\(\int \frac {((-x^2-x^3) \log (4)-x^2 \log ^2(4)) \log (5)+(-x^2+(1-x) \log (4)) \log (5) \log (\log (4))}{e^x (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4))+e^x (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)) \log (\log (4))+e^x (x^2+2 x \log (4)+\log ^2(4)) \log ^2(\log (4))} \, dx\) [7563]

Optimal result

Integrand size = 135, antiderivative size = 26 \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\frac {e^{-x} \log (5)}{(x+\log (4)) \left (\log (4)+\frac {\log (\log (4))}{x}\right )} \]

[Out]

ln(5)/(ln(2*ln(2))/x+2*ln(2))/(x+2*ln(2))/exp(x)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(63\) vs. \(2(26)=52\).

Time = 0.65 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.42, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6820, 12, 6874, 2208, 2209} \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\frac {e^{-x} \log (4) \log (5)}{\left (\log ^2(4)-\log (\log (4))\right ) (x+\log (4))}-\frac {e^{-x} \log (5) \log (\log (4))}{\left (\log ^2(4)-\log (\log (4))\right ) (x \log (4)+\log (\log (4)))} \]

[In]

Int[(((-x^2 - x^3)*Log[4] - x^2*Log[4]^2)*Log[5] + (-x^2 + (1 - x)*Log[4])*Log[5]*Log[Log[4]])/(E^x*(x^4*Log[4
]^2 + 2*x^3*Log[4]^3 + x^2*Log[4]^4) + E^x*(2*x^3*Log[4] + 4*x^2*Log[4]^2 + 2*x*Log[4]^3)*Log[Log[4]] + E^x*(x
^2 + 2*x*Log[4] + Log[4]^2)*Log[Log[4]]^2),x]

[Out]

(Log[4]*Log[5])/(E^x*(x + Log[4])*(Log[4]^2 - Log[Log[4]])) - (Log[5]*Log[Log[4]])/(E^x*(Log[4]^2 - Log[Log[4]
])*(x*Log[4] + Log[Log[4]]))

Rule 12

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

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\frac {e^{-x} x \log (5)}{(x+\log (4)) (x \log (4)+\log (\log (4)))} \]

[In]

Integrate[(((-x^2 - x^3)*Log[4] - x^2*Log[4]^2)*Log[5] + (-x^2 + (1 - x)*Log[4])*Log[5]*Log[Log[4]])/(E^x*(x^4
*Log[4]^2 + 2*x^3*Log[4]^3 + x^2*Log[4]^4) + E^x*(2*x^3*Log[4] + 4*x^2*Log[4]^2 + 2*x*Log[4]^3)*Log[Log[4]] +
E^x*(x^2 + 2*x*Log[4] + Log[4]^2)*Log[Log[4]]^2),x]

[Out]

(x*Log[5])/(E^x*(x + Log[4])*(x*Log[4] + Log[Log[4]]))

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62

method	result	size
gosper	\(\frac {\ln \left (5\right ) x \,{\mathrm e}^{-x}}{4 x \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )+2 \ln \left (2 \ln \left (2\right )\right ) \ln \left (2\right )+x \ln \left (2 \ln \left (2\right )\right )}\)	\(42\)
norman	\(\frac {\ln \left (5\right ) x \,{\mathrm e}^{-x}}{4 x \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )+2 \ln \left (2 \ln \left (2\right )\right ) \ln \left (2\right )+x \ln \left (2 \ln \left (2\right )\right )}\)	\(42\)
parallelrisch	\(\frac {\ln \left (5\right ) x \,{\mathrm e}^{-x}}{4 x \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )+2 \ln \left (2 \ln \left (2\right )\right ) \ln \left (2\right )+x \ln \left (2 \ln \left (2\right )\right )}\)	\(42\)
risch	\(\frac {\ln \left (5\right ) x \,{\mathrm e}^{-x}}{4 x \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )+2 \ln \left (2\right )^{2}+2 \ln \left (2\right ) \ln \left (\ln \left (2\right )\right )+x \ln \left (2\right )+x \ln \left (\ln \left (2\right )\right )}\)	\(48\)
default	\(\text {Expression too large to display}\)	\(2348\)

