3.87.54 \(\int \frac {1}{8} (36 x-4 e^4 x-4 x \log (4)+(72 x-8 e^4 x-8 x \log (4)) \log (x)+e^{e^{x/4}} (-4 x+(-8 x-e^{x/4} x^2) \log (x))) \, dx\)

Optimal. Leaf size=31 \[ \frac {1}{2} x^2 \left (9-e^4-e^{e^{x/4}}-\log (4)\right ) \log (x) \]

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Rubi [A]  time = 0.22, antiderivative size = 39, normalized size of antiderivative = 1.26, number of steps used = 12, number of rules used = 6, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {6, 12, 2304, 6742, 2288, 2554} \begin {gather*} \frac {1}{2} x^2 \left (9-e^4-\log (4)\right ) \log (x)-\frac {1}{2} e^{e^{x/4}} x^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(36*x - 4*E^4*x - 4*x*Log[4] + (72*x - 8*E^4*x - 8*x*Log[4])*Log[x] + E^E^(x/4)*(-4*x + (-8*x - E^(x/4)*x^
2)*Log[x]))/8,x]

[Out]

-1/2*(E^E^(x/4)*x^2*Log[x]) + (x^2*(9 - E^4 - Log[4])*Log[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 12

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{8} \left (\left (36-4 e^4\right ) x-4 x \log (4)+\left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x)+e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right )\right ) \, dx\\ &=\int \frac {1}{8} \left (x \left (36-4 e^4-4 \log (4)\right )+\left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x)+e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right )\right ) \, dx\\ &=\frac {1}{8} \int \left (x \left (36-4 e^4-4 \log (4)\right )+\left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x)+e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right )\right ) \, dx\\ &=\frac {1}{4} x^2 \left (9-e^4-\log (4)\right )+\frac {1}{8} \int \left (72 x-8 e^4 x-8 x \log (4)\right ) \log (x) \, dx+\frac {1}{8} \int e^{e^{x/4}} \left (-4 x+\left (-8 x-e^{x/4} x^2\right ) \log (x)\right ) \, dx\\ &=\frac {1}{4} x^2 \left (9-e^4-\log (4)\right )+\frac {1}{8} \int \left (\left (72-8 e^4\right ) x-8 x \log (4)\right ) \log (x) \, dx+\frac {1}{8} \int \left (-4 e^{e^{x/4}} x-e^{e^{x/4}} x \left (8+e^{x/4} x\right ) \log (x)\right ) \, dx\\ &=\frac {1}{4} x^2 \left (9-e^4-\log (4)\right )-\frac {1}{8} \int e^{e^{x/4}} x \left (8+e^{x/4} x\right ) \log (x) \, dx+\frac {1}{8} \int x \left (72-8 e^4-8 \log (4)\right ) \log (x) \, dx-\frac {1}{2} \int e^{e^{x/4}} x \, dx\\ &=\frac {1}{4} x^2 \left (9-e^4-\log (4)\right )-\frac {1}{2} e^{e^{x/4}} x^2 \log (x)+\frac {1}{8} \int 4 e^{e^{x/4}} x \, dx-\frac {1}{2} \int e^{e^{x/4}} x \, dx+\left (9-e^4-\log (4)\right ) \int x \log (x) \, dx\\ &=-\frac {1}{2} e^{e^{x/4}} x^2 \log (x)+\frac {1}{2} x^2 \left (9-e^4-\log (4)\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 25, normalized size = 0.81 \begin {gather*} -\frac {1}{2} x^2 \left (-9+e^4+e^{e^{x/4}}+\log (4)\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(36*x - 4*E^4*x - 4*x*Log[4] + (72*x - 8*E^4*x - 8*x*Log[4])*Log[x] + E^E^(x/4)*(-4*x + (-8*x - E^(x
/4)*x^2)*Log[x]))/8,x]

[Out]

