\(\int \frac {e (192 x+96 x^2+12 x^3)+(e^5 (-96-24 x)+e (192 x+48 x^2)) \log (e^4-2 x)}{(e^4 x^3-2 x^4) \log ^3(e^4-2 x)} \, dx\) [7464]

Optimal result

Integrand size = 73, antiderivative size = 21 \[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\frac {3 e (4+x)^2}{x^2 \log ^2\left (e^4-2 x\right )} \]

[Out]

3/x^2*exp(1)/ln(exp(4)-2*x)^2*(4+x)^2

Rubi [F]

\[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx \]

[In]

Int[(E*(192*x + 96*x^2 + 12*x^3) + (E^5*(-96 - 24*x) + E*(192*x + 48*x^2))*Log[E^4 - 2*x])/((E^4*x^3 - 2*x^4)*
Log[E^4 - 2*x]^3),x]

[Out]

(3*(8 + E^4)^2)/(E^7*Log[E^4 - 2*x]^2) + (192*Defer[Int][1/(x^2*Log[E^4 - 2*x]^3), x])/E^3 + (96*(4 + E^4)*Def
er[Int][1/(x*Log[E^4 - 2*x]^3), x])/E^7 - 96*E*Defer[Int][1/(x^3*Log[E^4 - 2*x]^2), x] - 24*E*Defer[Int][1/(x^
2*Log[E^4 - 2*x]^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4-2 x\right ) x^3 \log ^3\left (e^4-2 x\right )} \, dx \\ & = \int \frac {12 e (4+x) \left (x (4+x)-2 \left (e^4-2 x\right ) \log \left (e^4-2 x\right )\right )}{\left (e^4-2 x\right ) x^3 \log ^3\left (e^4-2 x\right )} \, dx \\ & = (12 e) \int \frac {(4+x) \left (x (4+x)-2 \left (e^4-2 x\right ) \log \left (e^4-2 x\right )\right )}{\left (e^4-2 x\right ) x^3 \log ^3\left (e^4-2 x\right )} \, dx \\ & = (12 e) \int \left (\frac {(4+x)^2}{\left (e^4-2 x\right ) x^2 \log ^3\left (e^4-2 x\right )}-\frac {2 (4+x)}{x^3 \log ^2\left (e^4-2 x\right )}\right ) \, dx \\ & = (12 e) \int \frac {(4+x)^2}{\left (e^4-2 x\right ) x^2 \log ^3\left (e^4-2 x\right )} \, dx-(24 e) \int \frac {4+x}{x^3 \log ^2\left (e^4-2 x\right )} \, dx \\ & = (12 e) \int \left (\frac {\left (8+e^4\right )^2}{e^8 \left (e^4-2 x\right ) \log ^3\left (e^4-2 x\right )}+\frac {16}{e^4 x^2 \log ^3\left (e^4-2 x\right )}+\frac {8 \left (4+e^4\right )}{e^8 x \log ^3\left (e^4-2 x\right )}\right ) \, dx-(24 e) \int \left (\frac {4}{x^3 \log ^2\left (e^4-2 x\right )}+\frac {1}{x^2 \log ^2\left (e^4-2 x\right )}\right ) \, dx \\ & = \frac {192 \int \frac {1}{x^2 \log ^3\left (e^4-2 x\right )} \, dx}{e^3}-(24 e) \int \frac {1}{x^2 \log ^2\left (e^4-2 x\right )} \, dx-(96 e) \int \frac {1}{x^3 \log ^2\left (e^4-2 x\right )} \, dx+\frac {\left (96 \left (4+e^4\right )\right ) \int \frac {1}{x \log ^3\left (e^4-2 x\right )} \, dx}{e^7}+\frac {\left (12 \left (8+e^4\right )^2\right ) \int \frac {1}{\left (e^4-2 x\right ) \log ^3\left (e^4-2 x\right )} \, dx}{e^7} \\ & = \frac {192 \int \frac {1}{x^2 \log ^3\left (e^4-2 x\right )} \, dx}{e^3}-(24 e) \int \frac {1}{x^2 \log ^2\left (e^4-2 x\right )} \, dx-(96 e) \int \frac {1}{x^3 \log ^2\left (e^4-2 x\right )} \, dx+\frac {\left (96 \left (4+e^4\right )\right ) \int \frac {1}{x \log ^3\left (e^4-2 x\right )} \, dx}{e^7}-\frac {\left (6 \left (8+e^4\right )^2\right ) \text {Subst}\left (\int \frac {1}{x \log ^3(x)} \, dx,x,e^4-2 x\right )}{e^7} \\ & = \frac {192 \int \frac {1}{x^2 \log ^3\left (e^4-2 x\right )} \, dx}{e^3}-(24 e) \int \frac {1}{x^2 \log ^2\left (e^4-2 x\right )} \, dx-(96 e) \int \frac {1}{x^3 \log ^2\left (e^4-2 x\right )} \, dx+\frac {\left (96 \left (4+e^4\right )\right ) \int \frac {1}{x \log ^3\left (e^4-2 x\right )} \, dx}{e^7}-\frac {\left (6 \left (8+e^4\right )^2\right ) \text {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log \left (e^4-2 x\right )\right )}{e^7} \\ & = \frac {3 \left (8+e^4\right )^2}{e^7 \log ^2\left (e^4-2 x\right )}+\frac {192 \int \frac {1}{x^2 \log ^3\left (e^4-2 x\right )} \, dx}{e^3}-(24 e) \int \frac {1}{x^2 \log ^2\left (e^4-2 x\right )} \, dx-(96 e) \int \frac {1}{x^3 \log ^2\left (e^4-2 x\right )} \, dx+\frac {\left (96 \left (4+e^4\right )\right ) \int \frac {1}{x \log ^3\left (e^4-2 x\right )} \, dx}{e^7} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\frac {3 e (4+x)^2}{x^2 \log ^2\left (e^4-2 x\right )} \]

