Integrand size = 55, antiderivative size = 29 \[ \int \frac {1}{4} e^{\frac {1}{4} \left (51 x-17 x \log (2)+e^x (-12 x+4 x \log (2))\right )} \left (51-17 \log (2)+e^x (-12-12 x+(4+4 x) \log (2))\right ) \, dx=1+e^{\frac {1}{4} \left (x+\left (4+4 \left (3-e^x\right )\right ) x\right ) (3-\log (2))} \] Output:
1+exp(((-4*exp(x)+16)*x+x)*(3/4-1/4*ln(2)))
Time = 0.37 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {1}{4} e^{\frac {1}{4} \left (51 x-17 x \log (2)+e^x (-12 x+4 x \log (2))\right )} \left (51-17 \log (2)+e^x (-12-12 x+(4+4 x) \log (2))\right ) \, dx=2^{-17 x/4} e^{\frac {51 x}{4}+e^x x (-3+\log (2))} \] Input:
Integrate[(E^((51*x - 17*x*Log[2] + E^x*(-12*x + 4*x*Log[2]))/4)*(51 - 17* Log[2] + E^x*(-12 - 12*x + (4 + 4*x)*Log[2])))/4,x]
Output:
E^((51*x)/4 + E^x*x*(-3 + Log[2]))/2^((17*x)/4)
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.
\(\displaystyle \int \frac {1}{4} \left (e^x (-12 x+(4 x+4) \log (2)-12)+51-17 \log (2)\right ) \exp \left (\frac {1}{4} \left (51 x-17 x \log (2)+e^x (4 x \log (2)-12 x)\right )\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int 2^{-17 x/4} e^{\frac {1}{4} \left (51 x-e^x x (12-\log (16))\right )} \left (17 (3-\log (2))-4 e^x (3 x-(x+1) \log (2)+3)\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {1}{4} \int 2^{-17 x/4} e^{\frac {1}{4} \left (51 x-e^x x (12-\log (16))\right )} \left (-4 e^x x-4 e^x+17\right ) (3-\log (2))dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} (3-\log (2)) \int 2^{-17 x/4} e^{\frac {1}{4} \left (51 x-e^x x (12-\log (16))\right )} \left (-4 e^x x-4 e^x+17\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{4} (3-\log (2)) \int \left (-2^{2-\frac {17 x}{4}} e^{x+\frac {1}{4} \left (51 x-e^x x (12-\log (16))\right )} x-2^{2-\frac {17 x}{4}} e^{x+\frac {1}{4} \left (51 x-e^x x (12-\log (16))\right )}+17\ 2^{-17 x/4} e^{\frac {1}{4} \left (51 x-e^x x (12-\log (16))\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} (3-\log (2)) \left (-\int 2^{2-\frac {17 x}{4}} e^{\frac {1}{4} x \left (55-12 e^x \left (1-\frac {\log (2)}{3}\right )\right )}dx+17 \int 2^{-17 x/4} e^{\frac {1}{4} \left (51 x-e^x x (12-\log (16))\right )}dx-\int 2^{2-\frac {17 x}{4}} e^{\frac {1}{4} x \left (55-12 e^x \left (1-\frac {\log (2)}{3}\right )\right )} xdx\right )\) |
Input:
Int[(E^((51*x - 17*x*Log[2] + E^x*(-12*x + 4*x*Log[2]))/4)*(51 - 17*Log[2] + E^x*(-12 - 12*x + (4 + 4*x)*Log[2])))/4,x]
Output:
$Aborted
Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {x \left (4 \,{\mathrm e}^{x} \ln \left (2\right )-12 \,{\mathrm e}^{x}-17 \ln \left (2\right )+51\right )}{4}}\) | \(21\) |
risch | \(2^{-\frac {17 x}{4}} 2^{{\mathrm e}^{x} x} {\mathrm e}^{-\frac {3 x \left (4 \,{\mathrm e}^{x}-17\right )}{4}}\) | \(23\) |
norman | \({\mathrm e}^{\frac {\left (4 x \ln \left (2\right )-12 x \right ) {\mathrm e}^{x}}{4}-\frac {17 x \ln \left (2\right )}{4}+\frac {51 x}{4}}\) | \(24\) |
Input:
int(1/4*(((4+4*x)*ln(2)-12*x-12)*exp(x)-17*ln(2)+51)*exp(1/4*(4*x*ln(2)-12 *x)*exp(x)-17/4*x*ln(2)+51/4*x),x,method=_RETURNVERBOSE)
Output:
exp(1/4*x*(4*exp(x)*ln(2)-12*exp(x)-17*ln(2)+51))
