\(\int \frac {-16 e^{2 x} \log (5)+e^x (-48 x^2+12 x^3) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx\) [7511]

Optimal result

Integrand size = 77, antiderivative size = 25 \[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=\frac {\log (5)}{x^2 \left (\frac {3}{4} e^{-x} x^2+\log (6 x)\right )} \]

[Out]

ln(5)/x^2/(3/4*x^2/exp(x)+ln(6*x))

Rubi [F]

\[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=\int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx \]

[In]

Int[(-16*E^(2*x)*Log[5] + E^x*(-48*x^2 + 12*x^3)*Log[5] - 32*E^(2*x)*Log[5]*Log[6*x])/(9*x^7 + 24*E^x*x^5*Log[
6*x] + 16*E^(2*x)*x^3*Log[6*x]^2),x]

[Out]

12*Log[5]*Defer[Int][E^x/(3*x^2 + 4*E^x*Log[6*x])^2, x] - 24*Log[5]*Defer[Int][E^x/(x*(3*x^2 + 4*E^x*Log[6*x])
^2), x] + 12*Log[5]*Defer[Int][E^x/(x*Log[6*x]*(3*x^2 + 4*E^x*Log[6*x])^2), x] - 8*Log[5]*Defer[Int][E^x/(x^3*
(3*x^2 + 4*E^x*Log[6*x])), x] - 4*Log[5]*Defer[Int][E^x/(x^3*Log[6*x]*(3*x^2 + 4*E^x*Log[6*x])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {4 e^x \log (5) \left (-4 e^x+3 (-4+x) x^2-8 e^x \log (6 x)\right )}{x^3 \left (3 x^2+4 e^x \log (6 x)\right )^2} \, dx \\ & = (4 \log (5)) \int \frac {e^x \left (-4 e^x+3 (-4+x) x^2-8 e^x \log (6 x)\right )}{x^3 \left (3 x^2+4 e^x \log (6 x)\right )^2} \, dx \\ & = (4 \log (5)) \int \left (-\frac {e^x (1+2 \log (6 x))}{x^3 \log (6 x) \left (3 x^2+4 e^x \log (6 x)\right )}+\frac {3 e^x (1-2 \log (6 x)+x \log (6 x))}{x \log (6 x) \left (3 x^2+4 e^x \log (6 x)\right )^2}\right ) \, dx \\ & = -\left ((4 \log (5)) \int \frac {e^x (1+2 \log (6 x))}{x^3 \log (6 x) \left (3 x^2+4 e^x \log (6 x)\right )} \, dx\right )+(12 \log (5)) \int \frac {e^x (1-2 \log (6 x)+x \log (6 x))}{x \log (6 x) \left (3 x^2+4 e^x \log (6 x)\right )^2} \, dx \\ & = -\left ((4 \log (5)) \int \left (\frac {2 e^x}{x^3 \left (3 x^2+4 e^x \log (6 x)\right )}+\frac {e^x}{x^3 \log (6 x) \left (3 x^2+4 e^x \log (6 x)\right )}\right ) \, dx\right )+(12 \log (5)) \int \frac {e^x (1+(-2+x) \log (6 x))}{x \log (6 x) \left (3 x^2+4 e^x \log (6 x)\right )^2} \, dx \\ & = -\left ((4 \log (5)) \int \frac {e^x}{x^3 \log (6 x) \left (3 x^2+4 e^x \log (6 x)\right )} \, dx\right )-(8 \log (5)) \int \frac {e^x}{x^3 \left (3 x^2+4 e^x \log (6 x)\right )} \, dx+(12 \log (5)) \int \left (\frac {e^x}{\left (3 x^2+4 e^x \log (6 x)\right )^2}-\frac {2 e^x}{x \left (3 x^2+4 e^x \log (6 x)\right )^2}+\frac {e^x}{x \log (6 x) \left (3 x^2+4 e^x \log (6 x)\right )^2}\right ) \, dx \\ & = -\left ((4 \log (5)) \int \frac {e^x}{x^3 \log (6 x) \left (3 x^2+4 e^x \log (6 x)\right )} \, dx\right )-(8 \log (5)) \int \frac {e^x}{x^3 \left (3 x^2+4 e^x \log (6 x)\right )} \, dx+(12 \log (5)) \int \frac {e^x}{\left (3 x^2+4 e^x \log (6 x)\right )^2} \, dx+(12 \log (5)) \int \frac {e^x}{x \log (6 x) \left (3 x^2+4 e^x \log (6 x)\right )^2} \, dx-(24 \log (5)) \int \frac {e^x}{x \left (3 x^2+4 e^x \log (6 x)\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=\frac {4 e^x \log (5)}{x^2 \left (3 x^2+4 e^x \log (6 x)\right )} \]

[In]

Integrate[(-16*E^(2*x)*Log[5] + E^x*(-48*x^2 + 12*x^3)*Log[5] - 32*E^(2*x)*Log[5]*Log[6*x])/(9*x^7 + 24*E^x*x^
5*Log[6*x] + 16*E^(2*x)*x^3*Log[6*x]^2),x]

[Out]

