\(\int \frac {3 x-4 e^{\frac {1}{4} (1+4 \log ^2(e^{\frac {1}{3} (-5-5 \log (2 x))} x))} \log (e^{\frac {1}{3} (-5-5 \log (2 x))} x)}{3 x} \, dx\) [5641]

Optimal result

Integrand size = 59, antiderivative size = 27 \[ \int \frac {3 x-4 e^{\frac {1}{4} \left (1+4 \log ^2\left (e^{\frac {1}{3} (-5-5 \log (2 x))} x\right )\right )} \log \left (e^{\frac {1}{3} (-5-5 \log (2 x))} x\right )}{3 x} \, dx=e^{\frac {1}{4}+\log ^2\left (e^{\frac {1}{3} (-5-5 \log (2 x))} x\right )}+x \]

[Out]

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

Rubi [A] (warning: unable to verify)

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {12, 14, 6838} \[ \int \frac {3 x-4 e^{\frac {1}{4} \left (1+4 \log ^2\left (e^{\frac {1}{3} (-5-5 \log (2 x))} x\right )\right )} \log \left (e^{\frac {1}{3} (-5-5 \log (2 x))} x\right )}{3 x} \, dx=e^{\log ^2\left (\frac {1}{2\ 2^{2/3} e^{5/3} x^{2/3}}\right )+\frac {1}{4}}+x \]

[In]

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

[Out]

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

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 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {3 x-4 e^{\frac {1}{4} \left (1+4 \log ^2\left (e^{\frac {1}{3} (-5-5 \log (2 x))} x\right )\right )} \log \left (e^{\frac {1}{3} (-5-5 \log (2 x))} x\right )}{3 x} \, dx=e^{\frac {1}{4}+\log ^2\left (e^{-\frac {5}{3} (1+\log (2 x))} x\right )}+x \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81

method	result	size
default	\(x +{\mathrm e}^{\ln \left (x \,{\mathrm e}^{-\frac {5 \ln \left (2 x \right )}{3}-\frac {5}{3}}\right )^{2}+\frac {1}{4}}\)	\(22\)
parallelrisch	\(x +{\mathrm e}^{\ln \left (x \,{\mathrm e}^{-\frac {5 \ln \left (2 x \right )}{3}-\frac {5}{3}}\right )^{2}+\frac {1}{4}}\)	\(22\)
parts	\(x +{\mathrm e}^{\ln \left (x \,{\mathrm e}^{-\frac {5 \ln \left (2 x \right )}{3}-\frac {5}{3}}\right )^{2}+\frac {1}{4}}\)	\(22\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {3 x-4 e^{\frac {1}{4} \left (1+4 \log ^2\left (e^{\frac {1}{3} (-5-5 \log (2 x))} x\right )\right )} \log \left (e^{\frac {1}{3} (-5-5 \log (2 x))} x\right )}{3 x} \, dx=x + e^{\left (\log \left (\frac {2^{\frac {1}{3}} e^{\left (-\frac {5}{3}\right )}}{4 \, x^{\frac {2}{3}}}\right )^{2} + \frac {1}{4}\right )} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {3 x-4 e^{\frac {1}{4} \left (1+4 \log ^2\left (e^{\frac {1}{3} (-5-5 \log (2 x))} x\right )\right )} \log \left (e^{\frac {1}{3} (-5-5 \log (2 x))} x\right )}{3 x} \, dx=x + e^{\left (\log \left (\frac {2^{\frac {1}{3}} e^{\left (-\frac {5}{3}\right )}}{4 \, x^{\frac {2}{3}}}\right )^{2} + \frac {1}{4}\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.71 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {3 x-4 e^{\frac {1}{4} \left (1+4 \log ^2\left (e^{\frac {1}{3} (-5-5 \log (2 x))} x\right )\right )} \log \left (e^{\frac {1}{3} (-5-5 \log (2 x))} x\right )}{3 x} \, dx=x + e^{\left (\frac {25}{9} \, \log \left (2\right )^{2} + \frac {20}{9} \, \log \left (2\right ) \log \left (x\right ) + \frac {4}{9} \, \log \left (x\right )^{2} + \frac {50}{9} \, \log \left (2\right ) + \frac {20}{9} \, \log \left (x\right ) + \frac {109}{36}\right )} \]

[In]

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

[Out]

x + e^(25/9*log(2)^2 + 20/9*log(2)*log(x) + 4/9*log(x)^2 + 50/9*log(2) + 20/9*log(x) + 109/36)

Mupad [B] (verification not implemented)

Time = 11.37 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {3 x-4 e^{\frac {1}{4} \left (1+4 \log ^2\left (e^{\frac {1}{3} (-5-5 \log (2 x))} x\right )\right )} \log \left (e^{\frac {1}{3} (-5-5 \log (2 x))} x\right )}{3 x} \, dx=x+32\,2^{5/9}\,x^{20/9}\,{\left ({\mathrm {e}}^{{\ln \left (2\right )}^2}\right )}^{25/9}\,{\mathrm {e}}^{{\ln \left (\frac {1}{x^{2/3}}\right )}^2}\,{\mathrm {e}}^{109/36}\,{\mathrm {e}}^{-\frac {10\,\ln \left (\frac {1}{x^{2/3}}\right )\,\ln \left (2\right )}{3}} \]

[In]

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

[Out]

x + 32*2^(5/9)*x^(20/9)*exp(log(2)^2)^(25/9)*exp(log(1/x^(2/3))^2)*exp(109/36)*exp(-(10*log(1/x^(2/3))*log(2))
/3)