3.60.19 \(\int \frac {250+400 x+180 x^2+32 x^3+2 x^4+e^{2 e^{2 x}} (-2+e^{2 x} (20+4 x))+(250+150 x+30 x^2+2 x^3) \log ^4(3)+e^{e^{2 x}} (-40-8 x+(10+2 x) \log ^4(3)+e^{2 x} (-100-140 x-44 x^2-4 x^3+(-100-40 x-4 x^2) \log ^4(3)))}{125+75 x+15 x^2+x^3} \, dx\)

Optimal. Leaf size=26 \[ \left (-1-x+\frac {e^{e^{2 x}}}{5+x}-\log ^4(3)\right )^2 \]

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Rubi [F]  time = 2.30, 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 {gather*} \int \frac {250+400 x+180 x^2+32 x^3+2 x^4+e^{2 e^{2 x}} \left (-2+e^{2 x} (20+4 x)\right )+\left (250+150 x+30 x^2+2 x^3\right ) \log ^4(3)+e^{e^{2 x}} \left (-40-8 x+(10+2 x) \log ^4(3)+e^{2 x} \left (-100-140 x-44 x^2-4 x^3+\left (-100-40 x-4 x^2\right ) \log ^4(3)\right )\right )}{125+75 x+15 x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(250 + 400*x + 180*x^2 + 32*x^3 + 2*x^4 + E^(2*E^(2*x))*(-2 + E^(2*x)*(20 + 4*x)) + (250 + 150*x + 30*x^2
+ 2*x^3)*Log[3]^4 + E^E^(2*x)*(-40 - 8*x + (10 + 2*x)*Log[3]^4 + E^(2*x)*(-100 - 140*x - 44*x^2 - 4*x^3 + (-10
0 - 40*x - 4*x^2)*Log[3]^4)))/(125 + 75*x + 15*x^2 + x^3),x]

[Out]