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {1}{4} e^{\frac {1}{4} \left (51 x-17 x \log (2)+e^x (-12 x+4 x \log (2))\right )} \left (51-17 \log (2)+e^x (-12-12 x+(4+4 x) \log (2))\right ) \, dx=e^{\left ({\left (x \log \left (2\right ) - 3 \, x\right )} e^{x} - \frac {17}{4} \, x \log \left (2\right ) + \frac {51}{4} \, x\right )} \] Input:
integrate(1/4*(((4+4*x)*log(2)-12*x-12)*exp(x)-17*log(2)+51)*exp(1/4*(4*x* log(2)-12*x)*exp(x)-17/4*x*log(2)+51/4*x),x, algorithm="fricas")
Output:
e^((x*log(2) - 3*x)*e^x - 17/4*x*log(2) + 51/4*x)
Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {1}{4} e^{\frac {1}{4} \left (51 x-17 x \log (2)+e^x (-12 x+4 x \log (2))\right )} \left (51-17 \log (2)+e^x (-12-12 x+(4+4 x) \log (2))\right ) \, dx=e^{- \frac {17 x \log {\left (2 \right )}}{4} + \frac {51 x}{4} + \left (- 3 x + x \log {\left (2 \right )}\right ) e^{x}} \] Input:
integrate(1/4*(((4+4*x)*ln(2)-12*x-12)*exp(x)-17*ln(2)+51)*exp(1/4*(4*x*ln (2)-12*x)*exp(x)-17/4*x*ln(2)+51/4*x),x)
Output:
exp(-17*x*log(2)/4 + 51*x/4 + (-3*x + x*log(2))*exp(x))
\[ \int \frac {1}{4} e^{\frac {1}{4} \left (51 x-17 x \log (2)+e^x (-12 x+4 x \log (2))\right )} \left (51-17 \log (2)+e^x (-12-12 x+(4+4 x) \log (2))\right ) \, dx=\int { \frac {1}{4} \, {\left (4 \, {\left ({\left (x + 1\right )} \log \left (2\right ) - 3 \, x - 3\right )} e^{x} - 17 \, \log \left (2\right ) + 51\right )} e^{\left ({\left (x \log \left (2\right ) - 3 \, x\right )} e^{x} - \frac {17}{4} \, x \log \left (2\right ) + \frac {51}{4} \, x\right )} \,d x } \] Input:
integrate(1/4*(((4+4*x)*log(2)-12*x-12)*exp(x)-17*log(2)+51)*exp(1/4*(4*x* log(2)-12*x)*exp(x)-17/4*x*log(2)+51/4*x),x, algorithm="maxima")
Output:
1/4*integrate((4*((x + 1)*log(2) - 3*x - 3)*e^x - 17*log(2) + 51)*e^((x*lo g(2) - 3*x)*e^x - 17/4*x*log(2) + 51/4*x), x)
Time = 0.12 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {1}{4} e^{\frac {1}{4} \left (51 x-17 x \log (2)+e^x (-12 x+4 x \log (2))\right )} \left (51-17 \log (2)+e^x (-12-12 x+(4+4 x) \log (2))\right ) \, dx=e^{\left (x e^{x} \log \left (2\right ) - 3 \, x e^{x} - \frac {17}{4} \, x \log \left (2\right ) + \frac {51}{4} \, x\right )} \] Input:
integrate(1/4*(((4+4*x)*log(2)-12*x-12)*exp(x)-17*log(2)+51)*exp(1/4*(4*x* log(2)-12*x)*exp(x)-17/4*x*log(2)+51/4*x),x, algorithm="giac")
Output:
e^(x*e^x*log(2) - 3*x*e^x - 17/4*x*log(2) + 51/4*x)
Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {1}{4} e^{\frac {1}{4} \left (51 x-17 x \log (2)+e^x (-12 x+4 x \log (2))\right )} \left (51-17 \log (2)+e^x (-12-12 x+(4+4 x) \log (2))\right ) \, dx=2^{x\,{\mathrm {e}}^x-\frac {17\,x}{4}}\,{\mathrm {e}}^{-3\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{\frac {51\,x}{4}} \] Input:
int(-(exp((51*x)/4 - (17*x*log(2))/4 - (exp(x)*(12*x - 4*x*log(2)))/4)*(17 *log(2) + exp(x)*(12*x - log(2)*(4*x + 4) + 12) - 51))/4,x)
Output:
2^(x*exp(x) - (17*x)/4)*exp(-3*x*exp(x))*exp((51*x)/4)
Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {1}{4} e^{\frac {1}{4} \left (51 x-17 x \log (2)+e^x (-12 x+4 x \log (2))\right )} \left (51-17 \log (2)+e^x (-12-12 x+(4+4 x) \log (2))\right ) \, dx=\frac {e^{e^{x} \mathrm {log}\left (2\right ) x +\frac {51 x}{4}}}{e^{3 e^{x} x} 2^{\frac {17 x}{4}}} \] Input:
int(1/4*(((4+4*x)*log(2)-12*x-12)*exp(x)-17*log(2)+51)*exp(1/4*(4*x*log(2) -12*x)*exp(x)-17/4*x*log(2)+51/4*x),x)
Output:
e**((4*e**x*log(2)*x + 51*x)/4)/(e**(3*e**x*x)*2**((17*x)/4))