\(\int \frac {(-4 x^2+x \log ^2(2)) \log (x)+e^{5+\log ^2(\frac {3}{\log (x)})} (16 x \log (x)+(-32 x+8 \log ^2(2)) \log (\frac {1}{4} (4 x-\log ^2(2))) \log (\frac {3}{\log (x)}))}{(-4 x^2+x \log ^2(2)) \log (x)} \, dx\) [6428]

Optimal result

Integrand size = 88, antiderivative size = 28 \[ \int \frac {\left (-4 x^2+x \log ^2(2)\right ) \log (x)+e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \left (16 x \log (x)+\left (-32 x+8 \log ^2(2)\right ) \log \left (\frac {1}{4} \left (4 x-\log ^2(2)\right )\right ) \log \left (\frac {3}{\log (x)}\right )\right )}{\left (-4 x^2+x \log ^2(2)\right ) \log (x)} \, dx=x-4 e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \log \left (x-\frac {\log ^2(2)}{4}\right ) \]

[Out]

x-4*ln(-1/4*ln(2)^2+x)*exp(ln(3/ln(x))^2+5)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(84\) vs. \(2(28)=56\).

Time = 1.38 (sec) , antiderivative size = 84, normalized size of antiderivative = 3.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {1607, 6874, 2326} \[ \int \frac {\left (-4 x^2+x \log ^2(2)\right ) \log (x)+e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \left (16 x \log (x)+\left (-32 x+8 \log ^2(2)\right ) \log \left (\frac {1}{4} \left (4 x-\log ^2(2)\right )\right ) \log \left (\frac {3}{\log (x)}\right )\right )}{\left (-4 x^2+x \log ^2(2)\right ) \log (x)} \, dx=x-\frac {4 e^{\log ^2\left (\frac {3}{\log (x)}\right )+5} \left (4 x \log \left (x-\frac {\log ^2(2)}{4}\right ) \log \left (\frac {3}{\log (x)}\right )-\log ^2(2) \log \left (x-\frac {\log ^2(2)}{4}\right ) \log \left (\frac {3}{\log (x)}\right )\right )}{\left (4 x-\log ^2(2)\right ) \log \left (\frac {3}{\log (x)}\right )} \]

[In]

Int[((-4*x^2 + x*Log[2]^2)*Log[x] + E^(5 + Log[3/Log[x]]^2)*(16*x*Log[x] + (-32*x + 8*Log[2]^2)*Log[(4*x - Log
[2]^2)/4]*Log[3/Log[x]]))/((-4*x^2 + x*Log[2]^2)*Log[x]),x]

[Out]

x - (4*E^(5 + Log[3/Log[x]]^2)*(4*x*Log[x - Log[2]^2/4]*Log[3/Log[x]] - Log[2]^2*Log[x - Log[2]^2/4]*Log[3/Log
[x]]))/((4*x - Log[2]^2)*Log[3/Log[x]])

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2326

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 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (-4 x^2+x \log ^2(2)\right ) \log (x)+e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \left (16 x \log (x)+\left (-32 x+8 \log ^2(2)\right ) \log \left (\frac {1}{4} \left (4 x-\log ^2(2)\right )\right ) \log \left (\frac {3}{\log (x)}\right )\right )}{x \left (-4 x+\log ^2(2)\right ) \log (x)} \, dx \\ & = \int \left (1-\frac {8 e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \left (2 x \log (x)-4 x \log \left (x-\frac {\log ^2(2)}{4}\right ) \log \left (\frac {3}{\log (x)}\right )+\log ^2(2) \log \left (x-\frac {\log ^2(2)}{4}\right ) \log \left (\frac {3}{\log (x)}\right )\right )}{x \left (4 x-\log ^2(2)\right ) \log (x)}\right ) \, dx \\ & = x-8 \int \frac {e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \left (2 x \log (x)-4 x \log \left (x-\frac {\log ^2(2)}{4}\right ) \log \left (\frac {3}{\log (x)}\right )+\log ^2(2) \log \left (x-\frac {\log ^2(2)}{4}\right ) \log \left (\frac {3}{\log (x)}\right )\right )}{x \left (4 x-\log ^2(2)\right ) \log (x)} \, dx \\ & = x-\frac {4 e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \left (4 x \log \left (x-\frac {\log ^2(2)}{4}\right ) \log \left (\frac {3}{\log (x)}\right )-\log ^2(2) \log \left (x-\frac {\log ^2(2)}{4}\right ) \log \left (\frac {3}{\log (x)}\right )\right )}{\left (4 x-\log ^2(2)\right ) \log \left (\frac {3}{\log (x)}\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-4 x^2+x \log ^2(2)\right ) \log (x)+e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \left (16 x \log (x)+\left (-32 x+8 \log ^2(2)\right ) \log \left (\frac {1}{4} \left (4 x-\log ^2(2)\right )\right ) \log \left (\frac {3}{\log (x)}\right )\right )}{\left (-4 x^2+x \log ^2(2)\right ) \log (x)} \, dx=x-4 e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \log \left (x-\frac {\log ^2(2)}{4}\right ) \]

