3.42.81 $\int \frac{e^{2e^{\frac{-5+10x}{2x}}}(-2x-5e^{\frac{-5+10x}{2x}}\log (4)+5e^{\frac{-5+10x}{2x}}\log (\frac{x}{5}))+e^{e^{\frac{-5+10x}{2x}}}(-30x\log (4)-75e^{\frac{-5+10x}{2x}}{\log }^2(4)+(30x+150e^{\frac{-5+10x}{2x}}\log (4))\log (\frac{x}{5})-75e^{\frac{-5+10x}{2x}}{\log }^2(\frac{x}{5}))}{-x^2{\log }^3(4)+3x^2{\log }^2(4)\log (\frac{x}{5})-3x^2\log (4){\log }^2(\frac{x}{5})+x^2{\log }^3(\frac{x}{5})}dx$

Optimal. Leaf size=32 $${(15-\frac{e^{e^{5-\frac{5}{2x}}}}{-\log (4)+\log \left(\frac{x}{5}\right)})}^2$$

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Rubi [F]  time = 6.30, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.000, Rules used = {} $$\begin{array}{c}\int \frac{e^{2e^{\frac{-5+10x}{2x}}}(-2x-5e^{\frac{-5+10x}{2x}}\log (4)+5e^{\frac{-5+10x}{2x}}\log \left(\frac{x}{5}\right))+e^{e^{\frac{-5+10x}{2x}}}(-30x\log (4)-75e^{\frac{-5+10x}{2x}}{\log }^2(4)+(30x+150e^{\frac{-5+10x}{2x}}\log (4))\log \left(\frac{x}{5}\right)-75e^{\frac{-5+10x}{2x}}{\log }^2\left(\frac{x}{5}\right))}{-x^2{\log }^3(4)+3x^2{\log }^2(4)\log \left(\frac{x}{5}\right)-3x^2\log (4){\log }^2\left(\frac{x}{5}\right)+x^2{\log }^3\left(\frac{x}{5}\right)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(E^(2*E^((-5 + 10*x)/(2*x)))*(-2*x - 5*E^((-5 + 10*x)/(2*x))*Log[4] + 5*E^((-5 + 10*x)/(2*x))*Log[x/5]) +
E^E^((-5 + 10*x)/(2*x))*(-30*x*Log[4] - 75*E^((-5 + 10*x)/(2*x))*Log[4]^2 + (30*x + 150*E^((-5 + 10*x)/(2*x))*
Log[4])*Log[x/5] - 75*E^((-5 + 10*x)/(2*x))*Log[x/5]^2))/(-(x^2*Log[4]^3) + 3*x^2*Log[4]^2*Log[x/5] - 3*x^2*Lo
g[4]*Log[x/5]^2 + x^2*Log[x/5]^3),x]

[Out]

-2*Defer[Int][E^(2*E^(5 - 5/(2*x)))/(x*Log[x/20]^3), x] + 5*Defer[Int][E^(5 + 2*E^(5 - 5/(2*x)) - 5/(2*x))/(x^
2*Log[x/20]^2), x] + 30*Defer[Int][E^E^(5 - 5/(2*x))/(x*Log[x/20]^2), x] - 75*Defer[Int][E^(5 + E^(5 - 5/(2*x)
) - 5/(2*x))/(x^2*Log[x/20]), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^{e^{5-\frac{5}{2x}}-\frac{5}{2x}}(e^{e^{5-\frac{5}{2x}}}+15\log (4)-15\log \left(\frac{x}{5}\right))(-2e^{\frac{5}{2}/x}x-5e^5\log (4)+5e^5\log \left(\frac{x}{5}\right))}{x^2{\log }^3\left(\frac{x}{20}\right)}dx\\ & =\int (-\frac{2e^{e^{5-\frac{5}{2x}}}(e^{e^{5-\frac{5}{2x}}}+15\log (4)-15\log \left(\frac{x}{5}\right))}{x{\log }^3\left(\frac{x}{20}\right)}-\frac{5e^{5+e^{5-\frac{5}{2x}}-\frac{5}{2x}}(e^{e^{5-\frac{5}{2x}}}+15\log (4)-15\log \left(\frac{x}{5}\right))(\log (20)-\log (x))}{x^2{\log }^3\left(\frac{x}{20}\right)})dx\\ & =-(2\int \frac{e^{e^{5-\frac{5}{2x}}}(e^{e^{5-\frac{5}{2x}}}+15\log (4)-15\log \left(\frac{x}{5}\right))}{x{\log }^3\left(\frac{x}{20}\right)}dx)-5\int \frac{e^{5+e^{5-\frac{5}{2x}}-\frac{5}{2x}}(e^{e^{5-\frac{5}{2x}}}+15\log (4)-15\log \left(\frac{x}{5}\right))(\log (20)-\log (x))}{x^2{\log }^3\left(\frac{x}{20}\right)}dx\\ & =-(2\int (\frac{e^{2e^{5-\frac{5}{2x}}}}{x{\log }^3\left(\frac{x}{20}\right)}-\frac{15e^{e^{5-\frac{5}{2x}}}}{x{\log }^2\left(\frac{x}{20}\right)})dx)-5\int (-\frac{e^{5+2e^{5-\frac{5}{2x}}-\frac{5}{2x}}}{x^2{\log }^2\left(\frac{x}{20}\right)}+\frac{15e^{5+e^{5-\frac{5}{2x}}-\frac{5}{2x}}}{x^2\log \left(\frac{x}{20}\right)})dx\\ & =-(2\int \frac{e^{2e^{5-\frac{5}{2x}}}}{x{\log }^3\left(\frac{x}{20}\right)}dx)+5\int \frac{e^{5+2e^{5-\frac{5}{2x}}-\frac{5}{2x}}}{x^2{\log }^2\left(\frac{x}{20}\right)}dx+30\int \frac{e^{e^{5-\frac{5}{2x}}}}{x{\log }^2\left(\frac{x}{20}\right)}dx-75\int \frac{e^{5+e^{5-\frac{5}{2x}}-\frac{5}{2x}}}{x^2\log \left(\frac{x}{20}\right)}dx\end{array}\end{array}$$

