3.11.65 \(\int \frac {-2 x-5 x^4-60 x^6+e^{2 x} (-5 x^2-60 x^4)+e^x (-1-x-10 x^3-120 x^5)+(5 e^{2 x} x^2+10 e^x x^3+5 x^4) \log (\log (5))}{5 e^{2 x} x^2+10 e^x x^3+5 x^4} \, dx\)

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

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Rubi [F]  time = 1.24, 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 {-2 x-5 x^4-60 x^6+e^{2 x} \left (-5 x^2-60 x^4\right )+e^x \left (-1-x-10 x^3-120 x^5\right )+\left (5 e^{2 x} x^2+10 e^x x^3+5 x^4\right ) \log (\log (5))}{5 e^{2 x} x^2+10 e^x x^3+5 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2*x - 5*x^4 - 60*x^6 + E^(2*x)*(-5*x^2 - 60*x^4) + E^x*(-1 - x - 10*x^3 - 120*x^5) + (5*E^(2*x)*x^2 + 10
*E^x*x^3 + 5*x^4)*Log[Log[5]])/(5*E^(2*x)*x^2 + 10*E^x*x^3 + 5*x^4),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 x-60 x^6-e^x \left (1+x+120 x^5-10 x^3 (-1+\log (\log (5)))\right )-5 e^{2 x} x^2 \left (1+12 x^2-\log (\log (5))\right )+5 x^4 (-1+\log (\log (5)))}{5 x^2 \left (e^x+x\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {-2 x-60 x^6-e^x \left (1+x+120 x^5-10 x^3 (-1+\log (\log (5)))\right )-5 e^{2 x} x^2 \left (1+12 x^2-\log (\log (5))\right )+5 x^4 (-1+\log (\log (5)))}{x^2 \left (e^x+x\right )^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {-1+x}{x \left (e^x+x\right )^2}-\frac {1+x}{x^2 \left (e^x+x\right )}-5 \left (1+12 x^2-\log (\log (5))\right )\right ) \, dx\\ &=\frac {1}{5} \int \frac {-1+x}{x \left (e^x+x\right )^2} \, dx-\frac {1}{5} \int \frac {1+x}{x^2 \left (e^x+x\right )} \, dx-\int \left (1+12 x^2-\log (\log (5))\right ) \, dx\\ &=-4 x^3-x (1-\log (\log (5)))+\frac {1}{5} \int \left (\frac {1}{\left (e^x+x\right )^2}-\frac {1}{x \left (e^x+x\right )^2}\right ) \, dx-\frac {1}{5} \int \left (\frac {1}{x^2 \left (e^x+x\right )}+\frac {1}{x \left (e^x+x\right )}\right ) \, dx\\ &=-4 x^3-x (1-\log (\log (5)))+\frac {1}{5} \int \frac {1}{\left (e^x+x\right )^2} \, dx-\frac {1}{5} \int \frac {1}{x \left (e^x+x\right )^2} \, dx-\frac {1}{5} \int \frac {1}{x^2 \left (e^x+x\right )} \, dx-\frac {1}{5} \int \frac {1}{x \left (e^x+x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 29, normalized size = 1.12 \begin {gather*} \frac {1}{5} \left (-20 x^3+\frac {1}{x \left (e^x+x\right )}+5 x (-1+\log (\log (5)))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x - 5*x^4 - 60*x^6 + E^(2*x)*(-5*x^2 - 60*x^4) + E^x*(-1 - x - 10*x^3 - 120*x^5) + (5*E^(2*x)*x^
2 + 10*E^x*x^3 + 5*x^4)*Log[Log[5]])/(5*E^(2*x)*x^2 + 10*E^x*x^3 + 5*x^4),x]

[Out]

(-20*x^3 + 1/(x*(E^x + x)) + 5*x*(-1 + Log[Log[5]]))/5

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fricas [A]  time = 0.59, size = 52, normalized size = 2.00 \begin {gather*} -\frac {20 \, x^{5} + 5 \, x^{3} + 5 \, {\left (4 \, x^{4} + x^{2}\right )} e^{x} - 5 \, {\left (x^{3} + x^{2} e^{x}\right )} \log \left (\log \relax (5)\right ) - 1}{5 \, {\left (x^{2} + x e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)^2*x^2+10*exp(x)*x^3+5*x^4)*log(log(5))+(-60*x^4-5*x^2)*exp(x)^2+(-120*x^5-10*x^3-x-1)*exp
(x)-60*x^6-5*x^4-2*x)/(5*exp(x)^2*x^2+10*exp(x)*x^3+5*x^4),x, algorithm="fricas")

[Out]

-1/5*(20*x^5 + 5*x^3 + 5*(4*x^4 + x^2)*e^x - 5*(x^3 + x^2*e^x)*log(log(5)) - 1)/(x^2 + x*e^x)

