3.59.59 \(\int (96 x^2-144 x^2 \log (x)+(48 x+96 x \log (x)) \log (\log (2))+e^{e^x} (6 x^2+3 e^x x^3+(-9 x^2-3 e^x x^3) \log (x)+(3 x+(6 x+3 e^x x^2) \log (x)) \log (\log (2)))) \, dx\)

Optimal. Leaf size=24 \[ 3 \left (16+e^{e^x}\right ) x^2 (x+\log (x) (-x+\log (\log (2)))) \]

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Rubi [B]  time = 0.81, antiderivative size = 63, normalized size of antiderivative = 2.62, number of steps used = 7, number of rules used = 4, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {2304, 6741, 12, 2288} \begin {gather*} 48 x^3-48 x^3 \log (x)+48 x^2 \log (\log (2)) \log (x)+3 e^{e^x-x} x \left (e^x x^2-e^x x^2 \log (x)+e^x x \log (\log (2)) \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[96*x^2 - 144*x^2*Log[x] + (48*x + 96*x*Log[x])*Log[Log[2]] + E^E^x*(6*x^2 + 3*E^x*x^3 + (-9*x^2 - 3*E^x*x^
3)*Log[x] + (3*x + (6*x + 3*E^x*x^2)*Log[x])*Log[Log[2]]),x]

[Out]

48*x^3 - 48*x^3*Log[x] + 48*x^2*Log[x]*Log[Log[2]] + 3*E^(E^x - x)*x*(E^x*x^2 - E^x*x^2*Log[x] + E^x*x*Log[x]*
Log[Log[2]])

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 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=32 x^3-144 \int x^2 \log (x) \, dx+\log (\log (2)) \int (48 x+96 x \log (x)) \, dx+\int e^{e^x} \left (6 x^2+3 e^x x^3+\left (-9 x^2-3 e^x x^3\right ) \log (x)+\left (3 x+\left (6 x+3 e^x x^2\right ) \log (x)\right ) \log (\log (2))\right ) \, dx\\ &=48 x^3-48 x^3 \log (x)+24 x^2 \log (\log (2))+(96 \log (\log (2))) \int x \log (x) \, dx+\int 3 e^{e^x} x \left (2 x+e^x x^2-3 x \log (x)-e^x x^2 \log (x)+\log (\log (2))+2 \log (x) \log (\log (2))+e^x x \log (x) \log (\log (2))\right ) \, dx\\ &=48 x^3-48 x^3 \log (x)+48 x^2 \log (x) \log (\log (2))+3 \int e^{e^x} x \left (2 x+e^x x^2-3 x \log (x)-e^x x^2 \log (x)+\log (\log (2))+2 \log (x) \log (\log (2))+e^x x \log (x) \log (\log (2))\right ) \, dx\\ &=48 x^3-48 x^3 \log (x)+48 x^2 \log (x) \log (\log (2))+3 e^{e^x-x} x \left (e^x x^2-e^x x^2 \log (x)+e^x x \log (x) \log (\log (2))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.63, size = 26, normalized size = 1.08 \begin {gather*} -3 \left (16+e^{e^x}\right ) x^2 (-x+\log (x) (x-\log (\log (2)))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[96*x^2 - 144*x^2*Log[x] + (48*x + 96*x*Log[x])*Log[Log[2]] + E^E^x*(6*x^2 + 3*E^x*x^3 + (-9*x^2 - 3*
E^x*x^3)*Log[x] + (3*x + (6*x + 3*E^x*x^2)*Log[x])*Log[Log[2]]),x]

[Out]

-3*(16 + E^E^x)*x^2*(-x + Log[x]*(x - Log[Log[2]]))

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fricas [B]  time = 1.88, size = 50, normalized size = 2.08 \begin {gather*} -48 \, x^{3} \log \relax (x) + 48 \, x^{2} \log \relax (x) \log \left (\log \relax (2)\right ) + 48 \, x^{3} - 3 \, {\left (x^{3} \log \relax (x) - x^{2} \log \relax (x) \log \left (\log \relax (2)\right ) - x^{3}\right )} e^{\left (e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*exp(x)*x^2+6*x)*log(x)+3*x)*log(log(2))+(-3*exp(x)*x^3-9*x^2)*log(x)+3*exp(x)*x^3+6*x^2)*exp(ex
p(x))+(96*x*log(x)+48*x)*log(log(2))-144*x^2*log(x)+96*x^2,x, algorithm="fricas")

[Out]

-48*x^3*log(x) + 48*x^2*log(x)*log(log(2)) + 48*x^3 - 3*(x^3*log(x) - x^2*log(x)*log(log(2)) - x^3)*e^(e^x)

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giac [B]  time = 0.19, size = 66, normalized size = 2.75 \begin {gather*} -48 \, x^{3} \log \relax (x) + 48 \, x^{2} \log \relax (x) \log \left (\log \relax (2)\right ) + 48 \, x^{3} - 3 \, {\left (x^{3} e^{\left (x + e^{x}\right )} \log \relax (x) - x^{2} e^{\left (x + e^{x}\right )} \log \relax (x) \log \left (\log \relax (2)\right ) - x^{3} e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*exp(x)*x^2+6*x)*log(x)+3*x)*log(log(2))+(-3*exp(x)*x^3-9*x^2)*log(x)+3*exp(x)*x^3+6*x^2)*exp(ex
p(x))+(96*x*log(x)+48*x)*log(log(2))-144*x^2*log(x)+96*x^2,x, algorithm="giac")

