3.86.58 \(\int \frac {e^{-\frac {-20 x-e^x x-2 x^2+2 x \log (\frac {5}{2})}{-10-x+\log (\frac {5}{2})}} (400-720 x-156 x^2-8 x^3+(-80+152 x+16 x^2) \log (\frac {5}{2})+(4-8 x) \log ^2(\frac {5}{2})+e^x (-40 x-40 x^2-4 x^3+(4 x+4 x^2) \log (\frac {5}{2})))}{1500+300 x+15 x^2+(-300-30 x) \log (\frac {5}{2})+15 \log ^2(\frac {5}{2})} \, dx\)

Optimal. Leaf size=27 \[ \frac {4}{15} e^{-x \left (2+\frac {e^x}{10+x-\log \left (\frac {5}{2}\right )}\right )} x \]

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Rubi [F]  time = 46.00, 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 {gather*} \int \frac {\exp \left (-\frac {-20 x-e^x x-2 x^2+2 x \log \left (\frac {5}{2}\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) \left (400-720 x-156 x^2-8 x^3+\left (-80+152 x+16 x^2\right ) \log \left (\frac {5}{2}\right )+(4-8 x) \log ^2\left (\frac {5}{2}\right )+e^x \left (-40 x-40 x^2-4 x^3+\left (4 x+4 x^2\right ) \log \left (\frac {5}{2}\right )\right )\right )}{1500+300 x+15 x^2+(-300-30 x) \log \left (\frac {5}{2}\right )+15 \log ^2\left (\frac {5}{2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(400 - 720*x - 156*x^2 - 8*x^3 + (-80 + 152*x + 16*x^2)*Log[5/2] + (4 - 8*x)*Log[5/2]^2 + E^x*(-40*x - 40*
x^2 - 4*x^3 + (4*x + 4*x^2)*Log[5/2]))/(E^((-20*x - E^x*x - 2*x^2 + 2*x*Log[5/2])/(-10 - x + Log[5/2]))*(1500
+ 300*x + 15*x^2 + (-300 - 30*x)*Log[5/2] + 15*Log[5/2]^2)),x]

[Out]