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giac [B]  time = 0.35, size = 56, normalized size = 2.15 \begin {gather*} -\frac {20 \, x^{5} + 20 \, x^{4} e^{x} - 5 \, x^{3} \log \left (\log \relax (5)\right ) - 5 \, x^{2} e^{x} \log \left (\log \relax (5)\right ) + 5 \, x^{3} + 5 \, x^{2} e^{x} - 1}{5 \, {\left (x^{2} + x e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)^2*x^2+10*exp(x)*x^3+5*x^4)*log(log(5))+(-60*x^4-5*x^2)*exp(x)^2+(-120*x^5-10*x^3-x-1)*exp
(x)-60*x^6-5*x^4-2*x)/(5*exp(x)^2*x^2+10*exp(x)*x^3+5*x^4),x, algorithm="giac")

[Out]

-1/5*(20*x^5 + 20*x^4*e^x - 5*x^3*log(log(5)) - 5*x^2*e^x*log(log(5)) + 5*x^3 + 5*x^2*e^x - 1)/(x^2 + x*e^x)

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maple [A]  time = 0.07, size = 26, normalized size = 1.00

method	result	size
risch	\(-4 x^{3}+x \ln \left (\ln \relax (5)\right )-x +\frac {1}{5 x \left ({\mathrm e}^{x}+x \right )}\)	\(26\)
norman	\(\frac {\frac {1}{5}+\left (-1+\ln \left (\ln \relax (5)\right )\right ) x^{3}+\left (-1+\ln \left (\ln \relax (5)\right )\right ) x^{2} {\mathrm e}^{x}-4 x^{5}-4 \,{\mathrm e}^{x} x^{4}}{\left ({\mathrm e}^{x}+x \right ) x}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*exp(x)^2*x^2+10*exp(x)*x^3+5*x^4)*ln(ln(5))+(-60*x^4-5*x^2)*exp(x)^2+(-120*x^5-10*x^3-x-1)*exp(x)-60*x
^6-5*x^4-2*x)/(5*exp(x)^2*x^2+10*exp(x)*x^3+5*x^4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.73, size = 49, normalized size = 1.88 \begin {gather*} -\frac {20 \, x^{5} - 5 \, x^{3} {\left (\log \left (\log \relax (5)\right ) - 1\right )} + 5 \, {\left (4 \, x^{4} - x^{2} {\left (\log \left (\log \relax (5)\right ) - 1\right )}\right )} e^{x} - 1}{5 \, {\left (x^{2} + x e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)^2*x^2+10*exp(x)*x^3+5*x^4)*log(log(5))+(-60*x^4-5*x^2)*exp(x)^2+(-120*x^5-10*x^3-x-1)*exp
(x)-60*x^6-5*x^4-2*x)/(5*exp(x)^2*x^2+10*exp(x)*x^3+5*x^4),x, algorithm="maxima")

[Out]

-1/5*(20*x^5 - 5*x^3*(log(log(5)) - 1) + 5*(4*x^4 - x^2*(log(log(5)) - 1))*e^x - 1)/(x^2 + x*e^x)

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mupad [B]  time = 0.83, size = 52, normalized size = 2.00 \begin {gather*} \frac {x^3\,\left (\ln \left ({\ln \relax (5)}^5\right )-5\right )-20\,x^4\,{\mathrm {e}}^x-20\,x^5+x^2\,{\mathrm {e}}^x\,\left (\ln \left ({\ln \relax (5)}^5\right )-5\right )+1}{5\,\left (x\,{\mathrm {e}}^x+x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x - log(log(5))*(10*x^3*exp(x) + 5*x^2*exp(2*x) + 5*x^4) + exp(2*x)*(5*x^2 + 60*x^4) + exp(x)*(x + 10*
x^3 + 120*x^5 + 1) + 5*x^4 + 60*x^6)/(10*x^3*exp(x) + 5*x^2*exp(2*x) + 5*x^4),x)

[Out]

(x^3*(log(log(5)^5) - 5) - 20*x^4*exp(x) - 20*x^5 + x^2*exp(x)*(log(log(5)^5) - 5) + 1)/(5*(x*exp(x) + x^2))

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sympy [A]  time = 0.14, size = 26, normalized size = 1.00 \begin {gather*} - 4 x^{3} + x \left (-1 + \log {\left (\log {\relax (5 )} \right )}\right ) + \frac {1}{5 x^{2} + 5 x e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)**2*x**2+10*exp(x)*x**3+5*x**4)*ln(ln(5))+(-60*x**4-5*x**2)*exp(x)**2+(-120*x**5-10*x**3-x
-1)*exp(x)-60*x**6-5*x**4-2*x)/(5*exp(x)**2*x**2+10*exp(x)*x**3+5*x**4),x)

[Out]

-4*x**3 + x*(-1 + log(log(5))) + 1/(5*x**2 + 5*x*exp(x))

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