\(\int \frac {e^{\frac {12}{(-15+20 x) \log (x \log ^2(2))}} (72-96 x-96 x \log (x \log ^2(2)))}{(45 x-120 x^2+80 x^3) \log ^2(x \log ^2(2))} \, dx\) [7422]

Optimal result

Integrand size = 61, antiderivative size = 28 \[ \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{\left (45 x-120 x^2+80 x^3\right ) \log ^2\left (x \log ^2(2)\right )} \, dx=-1+2 \left (-2+e^{\frac {3}{5 \left (-\frac {3}{4}+x\right ) \log \left (x \log ^2(2)\right )}}\right ) \]

[Out]

2*exp(3/5/ln(x*ln(2)^2)/(x-3/4))-5

Rubi [F]

\[ \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{\left (45 x-120 x^2+80 x^3\right ) \log ^2\left (x \log ^2(2)\right )} \, dx=\int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{\left (45 x-120 x^2+80 x^3\right ) \log ^2\left (x \log ^2(2)\right )} \, dx \]

[In]

Int[(E^(12/((-15 + 20*x)*Log[x*Log[2]^2]))*(72 - 96*x - 96*x*Log[x*Log[2]^2]))/((45*x - 120*x^2 + 80*x^3)*Log[
x*Log[2]^2]^2),x]

[Out]

(8*Defer[Int][E^(12/((-15 + 20*x)*Log[x*Log[2]^2]))/(x*Log[x*Log[2]^2]^2), x])/5 - (32*Defer[Int][E^(12/((-15
+ 20*x)*Log[x*Log[2]^2]))/((-3 + 4*x)*Log[x*Log[2]^2]^2), x])/5 - (96*Defer[Int][E^(12/((-15 + 20*x)*Log[x*Log
[2]^2]))/((-3 + 4*x)^2*Log[x*Log[2]^2]), x])/5

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{x \left (45-120 x+80 x^2\right ) \log ^2\left (x \log ^2(2)\right )} \, dx \\ & = \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{5 x (-3+4 x)^2 \log ^2\left (x \log ^2(2)\right )} \, dx \\ & = \frac {1}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{x (-3+4 x)^2 \log ^2\left (x \log ^2(2)\right )} \, dx \\ & = \frac {1}{5} \int \frac {24 e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (3-4 x-4 x \log \left (x \log ^2(2)\right )\right )}{(3-4 x)^2 x \log ^2\left (x \log ^2(2)\right )} \, dx \\ & = \frac {24}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (3-4 x-4 x \log \left (x \log ^2(2)\right )\right )}{(3-4 x)^2 x \log ^2\left (x \log ^2(2)\right )} \, dx \\ & = \frac {24}{5} \int \left (-\frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{x (-3+4 x) \log ^2\left (x \log ^2(2)\right )}-\frac {4 e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{(-3+4 x)^2 \log \left (x \log ^2(2)\right )}\right ) \, dx \\ & = -\left (\frac {24}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{x (-3+4 x) \log ^2\left (x \log ^2(2)\right )} \, dx\right )-\frac {96}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{(-3+4 x)^2 \log \left (x \log ^2(2)\right )} \, dx \\ & = -\left (\frac {24}{5} \int \left (-\frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{3 x \log ^2\left (x \log ^2(2)\right )}+\frac {4 e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{3 (-3+4 x) \log ^2\left (x \log ^2(2)\right )}\right ) \, dx\right )-\frac {96}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{(-3+4 x)^2 \log \left (x \log ^2(2)\right )} \, dx \\ & = \frac {8}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{x \log ^2\left (x \log ^2(2)\right )} \, dx-\frac {32}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{(-3+4 x) \log ^2\left (x \log ^2(2)\right )} \, dx-\frac {96}{5} \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}}}{(-3+4 x)^2 \log \left (x \log ^2(2)\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{\left (45 x-120 x^2+80 x^3\right ) \log ^2\left (x \log ^2(2)\right )} \, dx=2 e^{\frac {12}{5 (-3+4 x) \log \left (x \log ^2(2)\right )}} \]

[In]