[In]

int(((2*(1-x)*ln(2)-x^2)*ln(5)*ln(2*ln(2))+(-4*x^2*ln(2)^2+2*(-x^3-x^2)*ln(2))*ln(5))/((4*ln(2)^2+4*x*ln(2)+x^
2)*exp(x)*ln(2*ln(2))^2+(16*x*ln(2)^3+16*x^2*ln(2)^2+4*x^3*ln(2))*exp(x)*ln(2*ln(2))+(16*x^2*ln(2)^4+16*x^3*ln
(2)^3+4*x^4*ln(2)^2)*exp(x)),x,method=_RETURNVERBOSE)

[Out]

ln(5)*x/(4*x*ln(2)^2+2*x^2*ln(2)+2*ln(2*ln(2))*ln(2)+x*ln(2*ln(2)))/exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\frac {x \log \left (5\right )}{{\left (x + 2 \, \log \left (2\right )\right )} e^{x} \log \left (2 \, \log \left (2\right )\right ) + 2 \, {\left (x^{2} \log \left (2\right ) + 2 \, x \log \left (2\right )^{2}\right )} e^{x}} \]

[In]

integrate(((2*(1-x)*log(2)-x^2)*log(5)*log(2*log(2))+(-4*x^2*log(2)^2+2*(-x^3-x^2)*log(2))*log(5))/((4*log(2)^
2+4*x*log(2)+x^2)*exp(x)*log(2*log(2))^2+(16*x*log(2)^3+16*x^2*log(2)^2+4*x^3*log(2))*exp(x)*log(2*log(2))+(16
*x^2*log(2)^4+16*x^3*log(2)^3+4*x^4*log(2)^2)*exp(x)),x, algorithm="fricas")

[Out]

x*log(5)/((x + 2*log(2))*e^x*log(2*log(2)) + 2*(x^2*log(2) + 2*x*log(2)^2)*e^x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (26) = 52\).

Time = 0.17 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.04 \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\frac {x e^{- x} \log {\left (5 \right )}}{2 x^{2} \log {\left (2 \right )} + x \log {\left (\log {\left (2 \right )} \right )} + x \log {\left (2 \right )} + 4 x \log {\left (2 \right )}^{2} + 2 \log {\left (2 \right )} \log {\left (\log {\left (2 \right )} \right )} + 2 \log {\left (2 \right )}^{2}} \]

[In]

integrate(((2*(1-x)*ln(2)-x**2)*ln(5)*ln(2*ln(2))+(-4*x**2*ln(2)**2+2*(-x**3-x**2)*ln(2))*ln(5))/((4*ln(2)**2+
4*x*ln(2)+x**2)*exp(x)*ln(2*ln(2))**2+(16*x*ln(2)**3+16*x**2*ln(2)**2+4*x**3*ln(2))*exp(x)*ln(2*ln(2))+(16*x**
2*ln(2)**4+16*x**3*ln(2)**3+4*x**4*ln(2)**2)*exp(x)),x)

[Out]

x*exp(-x)*log(5)/(2*x**2*log(2) + x*log(log(2)) + x*log(2) + 4*x*log(2)**2 + 2*log(2)*log(log(2)) + 2*log(2)**
2)

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\frac {x e^{\left (-x\right )} \log \left (5\right )}{2 \, x^{2} \log \left (2\right ) + {\left (4 \, \log \left (2\right )^{2} + \log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )} x + 2 \, \log \left (2\right )^{2} + 2 \, \log \left (2\right ) \log \left (\log \left (2\right )\right )} \]

[In]

integrate(((2*(1-x)*log(2)-x^2)*log(5)*log(2*log(2))+(-4*x^2*log(2)^2+2*(-x^3-x^2)*log(2))*log(5))/((4*log(2)^
2+4*x*log(2)+x^2)*exp(x)*log(2*log(2))^2+(16*x*log(2)^3+16*x^2*log(2)^2+4*x^3*log(2))*exp(x)*log(2*log(2))+(16
*x^2*log(2)^4+16*x^3*log(2)^3+4*x^4*log(2)^2)*exp(x)),x, algorithm="maxima")

