\(\int \frac {e^x (-12-12 x) \log (4)+e^x (-12+12 x+12 x^2) \log (4) \log (x)+e^x (-3+3 x+3 x^2) \log ^2(x)}{x^2 \log ^2(x)} \, dx\) [3442]

Optimal result

Integrand size = 56, antiderivative size = 25 \[ \int \frac {e^x (-12-12 x) \log (4)+e^x \left (-12+12 x+12 x^2\right ) \log (4) \log (x)+e^x \left (-3+3 x+3 x^2\right ) \log ^2(x)}{x^2 \log ^2(x)} \, dx=\frac {3 e^x (2+2 x) \left (\frac {1}{2}+\frac {2 \log (4)}{\log (x)}\right )}{x} \]

[Out]

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

Rubi [F]

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

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {e^x (-12-12 x) \log (4)+e^x \left (-12+12 x+12 x^2\right ) \log (4) \log (x)+e^x \left (-3+3 x+3 x^2\right ) \log ^2(x)}{x^2 \log ^2(x)} \, dx=\frac {3 e^x (\log (256)+\log (x)+x \log (256 x))}{x \log (x)} \]

[In]

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

[Out]

(3*E^x*(Log[256] + Log[x] + x*Log[256*x]))/(x*Log[x])

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12

method	result	size
risch	\(\frac {3 \left (1+x \right ) {\mathrm e}^{x}}{x}+\frac {24 \ln \left (2\right ) {\mathrm e}^{x} \left (1+x \right )}{x \ln \left (x \right )}\)	\(28\)
norman	\(\frac {24 \,{\mathrm e}^{x} \ln \left (2\right )+3 \,{\mathrm e}^{x} \ln \left (x \right )+3 x \,{\mathrm e}^{x} \ln \left (x \right )+24 x \ln \left (2\right ) {\mathrm e}^{x}}{x \ln \left (x \right )}\)	\(36\)
parallelrisch	\(\frac {24 \,{\mathrm e}^{x} \ln \left (2\right )+3 \,{\mathrm e}^{x} \ln \left (x \right )+3 x \,{\mathrm e}^{x} \ln \left (x \right )+24 x \ln \left (2\right ) {\mathrm e}^{x}}{x \ln \left (x \right )}\)	\(36\)

[In]

int(((3*x^2+3*x-3)*exp(x)*ln(x)^2+2*(12*x^2+12*x-12)*ln(2)*exp(x)*ln(x)+2*(-12*x-12)*ln(2)*exp(x))/x^2/ln(x)^2
,x,method=_RETURNVERBOSE)

[Out]

3*(1+x)/x*exp(x)+24/x*ln(2)*exp(x)*(1+x)/ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^x (-12-12 x) \log (4)+e^x \left (-12+12 x+12 x^2\right ) \log (4) \log (x)+e^x \left (-3+3 x+3 x^2\right ) \log ^2(x)}{x^2 \log ^2(x)} \, dx=\frac {3 \, {\left (8 \, {\left (x + 1\right )} e^{x} \log \left (2\right ) + {\left (x + 1\right )} e^{x} \log \left (x\right )\right )}}{x \log \left (x\right )} \]

[In]

integrate(((3*x^2+3*x-3)*exp(x)*log(x)^2+2*(12*x^2+12*x-12)*log(2)*exp(x)*log(x)+2*(-12*x-12)*log(2)*exp(x))/x
^2/log(x)^2,x, algorithm="fricas")

[Out]

3*(8*(x + 1)*e^x*log(2) + (x + 1)*e^x*log(x))/(x*log(x))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^x (-12-12 x) \log (4)+e^x \left (-12+12 x+12 x^2\right ) \log (4) \log (x)+e^x \left (-3+3 x+3 x^2\right ) \log ^2(x)}{x^2 \log ^2(x)} \, dx=\frac {\left (3 x \log {\left (x \right )} + 24 x \log {\left (2 \right )} + 3 \log {\left (x \right )} + 24 \log {\left (2 \right )}\right ) e^{x}}{x \log {\left (x \right )}} \]

[In]

integrate(((3*x**2+3*x-3)*exp(x)*ln(x)**2+2*(12*x**2+12*x-12)*ln(2)*exp(x)*ln(x)+2*(-12*x-12)*ln(2)*exp(x))/x*
*2/ln(x)**2,x)

[Out]

(3*x*log(x) + 24*x*log(2) + 3*log(x) + 24*log(2))*exp(x)/(x*log(x))

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.32 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {e^x (-12-12 x) \log (4)+e^x \left (-12+12 x+12 x^2\right ) \log (4) \log (x)+e^x \left (-3+3 x+3 x^2\right ) \log ^2(x)}{x^2 \log ^2(x)} \, dx=\frac {24 \, {\left (x \log \left (2\right ) + \log \left (2\right )\right )} e^{x}}{x \log \left (x\right )} + 3 \, {\rm Ei}\left (x\right ) + 3 \, e^{x} - 3 \, \Gamma \left (-1, -x\right ) \]

[In]

integrate(((3*x^2+3*x-3)*exp(x)*log(x)^2+2*(12*x^2+12*x-12)*log(2)*exp(x)*log(x)+2*(-12*x-12)*log(2)*exp(x))/x
^2/log(x)^2,x, algorithm="maxima")

[Out]

24*(x*log(2) + log(2))*e^x/(x*log(x)) + 3*Ei(x) + 3*e^x - 3*gamma(-1, -x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {e^x (-12-12 x) \log (4)+e^x \left (-12+12 x+12 x^2\right ) \log (4) \log (x)+e^x \left (-3+3 x+3 x^2\right ) \log ^2(x)}{x^2 \log ^2(x)} \, dx=\frac {3 \, {\left (8 \, x e^{x} \log \left (2\right ) + x e^{x} \log \left (x\right ) + 8 \, e^{x} \log \left (2\right ) + e^{x} \log \left (x\right )\right )}}{x \log \left (x\right )} \]

[In]

integrate(((3*x^2+3*x-3)*exp(x)*log(x)^2+2*(12*x^2+12*x-12)*log(2)*exp(x)*log(x)+2*(-12*x-12)*log(2)*exp(x))/x
^2/log(x)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {e^x (-12-12 x) \log (4)+e^x \left (-12+12 x+12 x^2\right ) \log (4) \log (x)+e^x \left (-3+3 x+3 x^2\right ) \log ^2(x)}{x^2 \log ^2(x)} \, dx=3\,{\mathrm {e}}^x+\frac {3\,{\mathrm {e}}^x}{x}+\frac {24\,{\mathrm {e}}^x\,\ln \left (2\right )}{\ln \left (x\right )}+\frac {24\,{\mathrm {e}}^x\,\ln \left (2\right )}{x\,\ln \left (x\right )} \]

[In]

int((exp(x)*log(x)^2*(3*x + 3*x^2 - 3) - 2*exp(x)*log(2)*(12*x + 12) + 2*exp(x)*log(2)*log(x)*(12*x + 12*x^2 -
12))/(x^2*log(x)^2),x)

[Out]

3*exp(x) + (3*exp(x))/x + (24*exp(x)*log(2))/log(x) + (24*exp(x)*log(2))/(x*log(x))