Integrand size = 74, antiderivative size = 22 \[ \int \frac {e^{\frac {-10+\left (2 x^4+\log (x)\right ) \log (4 x \log (x))}{2 \log (4 x \log (x))}} \left (10+10 \log (x)+\left (1+8 x^4\right ) \log (x) \log ^2(4 x \log (x))\right )}{2 x \log (x) \log ^2(4 x \log (x))} \, dx=e^{x^4-\frac {5}{\log (4 x \log (x))}} \sqrt {x} \] Output:
exp(x^4-5/ln(4*x*ln(x))+1/2*ln(x))
Time = 0.45 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-10+\left (2 x^4+\log (x)\right ) \log (4 x \log (x))}{2 \log (4 x \log (x))}} \left (10+10 \log (x)+\left (1+8 x^4\right ) \log (x) \log ^2(4 x \log (x))\right )}{2 x \log (x) \log ^2(4 x \log (x))} \, dx=e^{x^4-\frac {5}{\log (4 x \log (x))}} \sqrt {x} \] Input:
Integrate[(E^((-10 + (2*x^4 + Log[x])*Log[4*x*Log[x]])/(2*Log[4*x*Log[x]]) )*(10 + 10*Log[x] + (1 + 8*x^4)*Log[x]*Log[4*x*Log[x]]^2))/(2*x*Log[x]*Log [4*x*Log[x]]^2),x]
Output:
E^(x^4 - 5/Log[4*x*Log[x]])*Sqrt[x]
Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(22)=44\).
Time = 2.75 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.59, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {27, 2726}
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.
\(\displaystyle \int \frac {\left (\left (8 x^4+1\right ) \log (x) \log ^2(4 x \log (x))+10 \log (x)+10\right ) \exp \left (\frac {\left (2 x^4+\log (x)\right ) \log (4 x \log (x))-10}{2 \log (4 x \log (x))}\right )}{2 x \log (x) \log ^2(4 x \log (x))} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {2^{\frac {2 x^4+\log (x)}{\log (4 x \log (x))}} e^{-\frac {5}{\log (4 x \log (x))}} (x \log (x))^{\frac {2 x^4+\log (x)}{2 \log (4 x \log (x))}} \left (\left (8 x^4+1\right ) \log (x) \log ^2(4 x \log (x))+10 \log (x)+10\right )}{x \log (x) \log ^2(4 x \log (x))}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle 2^{\frac {2 x^4+\log (x)}{\log (4 x \log (x))}} e^{-\frac {5}{\log (4 x \log (x))}} (x \log (x))^{\frac {2 x^4+\log (x)}{2 \log (4 x \log (x))}}\) |
Input:
Int[(E^((-10 + (2*x^4 + Log[x])*Log[4*x*Log[x]])/(2*Log[4*x*Log[x]]))*(10 + 10*Log[x] + (1 + 8*x^4)*Log[x]*Log[4*x*Log[x]]^2))/(2*x*Log[x]*Log[4*x*L og[x]]^2),x]
Output:
(2^((2*x^4 + Log[x])/Log[4*x*Log[x]])*(x*Log[x])^((2*x^4 + Log[x])/(2*Log[ 4*x*Log[x]])))/E^(5/Log[4*x*Log[x]])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 7.63 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left (\ln \left (x \right )+2 x^{4}\right ) \ln \left (4 x \ln \left (x \right )\right )-10}{2 \ln \left (4 x \ln \left (x \right )\right )}}\) | \(29\) |
