\(\int \frac {-1800 x+1800 \log (4)+e^{x/5} (-360 x+72 x^2+(360-72 x) \log (4))+(1800 x+360 e^{x/5} x-360 x^2) \log (\frac {-75-15 e^{x/5}+15 x}{x})}{25 x+5 e^{x/5} x-5 x^2} \, dx\) [4334]

Optimal result

Integrand size = 94, antiderivative size = 26 \[ \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx=72 (x-\log (4)) \log \left (\frac {15 \left (-5-e^{x/5}+x\right )}{x}\right ) \]

[Out]

72*(x-2*ln(2))*ln(15*(x-exp(1/5*x)-5)/x)

Rubi [F]

\[ \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx=\int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx \]

[In]

Int[(-1800*x + 1800*Log[4] + E^(x/5)*(-360*x + 72*x^2 + (360 - 72*x)*Log[4]) + (1800*x + 360*E^(x/5)*x - 360*x
^2)*Log[(-75 - 15*E^(x/5) + 15*x)/x])/(25*x + 5*E^(x/5)*x - 5*x^2),x]

[Out]

(-36*(5 - x)^2)/5 + (36*x^2)/5 - (72*x*(5 + Log[4]))/5 + 72*x*Log[(-15*(5 + E^(x/5) - x))/x] + 72*Log[4]*Log[x
] - (72*(10 + Log[4])*Defer[Int][x/(5 + E^(x/5) - x), x])/5 + (72*Defer[Int][x^2/(5 + E^(x/5) - x), x])/5 + 72
0*Log[4]*Defer[Subst][Defer[Int][(5 + E^x - 5*x)^(-1), x], x, x/5] + 3600*Defer[Subst][Defer[Int][x/(5 + E^x -
5*x), x], x, x/5] - 1800*Defer[Subst][Defer[Int][x^2/(5 + E^x - 5*x), x], x, x/5]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {72 \left (-25+e^{x/5} (-5+x)\right ) (x-\log (4))}{5 \left (5+e^{x/5}-x\right ) x}+72 \log \left (\frac {15 \left (-5-e^{x/5}+x\right )}{x}\right )\right ) \, dx \\ & = \frac {72}{5} \int \frac {\left (-25+e^{x/5} (-5+x)\right ) (x-\log (4))}{\left (5+e^{x/5}-x\right ) x} \, dx+72 \int \log \left (\frac {15 \left (-5-e^{x/5}+x\right )}{x}\right ) \, dx \\ & = 72 x \log \left (-\frac {15 \left (5+e^{x/5}-x\right )}{x}\right )+\frac {72}{5} \int \left (\frac {(-10+x) (x-\log (4))}{5+e^{x/5}-x}+\frac {(-5+x) (x-\log (4))}{x}\right ) \, dx-72 \int \frac {-25+e^{x/5} (-5+x)}{5 \left (5+e^{x/5}-x\right )} \, dx \\ & = 72 x \log \left (-\frac {15 \left (5+e^{x/5}-x\right )}{x}\right )-\frac {72}{5} \int \frac {-25+e^{x/5} (-5+x)}{5+e^{x/5}-x} \, dx+\frac {72}{5} \int \frac {(-10+x) (x-\log (4))}{5+e^{x/5}-x} \, dx+\frac {72}{5} \int \frac {(-5+x) (x-\log (4))}{x} \, dx \\ & = 72 x \log \left (-\frac {15 \left (5+e^{x/5}-x\right )}{x}\right )+\frac {72}{5} \int \left (-5+x-\log (4)+\frac {5 \log (4)}{x}\right ) \, dx+\frac {72}{5} \int \left (\frac {x^2}{5+e^{x/5}-x}+\frac {10 \log (4)}{5+e^{x/5}-x}-\frac {x (10+\log (4))}{5+e^{x/5}-x}\right ) \, dx-72 \text {Subst}\left (\int \frac {-25+e^x (-5+5 x)}{5+e^x-5 x} \, dx,x,\frac {x}{5}\right ) \\ & = \frac {36 x^2}{5}-\frac {72}{5} x (5+\log (4))+72 x \log \left (-\frac {15 \left (5+e^{x/5}-x\right )}{x}\right )+72 \log (4) \log (x)+\frac {72}{5} \int \frac {x^2}{5+e^{x/5}-x} \, dx-72 \text {Subst}\left (\int \left (5 (-1+x)+\frac {25 (-2+x) x}{5+e^x-5 x}\right ) \, dx,x,\frac {x}{5}\right )+(144 \log (4)) \int \frac {1}{5+e^{x/5}-x} \, dx-\frac {1}{5} (72 (10+\log (4))) \int \frac {x}{5+e^{x/5}-x} \, dx \\ & = -\frac {36}{5} (5-x)^2+\frac {36 x^2}{5}-\frac {72}{5} x (5+\log (4))+72 x \log \left (-\frac {15 \left (5+e^{x/5}-x\right )}{x}\right )+72 \log (4) \log (x)+\frac {72}{5} \int \frac {x^2}{5+e^{x/5}-x} \, dx-1800 \text {Subst}\left (\int \frac {(-2+x) x}{5+e^x-5 x} \, dx,x,\frac {x}{5}\right )+(720 \log (4)) \text {Subst}\left (\int \frac {1}{5+e^x-5 x} \, dx,x,\frac {x}{5}\right )-\frac {1}{5} (72 (10+\log (4))) \int \frac {x}{5+e^{x/5}-x} \, dx \\ & = -\frac {36}{5} (5-x)^2+\frac {36 x^2}{5}-\frac {72}{5} x (5+\log (4))+72 x \log \left (-\frac {15 \left (5+e^{x/5}-x\right )}{x}\right )+72 \log (4) \log (x)+\frac {72}{5} \int \frac {x^2}{5+e^{x/5}-x} \, dx-1800 \text {Subst}\left (\int \left (-\frac {2 x}{5+e^x-5 x}+\frac {x^2}{5+e^x-5 x}\right ) \, dx,x,\frac {x}{5}\right )+(720 \log (4)) \text {Subst}\left (\int \frac {1}{5+e^x-5 x} \, dx,x,\frac {x}{5}\right )-\frac {1}{5} (72 (10+\log (4))) \int \frac {x}{5+e^{x/5}-x} \, dx \\ & = -\frac {36}{5} (5-x)^2+\frac {36 x^2}{5}-\frac {72}{5} x (5+\log (4))+72 x \log \left (-\frac {15 \left (5+e^{x/5}-x\right )}{x}\right )+72 \log (4) \log (x)+\frac {72}{5} \int \frac {x^2}{5+e^{x/5}-x} \, dx-1800 \text {Subst}\left (\int \frac {x^2}{5+e^x-5 x} \, dx,x,\frac {x}{5}\right )+3600 \text {Subst}\left (\int \frac {x}{5+e^x-5 x} \, dx,x,\frac {x}{5}\right )+(720 \log (4)) \text {Subst}\left (\int \frac {1}{5+e^x-5 x} \, dx,x,\frac {x}{5}\right )-\frac {1}{5} (72 (10+\log (4))) \int \frac {x}{5+e^{x/5}-x} \, dx \\ \end{align*}

