3.40.56 \(\int \frac {-40 e^x x^2+e^{2 x} (100 x+140 x^2+40 x^3)+e^{1+x} (40 e^x x^2+e^{2 x} (-40 x-40 x^2))+(40 x^2+e^x (-200 x-180 x^2-40 x^3)+e^{1+x} (-40 x^2+e^x (80 x+40 x^2))) \log (10-4 e^{1+x}+4 x)+(100 x-40 e^{1+x} x+40 x^2) \log ^2(10-4 e^{1+x}+4 x)}{-125-50 x+e^{2 x} (-50 x^2-20 x^3)+e^{1+x} (50+20 e^{2 x} x^2+2 e^{4 x} x^4)+e^{4 x} (-5 x^4-2 x^5)+(e^x (100 x^2+40 x^3)+e^{1+x} (-40 e^x x^2-8 e^{3 x} x^4)+e^{3 x} (20 x^4+8 x^5)) \log (10-4 e^{1+x}+4 x)+(-50 x^2-20 x^3+e^{1+x} (20 x^2+12 e^{2 x} x^4)+e^{2 x} (-30 x^4-12 x^5)) \log ^2(10-4 e^{1+x}+4 x)+(-8 e^{1+2 x} x^4+e^x (20 x^4+8 x^5)) \log ^3(10-4 e^{1+x}+4 x)+(-5 x^4+2 e^{1+x} x^4-2 x^5) \log ^4(10-4 e^{1+x}+4 x)} \, dx\)

Optimal. Leaf size=35 \[ \frac {2}{1+\frac {1}{5} x^2 \left (-e^x+\log \left (10+4 \left (-e^{1+x}+x\right )\right )\right )^2} \]

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Rubi [F]  time = 180.00, 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*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-40*E^x*x^2 + E^(2*x)*(100*x + 140*x^2 + 40*x^3) + E^(1 + x)*(40*E^x*x^2 + E^(2*x)*(-40*x - 40*x^2)) + (4
0*x^2 + E^x*(-200*x - 180*x^2 - 40*x^3) + E^(1 + x)*(-40*x^2 + E^x*(80*x + 40*x^2)))*Log[10 - 4*E^(1 + x) + 4*
x] + (100*x - 40*E^(1 + x)*x + 40*x^2)*Log[10 - 4*E^(1 + x) + 4*x]^2)/(-125 - 50*x + E^(2*x)*(-50*x^2 - 20*x^3
) + E^(1 + x)*(50 + 20*E^(2*x)*x^2 + 2*E^(4*x)*x^4) + E^(4*x)*(-5*x^4 - 2*x^5) + (E^x*(100*x^2 + 40*x^3) + E^(
1 + x)*(-40*E^x*x^2 - 8*E^(3*x)*x^4) + E^(3*x)*(20*x^4 + 8*x^5))*Log[10 - 4*E^(1 + x) + 4*x] + (-50*x^2 - 20*x
^3 + E^(1 + x)*(20*x^2 + 12*E^(2*x)*x^4) + E^(2*x)*(-30*x^4 - 12*x^5))*Log[10 - 4*E^(1 + x) + 4*x]^2 + (-8*E^(
1 + 2*x)*x^4 + E^x*(20*x^4 + 8*x^5))*Log[10 - 4*E^(1 + x) + 4*x]^3 + (-5*x^4 + 2*E^(1 + x)*x^4 - 2*x^5)*Log[10
- 4*E^(1 + x) + 4*x]^4),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [A]  time = 0.28, size = 55, normalized size = 1.57 \begin {gather*} \frac {10}{5+e^{2 x} x^2-2 e^x x^2 \log \left (10-4 e^{1+x}+4 x\right )+x^2 \log ^2\left (10-4 e^{1+x}+4 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-40*E^x*x^2 + E^(2*x)*(100*x + 140*x^2 + 40*x^3) + E^(1 + x)*(40*E^x*x^2 + E^(2*x)*(-40*x - 40*x^2)
) + (40*x^2 + E^x*(-200*x - 180*x^2 - 40*x^3) + E^(1 + x)*(-40*x^2 + E^x*(80*x + 40*x^2)))*Log[10 - 4*E^(1 + x
) + 4*x] + (100*x - 40*E^(1 + x)*x + 40*x^2)*Log[10 - 4*E^(1 + x) + 4*x]^2)/(-125 - 50*x + E^(2*x)*(-50*x^2 -
20*x^3) + E^(1 + x)*(50 + 20*E^(2*x)*x^2 + 2*E^(4*x)*x^4) + E^(4*x)*(-5*x^4 - 2*x^5) + (E^x*(100*x^2 + 40*x^3)
+ E^(1 + x)*(-40*E^x*x^2 - 8*E^(3*x)*x^4) + E^(3*x)*(20*x^4 + 8*x^5))*Log[10 - 4*E^(1 + x) + 4*x] + (-50*x^2
- 20*x^3 + E^(1 + x)*(20*x^2 + 12*E^(2*x)*x^4) + E^(2*x)*(-30*x^4 - 12*x^5))*Log[10 - 4*E^(1 + x) + 4*x]^2 + (
-8*E^(1 + 2*x)*x^4 + E^x*(20*x^4 + 8*x^5))*Log[10 - 4*E^(1 + x) + 4*x]^3 + (-5*x^4 + 2*E^(1 + x)*x^4 - 2*x^5)*
Log[10 - 4*E^(1 + x) + 4*x]^4),x]

