3.25.55 \(\int \frac {25 x^2+10 x^5+x^8+e^5 (25 x+5 x^4)+(-25 x^2+15 e^5 x^4-10 x^5-x^8) \log (x)+(-25-10 x^3-x^6+(-25-10 x^3-x^6) \log (x)) \log (\log (3))}{(25 x^2+10 x^5+x^8) \log ^2(x)} \, dx\)

Optimal. Leaf size=33 \[ \frac {-x+\frac {-\frac {5 e^5}{\frac {5}{x}+x^2}+\log (\log (3))}{x}}{\log (x)} \]

________________________________________________________________________________________

Rubi [F]  time = 0.83, 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 {25 x^2+10 x^5+x^8+e^5 \left (25 x+5 x^4\right )+\left (-25 x^2+15 e^5 x^4-10 x^5-x^8\right ) \log (x)+\left (-25-10 x^3-x^6+\left (-25-10 x^3-x^6\right ) \log (x)\right ) \log (\log (3))}{\left (25 x^2+10 x^5+x^8\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(25*x^2 + 10*x^5 + x^8 + E^5*(25*x + 5*x^4) + (-25*x^2 + 15*E^5*x^4 - 10*x^5 - x^8)*Log[x] + (-25 - 10*x^3
- x^6 + (-25 - 10*x^3 - x^6)*Log[x])*Log[Log[3]])/((25*x^2 + 10*x^5 + x^8)*Log[x]^2),x]

[Out]

Defer[Int][(5*E^5*x + 5*x^2 + x^5 - 5*Log[Log[3]] - x^3*Log[Log[3]])/(x^2*(5 + x^3)*Log[x]^2), x] + Defer[Int]
[(-25*x^2 + 15*E^5*x^4 - 10*x^5 - x^8 - 25*Log[Log[3]] - 10*x^3*Log[Log[3]] - x^6*Log[Log[3]])/(x^2*(5 + x^3)^
2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 x^2+10 x^5+x^8+e^5 \left (25 x+5 x^4\right )+\left (-25 x^2+15 e^5 x^4-10 x^5-x^8\right ) \log (x)+\left (-25-10 x^3-x^6+\left (-25-10 x^3-x^6\right ) \log (x)\right ) \log (\log (3))}{x^2 \left (25+10 x^3+x^6\right ) \log ^2(x)} \, dx\\ &=\int \frac {25 x^2+10 x^5+x^8+e^5 \left (25 x+5 x^4\right )+\left (-25 x^2+15 e^5 x^4-10 x^5-x^8\right ) \log (x)+\left (-25-10 x^3-x^6+\left (-25-10 x^3-x^6\right ) \log (x)\right ) \log (\log (3))}{x^2 \left (5+x^3\right )^2 \log ^2(x)} \, dx\\ &=\int \left (\frac {5 e^5 x+5 x^2+x^5-5 \log (\log (3))-x^3 \log (\log (3))}{x^2 \left (5+x^3\right ) \log ^2(x)}+\frac {-25 x^2+15 e^5 x^4-10 x^5-x^8-25 \log (\log (3))-10 x^3 \log (\log (3))-x^6 \log (\log (3))}{x^2 \left (5+x^3\right )^2 \log (x)}\right ) \, dx\\ &=\int \frac {5 e^5 x+5 x^2+x^5-5 \log (\log (3))-x^3 \log (\log (3))}{x^2 \left (5+x^3\right ) \log ^2(x)} \, dx+\int \frac {-25 x^2+15 e^5 x^4-10 x^5-x^8-25 \log (\log (3))-10 x^3 \log (\log (3))-x^6 \log (\log (3))}{x^2 \left (5+x^3\right )^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 28, normalized size = 0.85 \begin {gather*} \frac {-x-\frac {5 e^5}{5+x^3}+\frac {\log (\log (3))}{x}}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*x^2 + 10*x^5 + x^8 + E^5*(25*x + 5*x^4) + (-25*x^2 + 15*E^5*x^4 - 10*x^5 - x^8)*Log[x] + (-25 -
10*x^3 - x^6 + (-25 - 10*x^3 - x^6)*Log[x])*Log[Log[3]])/((25*x^2 + 10*x^5 + x^8)*Log[x]^2),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.95, size = 39, normalized size = 1.18 \begin {gather*} -\frac {x^{5} + 5 \, x^{2} + 5 \, x e^{5} - {\left (x^{3} + 5\right )} \log \left (\log \relax (3)\right )}{{\left (x^{4} + 5 \, x\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^6-10*x^3-25)*log(x)-x^6-10*x^3-25)*log(log(3))+(15*x^4*exp(5)-x^8-10*x^5-25*x^2)*log(x)+(5*x^4
+25*x)*exp(5)+x^8+10*x^5+25*x^2)/(x^8+10*x^5+25*x^2)/log(x)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.22, size = 43, normalized size = 1.30 \begin {gather*} -\frac {x^{5} - x^{3} \log \left (\log \relax (3)\right ) + 5 \, x^{2} + 5 \, x e^{5} - 5 \, \log \left (\log \relax (3)\right )}{x^{4} \log \relax (x) + 5 \, x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^6-10*x^3-25)*log(x)-x^6-10*x^3-25)*log(log(3))+(15*x^4*exp(5)-x^8-10*x^5-25*x^2)*log(x)+(5*x^4
+25*x)*exp(5)+x^8+10*x^5+25*x^2)/(x^8+10*x^5+25*x^2)/log(x)^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.06, size = 44, normalized size = 1.33

