3.85.18 \(\int \frac {e^{-\frac {3600-2280 x+361 x^2}{400 \log ^2(x)}} (3600-2280 x+361 x^2+(1140 x-361 x^2) \log (x)+200 \log ^2(x)+200 \log ^3(x))}{100 \log ^2(x)} \, dx\)

Optimal. Leaf size=24 \[ -5+2 e^{-\frac {\left (3-\frac {19 x}{20}\right )^2}{\log ^2(x)}} x \log (x) \]

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Rubi [B]  time = 0.22, antiderivative size = 81, normalized size of antiderivative = 3.38, number of steps used = 2, number of rules used = 2, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {12, 2288} \begin {gather*} \frac {2 e^{-\frac {361 x^2-2280 x+3600}{400 \log ^2(x)}} \left (361 x^2+19 \left (60 x-19 x^2\right ) \log (x)-2280 x+3600\right )}{\left (\frac {361 x^2-2280 x+3600}{x \log ^3(x)}+\frac {19 (60-19 x)}{\log ^2(x)}\right ) \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3600 - 2280*x + 361*x^2 + (1140*x - 361*x^2)*Log[x] + 200*Log[x]^2 + 200*Log[x]^3)/(100*E^((3600 - 2280*x
+ 361*x^2)/(400*Log[x]^2))*Log[x]^2),x]

[Out]

(2*(3600 - 2280*x + 361*x^2 + 19*(60*x - 19*x^2)*Log[x]))/(E^((3600 - 2280*x + 361*x^2)/(400*Log[x]^2))*((3600
- 2280*x + 361*x^2)/(x*Log[x]^3) + (19*(60 - 19*x))/Log[x]^2)*Log[x]^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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{100} \int \frac {e^{-\frac {3600-2280 x+361 x^2}{400 \log ^2(x)}} \left (3600-2280 x+361 x^2+\left (1140 x-361 x^2\right ) \log (x)+200 \log ^2(x)+200 \log ^3(x)\right )}{\log ^2(x)} \, dx\\ &=\frac {2 e^{-\frac {3600-2280 x+361 x^2}{400 \log ^2(x)}} \left (3600-2280 x+361 x^2+19 \left (60 x-19 x^2\right ) \log (x)\right )}{\left (\frac {3600-2280 x+361 x^2}{x \log ^3(x)}+\frac {19 (60-19 x)}{\log ^2(x)}\right ) \log ^2(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 22, normalized size = 0.92 \begin {gather*} 2 e^{-\frac {(60-19 x)^2}{400 \log ^2(x)}} x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3600 - 2280*x + 361*x^2 + (1140*x - 361*x^2)*Log[x] + 200*Log[x]^2 + 200*Log[x]^3)/(100*E^((3600 -
2280*x + 361*x^2)/(400*Log[x]^2))*Log[x]^2),x]

[Out]

(2*x*Log[x])/E^((60 - 19*x)^2/(400*Log[x]^2))

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fricas [A]  time = 0.83, size = 22, normalized size = 0.92 \begin {gather*} 2 \, x e^{\left (-\frac {361 \, x^{2} - 2280 \, x + 3600}{400 \, \log \relax (x)^{2}}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/100*(200*log(x)^3+200*log(x)^2+(-361*x^2+1140*x)*log(x)+361*x^2-2280*x+3600)/log(x)^2/exp(1/400*(3
61*x^2-2280*x+3600)/log(x)^2),x, algorithm="fricas")

[Out]

2*x*e^(-1/400*(361*x^2 - 2280*x + 3600)/log(x)^2)*log(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(1/100*(200*log(x)^3+200*log(x)^2+(-361*x^2+1140*x)*log(x)+361*x^2-2280*x+3600)/log(x)^2/exp(1/400*(3
61*x^2-2280*x+3600)/log(x)^2),x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.03, size = 20, normalized size = 0.83

method	result	size
risch	\(2 \ln \relax (x ) x \,{\mathrm e}^{-\frac {\left (19 x -60\right )^{2}}{400 \ln \relax (x )^{2}}}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/100*(200*ln(x)^3+200*ln(x)^2+(-361*x^2+1140*x)*ln(x)+361*x^2-2280*x+3600)/ln(x)^2/exp(1/400*(361*x^2-228
0*x+3600)/ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

2*ln(x)*x*exp(-1/400*(19*x-60)^2/ln(x)^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{100} \, \int \frac {{\left (200 \, \log \relax (x)^{3} + 361 \, x^{2} - 19 \, {\left (19 \, x^{2} - 60 \, x\right )} \log \relax (x) + 200 \, \log \relax (x)^{2} - 2280 \, x + 3600\right )} e^{\left (-\frac {361 \, x^{2} - 2280 \, x + 3600}{400 \, \log \relax (x)^{2}}\right )}}{\log \relax (x)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/100*(200*log(x)^3+200*log(x)^2+(-361*x^2+1140*x)*log(x)+361*x^2-2280*x+3600)/log(x)^2/exp(1/400*(3
61*x^2-2280*x+3600)/log(x)^2),x, algorithm="maxima")

[Out]

1/100*integrate((200*log(x)^3 + 361*x^2 - 19*(19*x^2 - 60*x)*log(x) + 200*log(x)^2 - 2280*x + 3600)*e^(-1/400*
(361*x^2 - 2280*x + 3600)/log(x)^2)/log(x)^2, x)

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mupad [B]  time = 5.51, size = 22, normalized size = 0.92 \begin {gather*} 2\,x\,{\mathrm {e}}^{-\frac {361\,x^2-2280\,x+3600}{400\,{\ln \relax (x)}^2}}\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-((361*x^2)/400 - (57*x)/10 + 9)/log(x)^2)*(2*log(x)^2 - (114*x)/5 + 2*log(x)^3 + (log(x)*(1140*x - 3
61*x^2))/100 + (361*x^2)/100 + 36))/log(x)^2,x)

[Out]

2*x*exp(-(361*x^2 - 2280*x + 3600)/(400*log(x)^2))*log(x)

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sympy [A]  time = 4.01, size = 26, normalized size = 1.08 \begin {gather*} 2 x e^{- \frac {\frac {361 x^{2}}{400} - \frac {57 x}{10} + 9}{\log {\relax (x )}^{2}}} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/100*(200*ln(x)**3+200*ln(x)**2+(-361*x**2+1140*x)*ln(x)+361*x**2-2280*x+3600)/ln(x)**2/exp(1/400*(
361*x**2-2280*x+3600)/ln(x)**2),x)

[Out]

2*x*exp(-(361*x**2/400 - 57*x/10 + 9)/log(x)**2)*log(x)

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