3.79.3 \(\int \frac {-4 \log (e^{-x/4} x)+(e^5 (4 x^2-x^3)-4 e^5 x^2 \log (e^{-x/4} x)) \log (\frac {3 x}{\log (e^{-x/4} x)})}{2 e^5 x^3 \log (e^{-x/4} x)} \, dx\)

Optimal. Leaf size=31 \[ e^5+\frac {1}{e^5 x^2}-\log ^2\left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right ) \]

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Rubi [A]  time = 0.40, antiderivative size = 28, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 6, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 6688, 14, 6742, 6684, 6686} \begin {gather*} \frac {1}{e^5 x^2}-\log ^2\left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(E^5*x^2) - Log[(3*x)/Log[x/E^(x/4)]]^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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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} &=\frac {\int \frac {-4 \log \left (e^{-x/4} x\right )+\left (e^5 \left (4 x^2-x^3\right )-4 e^5 x^2 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{x^3 \log \left (e^{-x/4} x\right )} \, dx}{2 e^5}\\ &=\frac {\int \frac {-4-\frac {e^5 x^2 \left (-4+x+4 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{\log \left (e^{-x/4} x\right )}}{x^3} \, dx}{2 e^5}\\ &=\frac {\int \left (-\frac {4}{x^3}-\frac {e^5 \left (-4+x+4 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{x \log \left (e^{-x/4} x\right )}\right ) \, dx}{2 e^5}\\ &=\frac {1}{e^5 x^2}-\frac {1}{2} \int \frac {\left (-4+x+4 \log \left (e^{-x/4} x\right )\right ) \log \left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )}{x \log \left (e^{-x/4} x\right )} \, dx\\ &=\frac {1}{e^5 x^2}-\log ^2\left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 28, normalized size = 0.90 \begin {gather*} \frac {1}{e^5 x^2}-\log ^2\left (\frac {3 x}{\log \left (e^{-x/4} x\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.66, size = 30, normalized size = 0.97 \begin {gather*} -\frac {{\left (x^{2} e^{5} \log \left (\frac {3 \, x}{\log \left (x e^{\left (-\frac {1}{4} \, x\right )}\right )}\right )^{2} - 1\right )} e^{\left (-5\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.22, size = 456, normalized size = 14.71 \begin {gather*} -\frac {{\left (2 \, \pi ^{2} x^{2} e^{5} \mathrm {sgn}\left (-2 \, \pi + 2 \, \pi \mathrm {sgn}\relax (x)\right ) \mathrm {sgn}\left (x - 4 \, \log \left ({\left | x \right |}\right )\right ) \mathrm {sgn}\relax (x) - 10 \, \pi ^{2} x^{2} e^{5} \mathrm {sgn}\left (-2 \, \pi + 2 \, \pi \mathrm {sgn}\relax (x)\right ) \mathrm {sgn}\left (x - 4 \, \log \left ({\left | x \right |}\right )\right ) + 4 \, \pi x^{2} \arctan \left (-\frac {2 \, {\left (\pi - \pi \mathrm {sgn}\relax (x)\right )}}{x - 4 \, \log \left ({\left | x \right |}\right )}\right ) e^{5} \mathrm {sgn}\left (-2 \, \pi + 2 \, \pi \mathrm {sgn}\relax (x)\right ) \mathrm {sgn}\left (x - 4 \, \log \left ({\left | x \right |}\right )\right ) - 2 \, \pi ^{2} x^{2} e^{5} \mathrm {sgn}\left (-2 \, \pi + 2 \, \pi \mathrm {sgn}\relax (x)\right ) \mathrm {sgn}\relax (x) - 6 \, \pi ^{2} x^{2} e^{5} \mathrm {sgn}\left (-2 \, \pi + 2 \, \pi \mathrm {sgn}\relax (x)\right ) - 4 \, \pi x^{2} \arctan \left (-\frac {2 \, {\left (\pi - \pi \mathrm {sgn}\relax (x)\right )}}{x - 4 \, \log \left ({\left | x \right |}\right )}\right ) e^{5} \mathrm {sgn}\left (-2 \, \pi + 2 \, \pi \mathrm {sgn}\relax (x)\right ) + 2 \, \pi ^{2} x^{2} e^{5} \mathrm {sgn}\left (x - 4 \, \log \left ({\left | x \right |}\right )\right ) + 10 \, \pi ^{2} x^{2} e^{5} \mathrm {sgn}\relax (x) - 4 \, \pi x^{2} \arctan \left (-\frac {2 \, {\left (\pi - \pi \mathrm {sgn}\relax (x)\right )}}{x - 4 \, \log \left ({\left | x \right |}\right )}\right ) e^{5} \mathrm {sgn}\relax (x) - 12 \, \pi ^{2} x^{2} e^{5} + 20 \, \pi x^{2} \arctan \left (-\frac {2 \, {\left (\pi - \pi \mathrm {sgn}\relax (x)\right )}}{x - 4 \, \log \left ({\left | x \right |}\right )}\right ) e^{5} - 4 \, x^{2} \arctan \left (-\frac {2 \, {\left (\pi - \pi \mathrm {sgn}\relax (x)\right )}}{x - 4 \, \log \left ({\left | x \right |}\right )}\right )^{2} e^{5} - 4 \, x^{2} e^{5} \log \left (12\right ) \log \left (-8 \, \pi ^{2} \mathrm {sgn}\relax (x) + 8 \, \pi ^{2} + x^{2} - 8 \, x \log \left ({\left | x \right |}\right ) + 16 \, \log \left ({\left | x \right |}\right )^{2}\right ) + x^{2} e^{5} \log \left (-8 \, \pi ^{2} \mathrm {sgn}\relax (x) + 8 \, \pi ^{2} + x^{2} - 8 \, x \log \left ({\left | x \right |}\right ) + 16 \, \log \left ({\left | x \right |}\right )^{2}\right )^{2} + 8 \, x^{2} e^{5} \log \left (12\right ) \log \left ({\left | x \right |}\right ) - 4 \, x^{2} e^{5} \log \left (-8 \, \pi ^{2} \mathrm {sgn}\relax (x) + 8 \, \pi ^{2} + x^{2} - 8 \, x \log \left ({\left | x \right |}\right ) + 16 \, \log \left ({\left | x \right |}\right )^{2}\right ) \log \left ({\left | x \right |}\right ) + 4 \, x^{2} e^{5} \log \left ({\left | x \right |}\right )^{2} - 4\right )} e^{\left (-5\right )}}{4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*pi^2*x^2*e^5*sgn(-2*pi + 2*pi*sgn(x))*sgn(x - 4*log(abs(x)))*sgn(x) - 10*pi^2*x^2*e^5*sgn(-2*pi + 2*pi
*sgn(x))*sgn(x - 4*log(abs(x))) + 4*pi*x^2*arctan(-2*(pi - pi*sgn(x))/(x - 4*log(abs(x))))*e^5*sgn(-2*pi + 2*p
i*sgn(x))*sgn(x - 4*log(abs(x))) - 2*pi^2*x^2*e^5*sgn(-2*pi + 2*pi*sgn(x))*sgn(x) - 6*pi^2*x^2*e^5*sgn(-2*pi +
2*pi*sgn(x)) - 4*pi*x^2*arctan(-2*(pi - pi*sgn(x))/(x - 4*log(abs(x))))*e^5*sgn(-2*pi + 2*pi*sgn(x)) + 2*pi^2
*x^2*e^5*sgn(x - 4*log(abs(x))) + 10*pi^2*x^2*e^5*sgn(x) - 4*pi*x^2*arctan(-2*(pi - pi*sgn(x))/(x - 4*log(abs(
x))))*e^5*sgn(x) - 12*pi^2*x^2*e^5 + 20*pi*x^2*arctan(-2*(pi - pi*sgn(x))/(x - 4*log(abs(x))))*e^5 - 4*x^2*arc
tan(-2*(pi - pi*sgn(x))/(x - 4*log(abs(x))))^2*e^5 - 4*x^2*e^5*log(12)*log(-8*pi^2*sgn(x) + 8*pi^2 + x^2 - 8*x
*log(abs(x)) + 16*log(abs(x))^2) + x^2*e^5*log(-8*pi^2*sgn(x) + 8*pi^2 + x^2 - 8*x*log(abs(x)) + 16*log(abs(x)
)^2)^2 + 8*x^2*e^5*log(12)*log(abs(x)) - 4*x^2*e^5*log(-8*pi^2*sgn(x) + 8*pi^2 + x^2 - 8*x*log(abs(x)) + 16*lo
g(abs(x))^2)*log(abs(x)) + 4*x^2*e^5*log(abs(x))^2 - 4)*e^(-5)/x^2

