3.38.77 \(\int \frac {e^{\frac {x+(9-x) \log (\frac {1}{4} (x+x \log (2)))}{\log (\frac {1}{4} (x+x \log (2)))}} (-1+\log (\frac {1}{4} (x+x \log (2)))-\log ^2(\frac {1}{4} (x+x \log (2))))}{\log ^2(\frac {1}{4} (x+x \log (2)))} \, dx\)

Optimal. Leaf size=22 \[ e^{9-x+\frac {x}{\log \left (\frac {1}{4} (x+x \log (2))\right )}} \]

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Rubi [A]  time = 0.84, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 2, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {2444, 6706} \begin {gather*} e^{-x+\frac {x}{\log \left (\frac {1}{4} x (1+\log (2))\right )}+9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((x + (9 - x)*Log[(x + x*Log[2])/4])/Log[(x + x*Log[2])/4])*(-1 + Log[(x + x*Log[2])/4] - Log[(x + x*Lo
g[2])/4]^2))/Log[(x + x*Log[2])/4]^2,x]

[Out]

E^(9 - x + x/Log[(x*(1 + Log[2]))/4])

Rule 2444

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, 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 {\exp \left (\frac {x+(9-x) \log \left (\frac {1}{4} (x+x \log (2))\right )}{\log \left (\frac {1}{4} (x+x \log (2))\right )}\right ) \left (-1+\log \left (\frac {1}{4} (x+x \log (2))\right )-\log ^2\left (\frac {1}{4} (x+x \log (2))\right )\right )}{\log ^2\left (\frac {1}{4} x (1+\log (2))\right )} \, dx\\ &=4 \operatorname {Subst}\left (\int \frac {e^{9-4 x+\frac {4 x}{\log (x (1+\log (2)))}} \left (-1+\log (x (1+\log (2)))-\log ^2(x (1+\log (2)))\right )}{\log ^2(x (1+\log (2)))} \, dx,x,\frac {x}{4}\right )\\ &=e^{9-x+\frac {x}{\log \left (\frac {1}{4} x (1+\log (2))\right )}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.45, size = 21, normalized size = 0.95 \begin {gather*} e^{9-x+\frac {x}{\log \left (\frac {1}{4} x (1+\log (2))\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((x + (9 - x)*Log[(x + x*Log[2])/4])/Log[(x + x*Log[2])/4])*(-1 + Log[(x + x*Log[2])/4] - Log[(x
+ x*Log[2])/4]^2))/Log[(x + x*Log[2])/4]^2,x]

[Out]

E^(9 - x + x/Log[(x*(1 + Log[2]))/4])

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fricas [A]  time = 0.96, size = 33, normalized size = 1.50 \begin {gather*} e^{\left (-\frac {{\left (x - 9\right )} \log \left (\frac {1}{4} \, x \log \relax (2) + \frac {1}{4} \, x\right ) - x}{\log \left (\frac {1}{4} \, x \log \relax (2) + \frac {1}{4} \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(1/4*x*log(2)+1/4*x)^2+log(1/4*x*log(2)+1/4*x)-1)*exp(((9-x)*log(1/4*x*log(2)+1/4*x)+x)/log(1/4
*x*log(2)+1/4*x))/log(1/4*x*log(2)+1/4*x)^2,x, algorithm="fricas")

[Out]

e^(-((x - 9)*log(1/4*x*log(2) + 1/4*x) - x)/log(1/4*x*log(2) + 1/4*x))

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giac [A]  time = 1.11, size = 20, normalized size = 0.91 \begin {gather*} e^{\left (-x + \frac {x}{\log \left (\frac {1}{4} \, x \log \relax (2) + \frac {1}{4} \, x\right )} + 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(1/4*x*log(2)+1/4*x)^2+log(1/4*x*log(2)+1/4*x)-1)*exp(((9-x)*log(1/4*x*log(2)+1/4*x)+x)/log(1/4
*x*log(2)+1/4*x))/log(1/4*x*log(2)+1/4*x)^2,x, algorithm="giac")

[Out]

e^(-x + x/log(1/4*x*log(2) + 1/4*x) + 9)

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maple [A]  time = 0.07, size = 33, normalized size = 1.50

method	result	size
norman	\({\mathrm e}^{\frac {\left (9-x \right ) \ln \left (\frac {x \ln \relax (2)}{4}+\frac {x}{4}\right )+x}{\ln \left (\frac {x \ln \relax (2)}{4}+\frac {x}{4}\right )}}\)	\(33\)
risch	\({\mathrm e}^{-\frac {\ln \left (\frac {x \ln \relax (2)}{4}+\frac {x}{4}\right ) x -9 \ln \left (\frac {x \ln \relax (2)}{4}+\frac {x}{4}\right )-x}{\ln \left (\frac {x \ln \relax (2)}{4}+\frac {x}{4}\right )}}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(1/4*x*ln(2)+1/4*x)^2+ln(1/4*x*ln(2)+1/4*x)-1)*exp(((9-x)*ln(1/4*x*ln(2)+1/4*x)+x)/ln(1/4*x*ln(2)+1/4*
x))/ln(1/4*x*ln(2)+1/4*x)^2,x,method=_RETURNVERBOSE)

[Out]

exp(((9-x)*ln(1/4*x*ln(2)+1/4*x)+x)/ln(1/4*x*ln(2)+1/4*x))

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maxima [A]  time = 0.54, size = 27, normalized size = 1.23 \begin {gather*} e^{\left (-x - \frac {x}{2 \, \log \relax (2) - \log \relax (x) - \log \left (\log \relax (2) + 1\right )} + 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(1/4*x*log(2)+1/4*x)^2+log(1/4*x*log(2)+1/4*x)-1)*exp(((9-x)*log(1/4*x*log(2)+1/4*x)+x)/log(1/4
*x*log(2)+1/4*x))/log(1/4*x*log(2)+1/4*x)^2,x, algorithm="maxima")

[Out]

e^(-x - x/(2*log(2) - log(x) - log(log(2) + 1)) + 9)

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mupad [B]  time = 2.37, size = 22, normalized size = 1.00 \begin {gather*} {\mathrm {e}}^{-x}\,{\mathrm {e}}^9\,{\mathrm {e}}^{\frac {x}{\ln \left (\frac {x}{4}+\frac {x\,\ln \relax (2)}{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((x - log(x/4 + (x*log(2))/4)*(x - 9))/log(x/4 + (x*log(2))/4))*(log(x/4 + (x*log(2))/4)^2 - log(x/4
+ (x*log(2))/4) + 1))/log(x/4 + (x*log(2))/4)^2,x)

[Out]

exp(-x)*exp(9)*exp(x/log(x/4 + (x*log(2))/4))

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sympy [A]  time = 0.31, size = 29, normalized size = 1.32 \begin {gather*} e^{\frac {x + \left (9 - x\right ) \log {\left (\frac {x \log {\relax (2 )}}{4} + \frac {x}{4} \right )}}{\log {\left (\frac {x \log {\relax (2 )}}{4} + \frac {x}{4} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(1/4*x*ln(2)+1/4*x)**2+ln(1/4*x*ln(2)+1/4*x)-1)*exp(((9-x)*ln(1/4*x*ln(2)+1/4*x)+x)/ln(1/4*x*ln(
2)+1/4*x))/ln(1/4*x*ln(2)+1/4*x)**2,x)

[Out]

exp((x + (9 - x)*log(x*log(2)/4 + x/4))/log(x*log(2)/4 + x/4))

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