\(\int \frac {8 \log (5)+e^{\frac {e^{x/8}+16 \log (5)-3 e^x \log (5)}{\log (5)}} (-e^{x/8}+24 e^x \log (5))}{8 \log (5)} \, dx\) [1889]

Optimal result

Integrand size = 57, antiderivative size = 30 \[ \int \frac {8 \log (5)+e^{\frac {e^{x/8}+16 \log (5)-3 e^x \log (5)}{\log (5)}} \left (-e^{x/8}+24 e^x \log (5)\right )}{8 \log (5)} \, dx=-3-e^{-2+3 \left (6-e^x\right )+\frac {e^{x/8}}{\log (5)}}+x \]

[Out]

x-3-exp(-3*exp(x)+16+exp(1/8*x)/ln(5))

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 2306, 2320, 6838} \[ \int \frac {8 \log (5)+e^{\frac {e^{x/8}+16 \log (5)-3 e^x \log (5)}{\log (5)}} \left (-e^{x/8}+24 e^x \log (5)\right )}{8 \log (5)} \, dx=x-e^{-3 e^x+\frac {e^{x/8}}{\log (5)}+16} \]

[In]

Int[(8*Log[5] + E^((E^(x/8) + 16*Log[5] - 3*E^x*Log[5])/Log[5])*(-E^(x/8) + 24*E^x*Log[5]))/(8*Log[5]),x]

[Out]

-E^(16 - 3*E^x + E^(x/8)/Log[5]) + x

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2306

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

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]]]

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (8 \log (5)+e^{\frac {e^{x/8}+16 \log (5)-3 e^x \log (5)}{\log (5)}} \left (-e^{x/8}+24 e^x \log (5)\right )\right ) \, dx}{8 \log (5)} \\ & = x+\frac {\int e^{\frac {e^{x/8}+16 \log (5)-3 e^x \log (5)}{\log (5)}} \left (-e^{x/8}+24 e^x \log (5)\right ) \, dx}{8 \log (5)} \\ & = x+\frac {\int 5^{\frac {16}{\log (5)}} e^{\frac {e^{x/8}-3 e^x \log (5)}{\log (5)}} \left (-e^{x/8}+24 e^x \log (5)\right ) \, dx}{8 \log (5)} \\ & = x+\frac {e^{16} \int e^{\frac {e^{x/8}-3 e^x \log (5)}{\log (5)}} \left (-e^{x/8}+24 e^x \log (5)\right ) \, dx}{8 \log (5)} \\ & = x+\frac {e^{16} \text {Subst}\left (\int e^{-3 x^8+\frac {x}{\log (5)}} \left (-1+24 x^7 \log (5)\right ) \, dx,x,e^{x/8}\right )}{\log (5)} \\ & = -e^{16-3 e^x+\frac {e^{x/8}}{\log (5)}}+x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {8 \log (5)+e^{\frac {e^{x/8}+16 \log (5)-3 e^x \log (5)}{\log (5)}} \left (-e^{x/8}+24 e^x \log (5)\right )}{8 \log (5)} \, dx=-e^{16-3 e^x+\frac {e^{x/8}}{\log (5)}}+x \]

[In]

Integrate[(8*Log[5] + E^((E^(x/8) + 16*Log[5] - 3*E^x*Log[5])/Log[5])*(-E^(x/8) + 24*E^x*Log[5]))/(8*Log[5]),x
]

[Out]

-E^(16 - 3*E^x + E^(x/8)/Log[5]) + x

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97

method	result	size
risch	\(x -{\mathrm e}^{-\frac {3 \,{\mathrm e}^{x} \ln \left (5\right )-16 \ln \left (5\right )-{\mathrm e}^{\frac {x}{8}}}{\ln \left (5\right )}}\)	\(29\)
norman	\(x -{\mathrm e}^{\frac {-3 \,{\mathrm e}^{x} \ln \left (5\right )+{\mathrm e}^{\frac {x}{8}}+16 \ln \left (5\right )}{\ln \left (5\right )}}\)	\(30\)
parallelrisch	\(\frac {-8 \ln \left (5\right ) {\mathrm e}^{-\frac {3 \,{\mathrm e}^{x} \ln \left (5\right )-16 \ln \left (5\right )-{\mathrm e}^{\frac {x}{8}}}{\ln \left (5\right )}}+8 x \ln \left (5\right )}{8 \ln \left (5\right )}\)	\(41\)
default	\(\frac {-8 \ln \left (5\right ) {\mathrm e}^{\frac {-3 \,{\mathrm e}^{x} \ln \left (5\right )+{\mathrm e}^{\frac {x}{8}}+16 \ln \left (5\right )}{\ln \left (5\right )}}+8 x \ln \left (5\right )}{8 \ln \left (5\right )}\)	\(42\)