[Out]

10/(5 + E^(2*x)*x^2 - 2*E^x*x^2*Log[10 - 4*E^(1 + x) + 4*x] + x^2*Log[10 - 4*E^(1 + x) + 4*x]^2)

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fricas [B]  time = 0.75, size = 62, normalized size = 1.77 \begin {gather*} \frac {10 \, e^{2}}{x^{2} e^{2} \log \left (4 \, x - 4 \, e^{\left (x + 1\right )} + 10\right )^{2} - 2 \, x^{2} e^{\left (x + 2\right )} \log \left (4 \, x - 4 \, e^{\left (x + 1\right )} + 10\right ) + x^{2} e^{\left (2 \, x + 2\right )} + 5 \, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x*exp(x+1)+40*x^2+100*x)*log(-4*exp(x+1)+4*x+10)^2+(((40*x^2+80*x)*exp(x)-40*x^2)*exp(x+1)+(-4
0*x^3-180*x^2-200*x)*exp(x)+40*x^2)*log(-4*exp(x+1)+4*x+10)+((-40*x^2-40*x)*exp(x)^2+40*exp(x)*x^2)*exp(x+1)+(
40*x^3+140*x^2+100*x)*exp(x)^2-40*exp(x)*x^2)/((2*x^4*exp(x+1)-2*x^5-5*x^4)*log(-4*exp(x+1)+4*x+10)^4+(-8*x^4*
exp(x)*exp(x+1)+(8*x^5+20*x^4)*exp(x))*log(-4*exp(x+1)+4*x+10)^3+((12*exp(x)^2*x^4+20*x^2)*exp(x+1)+(-12*x^5-3
0*x^4)*exp(x)^2-20*x^3-50*x^2)*log(-4*exp(x+1)+4*x+10)^2+((-8*x^4*exp(x)^3-40*exp(x)*x^2)*exp(x+1)+(8*x^5+20*x
^4)*exp(x)^3+(40*x^3+100*x^2)*exp(x))*log(-4*exp(x+1)+4*x+10)+(2*x^4*exp(x)^4+20*exp(x)^2*x^2+50)*exp(x+1)+(-2
*x^5-5*x^4)*exp(x)^4+(-20*x^3-50*x^2)*exp(x)^2-50*x-125),x, algorithm="fricas")

[Out]