Mathematica [F]

\[ \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx=\int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx \]

[In]

Integrate[(-1800*x + 1800*Log[4] + E^(x/5)*(-360*x + 72*x^2 + (360 - 72*x)*Log[4]) + (1800*x + 360*E^(x/5)*x -
360*x^2)*Log[(-75 - 15*E^(x/5) + 15*x)/x])/(25*x + 5*E^(x/5)*x - 5*x^2),x]

[Out]

Integrate[(-1800*x + 1800*Log[4] + E^(x/5)*(-360*x + 72*x^2 + (360 - 72*x)*Log[4]) + (1800*x + 360*E^(x/5)*x -
360*x^2)*Log[(-75 - 15*E^(x/5) + 15*x)/x])/(25*x + 5*E^(x/5)*x - 5*x^2), x]

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50

method	result	size
parallelrisch	\(-144 \ln \left (-\frac {15 \left ({\mathrm e}^{\frac {x}{5}}-x +5\right )}{x}\right ) \ln \left (2\right )+72 \ln \left (-\frac {15 \left ({\mathrm e}^{\frac {x}{5}}-x +5\right )}{x}\right ) x\)	\(39\)
norman	\(-144 \ln \left (2\right ) \ln \left (\frac {-15 \,{\mathrm e}^{\frac {x}{5}}+15 x -75}{x}\right )+72 x \ln \left (\frac {-15 \,{\mathrm e}^{\frac {x}{5}}+15 x -75}{x}\right )\)	\(41\)
risch	\(72 x \ln \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )-72 x \ln \left (x \right )-36 i \pi x \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )\right ) \operatorname {csgn}\left (\frac {i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )}{x}\right )+36 i \pi x \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )}{x}\right )}^{2}+36 i \pi x \,\operatorname {csgn}\left (i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )}{x}\right )}^{2}-36 i \pi x {\operatorname {csgn}\left (\frac {i \left (x -{\mathrm e}^{\frac {x}{5}}-5\right )}{x}\right )}^{3}+72 x \ln \left (5\right )+72 x \ln \left (3\right )+144 \ln \left (2\right ) \ln \left (x \right )-144 \ln \left (2\right ) \ln \left ({\mathrm e}^{\frac {x}{5}}-x +5\right )\)	\(180\)

[In]

int(((360*x*exp(1/5*x)-360*x^2+1800*x)*ln((-15*exp(1/5*x)+15*x-75)/x)+(2*(-72*x+360)*ln(2)+72*x^2-360*x)*exp(1
/5*x)+3600*ln(2)-1800*x)/(5*x*exp(1/5*x)-5*x^2+25*x),x,method=_RETURNVERBOSE)

[Out]

