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

Optimal. Leaf size=32 \[ 2+\frac {x}{x+\log \left (\frac {\log (x)}{2+x-\frac {1-2 e^{-x}}{\log (5)}}\right )} \]

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Rubi [A]  time = 1.59, antiderivative size = 35, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, integrand size = 203, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {6688, 6711, 32} \begin {gather*} -\frac {1}{\frac {x}{\log \left (\frac {e^x \log (5) \log (x)}{e^x ((x+2) \log (5)-1)+2}\right )}+1} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 + x/Log[(E^x*Log[5]*Log[x])/(2 + E^x*(-1 + (2 + x)*Log[5]))])^(-1)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6688

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

Rule 6711

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2-e^x (-1+x \log (5)+\log (25))+\log (x) \left (x \left (-2+e^x \log (5)\right )+\left (2+e^x (-1+x \log (5)+\log (25))\right ) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )\right )}{\left (2+e^x (-1+x \log (5)+\log (25))\right ) \log (x) \left (x+\log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )\right )^2} \, dx\\ &=\operatorname {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,\frac {x}{\log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}\right )\\ &=-\frac {1}{1+\frac {x}{\log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.40, size = 31, normalized size = 0.97 \begin {gather*} \frac {x}{x+\log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/(x + Log[(E^x*Log[5]*Log[x])/(2 + E^x*(-1 + (2 + x)*Log[5]))])

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fricas [A]  time = 0.62, size = 29, normalized size = 0.91 \begin {gather*} \frac {x}{x + \log \left (\frac {e^{x} \log \relax (5) \log \relax (x)}{{\left ({\left (x + 2\right )} \log \relax (5) - 1\right )} e^{x} + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(x + log(e^x*log(5)*log(x)/(((x + 2)*log(5) - 1)*e^x + 2)))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [C]  time = 0.31, size = 430, normalized size = 13.44

method	result	size
risch	\(\frac {2 x}{-i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{x} \ln \relax (x )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \relax (5)-{\mathrm e}^{x}+2}\right )^{3}-i \pi \,\mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \relax (5)-{\mathrm e}^{x}+2}\right ) \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (\frac {i \ln \relax (x )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \relax (5)-{\mathrm e}^{x}+2}\right )-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (\frac {i \ln \relax (x )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \relax (5)-{\mathrm e}^{x}+2}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x} \ln \relax (x )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \relax (5)-{\mathrm e}^{x}+2}\right )+i \pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (\frac {i \ln \relax (x )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \relax (5)-{\mathrm e}^{x}+2}\right )^{2}+i \pi \,\mathrm {csgn}\left (\frac {i}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \relax (5)-{\mathrm e}^{x}+2}\right ) \mathrm {csgn}\left (\frac {i \ln \relax (x )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \relax (5)-{\mathrm e}^{x}+2}\right )^{2}+i \pi \,\mathrm {csgn}\left (\frac {i \ln \relax (x )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \relax (5)-{\mathrm e}^{x}+2}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x} \ln \relax (x )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \relax (5)-{\mathrm e}^{x}+2}\right )^{2}+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x} \ln \relax (x )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \relax (5)-{\mathrm e}^{x}+2}\right )^{2}-i \pi \mathrm {csgn}\left (\frac {i \ln \relax (x )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \relax (5)-{\mathrm e}^{x}+2}\right )^{3}+2 \ln \left (\ln \relax (5)\right )+2 x +2 \ln \left (\ln \relax (x )\right )-2 \ln \left (\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \relax (5)-{\mathrm e}^{x}+2\right )+2 \ln \left ({\mathrm e}^{x}\right )}\)	\(430\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x/(-I*Pi*csgn(I*exp(x)/((exp(x)*x+2*exp(x))*ln(5)-exp(x)+2)*ln(x))^3-I*Pi*csgn(I/((exp(x)*x+2*exp(x))*ln(5)-
exp(x)+2))*csgn(I*ln(x))*csgn(I*ln(x)/((exp(x)*x+2*exp(x))*ln(5)-exp(x)+2))-I*Pi*csgn(I*exp(x))*csgn(I*ln(x)/(
(exp(x)*x+2*exp(x))*ln(5)-exp(x)+2))*csgn(I*exp(x)/((exp(x)*x+2*exp(x))*ln(5)-exp(x)+2)*ln(x))+I*Pi*csgn(I*ln(
x))*csgn(I*ln(x)/((exp(x)*x+2*exp(x))*ln(5)-exp(x)+2))^2+I*Pi*csgn(I/((exp(x)*x+2*exp(x))*ln(5)-exp(x)+2))*csg
n(I*ln(x)/((exp(x)*x+2*exp(x))*ln(5)-exp(x)+2))^2+I*Pi*csgn(I*ln(x)/((exp(x)*x+2*exp(x))*ln(5)-exp(x)+2))*csgn
(I*exp(x)/((exp(x)*x+2*exp(x))*ln(5)-exp(x)+2)*ln(x))^2+I*Pi*csgn(I*exp(x))*csgn(I*exp(x)/((exp(x)*x+2*exp(x))
*ln(5)-exp(x)+2)*ln(x))^2-I*Pi*csgn(I*ln(x)/((exp(x)*x+2*exp(x))*ln(5)-exp(x)+2))^3+2*ln(ln(5))+2*x+2*ln(ln(x)
)-2*ln((exp(x)*x+2*exp(x))*ln(5)-exp(x)+2)+2*ln(exp(x)))