-2*E^E^(2*x) + x^2 + 2*x*(1 + Log[3]^4) - 2*Defer[Int][E^(2*E^(2*x))/(5 + x)^3, x] - 2*(4 - Log[3]^4)*Defer[In
t][E^E^(2*x)/(5 + x)^2, x] + 4*Defer[Int][E^(2*(E^(2*x) + x))/(5 + x)^2, x] + 4*(4 - Log[3]^4)*Defer[Int][E^(E
^(2*x) + 2*x)/(5 + x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (e^{e^{2 x}}-2 e^{e^{2 x}+2 x} (5+x)+(5+x)^2\right ) \left (-e^{e^{2 x}}+(5+x) \left (1+x+\log ^4(3)\right )\right )}{(5+x)^3} \, dx\\ &=2 \int \frac {\left (e^{e^{2 x}}-2 e^{e^{2 x}+2 x} (5+x)+(5+x)^2\right ) \left (-e^{e^{2 x}}+(5+x) \left (1+x+\log ^4(3)\right )\right )}{(5+x)^3} \, dx\\ &=2 \int \left (\frac {2 e^{e^{2 x}+2 x} \left (e^{e^{2 x}}-x^2-6 x \left (1+\frac {\log ^4(3)}{6}\right )-5 \left (1+\log ^4(3)\right )\right )}{(5+x)^2}+\frac {\left (25+e^{e^{2 x}}+10 x+x^2\right ) \left (-e^{e^{2 x}}+x^2+6 x \left (1+\frac {\log ^4(3)}{6}\right )+5 \left (1+\log ^4(3)\right )\right )}{(5+x)^3}\right ) \, dx\\ &=2 \int \frac {\left (25+e^{e^{2 x}}+10 x+x^2\right ) \left (-e^{e^{2 x}}+x^2+6 x \left (1+\frac {\log ^4(3)}{6}\right )+5 \left (1+\log ^4(3)\right )\right )}{(5+x)^3} \, dx+4 \int \frac {e^{e^{2 x}+2 x} \left (e^{e^{2 x}}-x^2-6 x \left (1+\frac {\log ^4(3)}{6}\right )-5 \left (1+\log ^4(3)\right )\right )}{(5+x)^2} \, dx\\ &=2 \int \frac {\left (e^{e^{2 x}}+(5+x)^2\right ) \left (-e^{e^{2 x}}+(5+x) \left (1+x+\log ^4(3)\right )\right )}{(5+x)^3} \, dx+4 \int \frac {e^{e^{2 x}+2 x} \left (e^{e^{2 x}}-(5+x) \left (1+x+\log ^4(3)\right )\right )}{(5+x)^2} \, dx\\ &=2 \int \left (1+x-\frac {e^{2 e^{2 x}}}{(5+x)^3}+\log ^4(3)+\frac {e^{e^{2 x}} \left (-4+\log ^4(3)\right )}{(5+x)^2}\right ) \, dx+4 \int \left (\frac {e^{2 e^{2 x}+2 x}}{(5+x)^2}+\frac {e^{e^{2 x}+2 x} \left (-1-x-\log ^4(3)\right )}{5+x}\right ) \, dx\\ &=x^2+2 x \left (1+\log ^4(3)\right )-2 \int \frac {e^{2 e^{2 x}}}{(5+x)^3} \, dx+4 \int \frac {e^{2 e^{2 x}+2 x}}{(5+x)^2} \, dx+4 \int \frac {e^{e^{2 x}+2 x} \left (-1-x-\log ^4(3)\right )}{5+x} \, dx-\left (2 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}}}{(5+x)^2} \, dx\\ &=x^2+2 x \left (1+\log ^4(3)\right )-2 \int \frac {e^{2 e^{2 x}}}{(5+x)^3} \, dx+4 \int \frac {e^{2 \left (e^{2 x}+x\right )}}{(5+x)^2} \, dx+4 \int \left (-e^{e^{2 x}+2 x}+\frac {e^{e^{2 x}+2 x} \left (4-\log ^4(3)\right )}{5+x}\right ) \, dx-\left (2 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}}}{(5+x)^2} \, dx\\ &=x^2+2 x \left (1+\log ^4(3)\right )-2 \int \frac {e^{2 e^{2 x}}}{(5+x)^3} \, dx-4 \int e^{e^{2 x}+2 x} \, dx+4 \int \frac {e^{2 \left (e^{2 x}+x\right )}}{(5+x)^2} \, dx-\left (2 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}}}{(5+x)^2} \, dx+\left (4 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}+2 x}}{5+x} \, dx\\ &=x^2+2 x \left (1+\log ^4(3)\right )-2 \int \frac {e^{2 e^{2 x}}}{(5+x)^3} \, dx-2 \operatorname {Subst}\left (\int e^x \, dx,x,e^{2 x}\right )+4 \int \frac {e^{2 \left (e^{2 x}+x\right )}}{(5+x)^2} \, dx-\left (2 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}}}{(5+x)^2} \, dx+\left (4 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}+2 x}}{5+x} \, dx\\ &=-2 e^{e^{2 x}}+x^2+2 x \left (1+\log ^4(3)\right )-2 \int \frac {e^{2 e^{2 x}}}{(5+x)^3} \, dx+4 \int \frac {e^{2 \left (e^{2 x}+x\right )}}{(5+x)^2} \, dx-\left (2 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}}}{(5+x)^2} \, dx+\left (4 \left (4-\log ^4(3)\right )\right ) \int \frac {e^{e^{2 x}+2 x}}{5+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.34, size = 44, normalized size = 1.69 \begin {gather*} \frac {\left (-e^{e^{2 x}}+x (5+x)\right ) \left (-e^{e^{2 x}}+(5+x) \left (2+x+2 \log ^4(3)\right )\right )}{(5+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(250 + 400*x + 180*x^2 + 32*x^3 + 2*x^4 + E^(2*E^(2*x))*(-2 + E^(2*x)*(20 + 4*x)) + (250 + 150*x + 3
0*x^2 + 2*x^3)*Log[3]^4 + E^E^(2*x)*(-40 - 8*x + (10 + 2*x)*Log[3]^4 + E^(2*x)*(-100 - 140*x - 44*x^2 - 4*x^3
+ (-100 - 40*x - 4*x^2)*Log[3]^4)))/(125 + 75*x + 15*x^2 + x^3),x]

[Out]

((-E^E^(2*x) + x*(5 + x))*(-E^E^(2*x) + (5 + x)*(2 + x + 2*Log[3]^4)))/(5 + x)^2

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fricas [B]  time = 0.69, size = 76, normalized size = 2.92 \begin {gather*} \frac {2 \, {\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} \log \relax (3)^{4} + x^{4} + 12 \, x^{3} + 45 \, x^{2} - 2 \, {\left ({\left (x + 5\right )} \log \relax (3)^{4} + x^{2} + 6 \, x + 5\right )} e^{\left (e^{\left (2 \, x\right )}\right )} + 50 \, x + e^{\left (2 \, e^{\left (2 \, x\right )}\right )}}{x^{2} + 10 \, x + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20+4*x)*exp(x)^2-2)*exp(exp(x)^2)^2+(((-4*x^2-40*x-100)*log(3)^4-4*x^3-44*x^2-140*x-100)*exp(x)^2
+(2*x+10)*log(3)^4-8*x-40)*exp(exp(x)^2)+(2*x^3+30*x^2+150*x+250)*log(3)^4+2*x^4+32*x^3+180*x^2+400*x+250)/(x^
3+15*x^2+75*x+125),x, algorithm="fricas")