10*e^2/(x^2*e^2*log(4*x - 4*e^(x + 1) + 10)^2 - 2*x^2*e^(x + 2)*log(4*x - 4*e^(x + 1) + 10) + x^2*e^(2*x + 2)
+ 5*e^2)

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giac [B]  time = 45.25, size = 90, normalized size = 2.57 \begin {gather*} -\frac {20}{2 \, x^{2} e^{x} \log \relax (2) - x^{2} \log \relax (2)^{2} + 2 \, x^{2} e^{x} \log \left (2 \, x - 2 \, e^{\left (x + 1\right )} + 5\right ) - 2 \, x^{2} \log \relax (2) \log \left (2 \, x - 2 \, e^{\left (x + 1\right )} + 5\right ) - x^{2} \log \left (2 \, x - 2 \, e^{\left (x + 1\right )} + 5\right )^{2} - x^{2} e^{\left (2 \, x\right )} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x*exp(x+1)+40*x^2+100*x)*log(-4*exp(x+1)+4*x+10)^2+(((40*x^2+80*x)*exp(x)-40*x^2)*exp(x+1)+(-4
0*x^3-180*x^2-200*x)*exp(x)+40*x^2)*log(-4*exp(x+1)+4*x+10)+((-40*x^2-40*x)*exp(x)^2+40*exp(x)*x^2)*exp(x+1)+(
40*x^3+140*x^2+100*x)*exp(x)^2-40*exp(x)*x^2)/((2*x^4*exp(x+1)-2*x^5-5*x^4)*log(-4*exp(x+1)+4*x+10)^4+(-8*x^4*
exp(x)*exp(x+1)+(8*x^5+20*x^4)*exp(x))*log(-4*exp(x+1)+4*x+10)^3+((12*exp(x)^2*x^4+20*x^2)*exp(x+1)+(-12*x^5-3
0*x^4)*exp(x)^2-20*x^3-50*x^2)*log(-4*exp(x+1)+4*x+10)^2+((-8*x^4*exp(x)^3-40*exp(x)*x^2)*exp(x+1)+(8*x^5+20*x
^4)*exp(x)^3+(40*x^3+100*x^2)*exp(x))*log(-4*exp(x+1)+4*x+10)+(2*x^4*exp(x)^4+20*exp(x)^2*x^2+50)*exp(x+1)+(-2
*x^5-5*x^4)*exp(x)^4+(-20*x^3-50*x^2)*exp(x)^2-50*x-125),x, algorithm="giac")

[Out]

-20/(2*x^2*e^x*log(2) - x^2*log(2)^2 + 2*x^2*e^x*log(2*x - 2*e^(x + 1) + 5) - 2*x^2*log(2)*log(2*x - 2*e^(x +
1) + 5) - x^2*log(2*x - 2*e^(x + 1) + 5)^2 - x^2*e^(2*x) - 5)

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maple [A]  time = 0.09, size = 52, normalized size = 1.49

method	result	size
risch	\(\frac {10}{{\mathrm e}^{2 x} x^{2}-2 \ln \left (-4 \,{\mathrm e}^{x +1}+4 x +10\right ) {\mathrm e}^{x} x^{2}+\ln \left (-4 \,{\mathrm e}^{x +1}+4 x +10\right )^{2} x^{2}+5}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-40*x*exp(x+1)+40*x^2+100*x)*ln(-4*exp(x+1)+4*x+10)^2+(((40*x^2+80*x)*exp(x)-40*x^2)*exp(x+1)+(-40*x^3-1
80*x^2-200*x)*exp(x)+40*x^2)*ln(-4*exp(x+1)+4*x+10)+((-40*x^2-40*x)*exp(x)^2+40*exp(x)*x^2)*exp(x+1)+(40*x^3+1
40*x^2+100*x)*exp(x)^2-40*exp(x)*x^2)/((2*x^4*exp(x+1)-2*x^5-5*x^4)*ln(-4*exp(x+1)+4*x+10)^4+(-8*x^4*exp(x)*ex
p(x+1)+(8*x^5+20*x^4)*exp(x))*ln(-4*exp(x+1)+4*x+10)^3+((12*exp(x)^2*x^4+20*x^2)*exp(x+1)+(-12*x^5-30*x^4)*exp
(x)^2-20*x^3-50*x^2)*ln(-4*exp(x+1)+4*x+10)^2+((-8*x^4*exp(x)^3-40*exp(x)*x^2)*exp(x+1)+(8*x^5+20*x^4)*exp(x)^
3+(40*x^3+100*x^2)*exp(x))*ln(-4*exp(x+1)+4*x+10)+(2*x^4*exp(x)^4+20*exp(x)^2*x^2+50)*exp(x+1)+(-2*x^5-5*x^4)*
exp(x)^4+(-20*x^3-50*x^2)*exp(x)^2-50*x-125),x,method=_RETURNVERBOSE)