[Out]

-48*x^3*log(x) + 48*x^2*log(x)*log(log(2)) + 48*x^3 - 3*(x^3*e^(x + e^x)*log(x) - x^2*e^(x + e^x)*log(x)*log(l
og(2)) - x^3*e^(x + e^x))*e^(-x)

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maple [B]  time = 0.10, size = 51, normalized size = 2.12

method	result	size
risch	\(\left (3 \ln \left (\ln \relax (2)\right ) x^{2} \ln \relax (x )-3 x^{3} \ln \relax (x )+3 x^{3}\right ) {\mathrm e}^{{\mathrm e}^{x}}+48 \ln \left (\ln \relax (2)\right ) x^{2} \ln \relax (x )-48 x^{3} \ln \relax (x )+48 x^{3}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3*exp(x)*x^2+6*x)*ln(x)+3*x)*ln(ln(2))+(-3*exp(x)*x^3-9*x^2)*ln(x)+3*exp(x)*x^3+6*x^2)*exp(exp(x))+(96*
x*ln(x)+48*x)*ln(ln(2))-144*x^2*ln(x)+96*x^2,x,method=_RETURNVERBOSE)

[Out]

(3*ln(ln(2))*x^2*ln(x)-3*x^3*ln(x)+3*x^3)*exp(exp(x))+48*ln(ln(2))*x^2*ln(x)-48*x^3*ln(x)+48*x^3

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maxima [A]  time = 0.44, size = 48, normalized size = 2.00 \begin {gather*} -48 \, x^{3} \log \relax (x) + 48 \, x^{2} \log \relax (x) \log \left (\log \relax (2)\right ) + 48 \, x^{3} + 3 \, {\left (x^{3} - {\left (x^{3} - x^{2} \log \left (\log \relax (2)\right )\right )} \log \relax (x)\right )} e^{\left (e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*exp(x)*x^2+6*x)*log(x)+3*x)*log(log(2))+(-3*exp(x)*x^3-9*x^2)*log(x)+3*exp(x)*x^3+6*x^2)*exp(ex
p(x))+(96*x*log(x)+48*x)*log(log(2))-144*x^2*log(x)+96*x^2,x, algorithm="maxima")

[Out]

-48*x^3*log(x) + 48*x^2*log(x)*log(log(2)) + 48*x^3 + 3*(x^3 - (x^3 - x^2*log(log(2)))*log(x))*e^(e^x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \ln \left (\ln \relax (2)\right )\,\left (48\,x+96\,x\,\ln \relax (x)\right )-144\,x^2\,\ln \relax (x)+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (3\,x^3\,{\mathrm {e}}^x+6\,x^2-\ln \relax (x)\,\left (3\,x^3\,{\mathrm {e}}^x+9\,x^2\right )+\ln \left (\ln \relax (2)\right )\,\left (3\,x+\ln \relax (x)\,\left (6\,x+3\,x^2\,{\mathrm {e}}^x\right )\right )\right )+96\,x^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(log(2))*(48*x + 96*x*log(x)) - 144*x^2*log(x) + exp(exp(x))*(3*x^3*exp(x) + 6*x^2 - log(x)*(3*x^3*exp(
x) + 9*x^2) + log(log(2))*(3*x + log(x)*(6*x + 3*x^2*exp(x)))) + 96*x^2,x)

[Out]

int(log(log(2))*(48*x + 96*x*log(x)) - 144*x^2*log(x) + exp(exp(x))*(3*x^3*exp(x) + 6*x^2 - log(x)*(3*x^3*exp(
x) + 9*x^2) + log(log(2))*(3*x + log(x)*(6*x + 3*x^2*exp(x)))) + 96*x^2, x)

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sympy [B]  time = 8.74, size = 54, normalized size = 2.25 \begin {gather*} 48 x^{3} + \left (- 48 x^{3} + 48 x^{2} \log {\left (\log {\relax (2 )} \right )}\right ) \log {\relax (x )} + \left (- 3 x^{3} \log {\relax (x )} + 3 x^{3} + 3 x^{2} \log {\relax (x )} \log {\left (\log {\relax (2 )} \right )}\right ) e^{e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*exp(x)*x**2+6*x)*ln(x)+3*x)*ln(ln(2))+(-3*exp(x)*x**3-9*x**2)*ln(x)+3*exp(x)*x**3+6*x**2)*exp(e
xp(x))+(96*x*ln(x)+48*x)*ln(ln(2))-144*x**2*ln(x)+96*x**2,x)

[Out]

48*x**3 + (-48*x**3 + 48*x**2*log(log(2)))*log(x) + (-3*x**3*log(x) + 3*x**3 + 3*x**2*log(x)*log(log(2)))*exp(
exp(x))

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