Integrand size = 53, antiderivative size = 24 \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=-e^{-1+\frac {e^4}{\log \left (4 e^{5+x}\right )}}-36 x \] Output:
-36*x-exp(exp(4)/ln(4*exp(5)*exp(x))-1)
Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=-e^{-1+\frac {e^4}{\log \left (4 e^{5+x}\right )}}-36 x \] Input:
Integrate[(E^(4 + (E^4 - Log[4*E^(5 + x)])/Log[4*E^(5 + x)]) - 36*Log[4*E^ (5 + x)]^2)/Log[4*E^(5 + x)]^2,x]
Output:
-E^(-1 + E^4/Log[4*E^(5 + x)]) - 36*x
Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {2720, 25, 2010, 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.
\(\displaystyle \int \frac {e^{\frac {e^4-\log \left (4 e^{x+5}\right )}{\log \left (4 e^{x+5}\right )}+4}-36 \log ^2\left (4 e^{x+5}\right )}{\log ^2\left (4 e^{x+5}\right )} \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \int -e^{-x-5} \left (36-\frac {e^{\frac {e^4}{\log \left (4 e^{x+5}\right )}+3}}{\log ^2\left (4 e^{x+5}\right )}\right )de^{x+5}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int e^{-x-5} \left (36-\frac {e^{3+\frac {e^4}{\log \left (4 e^{x+5}\right )}}}{\log ^2\left (4 e^{x+5}\right )}\right )de^{x+5}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\int \left (36 e^{-x-5}-\frac {e^{-x+\frac {e^4}{\log \left (4 e^{x+5}\right )}-2}}{\log ^2\left (4 e^{x+5}\right )}\right )de^{x+5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -36 \log \left (e^{x+5}\right )-e^{\frac {e^4}{\log \left (4 e^{x+5}\right )}-1}\) |
Input:
Int[(E^(4 + (E^4 - Log[4*E^(5 + x)])/Log[4*E^(5 + x)]) - 36*Log[4*E^(5 + x )]^2)/Log[4*E^(5 + x)]^2,x]
Output:
-E^(-1 + E^4/Log[4*E^(5 + x)]) - 36*Log[E^(5 + x)]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Time = 0.42 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17
method | result | size |
derivativedivides | \(-36 \ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )-{\mathrm e}^{\frac {{\mathrm e}^{4}}{\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )}-1}\) | \(28\) |
default | \(-36 \ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )-{\mathrm e}^{\frac {{\mathrm e}^{4}}{\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )}-1}\) | \(28\) |
parts | \(-36 x -{\mathrm e}^{\frac {-\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )+{\mathrm e}^{4}}{\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )}}\) | \(30\) |
parallelrisch | \(-36 x -{\mathrm e}^{-\frac {\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )-{\mathrm e}^{4}}{\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )}}\) | \(31\) |
risch | \(-36 x -{\mathrm e}^{\frac {-2 \ln \left (2\right )-5-\ln \left ({\mathrm e}^{x}\right )+{\mathrm e}^{4}}{2 \ln \left (2\right )+5+\ln \left ({\mathrm e}^{x}\right )}}\) | \(33\) |
norman | \(\frac {-36 \ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )^{2}-\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right ) {\mathrm e}^{\frac {-\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )+{\mathrm e}^{4}}{\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )}}}{\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )}\) | \(55\) |
Input:
int((exp(4)*exp((-ln(4*exp(5)*exp(x))+exp(4))/ln(4*exp(5)*exp(x)))-36*ln(4 *exp(5)*exp(x))^2)/ln(4*exp(5)*exp(x))^2,x,method=_RETURNVERBOSE)
Output:
-36*ln(4*exp(5)*exp(x))-exp(exp(4)/ln(4*exp(5)*exp(x))-1)
Time = 0.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=-{\left (36 \, x e^{4} + e^{\left (\frac {3 \, x + e^{4} + 6 \, \log \left (2\right ) + 15}{x + 2 \, \log \left (2\right ) + 5}\right )}\right )} e^{\left (-4\right )} \] Input:
integrate((exp(4)*exp((-log(4*exp(5)*exp(x))+exp(4))/log(4*exp(5)*exp(x))) -36*log(4*exp(5)*exp(x))^2)/log(4*exp(5)*exp(x))^2,x, algorithm="fricas")
Output:
-(36*x*e^4 + e^((3*x + e^4 + 6*log(2) + 15)/(x + 2*log(2) + 5)))*e^(-4)
Timed out. \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=\text {Timed out} \] Input:
integrate((exp(4)*exp((-ln(4*exp(5)*exp(x))+exp(4))/ln(4*exp(5)*exp(x)))-3 6*ln(4*exp(5)*exp(x))**2)/ln(4*exp(5)*exp(x))**2,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=-e^{\left (\frac {e^{4}}{\log \left (4 \, e^{\left (x + 5\right )}\right )} - 1\right )} - 36 \, \log \left (4 \, e^{\left (x + 5\right )}\right ) \] Input:
integrate((exp(4)*exp((-log(4*exp(5)*exp(x))+exp(4))/log(4*exp(5)*exp(x))) -36*log(4*exp(5)*exp(x))^2)/log(4*exp(5)*exp(x))^2,x, algorithm="maxima")
Output:
-e^(e^4/log(4*e^(x + 5)) - 1) - 36*log(4*e^(x + 5))
Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (21) = 42\).
