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

Optimal. Leaf size=24 \[ x+e^{2-x \left (1+\frac {4}{\log ^2(3)}\right )} \log (4+2 x) \]

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Rubi [A]  time = 0.80, antiderivative size = 40, normalized size of antiderivative = 1.67, 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 = {12, 6742, 2288} \begin {gather*} 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} \end {gather*}

Antiderivative was successfully verified.

[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 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} &=\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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 23, normalized size = 0.96 \begin {gather*} x+e^{2+x \left (-1-\frac {4}{\log ^2(3)}\right )} \log (2 (2+x)) \end {gather*}

Antiderivative was successfully verified.

[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)]

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fricas [A]  time = 0.83, size = 28, normalized size = 1.17 \begin {gather*} e^{\left (-\frac {{\left (x - 2\right )} \log \relax (3)^{2} + 4 \, x}{\log \relax (3)^{2}}\right )} \log \left (2 \, x + 4\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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maple [A]  time = 0.14, size = 33, normalized size = 1.38

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {e^{\left (\frac {2 \, {\left (\log \relax (3)^{2} + 4\right )}}{\log \relax (3)^{2}} + 2\right )} E_{1}\left (\frac {{\left (\log \relax (3)^{2} + 4\right )} {\left (x + 2\right )}}{\log \relax (3)^{2}}\right ) \log \relax (3)^{2} - e^{\left (-x - \frac {4 \, x}{\log \relax (3)^{2}} + 2\right )} \log \relax (3)^{2} \log \left (x + 2\right ) - {\left (x - 2 \, \log \left (x + 2\right )\right )} \log \relax (3)^{2} - 2 \, \log \relax (3)^{2} \log \left (x + 2\right ) + \mathit {undef}}{\log \relax (3)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\ln \relax (3)}^2\,\left (x+2\right )+{\mathrm {e}}^{-\frac {4\,x+{\ln \relax (3)}^2\,\left (x-2\right )}{{\ln \relax (3)}^2}}\,{\ln \relax (3)}^2-{\mathrm {e}}^{-\frac {4\,x+{\ln \relax (3)}^2\,\left (x-2\right )}{{\ln \relax (3)}^2}}\,\ln \left (2\,x+4\right )\,\left (4\,x+{\ln \relax (3)}^2\,\left (x+2\right )+8\right )}{{\ln \relax (3)}^2\,\left (x+2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.72, size = 26, normalized size = 1.08 \begin {gather*} x + e^{\frac {- 4 x + \left (2 - x\right ) \log {\relax (3 )}^{2}}{\log {\relax (3 )}^{2}}} \log {\left (2 x + 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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