\(\int \frac {e^4 (16-5 x)-e^4 x \log (e^{16/x} x)+e^4 x \log (\log (4))}{x} \, dx\) [925]

Optimal result

Integrand size = 38, antiderivative size = 22 \[ \int \frac {e^4 (16-5 x)-e^4 x \log \left (e^{16/x} x\right )+e^4 x \log (\log (4))}{x} \, dx=e^4 x \left (-4-\log \left (e^{16/x} x\right )+\log (\log (4))\right ) \]

[Out]

x*exp(4)*(ln(2*ln(2))-ln(x*exp(16/x))-4)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {14, 2628, 45} \[ \int \frac {e^4 (16-5 x)-e^4 x \log \left (e^{16/x} x\right )+e^4 x \log (\log (4))}{x} \, dx=e^4 x-e^4 x \log \left (e^{16/x} x\right )-e^4 x (5-\log (\log (4))) \]

[In]

Int[(E^4*(16 - 5*x) - E^4*x*Log[E^(16/x)*x] + E^4*x*Log[Log[4]])/x,x]

[Out]

E^4*x - E^4*x*Log[E^(16/x)*x] - E^4*x*(5 - Log[Log[4]])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2628

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*(D[u, x]/u), x], x] /; InverseFunctionFr
eeQ[u, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-e^4 \log \left (e^{16/x} x\right )+\frac {e^4 (16-x (5-\log (\log (4))))}{x}\right ) \, dx \\ & = -\left (e^4 \int \log \left (e^{16/x} x\right ) \, dx\right )+e^4 \int \frac {16-x (5-\log (\log (4)))}{x} \, dx \\ & = -e^4 x \log \left (e^{16/x} x\right )+e^4 \int \frac {-16+x}{x} \, dx+e^4 \int \left (-5+\frac {16}{x}+\log (\log (4))\right ) \, dx \\ & = 16 e^4 \log (x)-e^4 x \log \left (e^{16/x} x\right )-e^4 x (5-\log (\log (4)))+e^4 \int \left (1-\frac {16}{x}\right ) \, dx \\ & = e^4 x-e^4 x \log \left (e^{16/x} x\right )-e^4 x (5-\log (\log (4))) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^4 (16-5 x)-e^4 x \log \left (e^{16/x} x\right )+e^4 x \log (\log (4))}{x} \, dx=e^4 \left (16-x \log \left (e^{16/x} x\right )+x (-4+\log (\log (4)))\right ) \]

[In]

Integrate[(E^4*(16 - 5*x) - E^4*x*Log[E^(16/x)*x] + E^4*x*Log[Log[4]])/x,x]

[Out]

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55

method	result	size
norman	\(\left (-4 \,{\mathrm e}^{4}+{\mathrm e}^{4} \ln \left (2\right )+{\mathrm e}^{4} \ln \left (\ln \left (2\right )\right )\right ) x -x \,{\mathrm e}^{4} \ln \left (x \,{\mathrm e}^{\frac {16}{x}}\right )\)	\(34\)
parallelrisch	\(\frac {{\mathrm e}^{4} \ln \left (2 \ln \left (2\right )\right ) x^{2}-{\mathrm e}^{4} \ln \left (x \,{\mathrm e}^{\frac {16}{x}}\right ) x^{2}-4 x^{2} {\mathrm e}^{4}}{x}\)	\(40\)
default	\({\mathrm e}^{4} \left (x \ln \left (2 \ln \left (2\right )\right )-5 x +16 \ln \left (x \right )\right )-{\mathrm e}^{4} \left (\ln \left (x \,{\mathrm e}^{\frac {16}{x}}\right ) x -x -16 \ln \left (\frac {16}{x}\right )\right )\)	\(47\)
parts	\({\mathrm e}^{4} \left (x \ln \left (2 \ln \left (2\right )\right )-5 x +16 \ln \left (x \right )\right )-{\mathrm e}^{4} \left (\ln \left (x \,{\mathrm e}^{\frac {16}{x}}\right ) x -x -16 \ln \left (\frac {16}{x}\right )\right )\)	\(47\)
risch	\(-x \,{\mathrm e}^{4} \ln \left ({\mathrm e}^{\frac {16}{x}}\right )-x \,{\mathrm e}^{4} \ln \left (x \right )+\frac {i x \,{\mathrm e}^{4} \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {16}{x}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{\frac {16}{x}}\right )}{2}-\frac {i x \,{\mathrm e}^{4} \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{\frac {16}{x}}\right )^{2}}{2}-\frac {i x \,{\mathrm e}^{4} \pi \,\operatorname {csgn}\left (i {\mathrm e}^{\frac {16}{x}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{\frac {16}{x}}\right )^{2}}{2}+\frac {i x \,{\mathrm e}^{4} \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{\frac {16}{x}}\right )^{3}}{2}+x \,{\mathrm e}^{4} \ln \left (2\right )+x \,{\mathrm e}^{4} \ln \left (\ln \left (2\right )\right )-4 x \,{\mathrm e}^{4}\)	\(147\)

[In]

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

[Out]

(-4*exp(4)+exp(4)*ln(2)+exp(4)*ln(ln(2)))*x-x*exp(4)*ln(x*exp(16/x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {e^4 (16-5 x)-e^4 x \log \left (e^{16/x} x\right )+e^4 x \log (\log (4))}{x} \, dx=-x e^{4} \log \left (x e^{\frac {16}{x}}\right ) + x e^{4} \log \left (2 \, \log \left (2\right )\right ) - 4 \, x e^{4} \]

[In]

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

[Out]

-x*e^4*log(x*e^(16/x)) + x*e^4*log(2*log(2)) - 4*x*e^4

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {e^4 (16-5 x)-e^4 x \log \left (e^{16/x} x\right )+e^4 x \log (\log (4))}{x} \, dx=- x e^{4} \log {\left (x e^{\frac {16}{x}} \right )} + x \left (- 4 e^{4} + e^{4} \log {\left (\log {\left (2 \right )} \right )} + e^{4} \log {\left (2 \right )}\right ) \]

[In]

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

[Out]

-x*exp(4)*log(x*exp(16/x)) + x*(-4*exp(4) + exp(4)*log(log(2)) + exp(4)*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.00 \[ \int \frac {e^4 (16-5 x)-e^4 x \log \left (e^{16/x} x\right )+e^4 x \log (\log (4))}{x} \, dx=-x e^{4} \log \left (x e^{\frac {16}{x}}\right ) + x e^{4} \log \left (2 \, \log \left (2\right )\right ) + {\left (x - 16 \, \log \left (x\right )\right )} e^{4} - 5 \, x e^{4} + 16 \, e^{4} \log \left (x\right ) \]

[In]

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

[Out]

-x*e^4*log(x*e^(16/x)) + x*e^4*log(2*log(2)) + (x - 16*log(x))*e^4 - 5*x*e^4 + 16*e^4*log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^4 (16-5 x)-e^4 x \log \left (e^{16/x} x\right )+e^4 x \log (\log (4))}{x} \, dx=x e^{4} \log \left (2\right ) - x e^{4} \log \left (x\right ) + x e^{4} \log \left (\log \left (2\right )\right ) - 4 \, x e^{4} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.17 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {e^4 (16-5 x)-e^4 x \log \left (e^{16/x} x\right )+e^4 x \log (\log (4))}{x} \, dx=x\,{\mathrm {e}}^4\,\left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )-\ln \left (x\right )-4\right ) \]

[In]

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

[Out]

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