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Mathematica [F]  time = 2.11, size = 0, normalized size = 0.00 $$\begin{array}{c}\int \frac{e^{2e^{\frac{-5+10x}{2x}}}(-2x-5e^{\frac{-5+10x}{2x}}\log (4)+5e^{\frac{-5+10x}{2x}}\log \left(\frac{x}{5}\right))+e^{e^{\frac{-5+10x}{2x}}}(-30x\log (4)-75e^{\frac{-5+10x}{2x}}{\log }^2(4)+(30x+150e^{\frac{-5+10x}{2x}}\log (4))\log \left(\frac{x}{5}\right)-75e^{\frac{-5+10x}{2x}}{\log }^2\left(\frac{x}{5}\right))}{-x^2{\log }^3(4)+3x^2{\log }^2(4)\log \left(\frac{x}{5}\right)-3x^2\log (4){\log }^2\left(\frac{x}{5}\right)+x^2{\log }^3\left(\frac{x}{5}\right)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Integrate[(E^(2*E^((-5 + 10*x)/(2*x)))*(-2*x - 5*E^((-5 + 10*x)/(2*x))*Log[4] + 5*E^((-5 + 10*x)/(2*x))*Log[x/
5]) + E^E^((-5 + 10*x)/(2*x))*(-30*x*Log[4] - 75*E^((-5 + 10*x)/(2*x))*Log[4]^2 + (30*x + 150*E^((-5 + 10*x)/(
2*x))*Log[4])*Log[x/5] - 75*E^((-5 + 10*x)/(2*x))*Log[x/5]^2))/(-(x^2*Log[4]^3) + 3*x^2*Log[4]^2*Log[x/5] - 3*
x^2*Log[4]*Log[x/5]^2 + x^2*Log[x/5]^3),x]

[Out]

Integrate[(E^(2*E^((-5 + 10*x)/(2*x)))*(-2*x - 5*E^((-5 + 10*x)/(2*x))*Log[4] + 5*E^((-5 + 10*x)/(2*x))*Log[x/
5]) + E^E^((-5 + 10*x)/(2*x))*(-30*x*Log[4] - 75*E^((-5 + 10*x)/(2*x))*Log[4]^2 + (30*x + 150*E^((-5 + 10*x)/(
2*x))*Log[4])*Log[x/5] - 75*E^((-5 + 10*x)/(2*x))*Log[x/5]^2))/(-(x^2*Log[4]^3) + 3*x^2*Log[4]^2*Log[x/5] - 3*
x^2*Log[4]*Log[x/5]^2 + x^2*Log[x/5]^3), x]