risch | \({\mathrm e}^{\frac {2 i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2} x^{4}-2 i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \right ) x^{4}-2 i \pi \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{3} x^{4}+2 i \pi \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i x \right ) x^{4}+i \ln \left (x \right ) \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}-i \ln \left (x \right ) \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \right )-i \ln \left (x \right ) \pi \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{3}+i \ln \left (x \right ) \pi \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i x \right )+4 x^{4} \ln \left (x \right )+4 \ln \left (\ln \left (x \right )\right ) x^{4}+8 x^{4} \ln \left (2\right )+2 \ln \left (x \right )^{2}+2 \ln \left (x \right ) \ln \left (\ln \left (x \right )\right )+4 \ln \left (2\right ) \ln \left (x \right )-20}{2 i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}-2 i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \right )-2 i \pi \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{3}+2 i \pi \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i x \right )+4 \ln \left (x \right )+4 \ln \left (\ln \left (x \right )\right )+8 \ln \left (2\right )}}\) | \(299\) |
Input:
int(1/2*((8*x^4+1)*ln(x)*ln(4*x*ln(x))^2+10*ln(x)+10)*exp(1/2*((ln(x)+2*x^ 4)*ln(4*x*ln(x))-10)/ln(4*x*ln(x)))/x/ln(x)/ln(4*x*ln(x))^2,x,method=_RETU RNVERBOSE)
Output:
exp(1/2*((ln(x)+2*x^4)*ln(4*x*ln(x))-10)/ln(4*x*ln(x)))
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {e^{\frac {-10+\left (2 x^4+\log (x)\right ) \log (4 x \log (x))}{2 \log (4 x \log (x))}} \left (10+10 \log (x)+\left (1+8 x^4\right ) \log (x) \log ^2(4 x \log (x))\right )}{2 x \log (x) \log ^2(4 x \log (x))} \, dx=e^{\left (\frac {{\left (2 \, x^{4} + \log \left (x\right )\right )} \log \left (4 \, x \log \left (x\right )\right ) - 10}{2 \, \log \left (4 \, x \log \left (x\right )\right )}\right )} \] Input:
integrate(1/2*((8*x^4+1)*log(x)*log(4*x*log(x))^2+10*log(x)+10)*exp(1/2*(( log(x)+2*x^4)*log(4*x*log(x))-10)/log(4*x*log(x)))/x/log(x)/log(4*x*log(x) )^2,x, algorithm="fricas")
Output:
e^(1/2*((2*x^4 + log(x))*log(4*x*log(x)) - 10)/log(4*x*log(x)))
Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {e^{\frac {-10+\left (2 x^4+\log (x)\right ) \log (4 x \log (x))}{2 \log (4 x \log (x))}} \left (10+10 \log (x)+\left (1+8 x^4\right ) \log (x) \log ^2(4 x \log (x))\right )}{2 x \log (x) \log ^2(4 x \log (x))} \, dx=e^{\frac {\frac {\left (2 x^{4} + \log {\left (x \right )}\right ) \log {\left (4 x \log {\left (x \right )} \right )}}{2} - 5}{\log {\left (4 x \log {\left (x \right )} \right )}}} \] Input:
integrate(1/2*((8*x**4+1)*ln(x)*ln(4*x*ln(x))**2+10*ln(x)+10)*exp(1/2*((ln (x)+2*x**4)*ln(4*x*ln(x))-10)/ln(4*x*ln(x)))/x/ln(x)/ln(4*x*ln(x))**2,x)
Output:
exp(((2*x**4 + log(x))*log(4*x*log(x))/2 - 5)/log(4*x*log(x)))
\[ \int \frac {e^{\frac {-10+\left (2 x^4+\log (x)\right ) \log (4 x \log (x))}{2 \log (4 x \log (x))}} \left (10+10 \log (x)+\left (1+8 x^4\right ) \log (x) \log ^2(4 x \log (x))\right )}{2 x \log (x) \log ^2(4 x \log (x))} \, dx=\int { \frac {{\left ({\left (8 \, x^{4} + 1\right )} \log \left (4 \, x \log \left (x\right )\right )^{2} \log \left (x\right ) + 10 \, \log \left (x\right ) + 10\right )} e^{\left (\frac {{\left (2 \, x^{4} + \log \left (x\right )\right )} \log \left (4 \, x \log \left (x\right )\right ) - 10}{2 \, \log \left (4 \, x \log \left (x\right )\right )}\right )}}{2 \, x \log \left (4 \, x \log \left (x\right )\right )^{2} \log \left (x\right )} \,d x } \] Input:
