\(\int \frac {e^{\frac {(-29-2 x) \log (4)+(1-\log (4)) \log (x)}{\log (4)}} (1+(-1-2 x) \log (4))}{x \log (4)} \, dx\) [2648]

Optimal result

Integrand size = 43, antiderivative size = 18 \[ \int \frac {e^{\frac {(-29-2 x) \log (4)+(1-\log (4)) \log (x)}{\log (4)}} (1+(-1-2 x) \log (4))}{x \log (4)} \, dx=\frac {e^{-29-2 x+\frac {\log (x)}{\log (4)}}}{x} \]

[Out]

exp(1/2*ln(x)/ln(2)-29-2*x-ln(x))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 2306, 2228} \[ \int \frac {e^{\frac {(-29-2 x) \log (4)+(1-\log (4)) \log (x)}{\log (4)}} (1+(-1-2 x) \log (4))}{x \log (4)} \, dx=e^{-2 x-29} x^{\frac {1}{\log (4)}-1} \]

[In]

Int[(E^(((-29 - 2*x)*Log[4] + (1 - Log[4])*Log[x])/Log[4])*(1 + (-1 - 2*x)*Log[4]))/(x*Log[4]),x]

[Out]

E^(-29 - 2*x)*x^(-1 + Log[4]^(-1))

Rule 12

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

Rule 2228

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[g*u^(m + 1)*(F^(c*v)/(b*c*
e*Log[F])), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, 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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\exp \left (\frac {(-29-2 x) \log (4)+(1-\log (4)) \log (x)}{\log (4)}\right ) (1+(-1-2 x) \log (4))}{x} \, dx}{\log (4)} \\ & = \frac {\int e^{-29-2 x} x^{-1+\frac {1-\log (4)}{\log (4)}} (1+(-1-2 x) \log (4)) \, dx}{\log (4)} \\ & = e^{-29-2 x} x^{-1+\frac {1}{\log (4)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {e^{\frac {(-29-2 x) \log (4)+(1-\log (4)) \log (x)}{\log (4)}} (1+(-1-2 x) \log (4))}{x \log (4)} \, dx=\frac {e^{-29-2 x} x^{-1+\frac {1}{\log (4)}} \log (16)}{2 \log (4)} \]

[In]

Integrate[(E^(((-29 - 2*x)*Log[4] + (1 - Log[4])*Log[x])/Log[4])*(1 + (-1 - 2*x)*Log[4]))/(x*Log[4]),x]

[Out]

(E^(-29 - 2*x)*x^(-1 + Log[4]^(-1))*Log[16])/(2*Log[4])

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method	result	size
risch	\(\frac {x^{\frac {1}{2 \ln \left (2\right )}} {\mathrm e}^{-2 x -29}}{x}\)	\(19\)
norman	\({\mathrm e}^{\frac {\left (1-2 \ln \left (2\right )\right ) \ln \left (x \right )+2 \left (-2 x -29\right ) \ln \left (2\right )}{2 \ln \left (2\right )}}\)	\(27\)
parallelrisch	\({\mathrm e}^{\frac {\left (1-2 \ln \left (2\right )\right ) \ln \left (x \right )+2 \left (-2 x -29\right ) \ln \left (2\right )}{2 \ln \left (2\right )}}\)	\(27\)
gosper	\({\mathrm e}^{-\frac {2 \ln \left (2\right ) \ln \left (x \right )+4 x \ln \left (2\right )-\ln \left (x \right )+58 \ln \left (2\right )}{2 \ln \left (2\right )}}\)	\(28\)

[In]

int(1/2*(2*(-1-2*x)*ln(2)+1)*exp(1/2*((1-2*ln(2))*ln(x)+2*(-2*x-29)*ln(2))/ln(2))/x/ln(2),x,method=_RETURNVERB
OSE)

[Out]

x^(1/2/ln(2))/x*exp(-2*x-29)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {e^{\frac {(-29-2 x) \log (4)+(1-\log (4)) \log (x)}{\log (4)}} (1+(-1-2 x) \log (4))}{x \log (4)} \, dx=e^{\left (-\frac {2 \, {\left (2 \, x + 29\right )} \log \left (2\right ) + {\left (2 \, \log \left (2\right ) - 1\right )} \log \left (x\right )}{2 \, \log \left (2\right )}\right )} \]

[In]

integrate(1/2*(2*(-1-2*x)*log(2)+1)*exp(1/2*((1-2*log(2))*log(x)+2*(-2*x-29)*log(2))/log(2))/x/log(2),x, algor
ithm="fricas")

[Out]

e^(-1/2*(2*(2*x + 29)*log(2) + (2*log(2) - 1)*log(x))/log(2))

Sympy [A] (verification not implemented)

Time = 1.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {(-29-2 x) \log (4)+(1-\log (4)) \log (x)}{\log (4)}} (1+(-1-2 x) \log (4))}{x \log (4)} \, dx=\frac {x^{\frac {1}{2 \log {\left (2 \right )}}} e^{- 2 x}}{x e^{29}} \]

[In]

integrate(1/2*(2*(-1-2*x)*ln(2)+1)*exp(1/2*((1-2*ln(2))*ln(x)+2*(-2*x-29)*ln(2))/ln(2))/x/ln(2),x)

[Out]

x**(1/(2*log(2)))*exp(-29)*exp(-2*x)/x

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {(-29-2 x) \log (4)+(1-\log (4)) \log (x)}{\log (4)}} (1+(-1-2 x) \log (4))}{x \log (4)} \, dx=\frac {e^{\left (-2 \, x + \frac {\log \left (x\right )}{2 \, \log \left (2\right )} - 29\right )}}{x} \]

[In]

integrate(1/2*(2*(-1-2*x)*log(2)+1)*exp(1/2*((1-2*log(2))*log(x)+2*(-2*x-29)*log(2))/log(2))/x/log(2),x, algor
ithm="maxima")

[Out]

e^(-2*x + 1/2*log(x)/log(2) - 29)/x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {(-29-2 x) \log (4)+(1-\log (4)) \log (x)}{\log (4)}} (1+(-1-2 x) \log (4))}{x \log (4)} \, dx=e^{\left (-2 \, x + \frac {\log \left (x\right )}{2 \, \log \left (2\right )} - \log \left (x\right ) - 29\right )} \]

[In]

integrate(1/2*(2*(-1-2*x)*log(2)+1)*exp(1/2*((1-2*log(2))*log(x)+2*(-2*x-29)*log(2))/log(2))/x/log(2),x, algor
ithm="giac")

[Out]

e^(-2*x + 1/2*log(x)/log(2) - log(x) - 29)

Mupad [B] (verification not implemented)

Time = 8.44 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {(-29-2 x) \log (4)+(1-\log (4)) \log (x)}{\log (4)}} (1+(-1-2 x) \log (4))}{x \log (4)} \, dx=x^{\frac {1}{2\,\ln \left (2\right )}-1}\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{-29} \]

[In]

int(-(exp(-(log(2)*(2*x + 29) + (log(x)*(2*log(2) - 1))/2)/log(2))*(2*log(2)*(2*x + 1) - 1))/(2*x*log(2)),x)

[Out]

x^(1/(2*log(2)) - 1)*exp(-2*x)*exp(-29)