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

Optimal. Leaf size=28 \[ x-4 e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \log \left (x-\frac {\log ^2(2)}{4}\right ) \]

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Rubi [B]  time = 2.28, antiderivative size = 84, normalized size of antiderivative = 3.00, number of steps used = 4, number of rules used = 3, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {1593, 6742, 2288} \begin {gather*} 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 )} \end {gather*}

Antiderivative was successfully verified.

[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 1593

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 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 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 {\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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 28, normalized size = 1.00 \begin {gather*} x-4 e^{5+\log ^2\left (\frac {3}{\log (x)}\right )} \log \left (x-\frac {\log ^2(2)}{4}\right ) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.68, size = 25, normalized size = 0.89 \begin {gather*} -4 \, e^{\left (\log \left (\frac {3}{\log \relax (x)}\right )^{2} + 5\right )} \log \left (-\frac {1}{4} \, \log \relax (2)^{2} + x\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 2.09, size = 33, normalized size = 1.18

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

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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maxima [B]  time = 0.63, size = 57, normalized size = 2.04 \begin {gather*} 8 \, e^{\left (\log \relax (3)^{2} - 2 \, \log \relax (3) \log \left (\log \relax (x)\right ) + \log \left (\log \relax (x)\right )^{2} + 5\right )} \log \relax (2) - 4 \, e^{\left (\log \relax (3)^{2} - 2 \, \log \relax (3) \log \left (\log \relax (x)\right ) + \log \left (\log \relax (x)\right )^{2} + 5\right )} \log \left (-\log \relax (2)^{2} + 4 \, x\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 4.51, size = 37, normalized size = 1.32 \begin {gather*} x-4\,{\mathrm {e}}^{{\ln \relax (3)}^2}\,{\mathrm {e}}^5\,{\mathrm {e}}^{{\ln \left (\frac {1}{\ln \relax (x)}\right )}^2}\,\ln \left (x-\frac {{\ln \relax (2)}^2}{4}\right )\,{\left (\frac {1}{\ln \relax (x)}\right )}^{2\,\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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