-144*ln(-15*(exp(1/5*x)-x+5)/x)*ln(2)+72*ln(-15*(exp(1/5*x)-x+5)/x)*x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx=72 \, {\left (x - 2 \, \log \left (2\right )\right )} \log \left (\frac {15 \, {\left (x - e^{\left (\frac {1}{5} \, x\right )} - 5\right )}}{x}\right ) \]

[In]

integrate(((360*x*exp(1/5*x)-360*x^2+1800*x)*log((-15*exp(1/5*x)+15*x-75)/x)+(2*(-72*x+360)*log(2)+72*x^2-360*
x)*exp(1/5*x)+3600*log(2)-1800*x)/(5*x*exp(1/5*x)-5*x^2+25*x),x, algorithm="fricas")

[Out]

72*(x - 2*log(2))*log(15*(x - e^(1/5*x) - 5)/x)

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx=72 x \log {\left (\frac {15 x - 15 e^{\frac {x}{5}} - 75}{x} \right )} + 144 \log {\left (2 \right )} \log {\left (x \right )} - 144 \log {\left (2 \right )} \log {\left (- x + e^{\frac {x}{5}} + 5 \right )} \]

[In]

integrate(((360*x*exp(1/5*x)-360*x**2+1800*x)*ln((-15*exp(1/5*x)+15*x-75)/x)+(2*(-72*x+360)*ln(2)+72*x**2-360*
x)*exp(1/5*x)+3600*ln(2)-1800*x)/(5*x*exp(1/5*x)-5*x**2+25*x),x)

[Out]

72*x*log((15*x - 15*exp(x/5) - 75)/x) + 144*log(2)*log(x) - 144*log(2)*log(-x + exp(x/5) + 5)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx=72 \, {\left (i \, \pi + \log \left (5\right ) + \log \left (3\right )\right )} x - 72 \, {\left (x - 2 \, \log \left (2\right )\right )} \log \left (x\right ) + 72 \, {\left (x - 2 \, \log \left (2\right )\right )} \log \left (-x + e^{\left (\frac {1}{5} \, x\right )} + 5\right ) \]

[In]

integrate(((360*x*exp(1/5*x)-360*x^2+1800*x)*log((-15*exp(1/5*x)+15*x-75)/x)+(2*(-72*x+360)*log(2)+72*x^2-360*
x)*exp(1/5*x)+3600*log(2)-1800*x)/(5*x*exp(1/5*x)-5*x^2+25*x),x, algorithm="maxima")

[Out]

72*(I*pi + log(5) + log(3))*x - 72*(x - 2*log(2))*log(x) + 72*(x - 2*log(2))*log(-x + e^(1/5*x) + 5)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx=72 \, x \log \left (15 \, x - 15 \, e^{\left (\frac {1}{5} \, x\right )} - 75\right ) - 72 \, x \log \left (x\right ) + 144 \, \log \left (2\right ) \log \left (x\right ) - 144 \, \log \left (2\right ) \log \left (-x + e^{\left (\frac {1}{5} \, x\right )} + 5\right ) \]

[In]

integrate(((360*x*exp(1/5*x)-360*x^2+1800*x)*log((-15*exp(1/5*x)+15*x-75)/x)+(2*(-72*x+360)*log(2)+72*x^2-360*
x)*exp(1/5*x)+3600*log(2)-1800*x)/(5*x*exp(1/5*x)-5*x^2+25*x),x, algorithm="giac")

[Out]

72*x*log(15*x - 15*e^(1/5*x) - 75) - 72*x*log(x) + 144*log(2)*log(x) - 144*log(2)*log(-x + e^(1/5*x) + 5)

Mupad [B] (verification not implemented)

Time = 10.84 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {-1800 x+1800 \log (4)+e^{x/5} \left (-360 x+72 x^2+(360-72 x) \log (4)\right )+\left (1800 x+360 e^{x/5} x-360 x^2\right ) \log \left (\frac {-75-15 e^{x/5}+15 x}{x}\right )}{25 x+5 e^{x/5} x-5 x^2} \, dx=72\,x\,\ln \left (-\frac {15\,{\left ({\mathrm {e}}^x\right )}^{1/5}-15\,x+75}{x}\right )-72\,\ln \left (4\right )\,\ln \left ({\left ({\mathrm {e}}^x\right )}^{1/5}-x+5\right )+72\,\ln \left (4\right )\,\ln \left (x\right ) \]

[In]

int(-(1800*x - 3600*log(2) + exp(x/5)*(360*x + 2*log(2)*(72*x - 360) - 72*x^2) - log(-(15*exp(x/5) - 15*x + 75
)/x)*(1800*x + 360*x*exp(x/5) - 360*x^2))/(25*x + 5*x*exp(x/5) - 5*x^2),x)

[Out]

72*x*log(-(15*exp(x)^(1/5) - 15*x + 75)/x) - 72*log(4)*log(exp(x)^(1/5) - x + 5) + 72*log(4)*log(x)