\(\int \frac {(-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)) \log (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)})}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx\) [3985]

Optimal result

Integrand size = 68, antiderivative size = 26 \[ \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx=\log ^2\left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right ) \]

[Out]

ln(exp(25/16/x^2)/(2*ln(x)-x^2+x))^2

Rubi [F]

\[ \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx=\int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx \]

[In]

Int[((-25*x + 9*x^2 - 8*x^3 + 16*x^4 - 50*Log[x])*Log[E^(25/(16*x^2))/(x - x^2 + 2*Log[x])])/(4*x^4 - 4*x^5 +
8*x^3*Log[x]),x]

[Out]

2*Defer[Int][Log[E^(25/(16*x^2))/(x - x^2 + 2*Log[x])]/(-x + x^2 - 2*Log[x]), x] + (25*Defer[Int][Log[E^(25/(1
6*x^2))/(x - x^2 + 2*Log[x])]/(x^2*(-x + x^2 - 2*Log[x])), x])/4 - (9*Defer[Int][Log[E^(25/(16*x^2))/(x - x^2
+ 2*Log[x])]/(x*(-x + x^2 - 2*Log[x])), x])/4 - 4*Defer[Int][(x*Log[E^(25/(16*x^2))/(x - x^2 + 2*Log[x])])/(-x
+ x^2 - 2*Log[x]), x] + (25*Defer[Int][(Log[x]*Log[E^(25/(16*x^2))/(x - x^2 + 2*Log[x])])/(x^3*(-x + x^2 - 2*
Log[x])), x])/2

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^3 \left (x-x^2+2 \log (x)\right )} \, dx \\ & = \frac {1}{4} \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{x^3 \left (x-x^2+2 \log (x)\right )} \, dx \\ & = \frac {1}{4} \int \left (\frac {8 \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{-x+x^2-2 \log (x)}+\frac {25 \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{x^2 \left (-x+x^2-2 \log (x)\right )}-\frac {9 \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{x \left (-x+x^2-2 \log (x)\right )}-\frac {16 x \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{-x+x^2-2 \log (x)}+\frac {50 \log (x) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{x^3 \left (-x+x^2-2 \log (x)\right )}\right ) \, dx \\ & = 2 \int \frac {\log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{-x+x^2-2 \log (x)} \, dx-\frac {9}{4} \int \frac {\log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{x \left (-x+x^2-2 \log (x)\right )} \, dx-4 \int \frac {x \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{-x+x^2-2 \log (x)} \, dx+\frac {25}{4} \int \frac {\log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{x^2 \left (-x+x^2-2 \log (x)\right )} \, dx+\frac {25}{2} \int \frac {\log (x) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{x^3 \left (-x+x^2-2 \log (x)\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx=\log ^2\left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right ) \]

[In]

Integrate[((-25*x + 9*x^2 - 8*x^3 + 16*x^4 - 50*Log[x])*Log[E^(25/(16*x^2))/(x - x^2 + 2*Log[x])])/(4*x^4 - 4*
x^5 + 8*x^3*Log[x]),x]

[Out]

Log[E^(25/(16*x^2))/(x - x^2 + 2*Log[x])]^2

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 10.91 (sec) , antiderivative size = 569, normalized size of antiderivative = 21.88