(4*(10 - Log[5/2])*Defer[Int][E^(-((x*(E^x + x + 10*(1 - Log[5/2]/10)))/(10 + x - Log[5/2]))), x])/15 - (52*De
fer[Int][E^(-((-(E^x*x) - 2*x^2 - 20*x*(1 - Log[5/2]/10))/(-10 - x + Log[5/2]))), x])/5 + (16*(10 - Log[5/2])*
Defer[Int][E^(-((-(E^x*x) - 2*x^2 - 20*x*(1 - Log[5/2]/10))/(-10 - x + Log[5/2]))), x])/15 + (16*Log[5/2]*Defe
r[Int][E^(-((-(E^x*x) - 2*x^2 - 20*x*(1 - Log[5/2]/10))/(-10 - x + Log[5/2]))), x])/15 - (4*Defer[Int][x/E^((x
*(E^x + x + 10*(1 - Log[5/2]/10)))/(10 + x - Log[5/2])), x])/15 - (8*Defer[Int][x/E^((-(E^x*x) - 2*x^2 - 20*x*
(1 - Log[5/2]/10))/(-10 - x + Log[5/2])), x])/15 + (4*(10 - Log[5/2])^2*Defer[Int][1/(E^((x*(E^x + x + 10*(1 -
Log[5/2]/10)))/(10 + x - Log[5/2]))*(10 + x - Log[5/2])^2), x])/15 + (80*Defer[Int][1/(E^((-(E^x*x) - 2*x^2 -
20*x*(1 - Log[5/2]/10))/(-10 - x + Log[5/2]))*(10 + x - Log[5/2])^2), x])/3 + 48*(10 - Log[5/2])*Defer[Int][1
/(E^((-(E^x*x) - 2*x^2 - 20*x*(1 - Log[5/2]/10))/(-10 - x + Log[5/2]))*(10 + x - Log[5/2])^2), x] - (52*(10 -
Log[5/2])^2*Defer[Int][1/(E^((-(E^x*x) - 2*x^2 - 20*x*(1 - Log[5/2]/10))/(-10 - x + Log[5/2]))*(10 + x - Log[5
/2])^2), x])/5 + (8*(10 - Log[5/2])^3*Defer[Int][1/(E^((-(E^x*x) - 2*x^2 - 20*x*(1 - Log[5/2]/10))/(-10 - x +
Log[5/2]))*(10 + x - Log[5/2])^2), x])/15 - (4*Log[5/2]^2*(21 - Log[25/4])*Defer[Int][1/(E^((-(E^x*x) - 2*x^2
- 20*x*(1 - Log[5/2]/10))/(-10 - x + Log[5/2]))*(10 + x - Log[5/2])^2), x])/15 - (4*(10 - Log[5/2])*(11 - Log[
5/2])*Defer[Int][1/(E^((x*(E^x + x + 10*(1 - Log[5/2]/10)))/(10 + x - Log[5/2]))*(10 + x - Log[5/2])), x])/15
- 48*Defer[Int][1/(E^((-(E^x*x) - 2*x^2 - 20*x*(1 - Log[5/2]/10))/(-10 - x + Log[5/2]))*(10 + x - Log[5/2])),
x] + (104*(10 - Log[5/2])*Defer[Int][1/(E^((-(E^x*x) - 2*x^2 - 20*x*(1 - Log[5/2]/10))/(-10 - x + Log[5/2]))*(
10 + x - Log[5/2])), x])/5 - (8*(10 - Log[5/2])^2*Defer[Int][1/(E^((-(E^x*x) - 2*x^2 - 20*x*(1 - Log[5/2]/10))
/(-10 - x + Log[5/2]))*(10 + x - Log[5/2])), x])/5 - (8*(21 - 4*Log[5/2])*Log[5/2]*Defer[Int][1/(E^((-(E^x*x)
- 2*x^2 - 20*x*(1 - Log[5/2]/10))/(-10 - x + Log[5/2]))*(10 + x - Log[5/2])), x])/15 - (8*Log[5/2]^2*Defer[Int
][1/(E^((-(E^x*x) - 2*x^2 - 20*x*(1 - Log[5/2]/10))/(-10 - x + Log[5/2]))*(10 + x - Log[5/2])), x])/15