[Out]

10/(exp(2*x)*x^2-2*ln(-4*exp(x+1)+4*x+10)*exp(x)*x^2+ln(-4*exp(x+1)+4*x+10)^2*x^2+5)

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maxima [C]  time = 1.03, size = 99, normalized size = 2.83 \begin {gather*} -\frac {10}{2 \, {\left (i \, \pi + \log \relax (2)\right )} x^{2} e^{x} - x^{2} \log \left (-2 \, x + 2 \, e^{\left (x + 1\right )} - 5\right )^{2} + {\left (\pi ^{2} - 2 i \, \pi \log \relax (2) - \log \relax (2)^{2}\right )} x^{2} - x^{2} e^{\left (2 \, x\right )} + 2 \, {\left ({\left (-i \, \pi - \log \relax (2)\right )} x^{2} + x^{2} e^{x}\right )} \log \left (-2 \, x + 2 \, e^{\left (x + 1\right )} - 5\right ) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x*exp(x+1)+40*x^2+100*x)*log(-4*exp(x+1)+4*x+10)^2+(((40*x^2+80*x)*exp(x)-40*x^2)*exp(x+1)+(-4
0*x^3-180*x^2-200*x)*exp(x)+40*x^2)*log(-4*exp(x+1)+4*x+10)+((-40*x^2-40*x)*exp(x)^2+40*exp(x)*x^2)*exp(x+1)+(
40*x^3+140*x^2+100*x)*exp(x)^2-40*exp(x)*x^2)/((2*x^4*exp(x+1)-2*x^5-5*x^4)*log(-4*exp(x+1)+4*x+10)^4+(-8*x^4*
exp(x)*exp(x+1)+(8*x^5+20*x^4)*exp(x))*log(-4*exp(x+1)+4*x+10)^3+((12*exp(x)^2*x^4+20*x^2)*exp(x+1)+(-12*x^5-3
0*x^4)*exp(x)^2-20*x^3-50*x^2)*log(-4*exp(x+1)+4*x+10)^2+((-8*x^4*exp(x)^3-40*exp(x)*x^2)*exp(x+1)+(8*x^5+20*x
^4)*exp(x)^3+(40*x^3+100*x^2)*exp(x))*log(-4*exp(x+1)+4*x+10)+(2*x^4*exp(x)^4+20*exp(x)^2*x^2+50)*exp(x+1)+(-2
*x^5-5*x^4)*exp(x)^4+(-20*x^3-50*x^2)*exp(x)^2-50*x-125),x, algorithm="maxima")

[Out]