[In]

Integrate[((-4*x^2 + x*Log[2]^2)*Log[x] + E^(5 + Log[3/Log[x]]^2)*(16*x*Log[x] + (-32*x + 8*Log[2]^2)*Log[(4*x
- Log[2]^2)/4]*Log[3/Log[x]]))/((-4*x^2 + x*Log[2]^2)*Log[x]),x]

[Out]

x - 4*E^(5 + Log[3/Log[x]]^2)*Log[x - Log[2]^2/4]

Maple [A] (verified)

Time = 34.67 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14

method	result	size
parallelrisch	\(\frac {\ln \left (2\right )^{2}}{2}-4 \ln \left (-\frac {\ln \left (2\right )^{2}}{4}+x \right ) {\mathrm e}^{\ln \left (\frac {3}{\ln \left (x \right )}\right )^{2}+5}+x\)	\(32\)
risch	\(-4 \ln \left (-\frac {\ln \left (2\right )^{2}}{4}+x \right ) \ln \left (x \right )^{-2 \ln \left (3\right )} {\mathrm e}^{\ln \left (3\right )^{2}+5+\ln \left (\ln \left (x \right )\right )^{2}}+x\)	\(33\)

[In]

int((((8*ln(2)^2-32*x)*ln(-1/4*ln(2)^2+x)*ln(3/ln(x))+16*x*ln(x))*exp(ln(3/ln(x))^2+5)+(x*ln(2)^2-4*x^2)*ln(x)
)/(x*ln(2)^2-4*x^2)/ln(x),x,method=_RETURNVERBOSE)

[Out]

1/2*ln(2)^2-4*ln(-1/4*ln(2)^2+x)*exp(ln(3/ln(x))^2+5)+x

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {\left (-4 x^2+x \log ^2(2)\right ) \log (x)+e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \left (16 x \log (x)+\left (-32 x+8 \log ^2(2)\right ) \log \left (\frac {1}{4} \left (4 x-\log ^2(2)\right )\right ) \log \left (\frac {3}{\log (x)}\right )\right )}{\left (-4 x^2+x \log ^2(2)\right ) \log (x)} \, dx=-4 \, e^{\left (\log \left (\frac {3}{\log \left (x\right )}\right )^{2} + 5\right )} \log \left (-\frac {1}{4} \, \log \left (2\right )^{2} + x\right ) + x \]

[In]

integrate((((8*log(2)^2-32*x)*log(-1/4*log(2)^2+x)*log(3/log(x))+16*x*log(x))*exp(log(3/log(x))^2+5)+(x*log(2)
^2-4*x^2)*log(x))/(x*log(2)^2-4*x^2)/log(x),x, algorithm="fricas")

[Out]

-4*e^(log(3/log(x))^2 + 5)*log(-1/4*log(2)^2 + x) + x

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (-4 x^2+x \log ^2(2)\right ) \log (x)+e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \left (16 x \log (x)+\left (-32 x+8 \log ^2(2)\right ) \log \left (\frac {1}{4} \left (4 x-\log ^2(2)\right )\right ) \log \left (\frac {3}{\log (x)}\right )\right )}{\left (-4 x^2+x \log ^2(2)\right ) \log (x)} \, dx=\text {Timed out} \]

[In]

integrate((((8*ln(2)**2-32*x)*ln(-1/4*ln(2)**2+x)*ln(3/ln(x))+16*x*ln(x))*exp(ln(3/ln(x))**2+5)+(x*ln(2)**2-4*
x**2)*ln(x))/(x*ln(2)**2-4*x**2)/ln(x),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (25) = 50\).