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) \left (400-720 x-156 x^2-8 x^3+\left (-80+152 x+16 x^2\right ) \log \left (\frac {5}{2}\right )+(4-8 x) \log ^2\left (\frac {5}{2}\right )+e^x \left (-40 x-40 x^2-4 x^3+\left (4 x+4 x^2\right ) \log \left (\frac {5}{2}\right )\right )\right )}{15 x^2+30 x \left (10-\log \left (\frac {5}{2}\right )\right )+15 \left (10-\log \left (\frac {5}{2}\right )\right )^2} \, dx\\ &=\int \frac {\exp \left (-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) \left (400-720 x-156 x^2-8 x^3+\left (-80+152 x+16 x^2\right ) \log \left (\frac {5}{2}\right )+(4-8 x) \log ^2\left (\frac {5}{2}\right )+e^x \left (-40 x-40 x^2-4 x^3+\left (4 x+4 x^2\right ) \log \left (\frac {5}{2}\right )\right )\right )}{15 \left (10+x-\log \left (\frac {5}{2}\right )\right )^2} \, dx\\ &=\frac {1}{15} \int \frac {\exp \left (-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) \left (400-720 x-156 x^2-8 x^3+\left (-80+152 x+16 x^2\right ) \log \left (\frac {5}{2}\right )+(4-8 x) \log ^2\left (\frac {5}{2}\right )+e^x \left (-40 x-40 x^2-4 x^3+\left (4 x+4 x^2\right ) \log \left (\frac {5}{2}\right )\right )\right )}{\left (10+x-\log \left (\frac {5}{2}\right )\right )^2} \, dx\\ &=\frac {1}{15} \int \left (\frac {400 \exp \left (-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right )}{\left (10+x-\log \left (\frac {5}{2}\right )\right )^2}-\frac {720 \exp \left (-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) x}{\left (10+x-\log \left (\frac {5}{2}\right )\right )^2}-\frac {156 \exp \left (-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) x^2}{\left (10+x-\log \left (\frac {5}{2}\right )\right )^2}-\frac {8 \exp \left (-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) x^3}{\left (10+x-\log \left (\frac {5}{2}\right )\right )^2}+\frac {8 \exp \left (-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) (10+x) (-1+2 x) \log \left (\frac {5}{2}\right )}{\left (10+x-\log \left (\frac {5}{2}\right )\right )^2}-\frac {4 \exp \left (-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) (-1+2 x) \log ^2\left (\frac {5}{2}\right )}{\left (10+x-\log \left (\frac {5}{2}\right )\right )^2}+\frac {4 \exp \left (x-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) x \left (-10-x^2-x \left (10-\log \left (\frac {5}{2}\right )\right )+\log \left (\frac {5}{2}\right )\right )}{\left (10+x-\log \left (\frac {5}{2}\right )\right )^2}\right ) \, dx\\ &=\frac {4}{15} \int \frac {\exp \left (x-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) x \left (-10-x^2-x \left (10-\log \left (\frac {5}{2}\right )\right )+\log \left (\frac {5}{2}\right )\right )}{\left (10+x-\log \left (\frac {5}{2}\right )\right )^2} \, dx-\frac {8}{15} \int \frac {\exp \left (-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) x^3}{\left (10+x-\log \left (\frac {5}{2}\right )\right )^2} \, dx-\frac {52}{5} \int \frac {\exp \left (-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) x^2}{\left (10+x-\log \left (\frac {5}{2}\right )\right )^2} \, dx+\frac {80}{3} \int \frac {\exp \left (-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right )}{\left (10+x-\log \left (\frac {5}{2}\right )\right )^2} \, dx-48 \int \frac {\exp \left (-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) x}{\left (10+x-\log \left (\frac {5}{2}\right )\right )^2} \, dx+\frac {1}{15} \left (8 \log \left (\frac {5}{2}\right )\right ) \int \frac {\exp \left (-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) (10+x) (-1+2 x)}{\left (10+x-\log \left (\frac {5}{2}\right )\right )^2} \, dx-\frac {1}{15} \left (4 \log ^2\left (\frac {5}{2}\right )\right ) \int \frac {\exp \left (-\frac {-e^x x-2 x^2-20 x \left (1-\frac {1}{10} \log \left (\frac {5}{2}\right )\right )}{-10-x+\log \left (\frac {5}{2}\right )}\right ) (-1+2 x)}{\left (10+x-\log \left (\frac {5}{2}\right )\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [F]  time = 3.38, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{-\frac {-20 x-e^x x-2 x^2+2 x \log \left (\frac {5}{2}\right )}{-10-x+\log \left (\frac {5}{2}\right )}} \left (400-720 x-156 x^2-8 x^3+\left (-80+152 x+16 x^2\right ) \log \left (\frac {5}{2}\right )+(4-8 x) \log ^2\left (\frac {5}{2}\right )+e^x \left (-40 x-40 x^2-4 x^3+\left (4 x+4 x^2\right ) \log \left (\frac {5}{2}\right )\right )\right )}{1500+300 x+15 x^2+(-300-30 x) \log \left (\frac {5}{2}\right )+15 \log ^2\left (\frac {5}{2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(400 - 720*x - 156*x^2 - 8*x^3 + (-80 + 152*x + 16*x^2)*Log[5/2] + (4 - 8*x)*Log[5/2]^2 + E^x*(-40*x
- 40*x^2 - 4*x^3 + (4*x + 4*x^2)*Log[5/2]))/(E^((-20*x - E^x*x - 2*x^2 + 2*x*Log[5/2])/(-10 - x + Log[5/2]))*
(1500 + 300*x + 15*x^2 + (-300 - 30*x)*Log[5/2] + 15*Log[5/2]^2)),x]

