelrisch | \({\mathrm e}^{x \ln \left (-\frac {3 \,{\mathrm e}^{4}-6 x -7}{3 x}\right )+x^{2}+8 x +1}\) | \(27\) |
Input:
int(((3*exp(4)-6*x-7)*ln(1/3*(-3*exp(4)+6*x+7)/x)+(6*x+21)*exp(4)-12*x^2-6 2*x-49)*exp(x*ln(1/3*(-3*exp(4)+6*x+7)/x)+x^2+8*x+1)/(3*exp(4)-6*x-7),x,me thod=_RETURNVERBOSE)
Output:
exp(x*ln(1/3*(-3*exp(4)+6*x+7)/x)+x^2+8*x+1)
Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^{1+8 x+x^2+x \log \left (\frac {7-3 e^4+6 x}{3 x}\right )} \left (-49-62 x-12 x^2+e^4 (21+6 x)+\left (-7+3 e^4-6 x\right ) \log \left (\frac {7-3 e^4+6 x}{3 x}\right )\right )}{-7+3 e^4-6 x} \, dx=e^{\left (x^{2} + x \log \left (\frac {6 \, x - 3 \, e^{4} + 7}{3 \, x}\right ) + 8 \, x + 1\right )} \] Input:
integrate(((3*exp(4)-6*x-7)*log(1/3*(-3*exp(4)+6*x+7)/x)+(6*x+21)*exp(4)-1 2*x^2-62*x-49)*exp(x*log(1/3*(-3*exp(4)+6*x+7)/x)+x^2+8*x+1)/(3*exp(4)-6*x -7),x, algorithm="fricas")
Output:
e^(x^2 + x*log(1/3*(6*x - 3*e^4 + 7)/x) + 8*x + 1)
Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{1+8 x+x^2+x \log \left (\frac {7-3 e^4+6 x}{3 x}\right )} \left (-49-62 x-12 x^2+e^4 (21+6 x)+\left (-7+3 e^4-6 x\right ) \log \left (\frac {7-3 e^4+6 x}{3 x}\right )\right )}{-7+3 e^4-6 x} \, dx=e^{x^{2} + x \log {\left (\frac {2 x - e^{4} + \frac {7}{3}}{x} \right )} + 8 x + 1} \] Input:
integrate(((3*exp(4)-6*x-7)*ln(1/3*(-3*exp(4)+6*x+7)/x)+(6*x+21)*exp(4)-12 *x**2-62*x-49)*exp(x*ln(1/3*(-3*exp(4)+6*x+7)/x)+x**2+8*x+1)/(3*exp(4)-6*x -7),x)
Output:
exp(x**2 + x*log((2*x - exp(4) + 7/3)/x) + 8*x + 1)
Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{1+8 x+x^2+x \log \left (\frac {7-3 e^4+6 x}{3 x}\right )} \left (-49-62 x-12 x^2+e^4 (21+6 x)+\left (-7+3 e^4-6 x\right ) \log \left (\frac {7-3 e^4+6 x}{3 x}\right )\right )}{-7+3 e^4-6 x} \, dx=e^{\left (x^{2} - x \log \left (3\right ) + x \log \left (6 \, x - 3 \, e^{4} + 7\right ) - x \log \left (x\right ) + 8 \, x + 1\right )} \] Input:
integrate(((3*exp(4)-6*x-7)*log(1/3*(-3*exp(4)+6*x+7)/x)+(6*x+21)*exp(4)-1 2*x^2-62*x-49)*exp(x*log(1/3*(-3*exp(4)+6*x+7)/x)+x^2+8*x+1)/(3*exp(4)-6*x -7),x, algorithm="maxima")
Output:
e^(x^2 - x*log(3) + x*log(6*x - 3*e^4 + 7) - x*log(x) + 8*x + 1)