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maxima [A]  time = 1.06, size = 32, normalized size = 1.00 \begin {gather*} \frac {x}{2 \, x - \log \left ({\left (x \log \relax (5) + 2 \, \log \relax (5) - 1\right )} e^{x} + 2\right ) + \log \left (\log \relax (5)\right ) + \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(2*x - log((x*log(5) + 2*log(5) - 1)*e^x + 2) + log(log(5)) + log(log(x)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {{\mathrm {e}}^x\,\left (\ln \relax (5)\,\left (x+2\right )-1\right )+\ln \relax (x)\,\left (2\,x-x\,{\mathrm {e}}^x\,\ln \relax (5)\right )-\ln \relax (x)\,\ln \left (\frac {{\mathrm {e}}^x\,\ln \relax (5)\,\ln \relax (x)}{{\mathrm {e}}^x\,\left (\ln \relax (5)\,\left (x+2\right )-1\right )+2}\right )\,\left ({\mathrm {e}}^x\,\left (\ln \relax (5)\,\left (x+2\right )-1\right )+2\right )+2}{\ln \relax (x)\,\left ({\mathrm {e}}^x\,\left (\ln \relax (5)\,\left (x+2\right )-1\right )+2\right )\,{\ln \left (\frac {{\mathrm {e}}^x\,\ln \relax (5)\,\ln \relax (x)}{{\mathrm {e}}^x\,\left (\ln \relax (5)\,\left (x+2\right )-1\right )+2}\right )}^2+\ln \relax (x)\,\left (4\,x-{\mathrm {e}}^x\,\left (2\,x-\ln \relax (5)\,\left (2\,x^2+4\,x\right )\right )\right )\,\ln \left (\frac {{\mathrm {e}}^x\,\ln \relax (5)\,\ln \relax (x)}{{\mathrm {e}}^x\,\left (\ln \relax (5)\,\left (x+2\right )-1\right )+2}\right )+\ln \relax (x)\,\left ({\mathrm {e}}^x\,\left (\ln \relax (5)\,\left (x^3+2\,x^2\right )-x^2\right )+2\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.56, size = 27, normalized size = 0.84 \begin {gather*} \frac {x}{x + \log {\left (\frac {e^{x} \log {\relax (5 )} \log {\relax (x )}}{\left (\left (x + 2\right ) \log {\relax (5 )} - 1\right ) e^{x} + 2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(x + log(exp(x)*log(5)*log(x)/(((x + 2)*log(5) - 1)*exp(x) + 2)))

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