[Out]

Integrate[(400 - 720*x - 156*x^2 - 8*x^3 + (-80 + 152*x + 16*x^2)*Log[5/2] + (4 - 8*x)*Log[5/2]^2 + E^x*(-40*x
- 40*x^2 - 4*x^3 + (4*x + 4*x^2)*Log[5/2]))/(E^((-20*x - E^x*x - 2*x^2 + 2*x*Log[5/2])/(-10 - x + Log[5/2]))*
(1500 + 300*x + 15*x^2 + (-300 - 30*x)*Log[5/2] + 15*Log[5/2]^2)), x]

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fricas [A]  time = 0.60, size = 33, normalized size = 1.22 \begin {gather*} \frac {4}{15} \, x e^{\left (-\frac {2 \, x^{2} + x e^{x} - 2 \, x \log \left (\frac {5}{2}\right ) + 20 \, x}{x - \log \left (\frac {5}{2}\right ) + 10}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2+4*x)*log(5/2)-4*x^3-40*x^2-40*x)*exp(x)+(-8*x+4)*log(5/2)^2+(16*x^2+152*x-80)*log(5/2)-8*x^
3-156*x^2-720*x+400)/(15*log(5/2)^2+(-30*x-300)*log(5/2)+15*x^2+300*x+1500)/exp((-exp(x)*x+2*x*log(5/2)-2*x^2-
20*x)/(log(5/2)-x-10)),x, algorithm="fricas")

[Out]

4/15*x*e^(-(2*x^2 + x*e^x - 2*x*log(5/2) + 20*x)/(x - log(5/2) + 10))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {4 \, {\left (2 \, x^{3} + {\left (2 \, x - 1\right )} \log \left (\frac {5}{2}\right )^{2} + 39 \, x^{2} + {\left (x^{3} + 10 \, x^{2} - {\left (x^{2} + x\right )} \log \left (\frac {5}{2}\right ) + 10 \, x\right )} e^{x} - 2 \, {\left (2 \, x^{2} + 19 \, x - 10\right )} \log \left (\frac {5}{2}\right ) + 180 \, x - 100\right )} e^{\left (-\frac {2 \, x^{2} + x e^{x} - 2 \, x \log \left (\frac {5}{2}\right ) + 20 \, x}{x - \log \left (\frac {5}{2}\right ) + 10}\right )}}{15 \, {\left (x^{2} - 2 \, {\left (x + 10\right )} \log \left (\frac {5}{2}\right ) + \log \left (\frac {5}{2}\right )^{2} + 20 \, x + 100\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2+4*x)*log(5/2)-4*x^3-40*x^2-40*x)*exp(x)+(-8*x+4)*log(5/2)^2+(16*x^2+152*x-80)*log(5/2)-8*x^
3-156*x^2-720*x+400)/(15*log(5/2)^2+(-30*x-300)*log(5/2)+15*x^2+300*x+1500)/exp((-exp(x)*x+2*x*log(5/2)-2*x^2-
20*x)/(log(5/2)-x-10)),x, algorithm="giac")

[Out]

integrate(-4/15*(2*x^3 + (2*x - 1)*log(5/2)^2 + 39*x^2 + (x^3 + 10*x^2 - (x^2 + x)*log(5/2) + 10*x)*e^x - 2*(2
*x^2 + 19*x - 10)*log(5/2) + 180*x - 100)*e^(-(2*x^2 + x*e^x - 2*x*log(5/2) + 20*x)/(x - log(5/2) + 10))/(x^2
- 2*(x + 10)*log(5/2) + log(5/2)^2 + 20*x + 100), x)