Time = 0.39 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.04 \[ \int \frac {\left (-4 x^2+x \log ^2(2)\right ) \log (x)+e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \left (16 x \log (x)+\left (-32 x+8 \log ^2(2)\right ) \log \left (\frac {1}{4} \left (4 x-\log ^2(2)\right )\right ) \log \left (\frac {3}{\log (x)}\right )\right )}{\left (-4 x^2+x \log ^2(2)\right ) \log (x)} \, dx=8 \, e^{\left (\log \left (3\right )^{2} - 2 \, \log \left (3\right ) \log \left (\log \left (x\right )\right ) + \log \left (\log \left (x\right )\right )^{2} + 5\right )} \log \left (2\right ) - 4 \, e^{\left (\log \left (3\right )^{2} - 2 \, \log \left (3\right ) \log \left (\log \left (x\right )\right ) + \log \left (\log \left (x\right )\right )^{2} + 5\right )} \log \left (-\log \left (2\right )^{2} + 4 \, x\right ) + x \]

[In]

integrate((((8*log(2)^2-32*x)*log(-1/4*log(2)^2+x)*log(3/log(x))+16*x*log(x))*exp(log(3/log(x))^2+5)+(x*log(2)
^2-4*x^2)*log(x))/(x*log(2)^2-4*x^2)/log(x),x, algorithm="maxima")

[Out]

8*e^(log(3)^2 - 2*log(3)*log(log(x)) + log(log(x))^2 + 5)*log(2) - 4*e^(log(3)^2 - 2*log(3)*log(log(x)) + log(
log(x))^2 + 5)*log(-log(2)^2 + 4*x) + x

Giac [F]

\[ \int \frac {\left (-4 x^2+x \log ^2(2)\right ) \log (x)+e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \left (16 x \log (x)+\left (-32 x+8 \log ^2(2)\right ) \log \left (\frac {1}{4} \left (4 x-\log ^2(2)\right )\right ) \log \left (\frac {3}{\log (x)}\right )\right )}{\left (-4 x^2+x \log ^2(2)\right ) \log (x)} \, dx=\int { \frac {8 \, {\left ({\left (\log \left (2\right )^{2} - 4 \, x\right )} \log \left (-\frac {1}{4} \, \log \left (2\right )^{2} + x\right ) \log \left (\frac {3}{\log \left (x\right )}\right ) + 2 \, x \log \left (x\right )\right )} e^{\left (\log \left (\frac {3}{\log \left (x\right )}\right )^{2} + 5\right )} + {\left (x \log \left (2\right )^{2} - 4 \, x^{2}\right )} \log \left (x\right )}{{\left (x \log \left (2\right )^{2} - 4 \, x^{2}\right )} \log \left (x\right )} \,d x } \]

[In]

integrate((((8*log(2)^2-32*x)*log(-1/4*log(2)^2+x)*log(3/log(x))+16*x*log(x))*exp(log(3/log(x))^2+5)+(x*log(2)
^2-4*x^2)*log(x))/(x*log(2)^2-4*x^2)/log(x),x, algorithm="giac")

[Out]

undef

Mupad [B] (verification not implemented)

Time = 13.65 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {\left (-4 x^2+x \log ^2(2)\right ) \log (x)+e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \left (16 x \log (x)+\left (-32 x+8 \log ^2(2)\right ) \log \left (\frac {1}{4} \left (4 x-\log ^2(2)\right )\right ) \log \left (\frac {3}{\log (x)}\right )\right )}{\left (-4 x^2+x \log ^2(2)\right ) \log (x)} \, dx=x-4\,{\mathrm {e}}^{{\ln \left (3\right )}^2}\,{\mathrm {e}}^5\,{\mathrm {e}}^{{\ln \left (\frac {1}{\ln \left (x\right )}\right )}^2}\,\ln \left (x-\frac {{\ln \left (2\right )}^2}{4}\right )\,{\left (\frac {1}{\ln \left (x\right )}\right )}^{2\,\ln \left (3\right )} \]

[In]

int((log(x)*(x*log(2)^2 - 4*x^2) + exp(log(3/log(x))^2 + 5)*(16*x*log(x) - log(3/log(x))*log(x - log(2)^2/4)*(
32*x - 8*log(2)^2)))/(log(x)*(x*log(2)^2 - 4*x^2)),x)

[Out]

x - 4*exp(log(3)^2)*exp(5)*exp(log(1/log(x))^2)*log(x - log(2)^2/4)*(1/log(x))^(2*log(3))