Integrand size = 68, antiderivative size = 25 \[ \int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )}{2 \log (4)} \, dx=\frac {3}{2} e^{\frac {5 (3+e) (2-x)}{\log (4)}} x \log \left (x^2\right ) \]
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )}{2 \log (4)} \, dx=\frac {3 e^{-\frac {5 (3+e) (-2+x)}{\log (4)}} x \log (4) \log \left (x^2\right )}{\log (16)} \]
Integrate[(6*E^((30 + E*(10 - 5*x) - 15*x)/Log[4])*Log[4] + E^((30 + E*(10 - 5*x) - 15*x)/Log[4])*(-45*x - 15*E*x + 3*Log[4])*Log[x^2])/(2*Log[4]),x ]
Leaf count is larger than twice the leaf count of optimal. \(152\) vs. \(2(25)=50\).
Time = 0.50 (sec) , antiderivative size = 152, normalized size of antiderivative = 6.08, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {27, 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.
\(\displaystyle \int \frac {e^{\frac {e (10-5 x)-15 x+30}{\log (4)}} (-15 e x-45 x+3 \log (4)) \log \left (x^2\right )+6 \log (4) e^{\frac {e (10-5 x)-15 x+30}{\log (4)}}}{2 \log (4)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \left (6 e^{\frac {5 (e (2-x)-3 x+6)}{\log (4)}} \log (4)-e^{\frac {5 (e (2-x)-3 x+6)}{\log (4)}} (15 e x+45 x-\log (64)) \log \left (x^2\right )\right )dx}{2 \log (4)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {3 \log ^2(4) e^{\frac {5 (3+e) (2-x)}{\log (4)}} \log \left (x^2\right )}{5 (3+e)}+\frac {\log (4) e^{\frac {5 (3+e) (2-x)}{\log (4)}} (15 (3+e) x-\log (64)) \log \left (x^2\right )}{5 (3+e)}+\frac {6 \log ^2(4) e^{\frac {10 (3+e)}{\log (4)}-\frac {5 (3+e) x}{\log (4)}}}{5 (3+e)}-\frac {6 \log ^2(4) e^{\frac {5 (e (2-x)-3 x+6)}{\log (4)}}}{5 (3+e)}}{2 \log (4)}\) |
Int[(6*E^((30 + E*(10 - 5*x) - 15*x)/Log[4])*Log[4] + E^((30 + E*(10 - 5*x ) - 15*x)/Log[4])*(-45*x - 15*E*x + 3*Log[4])*Log[x^2])/(2*Log[4]),x]
((6*E^((10*(3 + E))/Log[4] - (5*(3 + E)*x)/Log[4])*Log[4]^2)/(5*(3 + E)) - (6*E^((5*(6 + E*(2 - x) - 3*x))/Log[4])*Log[4]^2)/(5*(3 + E)) + (3*E^((5* (3 + E)*(2 - x))/Log[4])*Log[4]^2*Log[x^2])/(5*(3 + E)) + (E^((5*(3 + E)*( 2 - x))/Log[4])*Log[4]*(15*(3 + E)*x - Log[64])*Log[x^2])/(5*(3 + E)))/(2* Log[4])
3.11.69.3.1 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 = 0.17 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12
method | result | size |
norman | \(\frac {3 x \,{\mathrm e}^{\frac {\left (-5 x +10\right ) {\mathrm e}-15 x +30}{2 \ln \left (2\right )}} \ln \left (x^{2}\right )}{2}\) | \(28\) |
parallelrisch | \(\frac {3 \,{\mathrm e}^{-\frac {5 \left (x \,{\mathrm e}-2 \,{\mathrm e}+3 x -6\right )}{2 \ln \left (2\right )}} \ln \left (x^{2}\right ) x}{2}\) | \(28\) |
risch | \(3 x \,{\mathrm e}^{-\frac {5 \left (-2+x \right ) \left (3+{\mathrm e}\right )}{2 \ln \left (2\right )}} \ln \left (x \right )-\frac {3 i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \left (\operatorname {csgn}\left (i x \right )^{2}-2 \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )+\operatorname {csgn}\left (i x^{2}\right )^{2}\right ) x \,{\mathrm e}^{-\frac {5 \left (-2+x \right ) \left (3+{\mathrm e}\right )}{2 \ln \left (2\right )}}}{4}\) | \(78\) |
