\(\int ((-2-x) \log (4)-\log (4) \log (e^x x^2)) \, dx\) [1352]

Optimal result

Integrand size = 21, antiderivative size = 16 \[ \int \left ((-2-x) \log (4)-\log (4) \log \left (e^x x^2\right )\right ) \, dx=5+e-x \log (4) \log \left (e^x x^2\right ) \]

[Out]

exp(1)+5-2*ln(exp(x)*x^2)*ln(2)*x

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(39\) vs. \(2(16)=32\).

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.44, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2628} \[ \int \left ((-2-x) \log (4)-\log (4) \log \left (e^x x^2\right )\right ) \, dx=\frac {1}{2} x^2 \log (4)-x \log (4) \log \left (e^x x^2\right )+2 x \log (4)-\frac {1}{2} (x+2)^2 \log (4) \]

[In]

Int[(-2 - x)*Log[4] - Log[4]*Log[E^x*x^2],x]

[Out]

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

Rule 2628

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*(D[u, x]/u), x], x] /; InverseFunctionFr
eeQ[u, x]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \left ((-2-x) \log (4)-\log (4) \log \left (e^x x^2\right )\right ) \, dx=-x \log (4) \log \left (e^x x^2\right ) \]

[In]

Integrate[(-2 - x)*Log[4] - Log[4]*Log[E^x*x^2],x]

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
norman	\(-2 \ln \left ({\mathrm e}^{x} x^{2}\right ) \ln \left (2\right ) x\)	\(13\)
parallelrisch	\(-2 \ln \left ({\mathrm e}^{x} x^{2}\right ) \ln \left (2\right ) x\)	\(13\)
default	\(2 \ln \left (2\right ) \left (-\frac {1}{2} x^{2}-2 x \right )-2 \ln \left (2\right ) \left (x \ln \left ({\mathrm e}^{x} x^{2}\right )-\frac {x^{2}}{2}-2 x \right )\)	\(37\)
parts	\(-2 \ln \left (2\right ) \left (2 x +\frac {1}{2} x^{2}\right )-2 \ln \left (2\right ) \left (x \ln \left ({\mathrm e}^{x} x^{2}\right )-\frac {x^{2}}{2}-2 x \right )\)	\(37\)
risch	\(-2 x \ln \left ({\mathrm e}^{x}\right ) \ln \left (2\right )-4 x \ln \left (2\right ) \ln \left (x \right )+i \ln \left (2\right ) \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 i \ln \left (2\right ) \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+i \ln \left (2\right ) \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}-i \ln \left (2\right ) \pi x \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} x^{2}\right )^{2}+i \ln \left (2\right ) \pi x \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} x^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )+i \ln \left (2\right ) \pi x \operatorname {csgn}\left (i {\mathrm e}^{x} x^{2}\right )^{3}-i \ln \left (2\right ) \pi x \operatorname {csgn}\left (i {\mathrm e}^{x} x^{2}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )\)	\(171\)

[In]

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

[Out]

-2*ln(exp(x)*x^2)*ln(2)*x

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \left ((-2-x) \log (4)-\log (4) \log \left (e^x x^2\right )\right ) \, dx=-2 \, x \log \left (2\right ) \log \left (x^{2} e^{x}\right ) \]

[In]

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

[Out]

-2*x*log(2)*log(x^2*e^x)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \left ((-2-x) \log (4)-\log (4) \log \left (e^x x^2\right )\right ) \, dx=- 2 x \log {\left (2 \right )} \log {\left (x^{2} e^{x} \right )} \]

[In]

integrate(-2*ln(2)*ln(exp(x)*x**2)+2*(-2-x)*ln(2),x)

[Out]

-2*x*log(2)*log(x**2*exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \left ((-2-x) \log (4)-\log (4) \log \left (e^x x^2\right )\right ) \, dx=-2 \, x \log \left (2\right ) \log \left (x^{2} e^{x}\right ) \]

[In]

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

[Out]

-2*x*log(2)*log(x^2*e^x)

Giac [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.00 \[ \int \left ((-2-x) \log (4)-\log (4) \log \left (e^x x^2\right )\right ) \, dx={\left (x^{2} - 2 \, x \log \left (x^{2} e^{x}\right ) + 4 \, x\right )} \log \left (2\right ) - {\left (x^{2} + 4 \, x\right )} \log \left (2\right ) \]

[In]

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

[Out]

(x^2 - 2*x*log(x^2*e^x) + 4*x)*log(2) - (x^2 + 4*x)*log(2)

Mupad [B] (verification not implemented)

Time = 9.39 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69 \[ \int \left ((-2-x) \log (4)-\log (4) \log \left (e^x x^2\right )\right ) \, dx=-2\,x\,\ln \left (2\right )\,\left (x+\ln \left (x^2\right )\right ) \]

[In]

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

[Out]

-2*x*log(2)*(x + log(x^2))