[In]

Integrate[(E*(192*x + 96*x^2 + 12*x^3) + (E^5*(-96 - 24*x) + E*(192*x + 48*x^2))*Log[E^4 - 2*x])/((E^4*x^3 - 2
*x^4)*Log[E^4 - 2*x]^3),x]

[Out]

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

Maple [A] (verified)

Time = 1.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19

method	result	size
risch	\(\frac {3 \,{\mathrm e} \left (x^{2}+8 x +16\right )}{x^{2} \ln \left ({\mathrm e}^{4}-2 x \right )^{2}}\)	\(25\)
norman	\(\frac {24 x \,{\mathrm e}+3 x^{2} {\mathrm e}+48 \,{\mathrm e}}{\ln \left ({\mathrm e}^{4}-2 x \right )^{2} x^{2}}\)	\(31\)
parallelrisch	\(\frac {12 x^{2} {\mathrm e}+96 x \,{\mathrm e}+192 \,{\mathrm e}}{4 \ln \left ({\mathrm e}^{4}-2 x \right )^{2} x^{2}}\)	\(32\)

[In]

int((((-24*x-96)*exp(1)*exp(4)+(48*x^2+192*x)*exp(1))*ln(exp(4)-2*x)+(12*x^3+96*x^2+192*x)*exp(1))/(x^3*exp(4)
-2*x^4)/ln(exp(4)-2*x)^3,x,method=_RETURNVERBOSE)

[Out]

3*exp(1)*(x^2+8*x+16)/x^2/ln(exp(4)-2*x)^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\frac {3 \, {\left (x^{2} + 8 \, x + 16\right )} e}{x^{2} \log \left (-2 \, x + e^{4}\right )^{2}} \]

[In]

integrate((((-24*x-96)*exp(1)*exp(4)+(48*x^2+192*x)*exp(1))*log(exp(4)-2*x)+(12*x^3+96*x^2+192*x)*exp(1))/(x^3
*exp(4)-2*x^4)/log(exp(4)-2*x)^3,x, algorithm="fricas")

[Out]

3*(x^2 + 8*x + 16)*e/(x^2*log(-2*x + e^4)^2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\frac {3 e x^{2} + 24 e x + 48 e}{x^{2} \log {\left (- 2 x + e^{4} \right )}^{2}} \]

[In]

integrate((((-24*x-96)*exp(1)*exp(4)+(48*x**2+192*x)*exp(1))*ln(exp(4)-2*x)+(12*x**3+96*x**2+192*x)*exp(1))/(x
**3*exp(4)-2*x**4)/ln(exp(4)-2*x)**3,x)

[Out]

(3*E*x**2 + 24*E*x + 48*E)/(x**2*log(-2*x + exp(4))**2)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\frac {3 \, {\left (x^{2} e + 8 \, x e + 16 \, e\right )}}{x^{2} \log \left (-2 \, x + e^{4}\right )^{2}} \]

[In]

integrate((((-24*x-96)*exp(1)*exp(4)+(48*x^2+192*x)*exp(1))*log(exp(4)-2*x)+(12*x^3+96*x^2+192*x)*exp(1))/(x^3
*exp(4)-2*x^4)/log(exp(4)-2*x)^3,x, algorithm="maxima")

[Out]

3*(x^2*e + 8*x*e + 16*e)/(x^2*log(-2*x + e^4)^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\frac {3 \, {\left (x^{2} e + 8 \, x e + 16 \, e\right )}}{x^{2} \log \left (-2 \, x + e^{4}\right )^{2}} \]

[In]

integrate((((-24*x-96)*exp(1)*exp(4)+(48*x^2+192*x)*exp(1))*log(exp(4)-2*x)+(12*x^3+96*x^2+192*x)*exp(1))/(x^3
*exp(4)-2*x^4)/log(exp(4)-2*x)^3,x, algorithm="giac")

[Out]

3*(x^2*e + 8*x*e + 16*e)/(x^2*log(-2*x + e^4)^2)

Mupad [B] (verification not implemented)

Time = 12.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e \left (192 x+96 x^2+12 x^3\right )+\left (e^5 (-96-24 x)+e \left (192 x+48 x^2\right )\right ) \log \left (e^4-2 x\right )}{\left (e^4 x^3-2 x^4\right ) \log ^3\left (e^4-2 x\right )} \, dx=\frac {3\,\mathrm {e}\,{\left (x+4\right )}^2}{x^2\,{\ln \left ({\mathrm {e}}^4-2\,x\right )}^2} \]

[In]

int((exp(1)*(192*x + 96*x^2 + 12*x^3) + log(exp(4) - 2*x)*(exp(1)*(192*x + 48*x^2) - exp(5)*(24*x + 96)))/(log
(exp(4) - 2*x)^3*(x^3*exp(4) - 2*x^4)),x)

[Out]

(3*exp(1)*(x + 4)^2)/(x^2*log(exp(4) - 2*x)^2)