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fricas [B]  time = 2.53, size = 64, normalized size = 2.00 $$\begin{array}{c}\frac{30(2\log (2)-\log \left(\frac{1}{5}x\right))e^{\left(e^{\left(\frac{5(2x-1)}{2x}\right)}\right)}+e^{\left(2e^{\left(\frac{5(2x-1)}{2x}\right)}\right)}}{4\log (2{)}^2-4\log (2)\log \left(\frac{1}{5}x\right)+\log {\left(\frac{1}{5}x\right)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(1/2*(10*x-5)/x)*log(1/5*x)-10*log(2)*exp(1/2*(10*x-5)/x)-2*x)*exp(exp(1/2*(10*x-5)/x))^2+(-7
5*exp(1/2*(10*x-5)/x)*log(1/5*x)^2+(300*log(2)*exp(1/2*(10*x-5)/x)+30*x)*log(1/5*x)-300*log(2)^2*exp(1/2*(10*x
-5)/x)-60*x*log(2))*exp(exp(1/2*(10*x-5)/x)))/(x^2*log(1/5*x)^3-6*x^2*log(2)*log(1/5*x)^2+12*x^2*log(2)^2*log(
1/5*x)-8*x^2*log(2)^3),x, algorithm="fricas")

[Out]

(30*(2*log(2) - log(1/5*x))*e^(e^(5/2*(2*x - 1)/x)) + e^(2*e^(5/2*(2*x - 1)/x)))/(4*log(2)^2 - 4*log(2)*log(1/
5*x) + log(1/5*x)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\int \frac{(10e^{\left(\frac{5(2x-1)}{2x}\right)}\log (2)-5e^{\left(\frac{5(2x-1)}{2x}\right)}\log \left(\frac{1}{5}x\right)+2x)e^{\left(2e^{\left(\frac{5(2x-1)}{2x}\right)}\right)}+15(20e^{\left(\frac{5(2x-1)}{2x}\right)}\log (2{)}^2+5e^{\left(\frac{5(2x-1)}{2x}\right)}\log {\left(\frac{1}{5}x\right)}^2+4x\log (2)-2(10e^{\left(\frac{5(2x-1)}{2x}\right)}\log (2)+x)\log \left(\frac{1}{5}x\right))e^{\left(e^{\left(\frac{5(2x-1)}{2x}\right)}\right)}}{8x^2\log (2{)}^3-12x^2\log (2{)}^2\log \left(\frac{1}{5}x\right)+6x^2\log (2)\log {\left(\frac{1}{5}x\right)}^2-x^2\log {\left(\frac{1}{5}x\right)}^3}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(1/2*(10*x-5)/x)*log(1/5*x)-10*log(2)*exp(1/2*(10*x-5)/x)-2*x)*exp(exp(1/2*(10*x-5)/x))^2+(-7
5*exp(1/2*(10*x-5)/x)*log(1/5*x)^2+(300*log(2)*exp(1/2*(10*x-5)/x)+30*x)*log(1/5*x)-300*log(2)^2*exp(1/2*(10*x
-5)/x)-60*x*log(2))*exp(exp(1/2*(10*x-5)/x)))/(x^2*log(1/5*x)^3-6*x^2*log(2)*log(1/5*x)^2+12*x^2*log(2)^2*log(
1/5*x)-8*x^2*log(2)^3),x, algorithm="giac")

[Out]

integrate(((10*e^(5/2*(2*x - 1)/x)*log(2) - 5*e^(5/2*(2*x - 1)/x)*log(1/5*x) + 2*x)*e^(2*e^(5/2*(2*x - 1)/x))
+ 15*(20*e^(5/2*(2*x - 1)/x)*log(2)^2 + 5*e^(5/2*(2*x - 1)/x)*log(1/5*x)^2 + 4*x*log(2) - 2*(10*e^(5/2*(2*x -
1)/x)*log(2) + x)*log(1/5*x))*e^(e^(5/2*(2*x - 1)/x)))/(8*x^2*log(2)^3 - 12*x^2*log(2)^2*log(1/5*x) + 6*x^2*lo
g(2)*log(1/5*x)^2 - x^2*log(1/5*x)^3), x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 $$\int \frac{(5{\mathrm{e}}^{\frac{10x-5}{2x}}\ln \left(\frac{x}{5}\right)-10\ln (2){\mathrm{e}}^{\frac{10x-5}{2x}}-2x){\mathrm{e}}^{2{\mathrm{e}}^{\frac{10x-5}{2x}}}+(-75{\mathrm{e}}^{\frac{10x-5}{2x}}\ln {\left(\frac{x}{5}\right)}^2+(300\ln (2){\mathrm{e}}^{\frac{10x-5}{2x}}+30x)\ln \left(\frac{x}{5}\right)-300\ln (2{)}^2{\mathrm{e}}^{\frac{10x-5}{2x}}-60x\ln (2)){\mathrm{e}}^{{\mathrm{e}}^{\frac{10x-5}{2x}}}}{x^2\ln {\left(\frac{x}{5}\right)}^3-6x^2\ln (2)\ln {\left(\frac{x}{5}\right)}^2+12x^2\ln (2{)}^2\ln \left(\frac{x}{5}\right)-8x^2\ln (2{)}^3}dx$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*exp(1/2*(10*x-5)/x)*ln(1/5*x)-10*ln(2)*exp(1/2*(10*x-5)/x)-2*x)*exp(exp(1/2*(10*x-5)/x))^2+(-75*exp(1/
2*(10*x-5)/x)*ln(1/5*x)^2+(300*ln(2)*exp(1/2*(10*x-5)/x)+30*x)*ln(1/5*x)-300*ln(2)^2*exp(1/2*(10*x-5)/x)-60*x*
ln(2))*exp(exp(1/2*(10*x-5)/x)))/(x^2*ln(1/5*x)^3-6*x^2*ln(2)*ln(1/5*x)^2+12*x^2*ln(2)^2*ln(1/5*x)-8*x^2*ln(2)
^3),x)

