\(\int \frac {e^{4+\frac {e^4-\log (4 e^{5+x})}{\log (4 e^{5+x})}}-36 \log ^2(4 e^{5+x})}{\log ^2(4 e^{5+x})} \, dx\) [7416]

Optimal result

Integrand size = 53, antiderivative size = 24 \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=-e^{-1+\frac {e^4}{\log \left (4 e^{5+x}\right )}}-36 x \]

[Out]

-36*x-exp(exp(4)/ln(4*exp(5)*exp(x))-1)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {2320, 14, 2240} \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=-36 x-e^{\frac {e^4}{\log \left (4 e^{x+5}\right )}-1} \]

[In]

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]

[Out]

-E^(-1 + E^4/Log[4*E^(5 + x)]) - 36*x

Rule 14

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]]

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {-36+\frac {e^{3+\frac {e^4}{\log (4 x)}}}{\log ^2(4 x)}}{x} \, dx,x,e^{5+x}\right ) \\ & = \text {Subst}\left (\int \left (-\frac {36}{x}+\frac {e^{3+\frac {e^4}{\log (4 x)}}}{x \log ^2(4 x)}\right ) \, dx,x,e^{5+x}\right ) \\ & = -36 x+\text {Subst}\left (\int \frac {e^{3+\frac {e^4}{\log (4 x)}}}{x \log ^2(4 x)} \, dx,x,e^{5+x}\right ) \\ & = -36 x+\text {Subst}\left (\int \frac {e^{3+\frac {e^4}{x}}}{x^2} \, dx,x,\log \left (4 e^{5+x}\right )\right ) \\ & = -e^{-1+\frac {e^4}{\log \left (4 e^{5+x}\right )}}-36 x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=-e^{-1+\frac {e^4}{\log \left (4 e^{5+x}\right )}}-36 x \]

[In]

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]

[Out]

-E^(-1 + E^4/Log[4*E^(5 + x)]) - 36*x

Maple [A] (verified)

Time = 0.49 (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	\(-36 \ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )-{\mathrm e}^{\frac {{\mathrm e}^{4}}{\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )}-1}\)	\(28\)
parallelrisch	\(-36 x -{\mathrm e}^{\frac {-\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )+{\mathrm e}^{4}}{\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )}}\)	\(30\)
parts	\(-36 x -{\mathrm e}^{\frac {-\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )+{\mathrm e}^{4}}{\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )}}\)	\(30\)
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	\(\frac {-36 \ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )^{2}-\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right ) {\mathrm e}^{\frac {-\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )+{\mathrm e}^{4}}{\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )}}}{\ln \left (4 \,{\mathrm e}^{5} {\mathrm e}^{x}\right )}\)	\(55\)

[In]

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)

[Out]

-36*ln(4*exp(5)*exp(x))-exp(exp(4)/ln(4*exp(5)*exp(x))-1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=-{\left (36 \, x e^{4} + e^{\left (\frac {3 \, x + e^{4} + 6 \, \log \left (2\right ) + 15}{x + 2 \, \log \left (2\right ) + 5}\right )}\right )} e^{\left (-4\right )} \]

[In]

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*ex
p(5)*exp(x))^2,x, algorithm="fricas")

[Out]

-(36*x*e^4 + e^((3*x + e^4 + 6*log(2) + 15)/(x + 2*log(2) + 5)))*e^(-4)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=\text {Timed out} \]

[In]

integrate((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)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=-e^{\left (\frac {e^{4}}{\log \left (4 \, e^{\left (x + 5\right )}\right )} - 1\right )} - 36 \, \log \left (4 \, e^{\left (x + 5\right )}\right ) \]

[In]

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*ex
p(5)*exp(x))^2,x, algorithm="maxima")

[Out]

-e^(e^4/log(4*e^(x + 5)) - 1) - 36*log(4*e^(x + 5))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=-36 \, x - e^{\left (\frac {e^{4}}{\log \left (4 \, e^{\left (x + 5\right )}\right )} - 1\right )} \]

[In]

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*ex
p(5)*exp(x))^2,x, algorithm="giac")

[Out]

-36*x - e^(e^4/log(4*e^(x + 5)) - 1)

Mupad [B] (verification not implemented)

Time = 12.45 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.12 \[ \int \frac {e^{4+\frac {e^4-\log \left (4 e^{5+x}\right )}{\log \left (4 e^{5+x}\right )}}-36 \log ^2\left (4 e^{5+x}\right )}{\log ^2\left (4 e^{5+x}\right )} \, dx=-36\,x-\frac {{\mathrm {e}}^{-\frac {5}{x+\ln \left (4\right )+5}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^4}{x+\ln \left (4\right )+5}}\,{\mathrm {e}}^{-\frac {x}{x+\ln \left (4\right )+5}}}{2^{\frac {2}{x+\ln \left (4\right )+5}}} \]

[In]

int((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)

[Out]

- 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
))