-10/(2*(I*pi + log(2))*x^2*e^x - x^2*log(-2*x + 2*e^(x + 1) - 5)^2 + (pi^2 - 2*I*pi*log(2) - log(2)^2)*x^2 - x
^2*e^(2*x) + 2*((-I*pi - log(2))*x^2 + x^2*e^x)*log(-2*x + 2*e^(x + 1) - 5) - 5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{2\,x}\,\left (40\,x^3+140\,x^2+100\,x\right )-{\mathrm {e}}^{x+1}\,\left ({\mathrm {e}}^{2\,x}\,\left (40\,x^2+40\,x\right )-40\,x^2\,{\mathrm {e}}^x\right )-40\,x^2\,{\mathrm {e}}^x+{\ln \left (4\,x-4\,{\mathrm {e}}^{x+1}+10\right )}^2\,\left (100\,x-40\,x\,{\mathrm {e}}^{x+1}+40\,x^2\right )+\ln \left (4\,x-4\,{\mathrm {e}}^{x+1}+10\right )\,\left (40\,x^2+{\mathrm {e}}^{x+1}\,\left ({\mathrm {e}}^x\,\left (40\,x^2+80\,x\right )-40\,x^2\right )-{\mathrm {e}}^x\,\left (40\,x^3+180\,x^2+200\,x\right )\right )}{\left (5\,x^4-2\,x^4\,{\mathrm {e}}^{x+1}+2\,x^5\right )\,{\ln \left (4\,x-4\,{\mathrm {e}}^{x+1}+10\right )}^4+\left (8\,x^4\,{\mathrm {e}}^{x+1}\,{\mathrm {e}}^x-{\mathrm {e}}^x\,\left (8\,x^5+20\,x^4\right )\right )\,{\ln \left (4\,x-4\,{\mathrm {e}}^{x+1}+10\right )}^3+\left ({\mathrm {e}}^{2\,x}\,\left (12\,x^5+30\,x^4\right )-{\mathrm {e}}^{x+1}\,\left (12\,x^4\,{\mathrm {e}}^{2\,x}+20\,x^2\right )+50\,x^2+20\,x^3\right )\,{\ln \left (4\,x-4\,{\mathrm {e}}^{x+1}+10\right )}^2+\left ({\mathrm {e}}^{x+1}\,\left (40\,x^2\,{\mathrm {e}}^x+8\,x^4\,{\mathrm {e}}^{3\,x}\right )-{\mathrm {e}}^{3\,x}\,\left (8\,x^5+20\,x^4\right )-{\mathrm {e}}^x\,\left (40\,x^3+100\,x^2\right )\right )\,\ln \left (4\,x-4\,{\mathrm {e}}^{x+1}+10\right )+50\,x+{\mathrm {e}}^{4\,x}\,\left (2\,x^5+5\,x^4\right )+{\mathrm {e}}^{2\,x}\,\left (20\,x^3+50\,x^2\right )-{\mathrm {e}}^{x+1}\,\left (20\,x^2\,{\mathrm {e}}^{2\,x}+2\,x^4\,{\mathrm {e}}^{4\,x}+50\right )+125} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x)*(100*x + 140*x^2 + 40*x^3) - exp(x + 1)*(exp(2*x)*(40*x + 40*x^2) - 40*x^2*exp(x)) - 40*x^2*exp
(x) + log(4*x - 4*exp(x + 1) + 10)^2*(100*x - 40*x*exp(x + 1) + 40*x^2) + log(4*x - 4*exp(x + 1) + 10)*(40*x^2
+ exp(x + 1)*(exp(x)*(80*x + 40*x^2) - 40*x^2) - exp(x)*(200*x + 180*x^2 + 40*x^3)))/(50*x + log(4*x - 4*exp(
x + 1) + 10)^4*(5*x^4 - 2*x^4*exp(x + 1) + 2*x^5) + exp(4*x)*(5*x^4 + 2*x^5) + exp(2*x)*(50*x^2 + 20*x^3) - lo
g(4*x - 4*exp(x + 1) + 10)^3*(exp(x)*(20*x^4 + 8*x^5) - 8*x^4*exp(x + 1)*exp(x)) + log(4*x - 4*exp(x + 1) + 10
)^2*(exp(2*x)*(30*x^4 + 12*x^5) - exp(x + 1)*(12*x^4*exp(2*x) + 20*x^2) + 50*x^2 + 20*x^3) - exp(x + 1)*(20*x^
2*exp(2*x) + 2*x^4*exp(4*x) + 50) - log(4*x - 4*exp(x + 1) + 10)*(exp(x)*(100*x^2 + 40*x^3) + exp(3*x)*(20*x^4
+ 8*x^5) - exp(x + 1)*(40*x^2*exp(x) + 8*x^4*exp(3*x))) + 125),x)