[Out]

int(((5*exp(1/2*(10*x-5)/x)*ln(1/5*x)-10*ln(2)*exp(1/2*(10*x-5)/x)-2*x)*exp(exp(1/2*(10*x-5)/x))^2+(-75*exp(1/
2*(10*x-5)/x)*ln(1/5*x)^2+(300*ln(2)*exp(1/2*(10*x-5)/x)+30*x)*ln(1/5*x)-300*ln(2)^2*exp(1/2*(10*x-5)/x)-60*x*
ln(2))*exp(exp(1/2*(10*x-5)/x)))/(x^2*ln(1/5*x)^3-6*x^2*ln(2)*ln(1/5*x)^2+12*x^2*ln(2)^2*ln(1/5*x)-8*x^2*ln(2)
^3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\int \frac{(10e^{\left(\frac{5(2x-1)}{2x}\right)}\log (2)-5e^{\left(\frac{5(2x-1)}{2x}\right)}\log \left(\frac{1}{5}x\right)+2x)e^{\left(2e^{\left(\frac{5(2x-1)}{2x}\right)}\right)}+15(20e^{\left(\frac{5(2x-1)}{2x}\right)}\log (2{)}^2+5e^{\left(\frac{5(2x-1)}{2x}\right)}\log {\left(\frac{1}{5}x\right)}^2+4x\log (2)-2(10e^{\left(\frac{5(2x-1)}{2x}\right)}\log (2)+x)\log \left(\frac{1}{5}x\right))e^{\left(e^{\left(\frac{5(2x-1)}{2x}\right)}\right)}}{8x^2\log (2{)}^3-12x^2\log (2{)}^2\log \left(\frac{1}{5}x\right)+6x^2\log (2)\log {\left(\frac{1}{5}x\right)}^2-x^2\log {\left(\frac{1}{5}x\right)}^3}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(1/2*(10*x-5)/x)*log(1/5*x)-10*log(2)*exp(1/2*(10*x-5)/x)-2*x)*exp(exp(1/2*(10*x-5)/x))^2+(-7
5*exp(1/2*(10*x-5)/x)*log(1/5*x)^2+(300*log(2)*exp(1/2*(10*x-5)/x)+30*x)*log(1/5*x)-300*log(2)^2*exp(1/2*(10*x
-5)/x)-60*x*log(2))*exp(exp(1/2*(10*x-5)/x)))/(x^2*log(1/5*x)^3-6*x^2*log(2)*log(1/5*x)^2+12*x^2*log(2)^2*log(
1/5*x)-8*x^2*log(2)^3),x, algorithm="maxima")

[Out]