[Out]

(2*(x^3 + 10*x^2 + 25*x)*log(3)^4 + x^4 + 12*x^3 + 45*x^2 - 2*((x + 5)*log(3)^4 + x^2 + 6*x + 5)*e^(e^(2*x)) +
50*x + e^(2*e^(2*x)))/(x^2 + 10*x + 25)

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giac [B]  time = 0.18, size = 108, normalized size = 4.15 \begin {gather*} \frac {2 \, x^{3} \log \relax (3)^{4} + 20 \, x^{2} \log \relax (3)^{4} - 2 \, x e^{\left (e^{\left (2 \, x\right )}\right )} \log \relax (3)^{4} + 50 \, x \log \relax (3)^{4} - 10 \, e^{\left (e^{\left (2 \, x\right )}\right )} \log \relax (3)^{4} + x^{4} + 12 \, x^{3} - 2 \, x^{2} e^{\left (e^{\left (2 \, x\right )}\right )} + 45 \, x^{2} - 12 \, x e^{\left (e^{\left (2 \, x\right )}\right )} + 50 \, x + e^{\left (2 \, e^{\left (2 \, x\right )}\right )} - 10 \, e^{\left (e^{\left (2 \, x\right )}\right )}}{x^{2} + 10 \, x + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20+4*x)*exp(x)^2-2)*exp(exp(x)^2)^2+(((-4*x^2-40*x-100)*log(3)^4-4*x^3-44*x^2-140*x-100)*exp(x)^2
+(2*x+10)*log(3)^4-8*x-40)*exp(exp(x)^2)+(2*x^3+30*x^2+150*x+250)*log(3)^4+2*x^4+32*x^3+180*x^2+400*x+250)/(x^
3+15*x^2+75*x+125),x, algorithm="giac")

[Out]

(2*x^3*log(3)^4 + 20*x^2*log(3)^4 - 2*x*e^(e^(2*x))*log(3)^4 + 50*x*log(3)^4 - 10*e^(e^(2*x))*log(3)^4 + x^4 +
12*x^3 - 2*x^2*e^(e^(2*x)) + 45*x^2 - 12*x*e^(e^(2*x)) + 50*x + e^(2*e^(2*x)) - 10*e^(e^(2*x)))/(x^2 + 10*x +
25)

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maple [B]  time = 0.10, size = 52, normalized size = 2.00

method	result	size
risch	\(2 x \ln \relax (3)^{4}+x^{2}+2 x +\frac {{\mathrm e}^{2 \,{\mathrm e}^{2 x}}}{x^{2}+10 x +25}-\frac {2 \left (\ln \relax (3)^{4}+x +1\right ) {\mathrm e}^{{\mathrm e}^{2 x}}}{5+x}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((20+4*x)*exp(x)^2-2)*exp(exp(x)^2)^2+(((-4*x^2-40*x-100)*ln(3)^4-4*x^3-44*x^2-140*x-100)*exp(x)^2+(2*x+1
0)*ln(3)^4-8*x-40)*exp(exp(x)^2)+(2*x^3+30*x^2+150*x+250)*ln(3)^4+2*x^4+32*x^3+180*x^2+400*x+250)/(x^3+15*x^2+
75*x+125),x,method=_RETURNVERBOSE)

[Out]

2*x*ln(3)^4+x^2+2*x+1/(x^2+10*x+25)*exp(2*exp(2*x))-2*(ln(3)^4+x+1)/(5+x)*exp(exp(2*x))

