3.75.20 \(\int (64 x \log (x)+64 x \log ^2(x)) \log (e^{16 x^2 \log ^2(x)} (25-5 \log (4))) \, dx\)

Optimal. Leaf size=21 \[ \log ^2\left (e^{16 x^2 \log ^2(x)} (25-5 \log (4))\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 22, normalized size of antiderivative = 1.05, number of steps used = 1, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2304, 2305, 6686} \begin {gather*} \log ^2\left (5 (5-\log (4)) e^{16 x^2 \log ^2(x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(64*x*Log[x] + 64*x*Log[x]^2)*Log[E^(16*x^2*Log[x]^2)*(25 - 5*Log[4])],x]

[Out]

Log[5*E^(16*x^2*Log[x]^2)*(5 - Log[4])]^2

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 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log ^2\left (5 e^{16 x^2 \log ^2(x)} (5-\log (4))\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 20, normalized size = 0.95 \begin {gather*} \log ^2\left (-5 e^{16 x^2 \log ^2(x)} (-5+\log (4))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(64*x*Log[x] + 64*x*Log[x]^2)*Log[E^(16*x^2*Log[x]^2)*(25 - 5*Log[4])],x]

[Out]

Log[-5*E^(16*x^2*Log[x]^2)*(-5 + Log[4])]^2

________________________________________________________________________________________

fricas [A]  time = 1.17, size = 26, normalized size = 1.24 \begin {gather*} 256 \, x^{4} \log \relax (x)^{4} + 32 \, x^{2} \log \relax (x)^{2} \log \left (-10 \, \log \relax (2) + 25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((64*x*log(x)^2+64*x*log(x))*log((-10*log(2)+25)*exp(16*x^2*log(x)^2)),x, algorithm="fricas")

[Out]

256*x^4*log(x)^4 + 32*x^2*log(x)^2*log(-10*log(2) + 25)

________________________________________________________________________________________

giac [A]  time = 0.28, size = 26, normalized size = 1.24 \begin {gather*} 256 \, x^{4} \log \relax (x)^{4} + 32 \, x^{2} \log \relax (x)^{2} \log \left (-10 \, \log \relax (2) + 25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((64*x*log(x)^2+64*x*log(x))*log((-10*log(2)+25)*exp(16*x^2*log(x)^2)),x, algorithm="giac")

[Out]

256*x^4*log(x)^4 + 32*x^2*log(x)^2*log(-10*log(2) + 25)

________________________________________________________________________________________

maple [B]  time = 0.19, size = 48, normalized size = 2.29

method	result	size
default	\(256 x^{4} \ln \relax (x )^{4}+32 \left (\ln \left (\left (-10 \ln \relax (2)+25\right ) {\mathrm e}^{16 x^{2} \ln \relax (x )^{2}}\right )-16 x^{2} \ln \relax (x )^{2}\right ) \ln \relax (x )^{2} x^{2}\)	\(48\)
risch	\(32 x^{2} \ln \relax (x )^{2} \ln \left ({\mathrm e}^{16 x^{2} \ln \relax (x )^{2}}\right )-256 x^{4} \ln \relax (x )^{4}+32 \ln \relax (5) x^{2} \ln \relax (x )^{2}+32 \ln \left (-2 \ln \relax (2)+5\right ) x^{2} \ln \relax (x )^{2}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((64*x*ln(x)^2+64*x*ln(x))*ln((-10*ln(2)+25)*exp(16*x^2*ln(x)^2)),x,method=_RETURNVERBOSE)

[Out]

256*x^4*ln(x)^4+32*(ln((-10*ln(2)+25)*exp(16*x^2*ln(x)^2))-16*x^2*ln(x)^2)*ln(x)^2*x^2

________________________________________________________________________________________

maxima [B]  time = 0.41, size = 60, normalized size = 2.86 \begin {gather*} -256 \, x^{4} \log \relax (x)^{4} + 16 \, {\left ({\left (2 \, \log \relax (x)^{2} - 2 \, \log \relax (x) + 1\right )} x^{2} + 2 \, x^{2} \log \relax (x) - x^{2}\right )} \log \left (-5 \, {\left (2 \, \log \relax (2) - 5\right )} e^{\left (16 \, x^{2} \log \relax (x)^{2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((64*x*log(x)^2+64*x*log(x))*log((-10*log(2)+25)*exp(16*x^2*log(x)^2)),x, algorithm="maxima")

[Out]

-256*x^4*log(x)^4 + 16*((2*log(x)^2 - 2*log(x) + 1)*x^2 + 2*x^2*log(x) - x^2)*log(-5*(2*log(2) - 5)*e^(16*x^2*
log(x)^2))

________________________________________________________________________________________

mupad [B]  time = 4.68, size = 37, normalized size = 1.76 \begin {gather*} 32\,x^2\,{\ln \relax (x)}^2\,\left (\ln \left ({\mathrm {e}}^{16\,x^2\,{\ln \relax (x)}^2}\right )+\ln \left (25-10\,\ln \relax (2)\right )-8\,x^2\,{\ln \relax (x)}^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(-exp(16*x^2*log(x)^2)*(10*log(2) - 25))*(64*x*log(x)^2 + 64*x*log(x)),x)

[Out]

32*x^2*log(x)^2*(log(exp(16*x^2*log(x)^2)) + log(25 - 10*log(2)) - 8*x^2*log(x)^2)

________________________________________________________________________________________

sympy [A]  time = 0.39, size = 36, normalized size = 1.71 \begin {gather*} 256 x^{4} \log {\relax (x )}^{4} + \left (32 x^{2} \log {\left (5 - 2 \log {\relax (2 )} \right )} + 32 x^{2} \log {\relax (5 )}\right ) \log {\relax (x )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((64*x*ln(x)**2+64*x*ln(x))*ln((-10*ln(2)+25)*exp(16*x**2*ln(x)**2)),x)

[Out]

256*x**4*log(x)**4 + (32*x**2*log(5 - 2*log(2)) + 32*x**2*log(5))*log(x)**2

________________________________________________________________________________________