\(\int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} (-20+40 \log ^2(x))}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+(145 x-15 x^2) \log ^2(x)} \, dx\) [5585]

Optimal result

Integrand size = 57, antiderivative size = 27 \[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\frac {1}{5} \log \left (x^2 \left (25+4 \left (1+5 e^{\frac {1}{\log (x)}}-x\right )+x\right )\right ) \]

[Out]

1/5*ln(x^2*(-3*x+20*exp(1/ln(x))+29))

Rubi [F]

\[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx \]

[In]

Int[((58 - 9*x)*Log[x]^2 + E^Log[x]^(-1)*(-20 + 40*Log[x]^2))/(100*E^Log[x]^(-1)*x*Log[x]^2 + (145*x - 15*x^2)
*Log[x]^2),x]

[Out]

1/(5*Log[x]) + (2*Log[x])/5 - (3*Defer[Int][(29 + 20*E^Log[x]^(-1) - 3*x)^(-1), x])/5 - (3*Defer[Int][1/((29 +
20*E^Log[x]^(-1) - 3*x)*Log[x]^2), x])/5 + (29*Defer[Int][1/((29 + 20*E^Log[x]^(-1) - 3*x)*x*Log[x]^2), x])/5

Rubi steps \begin{align*} \text {integral}& = \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{5 \left (29+20 e^{\frac {1}{\log (x)}}-3 x\right ) x \log ^2(x)} \, dx \\ & = \frac {1}{5} \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{\left (29+20 e^{\frac {1}{\log (x)}}-3 x\right ) x \log ^2(x)} \, dx \\ & = \frac {1}{5} \int \left (\frac {-1+2 \log ^2(x)}{x \log ^2(x)}-\frac {-29+3 x+3 x \log ^2(x)}{\left (29+20 e^{\frac {1}{\log (x)}}-3 x\right ) x \log ^2(x)}\right ) \, dx \\ & = \frac {1}{5} \int \frac {-1+2 \log ^2(x)}{x \log ^2(x)} \, dx-\frac {1}{5} \int \frac {-29+3 x+3 x \log ^2(x)}{\left (29+20 e^{\frac {1}{\log (x)}}-3 x\right ) x \log ^2(x)} \, dx \\ & = -\left (\frac {1}{5} \int \left (\frac {3}{29+20 e^{\frac {1}{\log (x)}}-3 x}+\frac {3}{\left (29+20 e^{\frac {1}{\log (x)}}-3 x\right ) \log ^2(x)}-\frac {29}{\left (29+20 e^{\frac {1}{\log (x)}}-3 x\right ) x \log ^2(x)}\right ) \, dx\right )+\frac {1}{5} \text {Subst}\left (\int \frac {-1+2 x^2}{x^2} \, dx,x,\log (x)\right ) \\ & = \frac {1}{5} \text {Subst}\left (\int \left (2-\frac {1}{x^2}\right ) \, dx,x,\log (x)\right )-\frac {3}{5} \int \frac {1}{29+20 e^{\frac {1}{\log (x)}}-3 x} \, dx-\frac {3}{5} \int \frac {1}{\left (29+20 e^{\frac {1}{\log (x)}}-3 x\right ) \log ^2(x)} \, dx+\frac {29}{5} \int \frac {1}{\left (29+20 e^{\frac {1}{\log (x)}}-3 x\right ) x \log ^2(x)} \, dx \\ & = \frac {1}{5 \log (x)}+\frac {2 \log (x)}{5}-\frac {3}{5} \int \frac {1}{29+20 e^{\frac {1}{\log (x)}}-3 x} \, dx-\frac {3}{5} \int \frac {1}{\left (29+20 e^{\frac {1}{\log (x)}}-3 x\right ) \log ^2(x)} \, dx+\frac {29}{5} \int \frac {1}{\left (29+20 e^{\frac {1}{\log (x)}}-3 x\right ) x \log ^2(x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\frac {1}{5} \left (\log \left (29+20 e^{\frac {1}{\log (x)}}-3 x\right )+2 \log (x)\right ) \]

[In]

Integrate[((58 - 9*x)*Log[x]^2 + E^Log[x]^(-1)*(-20 + 40*Log[x]^2))/(100*E^Log[x]^(-1)*x*Log[x]^2 + (145*x - 1
5*x^2)*Log[x]^2),x]

[Out]

