\(\int \frac {-24-12 x+\log (4)+(18+6 x) \log (2 e^x x)-3 \log ^2(2 e^x x)}{2 x^2} \, dx\) [5643]

Optimal result

Integrand size = 38, antiderivative size = 28 \[ \int \frac {-24-12 x+\log (4)+(18+6 x) \log \left (2 e^x x\right )-3 \log ^2\left (2 e^x x\right )}{2 x^2} \, dx=\frac {x-\log (4)+3 \left (2-\log \left (2 e^x x\right )\right )^2}{2 x} \]

[Out]

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

Rubi [F]

\[ \int \frac {-24-12 x+\log (4)+(18+6 x) \log \left (2 e^x x\right )-3 \log ^2\left (2 e^x x\right )}{2 x^2} \, dx=\int \frac {-24-12 x+\log (4)+(18+6 x) \log \left (2 e^x x\right )-3 \log ^2\left (2 e^x x\right )}{2 x^2} \, dx \]

[In]

Int[(-24 - 12*x + Log[4] + (18 + 6*x)*Log[2*E^x*x] - 3*Log[2*E^x*x]^2)/(2*x^2),x]

[Out]

-9/x + (24 - Log[4])/(2*x) + 3*Log[x] - (9*Log[2*E^x*x])/x + 3*Defer[Int][Log[2*E^x*x]/x, x] - (3*Defer[Int][L
og[2*E^x*x]^2/x^2, x])/2

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {-24-12 x+\log (4)+(18+6 x) \log \left (2 e^x x\right )-3 \log ^2\left (2 e^x x\right )}{x^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {-24-12 x+\log (4)}{x^2}+\frac {6 (3+x) \log \left (2 e^x x\right )}{x^2}-\frac {3 \log ^2\left (2 e^x x\right )}{x^2}\right ) \, dx \\ & = \frac {1}{2} \int \frac {-24-12 x+\log (4)}{x^2} \, dx-\frac {3}{2} \int \frac {\log ^2\left (2 e^x x\right )}{x^2} \, dx+3 \int \frac {(3+x) \log \left (2 e^x x\right )}{x^2} \, dx \\ & = \frac {1}{2} \int \left (-\frac {12}{x}+\frac {-24+\log (4)}{x^2}\right ) \, dx-\frac {3}{2} \int \frac {\log ^2\left (2 e^x x\right )}{x^2} \, dx+3 \int \left (\frac {3 \log \left (2 e^x x\right )}{x^2}+\frac {\log \left (2 e^x x\right )}{x}\right ) \, dx \\ & = \frac {24-\log (4)}{2 x}-6 \log (x)-\frac {3}{2} \int \frac {\log ^2\left (2 e^x x\right )}{x^2} \, dx+3 \int \frac {\log \left (2 e^x x\right )}{x} \, dx+9 \int \frac {\log \left (2 e^x x\right )}{x^2} \, dx \\ & = \frac {24-\log (4)}{2 x}-6 \log (x)-\frac {9 \log \left (2 e^x x\right )}{x}-\frac {3}{2} \int \frac {\log ^2\left (2 e^x x\right )}{x^2} \, dx+3 \int \frac {\log \left (2 e^x x\right )}{x} \, dx+9 \int \frac {1+x}{x^2} \, dx \\ & = \frac {24-\log (4)}{2 x}-6 \log (x)-\frac {9 \log \left (2 e^x x\right )}{x}-\frac {3}{2} \int \frac {\log ^2\left (2 e^x x\right )}{x^2} \, dx+3 \int \frac {\log \left (2 e^x x\right )}{x} \, dx+9 \int \left (\frac {1}{x^2}+\frac {1}{x}\right ) \, dx \\ & = -\frac {9}{x}+\frac {24-\log (4)}{2 x}+3 \log (x)-\frac {9 \log \left (2 e^x x\right )}{x}-\frac {3}{2} \int \frac {\log ^2\left (2 e^x x\right )}{x^2} \, dx+3 \int \frac {\log \left (2 e^x x\right )}{x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {-24-12 x+\log (4)+(18+6 x) \log \left (2 e^x x\right )-3 \log ^2\left (2 e^x x\right )}{2 x^2} \, dx=\frac {12-6 x+6 x^2-\log (4)+6 x \log (x)-6 (2+x) \log \left (2 e^x x\right )+3 \log ^2\left (2 e^x x\right )}{2 x} \]

