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

Optimal. Leaf size=22 \[ e^4 x \left (-4-\log \left (e^{16/x} x\right )+\log (\log (4))\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 35, normalized size of antiderivative = 1.59, number of steps used = 7, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {14, 2548, 43} \begin {gather*} e^4 x-e^4 x \log \left (e^{16/x} x\right )-e^4 x (5-\log (\log (4))) \end {gather*}

Antiderivative was successfully verified.

[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 43

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 2548

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 26, normalized size = 1.18 \begin {gather*} e^4 \left (16-x \log \left (e^{16/x} x\right )+x (-4+\log (\log (4)))\right ) \end {gather*}

Antiderivative was successfully verified.

[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]]))

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fricas [A]  time = 0.79, size = 29, normalized size = 1.32 \begin {gather*} -x e^{4} \log \left (x e^{\frac {16}{x}}\right ) + x e^{4} \log \left (2 \, \log \relax (2)\right ) - 4 \, x e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.30, size = 26, normalized size = 1.18 \begin {gather*} x e^{4} \log \relax (2) - x e^{4} \log \relax (x) + x e^{4} \log \left (\log \relax (2)\right ) - 4 \, x e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.09, size = 34, normalized size = 1.55

method	result	size
norman	\(\left (-4 \,{\mathrm e}^{4}+{\mathrm e}^{4} \ln \relax (2)+{\mathrm e}^{4} \ln \left (\ln \relax (2)\right )\right ) x -x \,{\mathrm e}^{4} \ln \left (x \,{\mathrm e}^{\frac {16}{x}}\right )\)	\(34\)
default	\(x \,{\mathrm e}^{4} \ln \left (2 \ln \relax (2)\right )+16 \,{\mathrm e}^{4} \ln \relax (x )-4 x \,{\mathrm e}^{4}-x \,{\mathrm e}^{4} \ln \left (x \,{\mathrm e}^{\frac {16}{x}}\right )+16 \,{\mathrm e}^{4} \ln \left (\frac {16}{x}\right )\)	\(46\)
risch	\(-x \,{\mathrm e}^{4} \ln \left ({\mathrm e}^{\frac {16}{x}}\right )-x \,{\mathrm e}^{4} \ln \relax (x )+\frac {i x \,{\mathrm e}^{4} \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {16}{x}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {16}{x}}\right )}{2}-\frac {i x \,{\mathrm e}^{4} \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {16}{x}}\right )^{2}}{2}-\frac {i x \,{\mathrm e}^{4} \pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {16}{x}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {16}{x}}\right )^{2}}{2}+\frac {i x \,{\mathrm e}^{4} \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {16}{x}}\right )^{3}}{2}+x \,{\mathrm e}^{4} \ln \relax (2)+x \,{\mathrm e}^{4} \ln \left (\ln \relax (2)\right )-4 x \,{\mathrm e}^{4}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.60, size = 44, normalized size = 2.00 \begin {gather*} -x e^{4} \log \left (x e^{\frac {16}{x}}\right ) + x e^{4} \log \left (2 \, \log \relax (2)\right ) + {\left (x - 16 \, \log \relax (x)\right )} e^{4} - 5 \, x e^{4} + 16 \, e^{4} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.72, size = 15, normalized size = 0.68 \begin {gather*} x\,{\mathrm {e}}^4\,\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )-\ln \relax (x)-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.33, size = 34, normalized size = 1.55 \begin {gather*} - x e^{4} \log {\left (x e^{\frac {16}{x}} \right )} + x \left (- 4 e^{4} + e^{4} \log {\left (\log {\relax (2 )} \right )} + e^{4} \log {\relax (2 )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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