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maxima [B]  time = 0.52, size = 234, normalized size = 9.00 \begin {gather*} {\left (2 \, x - \frac {25 \, {\left (6 \, x + 25\right )}}{x^{2} + 10 \, x + 25} - 30 \, \log \left (x + 5\right )\right )} \log \relax (3)^{4} + 15 \, {\left (\frac {5 \, {\left (4 \, x + 15\right )}}{x^{2} + 10 \, x + 25} + 2 \, \log \left (x + 5\right )\right )} \log \relax (3)^{4} - \frac {75 \, {\left (2 \, x + 5\right )} \log \relax (3)^{4}}{x^{2} + 10 \, x + 25} - \frac {125 \, \log \relax (3)^{4}}{x^{2} + 10 \, x + 25} + x^{2} + 2 \, x - \frac {2 \, {\left (5 \, \log \relax (3)^{4} + {\left (\log \relax (3)^{4} + 6\right )} x + x^{2} + 5\right )} e^{\left (e^{\left (2 \, x\right )}\right )} - e^{\left (2 \, e^{\left (2 \, x\right )}\right )}}{x^{2} + 10 \, x + 25} + \frac {125 \, {\left (8 \, x + 35\right )}}{x^{2} + 10 \, x + 25} - \frac {400 \, {\left (6 \, x + 25\right )}}{x^{2} + 10 \, x + 25} + \frac {450 \, {\left (4 \, x + 15\right )}}{x^{2} + 10 \, x + 25} - \frac {200 \, {\left (2 \, x + 5\right )}}{x^{2} + 10 \, x + 25} - \frac {125}{x^{2} + 10 \, x + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20+4*x)*exp(x)^2-2)*exp(exp(x)^2)^2+(((-4*x^2-40*x-100)*log(3)^4-4*x^3-44*x^2-140*x-100)*exp(x)^2
+(2*x+10)*log(3)^4-8*x-40)*exp(exp(x)^2)+(2*x^3+30*x^2+150*x+250)*log(3)^4+2*x^4+32*x^3+180*x^2+400*x+250)/(x^
3+15*x^2+75*x+125),x, algorithm="maxima")

[Out]

(2*x - 25*(6*x + 25)/(x^2 + 10*x + 25) - 30*log(x + 5))*log(3)^4 + 15*(5*(4*x + 15)/(x^2 + 10*x + 25) + 2*log(
x + 5))*log(3)^4 - 75*(2*x + 5)*log(3)^4/(x^2 + 10*x + 25) - 125*log(3)^4/(x^2 + 10*x + 25) + x^2 + 2*x - (2*(
5*log(3)^4 + (log(3)^4 + 6)*x + x^2 + 5)*e^(e^(2*x)) - e^(2*e^(2*x)))/(x^2 + 10*x + 25) + 125*(8*x + 35)/(x^2
+ 10*x + 25) - 400*(6*x + 25)/(x^2 + 10*x + 25) + 450*(4*x + 15)/(x^2 + 10*x + 25) - 200*(2*x + 5)/(x^2 + 10*x
+ 25) - 125/(x^2 + 10*x + 25)

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mupad [B]  time = 4.70, size = 55, normalized size = 2.12 \begin {gather*} x\,\left (2\,{\ln \relax (3)}^4+2\right )+\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}}}{x^2+10\,x+25}+x^2-\frac {{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\,\left (2\,x+2\,{\ln \relax (3)}^4+2\right )}{x+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((400*x - exp(exp(2*x))*(8*x - log(3)^4*(2*x + 10) + exp(2*x)*(140*x + log(3)^4*(40*x + 4*x^2 + 100) + 44*x
^2 + 4*x^3 + 100) + 40) + exp(2*exp(2*x))*(exp(2*x)*(4*x + 20) - 2) + log(3)^4*(150*x + 30*x^2 + 2*x^3 + 250)
+ 180*x^2 + 32*x^3 + 2*x^4 + 250)/(75*x + 15*x^2 + x^3 + 125),x)

[Out]

x*(2*log(3)^4 + 2) + exp(2*exp(2*x))/(10*x + x^2 + 25) + x^2 - (exp(exp(2*x))*(2*x + 2*log(3)^4 + 2))/(x + 5)

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sympy [B]  time = 0.29, size = 87, normalized size = 3.35 \begin {gather*} x^{2} + x \left (2 + 2 \log {\relax (3 )}^{4}\right ) + \frac {\left (x + 5\right ) e^{2 e^{2 x}} + \left (- 2 x^{3} - 22 x^{2} - 2 x^{2} \log {\relax (3 )}^{4} - 70 x - 20 x \log {\relax (3 )}^{4} - 50 \log {\relax (3 )}^{4} - 50\right ) e^{e^{2 x}}}{x^{3} + 15 x^{2} + 75 x + 125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20+4*x)*exp(x)**2-2)*exp(exp(x)**2)**2+(((-4*x**2-40*x-100)*ln(3)**4-4*x**3-44*x**2-140*x-100)*ex
p(x)**2+(2*x+10)*ln(3)**4-8*x-40)*exp(exp(x)**2)+(2*x**3+30*x**2+150*x+250)*ln(3)**4+2*x**4+32*x**3+180*x**2+4
00*x+250)/(x**3+15*x**2+75*x+125),x)

[Out]

x**2 + x*(2 + 2*log(3)**4) + ((x + 5)*exp(2*exp(2*x)) + (-2*x**3 - 22*x**2 - 2*x**2*log(3)**4 - 70*x - 20*x*lo
g(3)**4 - 50*log(3)**4 - 50)*exp(exp(2*x)))/(x**3 + 15*x**2 + 75*x + 125)

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