[In]

int(1/8*((24*exp(x)*ln(5)-exp(1/8*x))*exp((-3*exp(x)*ln(5)+exp(1/8*x)+16*ln(5))/ln(5))+8*ln(5))/ln(5),x,method
=_RETURNVERBOSE)

[Out]

x-exp(-(3*exp(x)*ln(5)-16*ln(5)-exp(1/8*x))/ln(5))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {8 \log (5)+e^{\frac {e^{x/8}+16 \log (5)-3 e^x \log (5)}{\log (5)}} \left (-e^{x/8}+24 e^x \log (5)\right )}{8 \log (5)} \, dx=x - e^{\left (-\frac {3 \, e^{x} \log \left (5\right ) - e^{\left (\frac {1}{8} \, x\right )} - 16 \, \log \left (5\right )}{\log \left (5\right )}\right )} \]

[In]

integrate(1/8*((24*exp(x)*log(5)-exp(1/8*x))*exp((-3*exp(x)*log(5)+exp(1/8*x)+16*log(5))/log(5))+8*log(5))/log
(5),x, algorithm="fricas")

[Out]

x - e^(-(3*e^x*log(5) - e^(1/8*x) - 16*log(5))/log(5))

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {8 \log (5)+e^{\frac {e^{x/8}+16 \log (5)-3 e^x \log (5)}{\log (5)}} \left (-e^{x/8}+24 e^x \log (5)\right )}{8 \log (5)} \, dx=x - e^{\frac {e^{\frac {x}{8}} - 3 e^{x} \log {\left (5 \right )} + 16 \log {\left (5 \right )}}{\log {\left (5 \right )}}} \]

[In]

integrate(1/8*((24*exp(x)*ln(5)-exp(1/8*x))*exp((-3*exp(x)*ln(5)+exp(1/8*x)+16*ln(5))/ln(5))+8*ln(5))/ln(5),x)

[Out]

x - exp((exp(x/8) - 3*exp(x)*log(5) + 16*log(5))/log(5))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {8 \log (5)+e^{\frac {e^{x/8}+16 \log (5)-3 e^x \log (5)}{\log (5)}} \left (-e^{x/8}+24 e^x \log (5)\right )}{8 \log (5)} \, dx=\frac {x \log \left (5\right ) - e^{\left (\frac {e^{\left (\frac {1}{8} \, x\right )}}{\log \left (5\right )} - 3 \, e^{x} + 16\right )} \log \left (5\right )}{\log \left (5\right )} \]

[In]

integrate(1/8*((24*exp(x)*log(5)-exp(1/8*x))*exp((-3*exp(x)*log(5)+exp(1/8*x)+16*log(5))/log(5))+8*log(5))/log
(5),x, algorithm="maxima")

[Out]

(x*log(5) - e^(e^(1/8*x)/log(5) - 3*e^x + 16)*log(5))/log(5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {8 \log (5)+e^{\frac {e^{x/8}+16 \log (5)-3 e^x \log (5)}{\log (5)}} \left (-e^{x/8}+24 e^x \log (5)\right )}{8 \log (5)} \, dx=\frac {x \log \left (5\right ) - e^{\left (\frac {e^{\left (\frac {1}{8} \, x\right )}}{\log \left (5\right )} - 3 \, e^{x} + 16\right )} \log \left (5\right )}{\log \left (5\right )} \]

[In]

integrate(1/8*((24*exp(x)*log(5)-exp(1/8*x))*exp((-3*exp(x)*log(5)+exp(1/8*x)+16*log(5))/log(5))+8*log(5))/log
(5),x, algorithm="giac")

[Out]

(x*log(5) - e^(e^(1/8*x)/log(5) - 3*e^x + 16)*log(5))/log(5)

Mupad [B] (verification not implemented)

Time = 9.40 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {8 \log (5)+e^{\frac {e^{x/8}+16 \log (5)-3 e^x \log (5)}{\log (5)}} \left (-e^{x/8}+24 e^x \log (5)\right )}{8 \log (5)} \, dx=x-{\mathrm {e}}^{16}\,{\mathrm {e}}^{-3\,{\mathrm {e}}^x}\,{\mathrm {e}}^{\frac {{\left ({\mathrm {e}}^x\right )}^{1/8}}{\ln \left (5\right )}} \]

[In]

int((log(5) - (exp((exp(x/8) + 16*log(5) - 3*exp(x)*log(5))/log(5))*(exp(x/8) - 24*exp(x)*log(5)))/8)/log(5),x
)

[Out]

x - exp(16)*exp(-3*exp(x))*exp(exp(x)^(1/8)/log(5))