method	result	size
risch	\(-\frac {\left (16 \ln \left (-2 \ln \left (x \right )+x^{2}-x \right ) x^{2}-25\right ) \ln \left ({\mathrm e}^{\frac {25}{16 x^{2}}}\right )}{8 x^{2}}+\frac {-400 i \pi \,x^{2} \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{3}-256 i \pi \ln \left (-\frac {x^{2}}{2}+\frac {x}{2}+\ln \left (x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {25}{16 x^{2}}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{2} x^{4}-400 i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{2 \ln \left (x \right )-x^{2}+x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {25}{16 x^{2}}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )+800 i \pi \,x^{2}-800 i \pi \,x^{2} \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{2}+400 i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{\frac {25}{16 x^{2}}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{2}-512 i \pi \ln \left (-\frac {x^{2}}{2}+\frac {x}{2}+\ln \left (x \right )\right ) x^{4}+256 i \pi \ln \left (-\frac {x^{2}}{2}+\frac {x}{2}+\ln \left (x \right )\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{3} x^{4}-400 i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{2 \ln \left (x \right )-x^{2}+x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{2}+512 i \pi \ln \left (-\frac {x^{2}}{2}+\frac {x}{2}+\ln \left (x \right )\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{2} x^{4}+256 i \pi \ln \left (-\frac {x^{2}}{2}+\frac {x}{2}+\ln \left (x \right )\right ) \operatorname {csgn}\left (\frac {i}{2 \ln \left (x \right )-x^{2}+x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {25}{16 x^{2}}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right ) x^{4}+256 i \pi \ln \left (-\frac {x^{2}}{2}+\frac {x}{2}+\ln \left (x \right )\right ) \operatorname {csgn}\left (\frac {i}{2 \ln \left (x \right )-x^{2}+x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {25}{16 x^{2}}}}{2 \ln \left (x \right )-x^{2}+x}\right )^{2} x^{4}-625+256 \ln \left (-2 \ln \left (x \right )+x^{2}-x \right )^{2} x^{4}}{256 x^{4}}\)	\(569\)

[In]

int((-50*ln(x)+16*x^4-8*x^3+9*x^2-25*x)*ln(exp(25/16/x^2)/(2*ln(x)-x^2+x))/(8*x^3*ln(x)-4*x^5+4*x^4),x,method=
_RETURNVERBOSE)

[Out]

-1/8*(16*ln(-2*ln(x)+x^2-x)*x^2-25)/x^2*ln(exp(25/16/x^2))+1/256*(-400*I*Pi*x^2*csgn(I*exp(25/16/x^2)/(2*ln(x)
-x^2+x))^3-256*I*Pi*ln(-1/2*x^2+1/2*x+ln(x))*csgn(I*exp(25/16/x^2))*csgn(I*exp(25/16/x^2)/(2*ln(x)-x^2+x))^2*x
^4-400*I*Pi*x^2*csgn(I/(2*ln(x)-x^2+x))*csgn(I*exp(25/16/x^2))*csgn(I*exp(25/16/x^2)/(2*ln(x)-x^2+x))+800*I*Pi
*x^2-800*I*Pi*x^2*csgn(I*exp(25/16/x^2)/(2*ln(x)-x^2+x))^2+400*I*Pi*x^2*csgn(I*exp(25/16/x^2))*csgn(I*exp(25/1
6/x^2)/(2*ln(x)-x^2+x))^2-512*I*Pi*ln(-1/2*x^2+1/2*x+ln(x))*x^4+256*I*Pi*ln(-1/2*x^2+1/2*x+ln(x))*csgn(I*exp(2
5/16/x^2)/(2*ln(x)-x^2+x))^3*x^4-400*I*Pi*x^2*csgn(I/(2*ln(x)-x^2+x))*csgn(I*exp(25/16/x^2)/(2*ln(x)-x^2+x))^2
+512*I*Pi*ln(-1/2*x^2+1/2*x+ln(x))*csgn(I*exp(25/16/x^2)/(2*ln(x)-x^2+x))^2*x^4+256*I*Pi*ln(-1/2*x^2+1/2*x+ln(
x))*csgn(I/(2*ln(x)-x^2+x))*csgn(I*exp(25/16/x^2))*csgn(I*exp(25/16/x^2)/(2*ln(x)-x^2+x))*x^4+256*I*Pi*ln(-1/2
*x^2+1/2*x+ln(x))*csgn(I/(2*ln(x)-x^2+x))*csgn(I*exp(25/16/x^2)/(2*ln(x)-x^2+x))^2*x^4-625+256*ln(-2*ln(x)+x^2
-x)^2*x^4)/x^4

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx=\log \left (-\frac {e^{\left (\frac {25}{16 \, x^{2}}\right )}}{x^{2} - x - 2 \, \log \left (x\right )}\right )^{2} \]

[In]

integrate((-50*log(x)+16*x^4-8*x^3+9*x^2-25*x)*log(exp(25/16/x^2)/(2*log(x)-x^2+x))/(8*x^3*log(x)-4*x^5+4*x^4)
,x, algorithm="fricas")