\[ \int \frac {e^{1+8 x+x^2+x \log \left (\frac {7-3 e^4+6 x}{3 x}\right )} \left (-49-62 x-12 x^2+e^4 (21+6 x)+\left (-7+3 e^4-6 x\right ) \log \left (\frac {7-3 e^4+6 x}{3 x}\right )\right )}{-7+3 e^4-6 x} \, dx=\int { \frac {{\left (12 \, x^{2} - 3 \, {\left (2 \, x + 7\right )} e^{4} + {\left (6 \, x - 3 \, e^{4} + 7\right )} \log \left (\frac {6 \, x - 3 \, e^{4} + 7}{3 \, x}\right ) + 62 \, x + 49\right )} e^{\left (x^{2} + x \log \left (\frac {6 \, x - 3 \, e^{4} + 7}{3 \, x}\right ) + 8 \, x + 1\right )}}{6 \, x - 3 \, e^{4} + 7} \,d x } \] Input:
integrate(((3*exp(4)-6*x-7)*log(1/3*(-3*exp(4)+6*x+7)/x)+(6*x+21)*exp(4)-1 2*x^2-62*x-49)*exp(x*log(1/3*(-3*exp(4)+6*x+7)/x)+x^2+8*x+1)/(3*exp(4)-6*x -7),x, algorithm="giac")
Output:
integrate((12*x^2 - 3*(2*x + 7)*e^4 + (6*x - 3*e^4 + 7)*log(1/3*(6*x - 3*e ^4 + 7)/x) + 62*x + 49)*e^(x^2 + x*log(1/3*(6*x - 3*e^4 + 7)/x) + 8*x + 1) /(6*x - 3*e^4 + 7), x)
Time = 0.89 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^{1+8 x+x^2+x \log \left (\frac {7-3 e^4+6 x}{3 x}\right )} \left (-49-62 x-12 x^2+e^4 (21+6 x)+\left (-7+3 e^4-6 x\right ) \log \left (\frac {7-3 e^4+6 x}{3 x}\right )\right )}{-7+3 e^4-6 x} \, dx={\mathrm {e}}^{8\,x}\,{\mathrm {e}}^{x^2}\,\mathrm {e}\,{\left (\frac {2\,x-{\mathrm {e}}^4+\frac {7}{3}}{x}\right )}^x \] Input:
int((exp(8*x + x*log((2*x - exp(4) + 7/3)/x) + x^2 + 1)*(62*x + 12*x^2 + l og((2*x - exp(4) + 7/3)/x)*(6*x - 3*exp(4) + 7) - exp(4)*(6*x + 21) + 49)) /(6*x - 3*exp(4) + 7),x)
Output:
exp(8*x)*exp(x^2)*exp(1)*((2*x - exp(4) + 7/3)/x)^x
\[ \int \frac {e^{1+8 x+x^2+x \log \left (\frac {7-3 e^4+6 x}{3 x}\right )} \left (-49-62 x-12 x^2+e^4 (21+6 x)+\left (-7+3 e^4-6 x\right ) \log \left (\frac {7-3 e^4+6 x}{3 x}\right )\right )}{-7+3 e^4-6 x} \, dx=\int \frac {\left (\left (3 \,{\mathrm e}^{4}-6 x -7\right ) \mathrm {log}\left (\frac {-3 \,{\mathrm e}^{4}+6 x +7}{3 x}\right )+\left (6 x +21\right ) {\mathrm e}^{4}-12 x^{2}-62 x -49\right ) {\mathrm e}^{x \,\mathrm {log}\left (\frac {-3 \,{\mathrm e}^{4}+6 x +7}{3 x}\right )+x^{2}+8 x +1}}{3 \,{\mathrm e}^{4}-6 x -7}d x \] Input:
int(((3*exp(4)-6*x-7)*log(1/3*(-3*exp(4)+6*x+7)/x)+(6*x+21)*exp(4)-12*x^2- 62*x-49)*exp(x*log(1/3*(-3*exp(4)+6*x+7)/x)+x^2+8*x+1)/(3*exp(4)-6*x-7),x)
Output:
int(((3*exp(4)-6*x-7)*log(1/3*(-3*exp(4)+6*x+7)/x)+(6*x+21)*exp(4)-12*x^2- 62*x-49)*exp(x*log(1/3*(-3*exp(4)+6*x+7)/x)+x^2+8*x+1)/(3*exp(4)-6*x-7),x)