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

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fricas [A]  time = 0.61, size = 36, normalized size = 1.16 \begin {gather*} -\frac {1}{2} \, x^{2} e^{\left (e^{\left (\frac {1}{4} \, x\right )}\right )} \log \relax (x) - \frac {1}{2} \, {\left (x^{2} e^{4} + 2 \, x^{2} \log \relax (2) - 9 \, x^{2}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((-x^2*exp(1/4*x)-8*x)*log(x)-4*x)*exp(exp(1/4*x))+1/8*(-16*x*log(2)-8*x*exp(4)+72*x)*log(x)-x*l
og(2)-1/2*x*exp(4)+9/2*x,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 38, normalized size = 1.23 \begin {gather*} -\frac {1}{2} \, x^{2} e^{4} \log \relax (x) - \frac {1}{2} \, x^{2} e^{\left (e^{\left (\frac {1}{4} \, x\right )}\right )} \log \relax (x) - x^{2} \log \relax (2) \log \relax (x) + \frac {9}{2} \, x^{2} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((-x^2*exp(1/4*x)-8*x)*log(x)-4*x)*exp(exp(1/4*x))+1/8*(-16*x*log(2)-8*x*exp(4)+72*x)*log(x)-x*l
og(2)-1/2*x*exp(4)+9/2*x,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 39, normalized size = 1.26

method	result	size
risch	\(-\frac {\ln \relax (x ) x^{2} {\mathrm e}^{{\mathrm e}^{\frac {x}{4}}}}{2}-x^{2} \ln \relax (2) \ln \relax (x )-\frac {x^{2} {\mathrm e}^{4} \ln \relax (x )}{2}+\frac {9 x^{2} \ln \relax (x )}{2}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*((-x^2*exp(1/4*x)-8*x)*ln(x)-4*x)*exp(exp(1/4*x))+1/8*(-16*x*ln(2)-8*x*exp(4)+72*x)*ln(x)-x*ln(2)-1/2*
x*exp(4)+9/2*x,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.38, size = 68, normalized size = 2.19 \begin {gather*} -\frac {1}{2} \, x^{2} e^{\left (e^{\left (\frac {1}{4} \, x\right )}\right )} \log \relax (x) + \frac {1}{4} \, x^{2} {\left (e^{4} + 2 \, \log \relax (2) - 9\right )} - \frac {1}{4} \, x^{2} e^{4} - \frac {1}{2} \, x^{2} \log \relax (2) + \frac {9}{4} \, x^{2} - \frac {1}{2} \, {\left (x^{2} e^{4} + 2 \, x^{2} \log \relax (2) - 9 \, x^{2}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((-x^2*exp(1/4*x)-8*x)*log(x)-4*x)*exp(exp(1/4*x))+1/8*(-16*x*log(2)-8*x*exp(4)+72*x)*log(x)-x*l
og(2)-1/2*x*exp(4)+9/2*x,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.52, size = 22, normalized size = 0.71 \begin {gather*} -\frac {x^2\,\ln \relax (x)\,\left (2\,{\mathrm {e}}^4+\ln \left (16\right )+2\,{\mathrm {e}}^{{\mathrm {e}}^{x/4}}-18\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*x)/2 - (exp(exp(x/4))*(4*x + log(x)*(8*x + x^2*exp(x/4))))/8 - (x*exp(4))/2 - x*log(2) - (log(x)*(8*x*e
xp(4) - 72*x + 16*x*log(2)))/8,x)

[Out]

-(x^2*log(x)*(2*exp(4) + log(16) + 2*exp(exp(x/4)) - 18))/4

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sympy [A]  time = 6.08, size = 39, normalized size = 1.26 \begin {gather*} - \frac {x^{2} e^{e^{\frac {x}{4}}} \log {\relax (x )}}{2} + \left (- \frac {x^{2} e^{4}}{2} - x^{2} \log {\relax (2 )} + \frac {9 x^{2}}{2}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((-x**2*exp(1/4*x)-8*x)*ln(x)-4*x)*exp(exp(1/4*x))+1/8*(-16*x*ln(2)-8*x*exp(4)+72*x)*ln(x)-x*ln(
2)-1/2*x*exp(4)+9/2*x,x)

[Out]

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

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