[In]

Integrate[(-24 - 12*x + Log[4] + (18 + 6*x)*Log[2*E^x*x] - 3*Log[2*E^x*x]^2)/(2*x^2),x]

[Out]

(12 - 6*x + 6*x^2 - Log[4] + 6*x*Log[x] - 6*(2 + x)*Log[2*E^x*x] + 3*Log[2*E^x*x]^2)/(2*x)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04

method	result	size
norman	\(\frac {6+\frac {3 \ln \left (2 \,{\mathrm e}^{x} x \right )^{2}}{2}-6 \ln \left (2 \,{\mathrm e}^{x} x \right )-\ln \left (2\right )}{x}\)	\(29\)
parallelrisch	\(\frac {12+3 \ln \left (2 \,{\mathrm e}^{x} x \right )^{2}-2 \ln \left (2\right )-12 \ln \left (2 \,{\mathrm e}^{x} x \right )}{2 x}\)	\(30\)
parts	\(3 \ln \left (x \right )-\frac {\ln \left (2\right )-12}{x}+\frac {3-3 \ln \left (2 \,{\mathrm e}^{x} x \right ) x -\frac {3 x^{3}}{2}+\frac {3 \ln \left (2 \,{\mathrm e}^{x} x \right )^{2}}{2}-\frac {3 x \ln \left (2 \,{\mathrm e}^{x} x \right )^{2}}{2}+3 x^{2} \ln \left (2 \,{\mathrm e}^{x} x \right )+3 \ln \left (2 \,{\mathrm e}^{x} x \right )}{x}+3 \ln \left (2 \,{\mathrm e}^{x} x \right ) \ln \left (x \right )-\frac {9 \ln \left (2 \,{\mathrm e}^{x} x \right )}{x}-3 x \ln \left (x \right )+3 x -\frac {3 \ln \left (x \right )^{2}}{2}-\frac {9}{x}\)	\(115\)
default	\(\frac {6-6 \ln \left (2 \,{\mathrm e}^{x} x \right ) x -3 x^{3}+3 \ln \left (2 \,{\mathrm e}^{x} x \right )^{2}-3 x \ln \left (2 \,{\mathrm e}^{x} x \right )^{2}+6 x^{2} \ln \left (2 \,{\mathrm e}^{x} x \right )+6 \ln \left (2 \,{\mathrm e}^{x} x \right )}{2 x}+3 \ln \left (x \right )-\frac {\ln \left (2\right )-12}{x}+3 \ln \left (2 \,{\mathrm e}^{x} x \right ) \ln \left (x \right )-\frac {9 \ln \left (2 \,{\mathrm e}^{x} x \right )}{x}-3 x \ln \left (x \right )+3 x -\frac {3 \ln \left (x \right )^{2}}{2}-\frac {9}{x}\)	\(116\)
risch	\(\text {Expression too large to display}\)	\(571\)

[In]

int(1/2*(-3*ln(2*exp(x)*x)^2+(18+6*x)*ln(2*exp(x)*x)+2*ln(2)-12*x-24)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {-24-12 x+\log (4)+(18+6 x) \log \left (2 e^x x\right )-3 \log ^2\left (2 e^x x\right )}{2 x^2} \, dx=\frac {3 \, \log \left (2 \, x e^{x}\right )^{2} - 2 \, \log \left (2\right ) - 12 \, \log \left (2 \, x e^{x}\right ) + 12}{2 \, x} \]

[In]

integrate(1/2*(-3*log(2*exp(x)*x)^2+(18+6*x)*log(2*exp(x)*x)+2*log(2)-12*x-24)/x^2,x, algorithm="fricas")

[Out]

1/2*(3*log(2*x*e^x)^2 - 2*log(2) - 12*log(2*x*e^x) + 12)/x

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-24-12 x+\log (4)+(18+6 x) \log \left (2 e^x x\right )-3 \log ^2\left (2 e^x x\right )}{2 x^2} \, dx=\frac {3 \log {\left (2 x e^{x} \right )}^{2}}{2 x} - \frac {6 \log {\left (2 x e^{x} \right )}}{x} - \frac {-6 + \log {\left (2 \right )}}{x} \]

[In]

integrate(1/2*(-3*ln(2*exp(x)*x)**2+(18+6*x)*ln(2*exp(x)*x)+2*ln(2)-12*x-24)/x**2,x)

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 113 vs. \(2 (23) = 46\).