(4*E^x*Log[5])/(x^2*(3*x^2 + 4*E^x*Log[6*x]))

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04

method	result	size
risch	\(\frac {4 \ln \left (5\right ) {\mathrm e}^{x}}{x^{2} \left (4 \,{\mathrm e}^{x} \ln \left (6 x \right )+3 x^{2}\right )}\)	\(26\)
parallelrisch	\(\frac {4 \ln \left (5\right ) {\mathrm e}^{x}}{x^{2} \left (4 \,{\mathrm e}^{x} \ln \left (6 x \right )+3 x^{2}\right )}\)	\(26\)

[In]

int((-32*ln(5)*exp(x)^2*ln(6*x)-16*ln(5)*exp(x)^2+(12*x^3-48*x^2)*ln(5)*exp(x))/(16*x^3*exp(x)^2*ln(6*x)^2+24*
x^5*exp(x)*ln(6*x)+9*x^7),x,method=_RETURNVERBOSE)

[Out]

4/x^2*ln(5)*exp(x)/(4*exp(x)*ln(6*x)+3*x^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=\frac {4 \, e^{x} \log \left (5\right )}{3 \, x^{4} + 4 \, x^{2} e^{x} \log \left (6 \, x\right )} \]

[In]

integrate((-32*log(5)*exp(x)^2*log(6*x)-16*log(5)*exp(x)^2+(12*x^3-48*x^2)*log(5)*exp(x))/(16*x^3*exp(x)^2*log
(6*x)^2+24*x^5*exp(x)*log(6*x)+9*x^7),x, algorithm="fricas")

[Out]

4*e^x*log(5)/(3*x^4 + 4*x^2*e^x*log(6*x))

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=- \frac {3 \log {\left (5 \right )}}{3 x^{2} \log {\left (6 x \right )} + 4 e^{x} \log {\left (6 x \right )}^{2}} + \frac {\log {\left (5 \right )}}{x^{2} \log {\left (6 x \right )}} \]

[In]

integrate((-32*ln(5)*exp(x)**2*ln(6*x)-16*ln(5)*exp(x)**2+(12*x**3-48*x**2)*ln(5)*exp(x))/(16*x**3*exp(x)**2*l
n(6*x)**2+24*x**5*exp(x)*ln(6*x)+9*x**7),x)

[Out]

-3*log(5)/(3*x**2*log(6*x) + 4*exp(x)*log(6*x)**2) + log(5)/(x**2*log(6*x))

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=\frac {4 \, e^{x} \log \left (5\right )}{3 \, x^{4} + 4 \, {\left (x^{2} {\left (\log \left (3\right ) + \log \left (2\right )\right )} + x^{2} \log \left (x\right )\right )} e^{x}} \]

[In]

integrate((-32*log(5)*exp(x)^2*log(6*x)-16*log(5)*exp(x)^2+(12*x^3-48*x^2)*log(5)*exp(x))/(16*x^3*exp(x)^2*log
(6*x)^2+24*x^5*exp(x)*log(6*x)+9*x^7),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=\frac {4 \, e^{x} \log \left (5\right )}{3 \, x^{4} + 4 \, x^{2} e^{x} \log \left (6 \, x\right )} \]

[In]

integrate((-32*log(5)*exp(x)^2*log(6*x)-16*log(5)*exp(x)^2+(12*x^3-48*x^2)*log(5)*exp(x))/(16*x^3*exp(x)^2*log
(6*x)^2+24*x^5*exp(x)*log(6*x)+9*x^7),x, algorithm="giac")

[Out]

4*e^x*log(5)/(3*x^4 + 4*x^2*e^x*log(6*x))

Mupad [B] (verification not implemented)

Time = 13.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.64 \[ \int \frac {-16 e^{2 x} \log (5)+e^x \left (-48 x^2+12 x^3\right ) \log (5)-32 e^{2 x} \log (5) \log (6 x)}{9 x^7+24 e^x x^5 \log (6 x)+16 e^{2 x} x^3 \log ^2(6 x)} \, dx=\frac {4\,{\mathrm {e}}^{2\,x}\,\ln \left (5\right )\,\left (4\,{\mathrm {e}}^x+6\,x^2-3\,x^3\right )}{x\,\left (4\,\ln \left (6\,x\right )\,{\mathrm {e}}^x+3\,x^2\right )\,\left (4\,x\,{\mathrm {e}}^{2\,x}+6\,x^3\,{\mathrm {e}}^x-3\,x^4\,{\mathrm {e}}^x\right )} \]

[In]

int(-(16*exp(2*x)*log(5) + 32*log(6*x)*exp(2*x)*log(5) + exp(x)*log(5)*(48*x^2 - 12*x^3))/(9*x^7 + 16*x^3*log(
6*x)^2*exp(2*x) + 24*x^5*log(6*x)*exp(x)),x)

[Out]

(4*exp(2*x)*log(5)*(4*exp(x) + 6*x^2 - 3*x^3))/(x*(4*log(6*x)*exp(x) + 3*x^2)*(4*x*exp(2*x) + 6*x^3*exp(x) - 3
*x^4*exp(x)))