Integrand size = 116, antiderivative size = 21 \[ \int \frac {x+\left (x+x^2+5\ 3^{5/x} \log (3)\right ) \log (x)}{\left (64 x^2+3^{10/x} x^2+16 x^3+x^4+3^{5/x} \left (-16 x^2-2 x^3\right )\right ) \log (x)+\left (16 x^2-2\ 3^{5/x} x^2+2 x^3\right ) \log (x) \log (x \log (x))+x^2 \log (x) \log ^2(x \log (x))} \, dx=\frac {1}{-8+3^{5/x}-x-\log (x \log (x))} \] Output:
1/(-8-x+exp(5*ln(3)/x)-ln(x*ln(x)))
Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {x+\left (x+x^2+5\ 3^{5/x} \log (3)\right ) \log (x)}{\left (64 x^2+3^{10/x} x^2+16 x^3+x^4+3^{5/x} \left (-16 x^2-2 x^3\right )\right ) \log (x)+\left (16 x^2-2\ 3^{5/x} x^2+2 x^3\right ) \log (x) \log (x \log (x))+x^2 \log (x) \log ^2(x \log (x))} \, dx=-\frac {1}{8-243^{\frac {1}{x}}+x+\log (x \log (x))} \] Input:
Integrate[(x + (x + x^2 + 5*3^(5/x)*Log[3])*Log[x])/((64*x^2 + 3^(10/x)*x^ 2 + 16*x^3 + x^4 + 3^(5/x)*(-16*x^2 - 2*x^3))*Log[x] + (16*x^2 - 2*3^(5/x) *x^2 + 2*x^3)*Log[x]*Log[x*Log[x]] + x^2*Log[x]*Log[x*Log[x]]^2),x]
Output:
-(8 - 243^x^(-1) + x + Log[x*Log[x]])^(-1)
Time = 0.73 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {7239, 7237}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (x^2+x+5\ 3^{5/x} \log (3)\right ) \log (x)+x}{x^2 \log (x) \log ^2(x \log (x))+\left (2 x^3-2\ 3^{5/x} x^2+16 x^2\right ) \log (x) \log (x \log (x))+\left (x^4+16 x^3+3^{10/x} x^2+64 x^2+3^{5/x} \left (-2 x^3-16 x^2\right )\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\left (x^2+x+5\ 243^{\frac {1}{x}} \log (3)\right ) \log (x)+x}{x^2 \log (x) \left (x-243^{\frac {1}{x}}+\log (x \log (x))+8\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle -\frac {1}{x-243^{\frac {1}{x}}+\log (x \log (x))+8}\) |
Input:
Int[(x + (x + x^2 + 5*3^(5/x)*Log[3])*Log[x])/((64*x^2 + 3^(10/x)*x^2 + 16 *x^3 + x^4 + 3^(5/x)*(-16*x^2 - 2*x^3))*Log[x] + (16*x^2 - 2*3^(5/x)*x^2 + 2*x^3)*Log[x]*Log[x*Log[x]] + x^2*Log[x]*Log[x*Log[x]]^2),x]
Output:
-(8 - 243^x^(-1) + x + Log[x*Log[x]])^(-1)
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 15.75 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10
method | result | size |
parallelrisch | \(-\frac {1}{x -{\mathrm e}^{\frac {5 \ln \left (3\right )}{x}}+\ln \left (x \ln \left (x \right )\right )+8}\) | \(23\) |
risch | \(-\frac {2 i}{\pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )-\pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}+\pi \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{3}+2 i x -2 i 243^{\frac {1}{x}}+2 i \ln \left (x \right )+2 i \ln \left (\ln \left (x \right )\right )+16 i}\) | \(98\) |