Time = 0.21 (sec) , antiderivative size = 113, normalized size of antiderivative = 4.04 \[ \int \frac {-24-12 x+\log (4)+(18+6 x) \log \left (2 e^x x\right )-3 \log ^2\left (2 e^x x\right )}{2 x^2} \, dx=3 \, {\left (\frac {1}{x} - \log \left (x\right )\right )} \log \left (2 \, x e^{x}\right ) - 3 \, {\left (x + \log \left (x\right )\right )} \log \left (x\right ) + 3 \, \log \left (2 \, x e^{x}\right ) \log \left (x\right ) + \frac {3}{2} \, \log \left (x\right )^{2} + 3 \, x + \frac {3 \, \log \left (2 \, x e^{x}\right )^{2}}{2 \, x} + \frac {3 \, {\left (x \log \left (x\right )^{2} - 2 \, x^{2} + 2 \, {\left (x^{2} - x\right )} \log \left (x\right ) + 2\right )}}{2 \, x} - \frac {\log \left (2\right )}{x} - \frac {9 \, \log \left (2 \, x e^{x}\right )}{x} + \frac {3}{x} + 3 \, \log \left (x\right ) \]

[In]

integrate(1/2*(-3*log(2*exp(x)*x)^2+(18+6*x)*log(2*exp(x)*x)+2*log(2)-12*x-24)/x^2,x, algorithm="maxima")

[Out]

3*(1/x - log(x))*log(2*x*e^x) - 3*(x + log(x))*log(x) + 3*log(2*x*e^x)*log(x) + 3/2*log(x)^2 + 3*x + 3/2*log(2
*x*e^x)^2/x + 3/2*(x*log(x)^2 - 2*x^2 + 2*(x^2 - x)*log(x) + 2)/x - log(2)/x - 9*log(2*x*e^x)/x + 3/x + 3*log(
x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.61 \[ \int \frac {-24-12 x+\log (4)+(18+6 x) \log \left (2 e^x x\right )-3 \log ^2\left (2 e^x x\right )}{2 x^2} \, dx=\frac {3}{2} \, x + \frac {3 \, {\left (\log \left (2\right ) - 2\right )} \log \left (x\right )}{x} + \frac {3 \, \log \left (x\right )^{2}}{2 \, x} + \frac {3 \, \log \left (2\right )^{2} - 14 \, \log \left (2\right ) + 12}{2 \, x} + 3 \, \log \left (x\right ) \]

[In]

integrate(1/2*(-3*log(2*exp(x)*x)^2+(18+6*x)*log(2*exp(x)*x)+2*log(2)-12*x-24)/x^2,x, algorithm="giac")

[Out]

3/2*x + 3*(log(2) - 2)*log(x)/x + 3/2*log(x)^2/x + 1/2*(3*log(2)^2 - 14*log(2) + 12)/x + 3*log(x)

Mupad [B] (verification not implemented)

Time = 11.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {-24-12 x+\log (4)+(18+6 x) \log \left (2 e^x x\right )-3 \log ^2\left (2 e^x x\right )}{2 x^2} \, dx=-\frac {-\frac {3\,{\ln \left (2\,x\,{\mathrm {e}}^x\right )}^2}{2}+6\,\ln \left (2\,x\,{\mathrm {e}}^x\right )+\ln \left (2\right )-6}{x} \]

[In]

int(-(6*x - log(2) + (3*log(2*x*exp(x))^2)/2 - (log(2*x*exp(x))*(6*x + 18))/2 + 12)/x^2,x)

[Out]

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