[Out]

int(-(exp(2*x)*(100*x + 140*x^2 + 40*x^3) - exp(x + 1)*(exp(2*x)*(40*x + 40*x^2) - 40*x^2*exp(x)) - 40*x^2*exp
(x) + log(4*x - 4*exp(x + 1) + 10)^2*(100*x - 40*x*exp(x + 1) + 40*x^2) + log(4*x - 4*exp(x + 1) + 10)*(40*x^2
+ exp(x + 1)*(exp(x)*(80*x + 40*x^2) - 40*x^2) - exp(x)*(200*x + 180*x^2 + 40*x^3)))/(50*x + log(4*x - 4*exp(
x + 1) + 10)^4*(5*x^4 - 2*x^4*exp(x + 1) + 2*x^5) + exp(4*x)*(5*x^4 + 2*x^5) + exp(2*x)*(50*x^2 + 20*x^3) - lo
g(4*x - 4*exp(x + 1) + 10)^3*(exp(x)*(20*x^4 + 8*x^5) - 8*x^4*exp(x + 1)*exp(x)) + log(4*x - 4*exp(x + 1) + 10
)^2*(exp(2*x)*(30*x^4 + 12*x^5) - exp(x + 1)*(12*x^4*exp(2*x) + 20*x^2) + 50*x^2 + 20*x^3) - exp(x + 1)*(20*x^
2*exp(2*x) + 2*x^4*exp(4*x) + 50) - log(4*x - 4*exp(x + 1) + 10)*(exp(x)*(100*x^2 + 40*x^3) + exp(3*x)*(20*x^4
+ 8*x^5) - exp(x + 1)*(40*x^2*exp(x) + 8*x^4*exp(3*x))) + 125), x)

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sympy [B]  time = 0.79, size = 54, normalized size = 1.54 \begin {gather*} \frac {10}{x^{2} e^{2 x} - 2 x^{2} e^{x} \log {\left (4 x - 4 e e^{x} + 10 \right )} + x^{2} \log {\left (4 x - 4 e e^{x} + 10 \right )}^{2} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x*exp(x+1)+40*x**2+100*x)*ln(-4*exp(x+1)+4*x+10)**2+(((40*x**2+80*x)*exp(x)-40*x**2)*exp(x+1)+
(-40*x**3-180*x**2-200*x)*exp(x)+40*x**2)*ln(-4*exp(x+1)+4*x+10)+((-40*x**2-40*x)*exp(x)**2+40*exp(x)*x**2)*ex
p(x+1)+(40*x**3+140*x**2+100*x)*exp(x)**2-40*exp(x)*x**2)/((2*x**4*exp(x+1)-2*x**5-5*x**4)*ln(-4*exp(x+1)+4*x+
10)**4+(-8*x**4*exp(x)*exp(x+1)+(8*x**5+20*x**4)*exp(x))*ln(-4*exp(x+1)+4*x+10)**3+((12*exp(x)**2*x**4+20*x**2
)*exp(x+1)+(-12*x**5-30*x**4)*exp(x)**2-20*x**3-50*x**2)*ln(-4*exp(x+1)+4*x+10)**2+((-8*x**4*exp(x)**3-40*exp(
x)*x**2)*exp(x+1)+(8*x**5+20*x**4)*exp(x)**3+(40*x**3+100*x**2)*exp(x))*ln(-4*exp(x+1)+4*x+10)+(2*x**4*exp(x)*
*4+20*exp(x)**2*x**2+50)*exp(x+1)+(-2*x**5-5*x**4)*exp(x)**4+(-20*x**3-50*x**2)*exp(x)**2-50*x-125),x)

[Out]

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

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