∫ e−75−29x−3x2+(25+10x+x2) log(3)+(−3+log(3)) log(12x) _____________________________________________________________________________ −3+log(3) (−29−6x+(10+2x) log(3)) ________________________________________________________________________________________________________

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Rubi [A] time = 0.56, antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 6, number of rules used = 5, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {6, 12, 2274, 2244, 2236} \begin {gather*} 12 e^{x^2+\frac {x (29-10 \log (3))}{3-\log (3)}+25} \end {gather*}

Int[(E^((-75 - 29*x - 3*x^2 + (25 + 10*x + x^2)*Log[3] + (-3 + Log[3])*Log[12*x])/(-3 + Log[3]))*(-29 - 6*x +

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}\right ) (-29-6 x+(10+2 x) \log (3))}{x (-3+\log (3))} \, dx\\ &=\frac {\int \frac {\exp \left (\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}\right ) (-29-6 x+(10+2 x) \log (3))}{x} \, dx}{-3+\log (3)}\\ &=\frac {\int 12 \exp \left (\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)}{-3+\log (3)}\right ) (-29-6 x+(10+2 x) \log (3)) \, dx}{-3+\log (3)}\\ &=-\frac {12 \int \exp \left (\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)}{-3+\log (3)}\right ) (-29-6 x+(10+2 x) \log (3)) \, dx}{3-\log (3)}\\ &=-\frac {12 \int e^{25+x^2+\frac {x (29-10 \log (3))}{3-\log (3)}} (-29-2 x (3-\log (3))+10 \log (3)) \, dx}{3-\log (3)}\\ &=12 e^{25+x^2+\frac {x (29-10 \log (3))}{3-\log (3)}}\\ \end {aligned} \end {gather*}

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Mathematica [F] time = 1.03, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx \end {gather*}

Integrate[(E^((-75 - 29*x - 3*x^2 + (25 + 10*x + x^2)*Log[3] + (-3 + Log[3])*Log[12*x])/(-3 + Log[3]))*(-29 -

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fricas [A] time = 0.52, size = 45, normalized size = 2.50 \begin {gather*} \frac {e^{\left (-\frac {3 \, x^{2} - {\left (x^{2} + 10 \, x + 25\right )} \log \relax (3) - {\left (\log \relax (3) - 3\right )} \log \left (12 \, x\right ) + 29 \, x + 75}{\log \relax (3) - 3}\right )}}{x} \end {gather*}

integrate(((2*x+10)*log(3)-6*x-29)*exp(((log(3)-3)*log(12*x)+(x^2+10*x+25)*log(3)-3*x^2-29*x-75)/(log(3)-3))/(

e^(-(3*x^2 - (x^2 + 10*x + 25)*log(3) - (log(3) - 3)*log(12*x) + 29*x + 75)/(log(3) - 3))/x

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giac [A] time = 0.57, size = 40, normalized size = 2.22 \begin {gather*} 4 \, e^{\left (\frac {x^{2} \log \relax (3) - 3 \, x^{2} + 10 \, x \log \relax (3) + \log \relax (3)^{2} - 29 \, x - 3 \, \log \relax (3)}{\log \relax (3) - 3} + 25\right )} \end {gather*}

integrate(((2*x+10)*log(3)-6*x-29)*exp(((log(3)-3)*log(12*x)+(x^2+10*x+25)*log(3)-3*x^2-29*x-75)/(log(3)-3))/(

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method	result	size

gosper	\(\frac {{\mathrm e}^{\frac {x^{2} \ln \relax (3)+\ln \left (12 x \right ) \ln \relax (3)+10 x \ln \relax (3)-3 x^{2}+25 \ln \relax (3)-3 \ln \left (12 x \right )-29 x -75}{\ln \relax (3)-3}}}{x}\)	\(51\)
risch	\(\frac {{\mathrm e}^{\frac {x^{2} \ln \relax (3)+\ln \left (12 x \right ) \ln \relax (3)+10 x \ln \relax (3)-3 x^{2}+25 \ln \relax (3)-3 \ln \left (12 x \right )-29 x -75}{\ln \relax (3)-3}}}{x}\)	\(51\)

int(((2*x+10)*ln(3)-6*x-29)*exp(((ln(3)-3)*ln(12*x)+(x^2+10*x+25)*ln(3)-3*x^2-29*x-75)/(ln(3)-3))/(x*ln(3)-3*x

exp((x^2*ln(3)+ln(12*x)*ln(3)+10*x*ln(3)-3*x^2+25*ln(3)-3*ln(12*x)-29*x-75)/(ln(3)-3))/x

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maxima [B] time = 0.65, size = 120, normalized size = 6.67 \begin {gather*} \frac {3^{\frac {22}{\log \relax (3) - 3}} 2^{\frac {2 \, \log \relax (3)}{\log \relax (3) - 3} - \frac {6}{\log \relax (3) - 3}} e^{\left (\frac {x^{2} \log \relax (3)}{\log \relax (3) - 3} - \frac {3 \, x^{2}}{\log \relax (3) - 3} + \frac {10 \, x \log \relax (3)}{\log \relax (3) - 3} + \frac {\log \relax (3)^{2}}{\log \relax (3) - 3} + \frac {\log \relax (3) \log \relax (x)}{\log \relax (3) - 3} - \frac {29 \, x}{\log \relax (3) - 3} - \frac {3 \, \log \relax (x)}{\log \relax (3) - 3} - \frac {75}{\log \relax (3) - 3}\right )}}{x} \end {gather*}

integrate(((2*x+10)*log(3)-6*x-29)*exp(((log(3)-3)*log(12*x)+(x^2+10*x+25)*log(3)-3*x^2-29*x-75)/(log(3)-3))/(

3^(22/(log(3) - 3))*2^(2*log(3)/(log(3) - 3) - 6/(log(3) - 3))*e^(x^2*log(3)/(log(3) - 3) - 3*x^2/(log(3) - 3)

+ 10*x*log(3)/(log(3) - 3) + log(3)^2/(log(3) - 3) + log(3)*log(x)/(log(3) - 3) - 29*x/(log(3) - 3) - 3*log(x

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mupad [B] time = 4.69, size = 59, normalized size = 3.28 \begin {gather*} {\left (\frac {1}{64}\right )}^{\frac {1}{\ln \relax (3)-3}}\,3^{\frac {x^2+10\,x+\ln \left (12\right )+22}{\ln \relax (3)-3}}\,{\mathrm {e}}^{-\frac {29\,x}{\ln \relax (3)-3}}\,{\mathrm {e}}^{-\frac {3\,x^2}{\ln \relax (3)-3}}\,{\mathrm {e}}^{-\frac {75}{\ln \relax (3)-3}} \end {gather*}

int((exp(-(29*x - log(12*x)*(log(3) - 3) + 3*x^2 - log(3)*(10*x + x^2 + 25) + 75)/(log(3) - 3))*(6*x - log(3)*

(1/64)^(1/(log(3) - 3))*3^((10*x + log(12) + x^2 + 22)/(log(3) - 3))*exp(-(29*x)/(log(3) - 3))*exp(-(3*x^2)/(l

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

integrate(((2*x+10)*ln(3)-6*x-29)*exp(((ln(3)-3)*ln(12*x)+(x**2+10*x+25)*ln(3)-3*x**2-29*x-75)/(ln(3)-3))/(x*l

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3.75.26 \(\int \frac {e^{\frac {-75-29 x-3 x^2+(25+10 x+x^2) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx\)