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (-4 x^{2} {\mathrm e}^{5} \ln \left (x \,{\mathrm e}^{-\frac {x}{4}}\right )+\left (-x^{3}+4 x^{2}\right ) {\mathrm e}^{5}\right ) \ln \left (\frac {3 x}{\ln \left (x \,{\mathrm e}^{-\frac {x}{4}}\right )}\right )-4 \ln \left (x \,{\mathrm e}^{-\frac {x}{4}}\right )\right ) {\mathrm e}^{-5}}{2 x^{3} \ln \left (x \,{\mathrm e}^{-\frac {x}{4}}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.54, size = 71, normalized size = 2.29 \begin {gather*} -{\left (2 \, {\left (\log \relax (3) + 2 \, \log \relax (2)\right )} e^{5} \log \relax (x) + e^{5} \log \relax (x)^{2} + e^{5} \log \left (-x + 4 \, \log \relax (x)\right )^{2} - 2 \, {\left ({\left (\log \relax (3) + 2 \, \log \relax (2)\right )} e^{5} + e^{5} \log \relax (x)\right )} \log \left (-x + 4 \, \log \relax (x)\right ) - \frac {1}{x^{2}}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.40, size = 23, normalized size = 0.74 \begin {gather*} \frac {{\mathrm {e}}^{-5}}{x^2}-{\ln \left (-\frac {12\,x}{x-4\,\ln \relax (x)}\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-5)*(2*log(x*exp(-x/4)) - (log((3*x)/log(x*exp(-x/4)))*(exp(5)*(4*x^2 - x^3) - 4*x^2*exp(5)*log(x*ex
p(-x/4))))/2))/(x^3*log(x*exp(-x/4))),x)

[Out]

exp(-5)/x^2 - log(-(12*x)/(x - 4*log(x)))^2

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sympy [A]  time = 0.44, size = 22, normalized size = 0.71 \begin {gather*} - \log {\left (\frac {3 x}{\log {\left (x e^{- \frac {x}{4}} \right )}} \right )}^{2} + \frac {1}{x^{2} e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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