default | \(\frac {\frac {24 \ln \left (2\right )^{2} {\mathrm e}^{\frac {\left (-5 x +10\right ) {\mathrm e}-15 x +30}{2 \ln \left (2\right )}}}{5 \left (3+{\mathrm e}\right )}+12 x \ln \left (2\right ) {\mathrm e}^{\frac {\left (-5 x +10\right ) {\mathrm e}-15 x +30}{2 \ln \left (2\right )}} \ln \left (x \right )+6 \ln \left (2\right ) \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right ) x \,{\mathrm e}^{\frac {\left (-5 x +10\right ) {\mathrm e}-15 x +30}{2 \ln \left (2\right )}}+\frac {24 \ln \left (2\right )^{2} {\mathrm e}^{\frac {\left (-5 \,{\mathrm e}-15\right ) x}{2 \ln \left (2\right )}+\frac {10 \,{\mathrm e}+30}{2 \ln \left (2\right )}}}{-5 \,{\mathrm e}-15}}{4 \ln \left (2\right )}\) | \(142\) |
int(1/4*((6*ln(2)-15*x*exp(1)-45*x)*exp(1/2*((-5*x+10)*exp(1)-15*x+30)/ln( 2))*ln(x^2)+12*ln(2)*exp(1/2*((-5*x+10)*exp(1)-15*x+30)/ln(2)))/ln(2),x,me thod=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )}{2 \log (4)} \, dx=\frac {3}{2} \, x e^{\left (-\frac {5 \, {\left ({\left (x - 2\right )} e + 3 \, x - 6\right )}}{2 \, \log \left (2\right )}\right )} \log \left (x^{2}\right ) \]
integrate(1/4*((6*log(2)-15*x*exp(1)-45*x)*exp(1/2*((-5*x+10)*exp(1)-15*x+ 30)/log(2))*log(x^2)+12*log(2)*exp(1/2*((-5*x+10)*exp(1)-15*x+30)/log(2))) /log(2),x, algorithm=\
Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )}{2 \log (4)} \, dx=\frac {3 x e^{\frac {- \frac {15 x}{2} + \frac {e \left (10 - 5 x\right )}{2} + 15}{\log {\left (2 \right )}}} \log {\left (x^{2} \right )}}{2} \]
integrate(1/4*((6*ln(2)-15*x*exp(1)-45*x)*exp(1/2*((-5*x+10)*exp(1)-15*x+3 0)/ln(2))*ln(x**2)+12*ln(2)*exp(1/2*((-5*x+10)*exp(1)-15*x+30)/ln(2)))/ln( 2),x)
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (21) = 42\).
Time = 0.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 5.20 \[ \int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )}{2 \log (4)} \, dx=\frac {3 \, {\left (5 \, x e^{\left (-\frac {5 \, x e}{2 \, \log \left (2\right )} - \frac {15 \, x}{2 \, \log \left (2\right )} + \frac {5 \, e}{\log \left (2\right )} + \frac {15}{\log \left (2\right )}\right )} \log \left (2\right ) \log \left (x^{2}\right ) - \frac {4 \, e^{\left (-\frac {5 \, {\left ({\left (x - 2\right )} e + 3 \, x - 6\right )}}{2 \, \log \left (2\right )}\right )} \log \left (2\right )^{2}}{e + 3} + \frac {4 \, e^{\left (-\frac {5 \, x e}{2 \, \log \left (2\right )} - \frac {15 \, x}{2 \, \log \left (2\right )} + \frac {5 \, e}{\log \left (2\right )} + \frac {15}{\log \left (2\right )}\right )} \log \left (2\right )}{\frac {e}{\log \left (2\right )} + \frac {3}{\log \left (2\right )}}\right )}}{10 \, \log \left (2\right )} \]
integrate(1/4*((6*log(2)-15*x*exp(1)-45*x)*exp(1/2*((-5*x+10)*exp(1)-15*x+ 30)/log(2))*log(x^2)+12*log(2)*exp(1/2*((-5*x+10)*exp(1)-15*x+30)/log(2))) /log(2),x, algorithm=\
3/10*(5*x*e^(-5/2*x*e/log(2) - 15/2*x/log(2) + 5*e/log(2) + 15/log(2))*log (2)*log(x^2) - 4*e^(-5/2*((x - 2)*e + 3*x - 6)/log(2))*log(2)^2/(e + 3) + 4*e^(-5/2*x*e/log(2) - 15/2*x/log(2) + 5*e/log(2) + 15/log(2))*log(2)/(e/l og(2) + 3/log(2)))/log(2)
Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (21) = 42\).