[Out]

x*e^(-x)*log(5)/(2*x^2*log(2) + (4*log(2)^2 + log(2) + log(log(2)))*x + 2*log(2)^2 + 2*log(2)*log(log(2)))

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81 \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\frac {x e^{\left (-x\right )} \log \left (5\right )}{2 \, x^{2} \log \left (2\right ) + 4 \, x \log \left (2\right )^{2} + x \log \left (2\right ) + 2 \, \log \left (2\right )^{2} + x \log \left (\log \left (2\right )\right ) + 2 \, \log \left (2\right ) \log \left (\log \left (2\right )\right )} \]

[In]

integrate(((2*(1-x)*log(2)-x^2)*log(5)*log(2*log(2))+(-4*x^2*log(2)^2+2*(-x^3-x^2)*log(2))*log(5))/((4*log(2)^
2+4*x*log(2)+x^2)*exp(x)*log(2*log(2))^2+(16*x*log(2)^3+16*x^2*log(2)^2+4*x^3*log(2))*exp(x)*log(2*log(2))+(16
*x^2*log(2)^4+16*x^3*log(2)^3+4*x^4*log(2)^2)*exp(x)),x, algorithm="giac")

[Out]

x*e^(-x)*log(5)/(2*x^2*log(2) + 4*x*log(2)^2 + x*log(2) + 2*log(2)^2 + x*log(log(2)) + 2*log(2)*log(log(2)))

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\int -\frac {\ln \left (5\right )\,\left (4\,x^2\,{\ln \left (2\right )}^2+2\,\ln \left (2\right )\,\left (x^3+x^2\right )\right )+\ln \left (2\,\ln \left (2\right )\right )\,\ln \left (5\right )\,\left (2\,\ln \left (2\right )\,\left (x-1\right )+x^2\right )}{{\mathrm {e}}^x\,\left (4\,{\ln \left (2\right )}^2\,x^4+16\,{\ln \left (2\right )}^3\,x^3+16\,{\ln \left (2\right )}^4\,x^2\right )+\ln \left (2\,\ln \left (2\right )\right )\,{\mathrm {e}}^x\,\left (4\,\ln \left (2\right )\,x^3+16\,{\ln \left (2\right )}^2\,x^2+16\,{\ln \left (2\right )}^3\,x\right )+{\ln \left (2\,\ln \left (2\right )\right )}^2\,{\mathrm {e}}^x\,\left (x^2+4\,\ln \left (2\right )\,x+4\,{\ln \left (2\right )}^2\right )} \,d x \]

[In]

int(-(log(5)*(4*x^2*log(2)^2 + 2*log(2)*(x^2 + x^3)) + log(2*log(2))*log(5)*(2*log(2)*(x - 1) + x^2))/(exp(x)*
(16*x^2*log(2)^4 + 16*x^3*log(2)^3 + 4*x^4*log(2)^2) + log(2*log(2))*exp(x)*(16*x^2*log(2)^2 + 16*x*log(2)^3 +
4*x^3*log(2)) + log(2*log(2))^2*exp(x)*(4*x*log(2) + 4*log(2)^2 + x^2)),x)

[Out]

int(-(log(5)*(4*x^2*log(2)^2 + 2*log(2)*(x^2 + x^3)) + log(2*log(2))*log(5)*(2*log(2)*(x - 1) + x^2))/(exp(x)*
(16*x^2*log(2)^4 + 16*x^3*log(2)^3 + 4*x^4*log(2)^2) + log(2*log(2))*exp(x)*(16*x^2*log(2)^2 + 16*x*log(2)^3 +
4*x^3*log(2)) + log(2*log(2))^2*exp(x)*(4*x*log(2) + 4*log(2)^2 + x^2)), x)