Integrate[(E^(12/((-15 + 20*x)*Log[x*Log[2]^2]))*(72 - 96*x - 96*x*Log[x*Log[2]^2]))/((45*x - 120*x^2 + 80*x^3
)*Log[x*Log[2]^2]^2),x]

[Out]

2*E^(12/(5*(-3 + 4*x)*Log[x*Log[2]^2]))

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79

method	result	size
risch	\(2 \,{\mathrm e}^{\frac {12}{5 \left (-3+4 x \right ) \ln \left (x \ln \left (2\right )^{2}\right )}}\)	\(22\)
parallelrisch	\(2 \,{\mathrm e}^{\frac {12}{\left (20 x -15\right ) \ln \left (x \ln \left (2\right )^{2}\right )}}\)	\(22\)

[In]

int((-96*x*ln(x*ln(2)^2)-96*x+72)*exp(12/(20*x-15)/ln(x*ln(2)^2))/(80*x^3-120*x^2+45*x)/ln(x*ln(2)^2)^2,x,meth
od=_RETURNVERBOSE)

[Out]

2*exp(12/5/(-3+4*x)/ln(x*ln(2)^2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{\left (45 x-120 x^2+80 x^3\right ) \log ^2\left (x \log ^2(2)\right )} \, dx=2 \, e^{\left (\frac {12}{5 \, {\left (4 \, x - 3\right )} \log \left (x \log \left (2\right )^{2}\right )}\right )} \]

[In]

integrate((-96*x*log(x*log(2)^2)-96*x+72)*exp(12/(20*x-15)/log(x*log(2)^2))/(80*x^3-120*x^2+45*x)/log(x*log(2)
^2)^2,x, algorithm="fricas")

[Out]

2*e^(12/5/((4*x - 3)*log(x*log(2)^2)))

Sympy [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{\left (45 x-120 x^2+80 x^3\right ) \log ^2\left (x \log ^2(2)\right )} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((-96*x*ln(x*ln(2)**2)-96*x+72)*exp(12/(20*x-15)/ln(x*ln(2)**2))/(80*x**3-120*x**2+45*x)/ln(x*ln(2)**
2)**2,x)

[Out]

Exception raised: TypeError >> '>' not supported between instances of 'Poly' and 'int'

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{\left (45 x-120 x^2+80 x^3\right ) \log ^2\left (x \log ^2(2)\right )} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((-96*x*log(x*log(2)^2)-96*x+72)*exp(12/(20*x-15)/log(x*log(2)^2))/(80*x^3-120*x^2+45*x)/log(x*log(2)
^2)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{\left (45 x-120 x^2+80 x^3\right ) \log ^2\left (x \log ^2(2)\right )} \, dx=2 \, e^{\left (\frac {12}{5 \, {\left (4 \, x \log \left (x \log \left (2\right )^{2}\right ) - 3 \, \log \left (x \log \left (2\right )^{2}\right )\right )}}\right )} \]

[In]

integrate((-96*x*log(x*log(2)^2)-96*x+72)*exp(12/(20*x-15)/log(x*log(2)^2))/(80*x^3-120*x^2+45*x)/log(x*log(2)
^2)^2,x, algorithm="giac")

[Out]

2*e^(12/5/(4*x*log(x*log(2)^2) - 3*log(x*log(2)^2)))

Mupad [B] (verification not implemented)

Time = 12.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {12}{(-15+20 x) \log \left (x \log ^2(2)\right )}} \left (72-96 x-96 x \log \left (x \log ^2(2)\right )\right )}{\left (45 x-120 x^2+80 x^3\right ) \log ^2\left (x \log ^2(2)\right )} \, dx=2\,{\mathrm {e}}^{-\frac {12}{30\,\ln \left (\ln \left (2\right )\right )+15\,\ln \left (x\right )-40\,x\,\ln \left (\ln \left (2\right )\right )-20\,x\,\ln \left (x\right )}} \]

[In]

int(-(exp(12/(log(x*log(2)^2)*(20*x - 15)))*(96*x + 96*x*log(x*log(2)^2) - 72))/(log(x*log(2)^2)^2*(45*x - 120
*x^2 + 80*x^3)),x)

[Out]

2*exp(-12/(30*log(log(2)) + 15*log(x) - 40*x*log(log(2)) - 20*x*log(x)))