\(\int \frac {216 e^x x^3+e^x (-864 x^3-216 x^4) \log (x)+e^x \log ^2(x)}{\log ^2(x)} \, dx\) [8627]

Optimal result

Integrand size = 39, antiderivative size = 16 \[ \int \frac {216 e^x x^3+e^x \left (-864 x^3-216 x^4\right ) \log (x)+e^x \log ^2(x)}{\log ^2(x)} \, dx=e^x-\frac {216 e^x x^4}{\log (x)} \]

[Out]

exp(x)-216*exp(x)*x^4/ln(x)

Rubi [F]

\[ \int \frac {216 e^x x^3+e^x \left (-864 x^3-216 x^4\right ) \log (x)+e^x \log ^2(x)}{\log ^2(x)} \, dx=\int \frac {216 e^x x^3+e^x \left (-864 x^3-216 x^4\right ) \log (x)+e^x \log ^2(x)}{\log ^2(x)} \, dx \]

[In]

Int[(216*E^x*x^3 + E^x*(-864*x^3 - 216*x^4)*Log[x] + E^x*Log[x]^2)/Log[x]^2,x]

[Out]

E^x + 216*Defer[Int][(E^x*x^3)/Log[x]^2, x] - 864*Defer[Int][(E^x*x^3)/Log[x], x] - 216*Defer[Int][(E^x*x^4)/L
og[x], x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (216 x^3-216 x^3 (4+x) \log (x)+\log ^2(x)\right )}{\log ^2(x)} \, dx \\ & = \int \left (e^x+\frac {216 e^x x^3}{\log ^2(x)}-\frac {216 e^x x^3 (4+x)}{\log (x)}\right ) \, dx \\ & = 216 \int \frac {e^x x^3}{\log ^2(x)} \, dx-216 \int \frac {e^x x^3 (4+x)}{\log (x)} \, dx+\int e^x \, dx \\ & = e^x-216 \int \left (\frac {4 e^x x^3}{\log (x)}+\frac {e^x x^4}{\log (x)}\right ) \, dx+216 \int \frac {e^x x^3}{\log ^2(x)} \, dx \\ & = e^x+216 \int \frac {e^x x^3}{\log ^2(x)} \, dx-216 \int \frac {e^x x^4}{\log (x)} \, dx-864 \int \frac {e^x x^3}{\log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {216 e^x x^3+e^x \left (-864 x^3-216 x^4\right ) \log (x)+e^x \log ^2(x)}{\log ^2(x)} \, dx=e^x-\frac {216 e^x x^4}{\log (x)} \]

[In]

Integrate[(216*E^x*x^3 + E^x*(-864*x^3 - 216*x^4)*Log[x] + E^x*Log[x]^2)/Log[x]^2,x]

[Out]

E^x - (216*E^x*x^4)/Log[x]

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
risch	\({\mathrm e}^{x}-\frac {216 \,{\mathrm e}^{x} x^{4}}{\ln \left (x \right )}\)	\(15\)
parallelrisch	\(-\frac {216 \,{\mathrm e}^{x} x^{4}-{\mathrm e}^{x} \ln \left (x \right )}{\ln \left (x \right )}\)	\(21\)

[In]

int((exp(x)*ln(x)^2+(-216*x^4-864*x^3)*exp(x)*ln(x)+216*exp(x)*x^3)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

exp(x)-216*exp(x)*x^4/ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25 \[ \int \frac {216 e^x x^3+e^x \left (-864 x^3-216 x^4\right ) \log (x)+e^x \log ^2(x)}{\log ^2(x)} \, dx=-\frac {216 \, x^{4} e^{x} - e^{x} \log \left (x\right )}{\log \left (x\right )} \]

[In]

integrate((exp(x)*log(x)^2+(-216*x^4-864*x^3)*exp(x)*log(x)+216*exp(x)*x^3)/log(x)^2,x, algorithm="fricas")

[Out]

-(216*x^4*e^x - e^x*log(x))/log(x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {216 e^x x^3+e^x \left (-864 x^3-216 x^4\right ) \log (x)+e^x \log ^2(x)}{\log ^2(x)} \, dx=\frac {\left (- 216 x^{4} + \log {\left (x \right )}\right ) e^{x}}{\log {\left (x \right )}} \]

[In]

integrate((exp(x)*ln(x)**2+(-216*x**4-864*x**3)*exp(x)*ln(x)+216*exp(x)*x**3)/ln(x)**2,x)

[Out]

(-216*x**4 + log(x))*exp(x)/log(x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {216 e^x x^3+e^x \left (-864 x^3-216 x^4\right ) \log (x)+e^x \log ^2(x)}{\log ^2(x)} \, dx=-\frac {216 \, x^{4} e^{x}}{\log \left (x\right )} + e^{x} \]

[In]

integrate((exp(x)*log(x)^2+(-216*x^4-864*x^3)*exp(x)*log(x)+216*exp(x)*x^3)/log(x)^2,x, algorithm="maxima")

[Out]

-216*x^4*e^x/log(x) + e^x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25 \[ \int \frac {216 e^x x^3+e^x \left (-864 x^3-216 x^4\right ) \log (x)+e^x \log ^2(x)}{\log ^2(x)} \, dx=-\frac {216 \, x^{4} e^{x} - e^{x} \log \left (x\right )}{\log \left (x\right )} \]

[In]

integrate((exp(x)*log(x)^2+(-216*x^4-864*x^3)*exp(x)*log(x)+216*exp(x)*x^3)/log(x)^2,x, algorithm="giac")

[Out]

-(216*x^4*e^x - e^x*log(x))/log(x)

Mupad [B] (verification not implemented)

Time = 12.46 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {216 e^x x^3+e^x \left (-864 x^3-216 x^4\right ) \log (x)+e^x \log ^2(x)}{\log ^2(x)} \, dx={\mathrm {e}}^x-\frac {216\,x^4\,{\mathrm {e}}^x}{\ln \left (x\right )} \]

[In]

int((216*x^3*exp(x) + exp(x)*log(x)^2 - exp(x)*log(x)*(864*x^3 + 216*x^4))/log(x)^2,x)

[Out]

exp(x) - (216*x^4*exp(x))/log(x)