[Out]

log(-e^(25/16/x^2)/(x^2 - x - 2*log(x)))^2

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx=\log {\left (\frac {e^{\frac {25}{16 x^{2}}}}{- x^{2} + x + 2 \log {\left (x \right )}} \right )}^{2} \]

[In]

integrate((-50*ln(x)+16*x**4-8*x**3+9*x**2-25*x)*ln(exp(25/16/x**2)/(2*ln(x)-x**2+x))/(8*x**3*ln(x)-4*x**5+4*x
**4),x)

[Out]

log(exp(25/(16*x**2))/(-x**2 + x + 2*log(x)))**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (24) = 48\).

Time = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 4.04 \[ \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx=\frac {1}{8} \, {\left (\frac {25}{x^{2}} - 16 \, \log \left (-\frac {1}{2} \, x^{2} + \frac {1}{2} \, x + \log \left (x\right )\right )\right )} \log \left (-\frac {e^{\left (\frac {25}{16 \, x^{2}}\right )}}{x^{2} - x - 2 \, \log \left (x\right )}\right ) - \frac {256 \, x^{4} \log \left (-x^{2} + x + 2 \, \log \left (x\right )\right )^{2} + 800 \, x^{2} \log \left (2\right ) - 32 \, {\left (16 \, x^{4} \log \left (2\right ) + 25 \, x^{2}\right )} \log \left (-x^{2} + x + 2 \, \log \left (x\right )\right ) + 625}{256 \, x^{4}} \]

[In]

integrate((-50*log(x)+16*x^4-8*x^3+9*x^2-25*x)*log(exp(25/16/x^2)/(2*log(x)-x^2+x))/(8*x^3*log(x)-4*x^5+4*x^4)
,x, algorithm="maxima")

[Out]

1/8*(25/x^2 - 16*log(-1/2*x^2 + 1/2*x + log(x)))*log(-e^(25/16/x^2)/(x^2 - x - 2*log(x))) - 1/256*(256*x^4*log
(-x^2 + x + 2*log(x))^2 + 800*x^2*log(2) - 32*(16*x^4*log(2) + 25*x^2)*log(-x^2 + x + 2*log(x)) + 625)/x^4

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.31 \[ \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx=\log \left (x^{2} - x - 2 \, \log \left (x\right )\right )^{2} - 2 i \, \pi \log \left (-x^{2} + x + 2 \, \log \left (x\right )\right ) - \frac {25 \, \log \left (x^{2} - x - 2 \, \log \left (x\right )\right )}{8 \, x^{2}} + \frac {25 \, {\left (32 i \, \pi x^{2} + 25\right )}}{256 \, x^{4}} \]

[In]

integrate((-50*log(x)+16*x^4-8*x^3+9*x^2-25*x)*log(exp(25/16/x^2)/(2*log(x)-x^2+x))/(8*x^3*log(x)-4*x^5+4*x^4)
,x, algorithm="giac")

[Out]

log(x^2 - x - 2*log(x))^2 - 2*I*pi*log(-x^2 + x + 2*log(x)) - 25/8*log(x^2 - x - 2*log(x))/x^2 + 25/256*(32*I*
pi*x^2 + 25)/x^4

Mupad [B] (verification not implemented)

Time = 9.36 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {\left (-25 x+9 x^2-8 x^3+16 x^4-50 \log (x)\right ) \log \left (\frac {e^{\frac {25}{16 x^2}}}{x-x^2+2 \log (x)}\right )}{4 x^4-4 x^5+8 x^3 \log (x)} \, dx=\frac {{\left (16\,x^2\,\ln \left (\frac {1}{x+2\,\ln \left (x\right )-x^2}\right )+25\right )}^2}{256\,x^4} \]

[In]

int(-(log(exp(25/(16*x^2))/(x + 2*log(x) - x^2))*(25*x + 50*log(x) - 9*x^2 + 8*x^3 - 16*x^4))/(8*x^3*log(x) +
4*x^4 - 4*x^5),x)

[Out]

(16*x^2*log(1/(x + 2*log(x) - x^2)) + 25)^2/(256*x^4)