∫ e x ____________________________________ 2+log(−18−2x+log(4)+3 log(25)) (−36−2x+2 log(4)+6 log(25)+(−18−2x+log(4)+3 log(25)) log(−18−2x+log(4)+3 log(25))) ___________________________________________________________________________________________________________________________________________________________________________________________________________−72−8x+4 log (4)+12 log (25)+(−72−8x+4 log (4)+12 log (25)) log (−18−2x+log (4)+3 log (25))+(−18−2x+log (4)+3 log (25)) log 2 (−18−2x+log(4)+3 log(25)) dx

3.89.100 \(\int \frac {e^{\frac {x}{2+\log (-18-2 x+\log (4)+3 \log (25))}} (-36-2 x+2 \log (4)+6 \log (25)+(-18-2 x+\log (4)+3 \log (25)) \log (-18-2 x+\log (4)+3 \log (25)))}{-72-8 x+4 \log (4)+12 \log (25)+(-72-8 x+4 \log (4)+12 \log (25)) \log (-18-2 x+\log (4)+3 \log (25))+(-18-2 x+\log (4)+3 \log (25)) \log ^2(-18-2 x+\log (4)+3 \log (25))} \, dx\)

Optimal. Leaf size=23 \[ e^{\frac {x}{2+\log (3+x+\log (4)-3 (7+x-\log (25)))}} \]

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Rubi [A] time = 0.74, antiderivative size = 16, normalized size of antiderivative = 0.70, number of steps used = 2, number of rules used = 2, integrand size = 125, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {6688, 6706} \begin {gather*} e^{\frac {x}{\log (-2 x-18+\log (62500))+2}} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[(E^(x/(2 + Log[-18 - 2*x + Log[4] + 3*Log[25]]))*(-36 - 2*x + 2*Log[4] + 6*Log[25] + (-18 - 2*x + Log[4] +

3*Log[25])*Log[-18 - 2*x + Log[4] + 3*Log[25]]))/(-72 - 8*x + 4*Log[4] + 12*Log[25] + (-72 - 8*x + 4*Log[4] +

12*Log[25])*Log[-18 - 2*x + Log[4] + 3*Log[25]] + (-18 - 2*x + Log[4] + 3*Log[25])*Log[-18 - 2*x + Log[4] + 3

*Log[25]]^2),x]

[Out]

E^(x/(2 + Log[-18 - 2*x + Log[62500]]))

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6706

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {x}{2+\log (-18-2 x+\log (62500))}} (2 (18+x-\log (62500))+(18+2 x-\log (62500)) \log (-18-2 x+\log (62500)))}{(18+2 x-\log (62500)) (2+\log (-18-2 x+\log (62500)))^2} \, dx\\ &=e^{\frac {x}{2+\log (-18-2 x+\log (62500))}}\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.22, size = 16, normalized size = 0.70 \begin {gather*} e^{\frac {x}{2+\log (-18-2 x+\log (62500))}} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(E^(x/(2 + Log[-18 - 2*x + Log[4] + 3*Log[25]]))*(-36 - 2*x + 2*Log[4] + 6*Log[25] + (-18 - 2*x + Lo

g[4] + 3*Log[25])*Log[-18 - 2*x + Log[4] + 3*Log[25]]))/(-72 - 8*x + 4*Log[4] + 12*Log[25] + (-72 - 8*x + 4*Lo

g[4] + 12*Log[25])*Log[-18 - 2*x + Log[4] + 3*Log[25]] + (-18 - 2*x + Log[4] + 3*Log[25])*Log[-18 - 2*x + Log[

4] + 3*Log[25]]^2),x]

[Out]

E^(x/(2 + Log[-18 - 2*x + Log[62500]]))

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fricas [A] time = 0.52, size = 21, normalized size = 0.91 \begin {gather*} e^{\left (\frac {x}{\log \left (-2 \, x + 6 \, \log \relax (5) + 2 \, \log \relax (2) - 18\right ) + 2}\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*log(5)+2*log(2)-2*x-18)*log(6*log(5)+2*log(2)-2*x-18)+12*log(5)+4*log(2)-2*x-36)*exp(x/(log(6*lo

g(5)+2*log(2)-2*x-18)+2))/((6*log(5)+2*log(2)-2*x-18)*log(6*log(5)+2*log(2)-2*x-18)^2+(24*log(5)+8*log(2)-8*x-

72)*log(6*log(5)+2*log(2)-2*x-18)+24*log(5)+8*log(2)-8*x-72),x, algorithm="fricas")

[Out]

e^(x/(log(-2*x + 6*log(5) + 2*log(2) - 18) + 2))