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maple [A]  time = 0.53, size = 34, normalized size = 1.26

method	result	size
risch	\(\frac {4 x \,{\mathrm e}^{-\frac {x \left ({\mathrm e}^{x}-2 \ln \relax (5)+2 \ln \relax (2)+2 x +20\right )}{-\ln \relax (5)+\ln \relax (2)+x +10}}}{15}\)	\(34\)
norman	\(\frac {\left (\left (-\frac {8}{3}+\frac {4 \ln \relax (5)}{15}-\frac {4 \ln \relax (2)}{15}\right ) x -\frac {4 x^{2}}{15}\right ) {\mathrm e}^{-\frac {-{\mathrm e}^{x} x +2 x \ln \left (\frac {5}{2}\right )-2 x^{2}-20 x}{\ln \left (\frac {5}{2}\right )-x -10}}}{\ln \left (\frac {5}{2}\right )-x -10}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x^2+4*x)*ln(5/2)-4*x^3-40*x^2-40*x)*exp(x)+(-8*x+4)*ln(5/2)^2+(16*x^2+152*x-80)*ln(5/2)-8*x^3-156*x^2
-720*x+400)/(15*ln(5/2)^2+(-30*x-300)*ln(5/2)+15*x^2+300*x+1500)/exp((-exp(x)*x+2*x*ln(5/2)-2*x^2-20*x)/(ln(5/
2)-x-10)),x,method=_RETURNVERBOSE)

[Out]

4/15*x*exp(-x*(exp(x)-2*ln(5)+2*ln(2)+2*x+20)/(-ln(5)+ln(2)+x+10))

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maxima [B]  time = 0.90, size = 60, normalized size = 2.22 \begin {gather*} \frac {4}{15} \, x e^{\left (-2 \, x - \frac {e^{x} \log \relax (5)}{x - \log \relax (5) + \log \relax (2) + 10} + \frac {e^{x} \log \relax (2)}{x - \log \relax (5) + \log \relax (2) + 10} + \frac {10 \, e^{x}}{x - \log \relax (5) + \log \relax (2) + 10} - e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2+4*x)*log(5/2)-4*x^3-40*x^2-40*x)*exp(x)+(-8*x+4)*log(5/2)^2+(16*x^2+152*x-80)*log(5/2)-8*x^
3-156*x^2-720*x+400)/(15*log(5/2)^2+(-30*x-300)*log(5/2)+15*x^2+300*x+1500)/exp((-exp(x)*x+2*x*log(5/2)-2*x^2-
20*x)/(log(5/2)-x-10)),x, algorithm="maxima")

[Out]

4/15*x*e^(-2*x - e^x*log(5)/(x - log(5) + log(2) + 10) + e^x*log(2)/(x - log(5) + log(2) + 10) + 10*e^x/(x - l
og(5) + log(2) + 10) - e^x)

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(20*x - 2*x*log(5/2) + x*exp(x) + 2*x^2)/(x - log(5/2) + 10))*(720*x + exp(x)*(40*x - log(5/2)*(4*x
+ 4*x^2) + 40*x^2 + 4*x^3) - log(5/2)*(152*x + 16*x^2 - 80) + log(5/2)^2*(8*x - 4) + 156*x^2 + 8*x^3 - 400))/
(300*x - log(5/2)*(30*x + 300) + 15*log(5/2)^2 + 15*x^2 + 1500),x)

[Out]

\text{Hanged}

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sympy [A]  time = 124.09, size = 36, normalized size = 1.33 \begin {gather*} \frac {4 x e^{- \frac {- 2 x^{2} - x e^{x} - 20 x + 2 x \log {\left (\frac {5}{2} \right )}}{- x - 10 + \log {\left (\frac {5}{2} \right )}}}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x**2+4*x)*ln(5/2)-4*x**3-40*x**2-40*x)*exp(x)+(-8*x+4)*ln(5/2)**2+(16*x**2+152*x-80)*ln(5/2)-8*
x**3-156*x**2-720*x+400)/(15*ln(5/2)**2+(-30*x-300)*ln(5/2)+15*x**2+300*x+1500)/exp((-exp(x)*x+2*x*ln(5/2)-2*x
**2-20*x)/(ln(5/2)-x-10)),x)

[Out]

4*x*exp(-(-2*x**2 - x*exp(x) - 20*x + 2*x*log(5/2))/(-x - 10 + log(5/2)))/15

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