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

Optimal. Leaf size=27 \[ \log \left (e^{-e^{\frac {x}{\log \left (1+e^5-2 x\right )}}} \log \left (\frac {3 x}{2}\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 1.71, antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 6, integrand size = 114, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {6, 1593, 6688, 6706, 2302, 29} \begin {gather*} \log \left (\log \left (\frac {3 x}{2}\right )\right )-e^{\frac {x}{\log \left (-2 x+e^5+1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6688

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

Rule 6706

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (\left (1+e^5\right ) x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx\\ &=\int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (1+e^5-2 x\right ) x \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx\\ &=\int \left (-\frac {e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (2 x+\left (1+e^5-2 x\right ) \log \left (1+e^5-2 x\right )\right )}{\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )}+\frac {1}{x \log \left (\frac {3 x}{2}\right )}\right ) \, dx\\ &=-\int \frac {e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (2 x+\left (1+e^5-2 x\right ) \log \left (1+e^5-2 x\right )\right )}{\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )} \, dx+\int \frac {1}{x \log \left (\frac {3 x}{2}\right )} \, dx\\ &=-e^{\frac {x}{\log \left (1+e^5-2 x\right )}}+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\frac {3 x}{2}\right )\right )\\ &=-e^{\frac {x}{\log \left (1+e^5-2 x\right )}}+\log \left (\log \left (\frac {3 x}{2}\right )\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.43, size = 25, normalized size = 0.93 \begin {gather*} -e^{\frac {x}{\log \left (1+e^5-2 x\right )}}+\log \left (\log \left (\frac {3 x}{2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.96, size = 21, normalized size = 0.78 \begin {gather*} -e^{\left (\frac {x}{\log \left (-2 \, x + e^{5} + 1\right )}\right )} + \log \left (\log \left (\frac {3}{2} \, x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^(x/log(-2*x + e^5 + 1)) + log(log(3/2*x))

________________________________________________________________________________________

giac [A]  time = 0.54, size = 26, normalized size = 0.96 \begin {gather*} -e^{\left (\frac {x}{\log \left (-2 \, x + e^{5} + 1\right )}\right )} + \log \left (-\log \relax (2) + \log \left (3 \, x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^(x/log(-2*x + e^5 + 1)) + log(-log(2) + log(3*x))

________________________________________________________________________________________

maple [A]  time = 0.11, size = 22, normalized size = 0.81

method	result	size
default	\(\ln \left (\ln \left (\frac {3 x}{2}\right )\right )-{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{5}+1-2 x \right )}}\)	\(22\)
risch	\(\ln \left (\ln \left (\frac {3 x}{2}\right )\right )-{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{5}+1-2 x \right )}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.49, size = 26, normalized size = 0.96 \begin {gather*} -e^{\left (\frac {x}{\log \left (-2 \, x + e^{5} + 1\right )}\right )} + \log \left (\log \relax (3) - \log \relax (2) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^(x/log(-2*x + e^5 + 1)) + log(log(3) - log(2) + log(x))

________________________________________________________________________________________

mupad [B]  time = 2.62, size = 21, normalized size = 0.78 \begin {gather*} \ln \left (\ln \left (\frac {3\,x}{2}\right )\right )-{\mathrm {e}}^{\frac {x}{\ln \left ({\mathrm {e}}^5-2\,x+1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log((3*x)/2)) - exp(x/log(exp(5) - 2*x + 1))

________________________________________________________________________________________

sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: CoercionFailed} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: CoercionFailed

________________________________________________________________________________________