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giac [A] time = 0.49, size = 21, normalized size = 0.91 \begin {gather*} e^{\left (\frac {x}{\log \left (-2 \, x + 6 \, \log \relax (5) + 2 \, \log \relax (2) - 18\right ) + 2}\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*log(5)+2*log(2)-2*x-18)*log(6*log(5)+2*log(2)-2*x-18)+12*log(5)+4*log(2)-2*x-36)*exp(x/(log(6*lo

g(5)+2*log(2)-2*x-18)+2))/((6*log(5)+2*log(2)-2*x-18)*log(6*log(5)+2*log(2)-2*x-18)^2+(24*log(5)+8*log(2)-8*x-

72)*log(6*log(5)+2*log(2)-2*x-18)+24*log(5)+8*log(2)-8*x-72),x, algorithm="giac")

[Out]

e^(x/(log(-2*x + 6*log(5) + 2*log(2) - 18) + 2))

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maple [A] time = 0.21, size = 22, normalized size = 0.96


method	result	size

risch	\({\mathrm e}^{\frac {x}{\ln \left (6 \ln \relax (5)+2 \ln \relax (2)-2 x -18\right )+2}}\)	\(22\)
norman	\(\frac {\ln \left (6 \ln \relax (5)+2 \ln \relax (2)-2 x -18\right ) {\mathrm e}^{\frac {x}{\ln \left (6 \ln \relax (5)+2 \ln \relax (2)-2 x -18\right )+2}}+2 \,{\mathrm e}^{\frac {x}{\ln \left (6 \ln \relax (5)+2 \ln \relax (2)-2 x -18\right )+2}}}{\ln \left (6 \ln \relax (5)+2 \ln \relax (2)-2 x -18\right )+2}\)	\(80\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((6*ln(5)+2*ln(2)-2*x-18)*ln(6*ln(5)+2*ln(2)-2*x-18)+12*ln(5)+4*ln(2)-2*x-36)*exp(x/(ln(6*ln(5)+2*ln(2)-2*

x-18)+2))/((6*ln(5)+2*ln(2)-2*x-18)*ln(6*ln(5)+2*ln(2)-2*x-18)^2+(24*ln(5)+8*ln(2)-8*x-72)*ln(6*ln(5)+2*ln(2)-

2*x-18)+24*ln(5)+8*ln(2)-8*x-72),x,method=_RETURNVERBOSE)

[Out]

exp(x/(ln(6*ln(5)+2*ln(2)-2*x-18)+2))

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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*log(5)+2*log(2)-2*x-18)*log(6*log(5)+2*log(2)-2*x-18)+12*log(5)+4*log(2)-2*x-36)*exp(x/(log(6*lo

g(5)+2*log(2)-2*x-18)+2))/((6*log(5)+2*log(2)-2*x-18)*log(6*log(5)+2*log(2)-2*x-18)^2+(24*log(5)+8*log(2)-8*x-

72)*log(6*log(5)+2*log(2)-2*x-18)+24*log(5)+8*log(2)-8*x-72),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is 0which is not

of the expected type LIST

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mupad [B] time = 0.78, size = 17, normalized size = 0.74 \begin {gather*} {\mathrm {e}}^{\frac {x}{\ln \left (2\,\ln \left (250\right )-2\,x-18\right )+2}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x/(log(2*log(2) - 2*x + 6*log(5) - 18) + 2))*(2*x - 4*log(2) - 12*log(5) + log(2*log(2) - 2*x + 6*log

(5) - 18)*(2*x - 2*log(2) - 6*log(5) + 18) + 36))/(8*x - 8*log(2) - 24*log(5) + log(2*log(2) - 2*x + 6*log(5)

- 18)*(8*x - 8*log(2) - 24*log(5) + 72) + log(2*log(2) - 2*x + 6*log(5) - 18)^2*(2*x - 2*log(2) - 6*log(5) + 1

8) + 72),x)

[Out]

exp(x/(log(2*log(250) - 2*x - 18) + 2))

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sympy [A] time = 0.67, size = 20, normalized size = 0.87 \begin {gather*} e^{\frac {x}{\log {\left (- 2 x - 18 + 2 \log {\relax (2 )} + 6 \log {\relax (5 )} \right )} + 2}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*ln(5)+2*ln(2)-2*x-18)*ln(6*ln(5)+2*ln(2)-2*x-18)+12*ln(5)+4*ln(2)-2*x-36)*exp(x/(ln(6*ln(5)+2*ln

(2)-2*x-18)+2))/((6*ln(5)+2*ln(2)-2*x-18)*ln(6*ln(5)+2*ln(2)-2*x-18)**2+(24*ln(5)+8*ln(2)-8*x-72)*ln(6*ln(5)+2

*ln(2)-2*x-18)+24*ln(5)+8*ln(2)-8*x-72),x)

[Out]

exp(x/(log(-2*x - 18 + 2*log(2) + 6*log(5)) + 2))

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