integrate(((10*e^(5/2*(2*x - 1)/x)*log(2) - 5*e^(5/2*(2*x - 1)/x)*log(1/5*x) + 2*x)*e^(2*e^(5/2*(2*x - 1)/x))
+ 15*(20*e^(5/2*(2*x - 1)/x)*log(2)^2 + 5*e^(5/2*(2*x - 1)/x)*log(1/5*x)^2 + 4*x*log(2) - 2*(10*e^(5/2*(2*x -
1)/x)*log(2) + x)*log(1/5*x))*e^(e^(5/2*(2*x - 1)/x)))/(8*x^2*log(2)^3 - 12*x^2*log(2)^2*log(1/5*x) + 6*x^2*lo
g(2)*log(1/5*x)^2 - x^2*log(1/5*x)^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 $$\begin{array}{c}-\int -\frac{{\mathrm{e}}^{2{\mathrm{e}}^{\frac{5x-\frac{5}{2}}{x}}}(2x-5\ln \left(\frac{x}{5}\right){\mathrm{e}}^{\frac{5x-\frac{5}{2}}{x}}+10{\mathrm{e}}^{\frac{5x-\frac{5}{2}}{x}}\ln (2))+{\mathrm{e}}^{{\mathrm{e}}^{\frac{5x-\frac{5}{2}}{x}}}(75{\mathrm{e}}^{\frac{5x-\frac{5}{2}}{x}}{\ln \left(\frac{x}{5}\right)}^2+(-30x-300{\mathrm{e}}^{\frac{5x-\frac{5}{2}}{x}}\ln (2))\ln \left(\frac{x}{5}\right)+60x\ln (2)+300{\mathrm{e}}^{\frac{5x-\frac{5}{2}}{x}}{\ln (2)}^2)}{-x^2{\ln \left(\frac{x}{5}\right)}^3+6\ln (2)x^2{\ln \left(\frac{x}{5}\right)}^2-12{\ln (2)}^2{x}^2\ln \left(\frac{x}{5}\right)+8{\ln (2)}^3{x}^2}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*exp((5*x - 5/2)/x))*(2*x - 5*log(x/5)*exp((5*x - 5/2)/x) + 10*exp((5*x - 5/2)/x)*log(2)) + exp(exp(
(5*x - 5/2)/x))*(60*x*log(2) - log(x/5)*(30*x + 300*exp((5*x - 5/2)/x)*log(2)) + 75*log(x/5)^2*exp((5*x - 5/2)
/x) + 300*exp((5*x - 5/2)/x)*log(2)^2))/(8*x^2*log(2)^3 - x^2*log(x/5)^3 - 12*x^2*log(x/5)*log(2)^2 + 6*x^2*lo
g(x/5)^2*log(2)),x)

[Out]

-int(-(exp(2*exp((5*x - 5/2)/x))*(2*x - 5*log(x/5)*exp((5*x - 5/2)/x) + 10*exp((5*x - 5/2)/x)*log(2)) + exp(ex
p((5*x - 5/2)/x))*(60*x*log(2) - log(x/5)*(30*x + 300*exp((5*x - 5/2)/x)*log(2)) + 75*log(x/5)^2*exp((5*x - 5/
2)/x) + 300*exp((5*x - 5/2)/x)*log(2)^2))/(8*x^2*log(2)^3 - x^2*log(x/5)^3 - 12*x^2*log(x/5)*log(2)^2 + 6*x^2*
log(x/5)^2*log(2)), x)

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sympy [B]  time = 0.60, size = 97, normalized size = 3.03 $$\begin{array}{c}\frac{(\log \left(\frac{x}{5}\right)-2\log (2))e^{2e^{\frac{5x-\frac{5}{2}}{x}}}+(-30\log {\left(\frac{x}{5}\right)}^2+120\log (2)\log \left(\frac{x}{5}\right)-120\log {(2)}^2)e^{e^{\frac{5x-\frac{5}{2}}{x}}}}{\log {\left(\frac{x}{5}\right)}^3-6\log (2)\log {\left(\frac{x}{5}\right)}^2+12\log {(2)}^2\log \left(\frac{x}{5}\right)-8\log {(2)}^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(1/2*(10*x-5)/x)*ln(1/5*x)-10*ln(2)*exp(1/2*(10*x-5)/x)-2*x)*exp(exp(1/2*(10*x-5)/x))**2+(-75
*exp(1/2*(10*x-5)/x)*ln(1/5*x)**2+(300*ln(2)*exp(1/2*(10*x-5)/x)+30*x)*ln(1/5*x)-300*ln(2)**2*exp(1/2*(10*x-5)
/x)-60*x*ln(2))*exp(exp(1/2*(10*x-5)/x)))/(x**2*ln(1/5*x)**3-6*x**2*ln(2)*ln(1/5*x)**2+12*x**2*ln(2)**2*ln(1/5
*x)-8*x**2*ln(2)**3),x)

[Out]

((log(x/5) - 2*log(2))*exp(2*exp((5*x - 5/2)/x)) + (-30*log(x/5)**2 + 120*log(2)*log(x/5) - 120*log(2)**2)*exp
(exp((5*x - 5/2)/x)))/(log(x/5)**3 - 6*log(2)*log(x/5)**2 + 12*log(2)**2*log(x/5) - 8*log(2)**3)

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