method	result	size
risch	\(-\frac {x^{5}-x^{3} \ln \left (\ln \relax (3)\right )+5 x \,{\mathrm e}^{5}+5 x^{2}-5 \ln \left (\ln \relax (3)\right )}{x \left (x^{3}+5\right ) \ln \relax (x )}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^6-10*x^3-25)*ln(x)-x^6-10*x^3-25)*ln(ln(3))+(15*x^4*exp(5)-x^8-10*x^5-25*x^2)*ln(x)+(5*x^4+25*x)*exp
(5)+x^8+10*x^5+25*x^2)/(x^8+10*x^5+25*x^2)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.73, size = 42, normalized size = 1.27 \begin {gather*} -\frac {x^{5} - x^{3} \log \left (\log \relax (3)\right ) + 5 \, x^{2} + 5 \, x e^{5} - 5 \, \log \left (\log \relax (3)\right )}{{\left (x^{4} + 5 \, x\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^6-10*x^3-25)*log(x)-x^6-10*x^3-25)*log(log(3))+(15*x^4*exp(5)-x^8-10*x^5-25*x^2)*log(x)+(5*x^4
+25*x)*exp(5)+x^8+10*x^5+25*x^2)/(x^8+10*x^5+25*x^2)/log(x)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.55, size = 43, normalized size = 1.30 \begin {gather*} -\frac {x^5-\ln \left (\ln \relax (3)\right )\,x^3+5\,x^2+5\,{\mathrm {e}}^5\,x-5\,\ln \left (\ln \relax (3)\right )}{x\,\ln \relax (x)\,\left (x^3+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5)*(25*x + 5*x^4) - log(log(3))*(log(x)*(10*x^3 + x^6 + 25) + 10*x^3 + x^6 + 25) - log(x)*(25*x^2 - 1
5*x^4*exp(5) + 10*x^5 + x^8) + 25*x^2 + 10*x^5 + x^8)/(log(x)^2*(25*x^2 + 10*x^5 + x^8)),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.23, size = 39, normalized size = 1.18 \begin {gather*} \frac {- x^{5} + x^{3} \log {\left (\log {\relax (3 )} \right )} - 5 x^{2} - 5 x e^{5} + 5 \log {\left (\log {\relax (3 )} \right )}}{\left (x^{4} + 5 x\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**6-10*x**3-25)*ln(x)-x**6-10*x**3-25)*ln(ln(3))+(15*x**4*exp(5)-x**8-10*x**5-25*x**2)*ln(x)+(5
*x**4+25*x)*exp(5)+x**8+10*x**5+25*x**2)/(x**8+10*x**5+25*x**2)/ln(x)**2,x)

[Out]

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

________________________________________________________________________________________