Input:
int(((5*ln(3)*exp(5*ln(3)/x)+x^2+x)*ln(x)+x)/(x^2*ln(x)*ln(x*ln(x))^2+(-2* x^2*exp(5*ln(3)/x)+2*x^3+16*x^2)*ln(x)*ln(x*ln(x))+(x^2*exp(5*ln(3)/x)^2+( -2*x^3-16*x^2)*exp(5*ln(3)/x)+x^4+16*x^3+64*x^2)*ln(x)),x,method=_RETURNVE RBOSE)
Output:
-1/(x-exp(5*ln(3)/x)+ln(x*ln(x))+8)
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {x+\left (x+x^2+5\ 3^{5/x} \log (3)\right ) \log (x)}{\left (64 x^2+3^{10/x} x^2+16 x^3+x^4+3^{5/x} \left (-16 x^2-2 x^3\right )\right ) \log (x)+\left (16 x^2-2\ 3^{5/x} x^2+2 x^3\right ) \log (x) \log (x \log (x))+x^2 \log (x) \log ^2(x \log (x))} \, dx=\frac {1}{3^{\frac {5}{x}} - x - \log \left (x \log \left (x\right )\right ) - 8} \] Input:
integrate(((5*log(3)*exp(5*log(3)/x)+x^2+x)*log(x)+x)/(x^2*log(x)*log(x*lo g(x))^2+(-2*x^2*exp(5*log(3)/x)+2*x^3+16*x^2)*log(x)*log(x*log(x))+(x^2*ex p(5*log(3)/x)^2+(-2*x^3-16*x^2)*exp(5*log(3)/x)+x^4+16*x^3+64*x^2)*log(x)) ,x, algorithm="fricas")
Output:
1/(3^(5/x) - x - log(x*log(x)) - 8)
Time = 0.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {x+\left (x+x^2+5\ 3^{5/x} \log (3)\right ) \log (x)}{\left (64 x^2+3^{10/x} x^2+16 x^3+x^4+3^{5/x} \left (-16 x^2-2 x^3\right )\right ) \log (x)+\left (16 x^2-2\ 3^{5/x} x^2+2 x^3\right ) \log (x) \log (x \log (x))+x^2 \log (x) \log ^2(x \log (x))} \, dx=\frac {1}{- x + e^{\frac {5 \log {\left (3 \right )}}{x}} - \log {\left (x \log {\left (x \right )} \right )} - 8} \] Input:
integrate(((5*ln(3)*exp(5*ln(3)/x)+x**2+x)*ln(x)+x)/(x**2*ln(x)*ln(x*ln(x) )**2+(-2*x**2*exp(5*ln(3)/x)+2*x**3+16*x**2)*ln(x)*ln(x*ln(x))+(x**2*exp(5 *ln(3)/x)**2+(-2*x**3-16*x**2)*exp(5*ln(3)/x)+x**4+16*x**3+64*x**2)*ln(x)) ,x)
Output:
1/(-x + exp(5*log(3)/x) - log(x*log(x)) - 8)
Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {x+\left (x+x^2+5\ 3^{5/x} \log (3)\right ) \log (x)}{\left (64 x^2+3^{10/x} x^2+16 x^3+x^4+3^{5/x} \left (-16 x^2-2 x^3\right )\right ) \log (x)+\left (16 x^2-2\ 3^{5/x} x^2+2 x^3\right ) \log (x) \log (x \log (x))+x^2 \log (x) \log ^2(x \log (x))} \, dx=\frac {1}{3^{\frac {5}{x}} - x - \log \left (x\right ) - \log \left (\log \left (x\right )\right ) - 8} \] Input:
integrate(((5*log(3)*exp(5*log(3)/x)+x^2+x)*log(x)+x)/(x^2*log(x)*log(x*lo g(x))^2+(-2*x^2*exp(5*log(3)/x)+2*x^3+16*x^2)*log(x)*log(x*log(x))+(x^2*ex p(5*log(3)/x)^2+(-2*x^3-16*x^2)*exp(5*log(3)/x)+x^4+16*x^3+64*x^2)*log(x)) ,x, algorithm="maxima")
Output:
1/(3^(5/x) - x - log(x) - log(log(x)) - 8)