Time = 0.34 (sec) , antiderivative size = 198, normalized size of antiderivative = 7.92 \[ \int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )}{2 \log (4)} \, dx=\frac {3 \, {\left (\frac {4 \, e^{\left (-\frac {5 \, x e}{2 \, \log \left (2\right )} - \frac {15 \, x}{2 \, \log \left (2\right )} + \frac {5 \, e}{\log \left (2\right )} + \frac {15}{\log \left (2\right )} + 1\right )} \log \left (2\right )^{2}}{e^{2} + 3 \, e} - \frac {4 \, e^{\left (-\frac {5 \, {\left ({\left (x - 2\right )} e + 3 \, x - 6\right )}}{2 \, \log \left (2\right )}\right )} \log \left (2\right )^{2}}{e + 3} + {\left (\frac {{\left (5 \, x e \log \left (2\right ) + 15 \, x \log \left (2\right ) + 2 \, \log \left (2\right )^{2}\right )} e^{\left (-\frac {5 \, x e + 15 \, x - 10 \, e - 2 \, \log \left (2\right ) - 30}{2 \, \log \left (2\right )}\right )}}{e^{2} + 6 \, e + 9} + \frac {{\left (15 \, x e \log \left (2\right ) - 2 \, e \log \left (2\right )^{2} + 45 \, x \log \left (2\right )\right )} e^{\left (-\frac {5 \, {\left (x e + 3 \, x - 2 \, e - 6\right )}}{2 \, \log \left (2\right )}\right )}}{e^{2} + 6 \, e + 9}\right )} \log \left (x^{2}\right )\right )}}{10 \, \log \left (2\right )} \]
integrate(1/4*((6*log(2)-15*x*exp(1)-45*x)*exp(1/2*((-5*x+10)*exp(1)-15*x+ 30)/log(2))*log(x^2)+12*log(2)*exp(1/2*((-5*x+10)*exp(1)-15*x+30)/log(2))) /log(2),x, algorithm=\
3/10*(4*e^(-5/2*x*e/log(2) - 15/2*x/log(2) + 5*e/log(2) + 15/log(2) + 1)*l og(2)^2/(e^2 + 3*e) - 4*e^(-5/2*((x - 2)*e + 3*x - 6)/log(2))*log(2)^2/(e + 3) + ((5*x*e*log(2) + 15*x*log(2) + 2*log(2)^2)*e^(-1/2*(5*x*e + 15*x - 10*e - 2*log(2) - 30)/log(2))/(e^2 + 6*e + 9) + (15*x*e*log(2) - 2*e*log(2 )^2 + 45*x*log(2))*e^(-5/2*(x*e + 3*x - 2*e - 6)/log(2))/(e^2 + 6*e + 9))* log(x^2))/log(2)
Time = 8.45 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {6 e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} \log (4)+e^{\frac {30+e (10-5 x)-15 x}{\log (4)}} (-45 x-15 e x+3 \log (4)) \log \left (x^2\right )}{2 \log (4)} \, dx=\frac {3\,x\,\ln \left (x^2\right )\,{\mathrm {e}}^{-\frac {15\,x}{2\,\ln \left (2\right )}}\,{\mathrm {e}}^{-\frac {5\,x\,\mathrm {e}}{2\,\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {15}{\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {5\,\mathrm {e}}{\ln \left (2\right )}}}{2} \]
int((3*exp(-((15*x)/2 + (exp(1)*(5*x - 10))/2 - 15)/log(2))*log(2) - (log( x^2)*exp(-((15*x)/2 + (exp(1)*(5*x - 10))/2 - 15)/log(2))*(45*x - 6*log(2) + 15*x*exp(1)))/4)/log(2),x)