\(\int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} (-8-4 x+(-2-x) \log ^2(3)) \log (4+2 x)}{(2+x) \log ^2(3)} \, dx\) [6908]

Optimal result

Integrand size = 88, antiderivative size = 24 \[ \int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \left (-8-4 x+(-2-x) \log ^2(3)\right ) \log (4+2 x)}{(2+x) \log ^2(3)} \, dx=x+e^{2-x \left (1+\frac {4}{\log ^2(3)}\right )} \log (4+2 x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {12, 6874, 2326} \[ \int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \left (-8-4 x+(-2-x) \log ^2(3)\right ) \log (4+2 x)}{(2+x) \log ^2(3)} \, dx=x+\frac {e^{2-x \left (1+\frac {4}{\log ^2(3)}\right )} (x \log (2 (x+2))+2 \log (2 (x+2)))}{x+2} \]

[In]

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

[Out]

x + (E^(2 - x*(1 + 4/Log[3]^2))*(2*Log[2*(2 + x)] + x*Log[2*(2 + x)]))/(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 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}& = \frac {\int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \left (-8-4 x+(-2-x) \log ^2(3)\right ) \log (4+2 x)}{2+x} \, dx}{\log ^2(3)} \\ & = \frac {\int \left (\log ^2(3)+\frac {e^{2-x \left (1+\frac {4}{\log ^2(3)}\right )} \left (\log ^2(3)-8 \left (1+\frac {\log ^2(3)}{4}\right ) \log (2 (2+x))-4 x \left (1+\frac {\log ^2(3)}{4}\right ) \log (2 (2+x))\right )}{2+x}\right ) \, dx}{\log ^2(3)} \\ & = x+\frac {\int \frac {e^{2-x \left (1+\frac {4}{\log ^2(3)}\right )} \left (\log ^2(3)-8 \left (1+\frac {\log ^2(3)}{4}\right ) \log (2 (2+x))-4 x \left (1+\frac {\log ^2(3)}{4}\right ) \log (2 (2+x))\right )}{2+x} \, dx}{\log ^2(3)} \\ & = x+\frac {e^{2-x \left (1+\frac {4}{\log ^2(3)}\right )} (2 \log (2 (2+x))+x \log (2 (2+x)))}{2+x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \left (-8-4 x+(-2-x) \log ^2(3)\right ) \log (4+2 x)}{(2+x) \log ^2(3)} \, dx=x+e^{2+x \left (-1-\frac {4}{\log ^2(3)}\right )} \log (2 (2+x)) \]

[In]

Integrate[(E^((-4*x + (2 - x)*Log[3]^2)/Log[3]^2)*Log[3]^2 + (2 + x)*Log[3]^2 + E^((-4*x + (2 - x)*Log[3]^2)/L
og[3]^2)*(-8 - 4*x + (-2 - x)*Log[3]^2)*Log[4 + 2*x])/((2 + x)*Log[3]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25

method	result	size
parts	\(x +{\mathrm e}^{\frac {\left (2-x \right ) \ln \left (3\right )^{2}-4 x}{\ln \left (3\right )^{2}}} \ln \left (4+2 x \right )\)	\(30\)
risch	\(x +{\mathrm e}^{-\frac {x \ln \left (3\right )^{2}-2 \ln \left (3\right )^{2}+4 x}{\ln \left (3\right )^{2}}} \ln \left (4+2 x \right )\)	\(33\)
norman	\(\frac {x \ln \left (3\right )+\ln \left (3\right ) {\mathrm e}^{\frac {\left (2-x \right ) \ln \left (3\right )^{2}-4 x}{\ln \left (3\right )^{2}}} \ln \left (4+2 x \right )}{\ln \left (3\right )}\)	\(40\)
default	\(\frac {x \ln \left (3\right )^{2}+\ln \left (3\right )^{2} {\mathrm e}^{\frac {\left (2-x \right ) \ln \left (3\right )^{2}-4 x}{\ln \left (3\right )^{2}}} \ln \left (4+2 x \right )}{\ln \left (3\right )^{2}}\)	\(44\)
parallelrisch	\(\frac {\ln \left (3\right )^{2} {\mathrm e}^{\frac {\left (2-x \right ) \ln \left (3\right )^{2}-4 x}{\ln \left (3\right )^{2}}} \ln \left (4+2 x \right )+x \ln \left (3\right )^{2}-4 \ln \left (3\right )^{2}}{\ln \left (3\right )^{2}}\)	\(50\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \left (-8-4 x+(-2-x) \log ^2(3)\right ) \log (4+2 x)}{(2+x) \log ^2(3)} \, dx=e^{\left (-\frac {{\left (x - 2\right )} \log \left (3\right )^{2} + 4 \, x}{\log \left (3\right )^{2}}\right )} \log \left (2 \, x + 4\right ) + x \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \left (-8-4 x+(-2-x) \log ^2(3)\right ) \log (4+2 x)}{(2+x) \log ^2(3)} \, dx=x + e^{\frac {- 4 x + \left (2 - x\right ) \log {\left (3 \right )}^{2}}{\log {\left (3 \right )}^{2}}} \log {\left (2 x + 4 \right )} \]

[In]

integrate((((-2-x)*ln(3)**2-4*x-8)*exp(((2-x)*ln(3)**2-4*x)/ln(3)**2)*ln(4+2*x)+ln(3)**2*exp(((2-x)*ln(3)**2-4
*x)/ln(3)**2)+(2+x)*ln(3)**2)/(2+x)/ln(3)**2,x)

[Out]

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

Maxima [F]

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

[In]

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

[Out]

-(e^(2*(log(3)^2 + 4)/log(3)^2 + 2)*exp_integral_e(1, (log(3)^2 + 4)*(x + 2)/log(3)^2)*log(3)^2 - e^(-x - 4*x/
log(3)^2 + 2)*log(3)^2*log(x + 2) - (x - 2*log(x + 2))*log(3)^2 - 2*log(3)^2*log(x + 2) + integrate(((log(3)^2
+ 4)*x*e^2*log(2) + (log(3)^2 + 2*(log(3)^2 + 4)*log(2))*e^2)*e^(-x - 4*x/log(3)^2)/(x + 2), x))/log(3)^2

Giac [F]

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

[In]

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

[Out]

undef

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \left (-8-4 x+(-2-x) \log ^2(3)\right ) \log (4+2 x)}{(2+x) \log ^2(3)} \, dx=\int \frac {{\ln \left (3\right )}^2\,\left (x+2\right )+{\mathrm {e}}^{-\frac {4\,x+{\ln \left (3\right )}^2\,\left (x-2\right )}{{\ln \left (3\right )}^2}}\,{\ln \left (3\right )}^2-{\mathrm {e}}^{-\frac {4\,x+{\ln \left (3\right )}^2\,\left (x-2\right )}{{\ln \left (3\right )}^2}}\,\ln \left (2\,x+4\right )\,\left (4\,x+{\ln \left (3\right )}^2\,\left (x+2\right )+8\right )}{{\ln \left (3\right )}^2\,\left (x+2\right )} \,d x \]

[In]

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

[Out]

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