integrate(1/2*((8*x^4+1)*log(x)*log(4*x*log(x))^2+10*log(x)+10)*exp(1/2*(( log(x)+2*x^4)*log(4*x*log(x))-10)/log(4*x*log(x)))/x/log(x)/log(4*x*log(x) )^2,x, algorithm="maxima")
Output:
1/2*integrate(((8*x^4 + 1)*log(4*x*log(x))^2*log(x) + 10*log(x) + 10)*e^(1 /2*((2*x^4 + log(x))*log(4*x*log(x)) - 10)/log(4*x*log(x)))/(x*log(4*x*log (x))^2*log(x)), x)
Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\frac {-10+\left (2 x^4+\log (x)\right ) \log (4 x \log (x))}{2 \log (4 x \log (x))}} \left (10+10 \log (x)+\left (1+8 x^4\right ) \log (x) \log ^2(4 x \log (x))\right )}{2 x \log (x) \log ^2(4 x \log (x))} \, dx=e^{\left (x^{4} - \frac {5}{\log \left (4 \, x \log \left (x\right )\right )} + \frac {1}{2} \, \log \left (x\right )\right )} \] Input:
integrate(1/2*((8*x^4+1)*log(x)*log(4*x*log(x))^2+10*log(x)+10)*exp(1/2*(( log(x)+2*x^4)*log(4*x*log(x))-10)/log(4*x*log(x)))/x/log(x)/log(4*x*log(x) )^2,x, algorithm="giac")
Output:
e^(x^4 - 5/log(4*x*log(x)) + 1/2*log(x))
Time = 0.68 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.82 \[ \int \frac {e^{\frac {-10+\left (2 x^4+\log (x)\right ) \log (4 x \log (x))}{2 \log (4 x \log (x))}} \left (10+10 \log (x)+\left (1+8 x^4\right ) \log (x) \log ^2(4 x \log (x))\right )}{2 x \log (x) \log ^2(4 x \log (x))} \, dx=x^{\frac {\ln \left (2\right )}{\ln \left (4\,x\,\ln \left (x\right )\right )}}\,{\mathrm {e}}^{\frac {\ln \left (x\,\ln \left (x\right )\right )\,\ln \left (\sqrt {x}\right )}{\ln \left (4\,x\,\ln \left (x\right )\right )}-\frac {5}{\ln \left (4\,x\,\ln \left (x\right )\right )}}\,{\left (4\,x\,\ln \left (x\right )\right )}^{\frac {x^4}{\ln \left (4\,x\,\ln \left (x\right )\right )}} \] Input:
int((exp(((log(4*x*log(x))*(log(x) + 2*x^4))/2 - 5)/log(4*x*log(x)))*(10*l og(x) + log(4*x*log(x))^2*log(x)*(8*x^4 + 1) + 10))/(2*x*log(4*x*log(x))^2 *log(x)),x)
Output:
x^(log(2)/log(4*x*log(x)))*exp((log(x*log(x))*log(x^(1/2)))/log(4*x*log(x) ) - 5/log(4*x*log(x)))*(4*x*log(x))^(x^4/log(4*x*log(x)))
Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-10+\left (2 x^4+\log (x)\right ) \log (4 x \log (x))}{2 \log (4 x \log (x))}} \left (10+10 \log (x)+\left (1+8 x^4\right ) \log (x) \log ^2(4 x \log (x))\right )}{2 x \log (x) \log ^2(4 x \log (x))} \, dx=\frac {\sqrt {x}\, e^{x^{4}}}{e^{\frac {5}{\mathrm {log}\left (4 \,\mathrm {log}\left (x \right ) x \right )}}} \] Input:
int(1/2*((8*x^4+1)*log(x)*log(4*x*log(x))^2+10*log(x)+10)*exp(1/2*((log(x) +2*x^4)*log(4*x*log(x))-10)/log(4*x*log(x)))/x/log(x)/log(4*x*log(x))^2,x)
Output:
(sqrt(x)*e**(x**4))/e**(5/log(4*log(x)*x))