\(\int (64 x \log (x)+64 x \log ^2(x)) \log (e^{16 x^2 \log ^2(x)} (25-5 \log (4))) \, dx\) [7420]

Optimal result

Integrand size = 33, antiderivative size = 21 \[ \int \left (64 x \log (x)+64 x \log ^2(x)\right ) \log \left (e^{16 x^2 \log ^2(x)} (25-5 \log (4))\right ) \, dx=\log ^2\left (e^{16 x^2 \log ^2(x)} (25-5 \log (4))\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05, number of steps used = 1, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2341, 2342, 6818} \[ \int \left (64 x \log (x)+64 x \log ^2(x)\right ) \log \left (e^{16 x^2 \log ^2(x)} (25-5 \log (4))\right ) \, dx=\log ^2\left (5 (5-\log (4)) e^{16 x^2 \log ^2(x)}\right ) \]

[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 2341

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 2342

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 6818

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \left (64 x \log (x)+64 x \log ^2(x)\right ) \log \left (e^{16 x^2 \log ^2(x)} (25-5 \log (4))\right ) \, dx=\log ^2\left (-5 e^{16 x^2 \log ^2(x)} (-5+\log (4))\right ) \]

[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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05

method	result	size
parallelrisch	\(\ln \left (-5 \left (2 \ln \left (2\right )-5\right ) {\mathrm e}^{16 x^{2} \ln \left (x \right )^{2}}\right )^{2}\)	\(22\)
risch	\(32 x^{2} \ln \left (x \right )^{2} \ln \left ({\mathrm e}^{16 x^{2} \ln \left (x \right )^{2}}\right )-256 x^{4} \ln \left (x \right )^{4}+32 \ln \left (5\right ) x^{2} \ln \left (x \right )^{2}+32 \ln \left (-2 \ln \left (2\right )+5\right ) x^{2} \ln \left (x \right )^{2}\)	\(58\)
default	\(64 \left (\ln \left (\left (-10 \ln \left (2\right )+25\right ) {\mathrm e}^{16 x^{2} \ln \left (x \right )^{2}}\right )-16 x^{2} \ln \left (x \right )^{2}\right ) \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+256 x^{4} \ln \left (x \right )^{4}+64 \left (\ln \left (\left (-10 \ln \left (2\right )+25\right ) {\mathrm e}^{16 x^{2} \ln \left (x \right )^{2}}\right )-16 x^{2} \ln \left (x \right )^{2}\right ) \left (\frac {x^{2} \ln \left (x \right )^{2}}{2}-\frac {x^{2} \ln \left (x \right )}{2}+\frac {x^{2}}{4}\right )\)	\(106\)
parts	\(64 \left (\ln \left (\left (-10 \ln \left (2\right )+25\right ) {\mathrm e}^{16 x^{2} \ln \left (x \right )^{2}}\right )-16 x^{2} \ln \left (x \right )^{2}\right ) \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+256 x^{4} \ln \left (x \right )^{4}+64 \left (\ln \left (\left (-10 \ln \left (2\right )+25\right ) {\mathrm e}^{16 x^{2} \ln \left (x \right )^{2}}\right )-16 x^{2} \ln \left (x \right )^{2}\right ) \left (\frac {x^{2} \ln \left (x \right )^{2}}{2}-\frac {x^{2} \ln \left (x \right )}{2}+\frac {x^{2}}{4}\right )\)	\(106\)

[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]

ln(-5*(2*ln(2)-5)*exp(16*x^2*ln(x)^2))^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \left (64 x \log (x)+64 x \log ^2(x)\right ) \log \left (e^{16 x^2 \log ^2(x)} (25-5 \log (4))\right ) \, dx=256 \, x^{4} \log \left (x\right )^{4} + 32 \, x^{2} \log \left (x\right )^{2} \log \left (-10 \, \log \left (2\right ) + 25\right ) \]

[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)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \left (64 x \log (x)+64 x \log ^2(x)\right ) \log \left (e^{16 x^2 \log ^2(x)} (25-5 \log (4))\right ) \, dx=256 x^{4} \log {\left (x \right )}^{4} + \left (32 x^{2} \log {\left (5 - 2 \log {\left (2 \right )} \right )} + 32 x^{2} \log {\left (5 \right )}\right ) \log {\left (x \right )}^{2} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (21) = 42\).

Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.86 \[ \int \left (64 x \log (x)+64 x \log ^2(x)\right ) \log \left (e^{16 x^2 \log ^2(x)} (25-5 \log (4))\right ) \, dx=-256 \, x^{4} \log \left (x\right )^{4} + 16 \, {\left ({\left (2 \, \log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 1\right )} x^{2} + 2 \, x^{2} \log \left (x\right ) - x^{2}\right )} \log \left (-5 \, {\left (2 \, \log \left (2\right ) - 5\right )} e^{\left (16 \, x^{2} \log \left (x\right )^{2}\right )}\right ) \]

[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))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (21) = 42\).

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.24 \[ \int \left (64 x \log (x)+64 x \log ^2(x)\right ) \log \left (e^{16 x^2 \log ^2(x)} (25-5 \log (4))\right ) \, dx=-256 \, x^{4} \log \left (x\right )^{4} + 32 \, x^{2} \log \left (-10 \, e^{\left (16 \, x^{2} \log \left (x\right )^{2}\right )} \log \left (2\right ) + 25 \, e^{\left (16 \, x^{2} \log \left (x\right )^{2}\right )}\right ) \log \left (x\right )^{2} \]

[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(-10*e^(16*x^2*log(x)^2)*log(2) + 25*e^(16*x^2*log(x)^2))*log(x)^2

Mupad [B] (verification not implemented)

Time = 11.36 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.76 \[ \int \left (64 x \log (x)+64 x \log ^2(x)\right ) \log \left (e^{16 x^2 \log ^2(x)} (25-5 \log (4))\right ) \, dx=32\,x^2\,{\ln \left (x\right )}^2\,\left (\ln \left ({\mathrm {e}}^{16\,x^2\,{\ln \left (x\right )}^2}\right )+\ln \left (25-10\,\ln \left (2\right )\right )-8\,x^2\,{\ln \left (x\right )}^2\right ) \]

[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)