Time = 0.13 (sec) , antiderivative size = 105, normalized size of antiderivative = 4.38 \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=-\frac {{\left (\frac {{\left (e^{4} + 3 \, \log \left (4 \, e^{\left (x + 5\right )}\right )\right )} e^{\left (\frac {e^{4} + 3 \, \log \left (4 \, e^{\left (x + 5\right )}\right )}{\log \left (4 \, e^{\left (x + 5\right )}\right )}\right )}}{\log \left (4 \, e^{\left (x + 5\right )}\right )} + 36 \, e^{8} - 3 \, e^{\left (\frac {e^{4} + 3 \, \log \left (4 \, e^{\left (x + 5\right )}\right )}{\log \left (4 \, e^{\left (x + 5\right )}\right )}\right )}\right )} e^{\left (-4\right )}}{\frac {e^{4} + 3 \, \log \left (4 \, e^{\left (x + 5\right )}\right )}{\log \left (4 \, e^{\left (x + 5\right )}\right )} - 3} \] Input:
integrate((exp(4)*exp((-log(4*exp(5)*exp(x))+exp(4))/log(4*exp(5)*exp(x))) -36*log(4*exp(5)*exp(x))^2)/log(4*exp(5)*exp(x))^2,x, algorithm="giac")
Output:
-((e^4 + 3*log(4*e^(x + 5)))*e^((e^4 + 3*log(4*e^(x + 5)))/log(4*e^(x + 5) ))/log(4*e^(x + 5)) + 36*e^8 - 3*e^((e^4 + 3*log(4*e^(x + 5)))/log(4*e^(x + 5))))*e^(-4)/((e^4 + 3*log(4*e^(x + 5)))/log(4*e^(x + 5)) - 3)
Time = 3.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.12 \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=-36\,x-\frac {{\mathrm {e}}^{-\frac {5}{x+\ln \left (4\right )+5}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^4}{x+\ln \left (4\right )+5}}\,{\mathrm {e}}^{-\frac {x}{x+\ln \left (4\right )+5}}}{2^{\frac {2}{x+\ln \left (4\right )+5}}} \] Input:
int((exp(4)*exp(-(log(4*exp(5)*exp(x)) - exp(4))/log(4*exp(5)*exp(x))) - 3 6*log(4*exp(5)*exp(x))^2)/log(4*exp(5)*exp(x))^2,x)
Output:
- 36*x - (exp(-5/(x + log(4) + 5))*exp(exp(4)/(x + log(4) + 5))*exp(-x/(x + log(4) + 5)))/2^(2/(x + log(4) + 5))
Time = 0.15 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=\frac {-e^{\frac {e^{4}}{\mathrm {log}\left (4 e^{x} e^{5}\right )}}-36 e x}{e} \] Input:
int((exp(4)*exp((-log(4*exp(5)*exp(x))+exp(4))/log(4*exp(5)*exp(x)))-36*lo g(4*exp(5)*exp(x))^2)/log(4*exp(5)*exp(x))^2,x)
Output:
( - e**(e**4/log(4*e**x*e**5)) - 36*e*x)/e