\[ \int \frac {x+\left (x+x^2+5\ 3^{5/x} \log (3)\right ) \log (x)}{\left (64 x^2+3^{10/x} x^2+16 x^3+x^4+3^{5/x} \left (-16 x^2-2 x^3\right )\right ) \log (x)+\left (16 x^2-2\ 3^{5/x} x^2+2 x^3\right ) \log (x) \log (x \log (x))+x^2 \log (x) \log ^2(x \log (x))} \, dx=\int { \frac {{\left (x^{2} + 5 \cdot 3^{\frac {5}{x}} \log \left (3\right ) + x\right )} \log \left (x\right ) + x}{x^{2} \log \left (x \log \left (x\right )\right )^{2} \log \left (x\right ) - 2 \, {\left (3^{\frac {5}{x}} x^{2} - x^{3} - 8 \, x^{2}\right )} \log \left (x \log \left (x\right )\right ) \log \left (x\right ) + {\left (x^{4} + 3^{\frac {10}{x}} x^{2} + 16 \, x^{3} - 2 \, {\left (x^{3} + 8 \, x^{2}\right )} 3^{\frac {5}{x}} + 64 \, x^{2}\right )} \log \left (x\right )} \,d x } \] Input:
integrate(((5*log(3)*exp(5*log(3)/x)+x^2+x)*log(x)+x)/(x^2*log(x)*log(x*lo g(x))^2+(-2*x^2*exp(5*log(3)/x)+2*x^3+16*x^2)*log(x)*log(x*log(x))+(x^2*ex p(5*log(3)/x)^2+(-2*x^3-16*x^2)*exp(5*log(3)/x)+x^4+16*x^3+64*x^2)*log(x)) ,x, algorithm="giac")
Output:
integrate(((x^2 + 5*3^(5/x)*log(3) + x)*log(x) + x)/(x^2*log(x*log(x))^2*l og(x) - 2*(3^(5/x)*x^2 - x^3 - 8*x^2)*log(x*log(x))*log(x) + (x^4 + 3^(10/ x)*x^2 + 16*x^3 - 2*(x^3 + 8*x^2)*3^(5/x) + 64*x^2)*log(x)), x)
Time = 2.92 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {x+\left (x+x^2+5\ 3^{5/x} \log (3)\right ) \log (x)}{\left (64 x^2+3^{10/x} x^2+16 x^3+x^4+3^{5/x} \left (-16 x^2-2 x^3\right )\right ) \log (x)+\left (16 x^2-2\ 3^{5/x} x^2+2 x^3\right ) \log (x) \log (x \log (x))+x^2 \log (x) \log ^2(x \log (x))} \, dx=-\frac {1}{x+\ln \left (x\,\ln \left (x\right )\right )-3^{5/x}+8} \] Input:
int((x + log(x)*(x + 5*exp((5*log(3))/x)*log(3) + x^2))/(log(x)*(x^2*exp(( 10*log(3))/x) - exp((5*log(3))/x)*(16*x^2 + 2*x^3) + 64*x^2 + 16*x^3 + x^4 ) + log(x*log(x))*log(x)*(16*x^2 - 2*x^2*exp((5*log(3))/x) + 2*x^3) + x^2* log(x*log(x))^2*log(x)),x)
Output:
-1/(x + log(x*log(x)) - 3^(5/x) + 8)
Time = 0.56 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {x+\left (x+x^2+5\ 3^{5/x} \log (3)\right ) \log (x)}{\left (64 x^2+3^{10/x} x^2+16 x^3+x^4+3^{5/x} \left (-16 x^2-2 x^3\right )\right ) \log (x)+\left (16 x^2-2\ 3^{5/x} x^2+2 x^3\right ) \log (x) \log (x \log (x))+x^2 \log (x) \log ^2(x \log (x))} \, dx=\frac {1}{e^{\frac {5 \,\mathrm {log}\left (3\right )}{x}}-\mathrm {log}\left (\mathrm {log}\left (x \right ) x \right )-x -8} \] Input:
int(((5*log(3)*exp(5*log(3)/x)+x^2+x)*log(x)+x)/(x^2*log(x)*log(x*log(x))^ 2+(-2*x^2*exp(5*log(3)/x)+2*x^3+16*x^2)*log(x)*log(x*log(x))+(x^2*exp(5*lo g(3)/x)^2+(-2*x^3-16*x^2)*exp(5*log(3)/x)+x^4+16*x^3+64*x^2)*log(x)),x)
Output:
1/(e**((5*log(3))/x) - log(log(x)*x) - x - 8)