(Log[29 + 20*E^Log[x]^(-1) - 3*x] + 2*Log[x])/5

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70

method	result	size
risch	\(\frac {2 \ln \left (x \right )}{5}+\frac {\ln \left (-\frac {3 x}{20}+{\mathrm e}^{\frac {1}{\ln \left (x \right )}}+\frac {29}{20}\right )}{5}\)	\(19\)
parallelrisch	\(\frac {2 \ln \left (x \right )}{5}+\frac {\ln \left (x -\frac {20 \,{\mathrm e}^{\frac {1}{\ln \left (x \right )}}}{3}-\frac {29}{3}\right )}{5}\)	\(19\)
norman	\(\frac {2 \ln \left (x \right )}{5}+\frac {\ln \left (3 x -20 \,{\mathrm e}^{\frac {1}{\ln \left (x \right )}}-29\right )}{5}\)	\(21\)

[In]

int(((40*ln(x)^2-20)*exp(1/ln(x))+(-9*x+58)*ln(x)^2)/(100*x*ln(x)^2*exp(1/ln(x))+(-15*x^2+145*x)*ln(x)^2),x,me
thod=_RETURNVERBOSE)

[Out]

2/5*ln(x)+1/5*ln(-3/20*x+exp(1/ln(x))+29/20)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\frac {2}{5} \, \log \left (x\right ) + \frac {1}{5} \, \log \left (-3 \, x + 20 \, e^{\frac {1}{\log \left (x\right )}} + 29\right ) \]

[In]

integrate(((40*log(x)^2-20)*exp(1/log(x))+(-9*x+58)*log(x)^2)/(100*x*log(x)^2*exp(1/log(x))+(-15*x^2+145*x)*lo
g(x)^2),x, algorithm="fricas")

[Out]

2/5*log(x) + 1/5*log(-3*x + 20*e^(1/log(x)) + 29)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\frac {2 \log {\left (x \right )}}{5} + \frac {\log {\left (- \frac {3 x}{20} + e^{\frac {1}{\log {\left (x \right )}}} + \frac {29}{20} \right )}}{5} \]

[In]

integrate(((40*ln(x)**2-20)*exp(1/ln(x))+(-9*x+58)*ln(x)**2)/(100*x*ln(x)**2*exp(1/ln(x))+(-15*x**2+145*x)*ln(
x)**2),x)

[Out]

2*log(x)/5 + log(-3*x/20 + exp(1/log(x)) + 29/20)/5

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\frac {2}{5} \, \log \left (x\right ) + \frac {1}{5} \, \log \left (-\frac {3}{20} \, x + e^{\frac {1}{\log \left (x\right )}} + \frac {29}{20}\right ) \]

[In]

integrate(((40*log(x)^2-20)*exp(1/log(x))+(-9*x+58)*log(x)^2)/(100*x*log(x)^2*exp(1/log(x))+(-15*x^2+145*x)*lo
g(x)^2),x, algorithm="maxima")

[Out]

2/5*log(x) + 1/5*log(-3/20*x + e^(1/log(x)) + 29/20)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\frac {2}{5} \, \log \left (x\right ) + \frac {1}{5} \, \log \left (-3 \, x + 20 \, e^{\frac {1}{\log \left (x\right )}} + 29\right ) \]

[In]

integrate(((40*log(x)^2-20)*exp(1/log(x))+(-9*x+58)*log(x)^2)/(100*x*log(x)^2*exp(1/log(x))+(-15*x^2+145*x)*lo
g(x)^2),x, algorithm="giac")

[Out]

2/5*log(x) + 1/5*log(-3*x + 20*e^(1/log(x)) + 29)

Mupad [B] (verification not implemented)

Time = 11.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {(58-9 x) \log ^2(x)+e^{\frac {1}{\log (x)}} \left (-20+40 \log ^2(x)\right )}{100 e^{\frac {1}{\log (x)}} x \log ^2(x)+\left (145 x-15 x^2\right ) \log ^2(x)} \, dx=\frac {\ln \left ({\mathrm {e}}^{\frac {1}{\ln \left (x\right )}}-\frac {3\,x}{20}+\frac {29}{20}\right )}{5}+\frac {2\,\ln \left (x\right )}{5} \]

[In]

int((exp(1/log(x))*(40*log(x)^2 - 20) - log(x)^2*(9*x - 58))/(log(x)^2*(145*x - 15*x^2) + 100*x*exp(1/log(x))*
log(x)^2),x)

[Out]

log(exp(1/log(x)) - (3*x)/20 + 29/20)/5 + (2*log(x))/5