\(\int \frac {3+2 x+(-9-12 x+(3+4 x) \log (x)) \log (3-\log (x))}{-6+2 \log (x)} \, dx\) [5243]

Optimal result

Integrand size = 35, antiderivative size = 14 \[ \int \frac {3+2 x+(-9-12 x+(3+4 x) \log (x)) \log (3-\log (x))}{-6+2 \log (x)} \, dx=x \left (\frac {3}{2}+x\right ) \log (3-\log (x)) \]

[Out]

x*(x+3/2)*ln(3-ln(x))

Rubi [F]

\[ \int \frac {3+2 x+(-9-12 x+(3+4 x) \log (x)) \log (3-\log (x))}{-6+2 \log (x)} \, dx=\int \frac {3+2 x+(-9-12 x+(3+4 x) \log (x)) \log (3-\log (x))}{-6+2 \log (x)} \, dx \]

[In]

Int[(3 + 2*x + (-9 - 12*x + (3 + 4*x)*Log[x])*Log[3 - Log[x]])/(-6 + 2*Log[x]),x]

[Out]

E^6*ExpIntegralEi[-2*(3 - Log[x])] + (3*E^3*ExpIntegralEi[-3 + Log[x]])/2 + (3*Defer[Int][Log[3 - Log[x]], x])
/2 + 2*Defer[Int][x*Log[3 - Log[x]], x]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.21 \[ \int \frac {3+2 x+(-9-12 x+(3+4 x) \log (x)) \log (3-\log (x))}{-6+2 \log (x)} \, dx=\frac {1}{2} x (3+2 x) \log (3-\log (x)) \]

[In]

Integrate[(3 + 2*x + (-9 - 12*x + (3 + 4*x)*Log[x])*Log[3 - Log[x]])/(-6 + 2*Log[x]),x]

[Out]

(x*(3 + 2*x)*Log[3 - Log[x]])/2

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14

method	result	size
risch	\(\left (x^{2}+\frac {3}{2} x \right ) \ln \left (3-\ln \left (x \right )\right )\)	\(16\)
default	\(\ln \left (3-\ln \left (x \right )\right ) x^{2}+\frac {3 \ln \left (3-\ln \left (x \right )\right ) x}{2}\)	\(23\)
norman	\(\ln \left (3-\ln \left (x \right )\right ) x^{2}+\frac {3 \ln \left (3-\ln \left (x \right )\right ) x}{2}\)	\(23\)
parallelrisch	\(\ln \left (3-\ln \left (x \right )\right ) x^{2}+\frac {3 \ln \left (3-\ln \left (x \right )\right ) x}{2}\)	\(23\)

[In]

int((((3+4*x)*ln(x)-12*x-9)*ln(3-ln(x))+2*x+3)/(2*ln(x)-6),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.29 \[ \int \frac {3+2 x+(-9-12 x+(3+4 x) \log (x)) \log (3-\log (x))}{-6+2 \log (x)} \, dx=\frac {1}{2} \, {\left (2 \, x^{2} + 3 \, x\right )} \log \left (-\log \left (x\right ) + 3\right ) \]

[In]

integrate((((3+4*x)*log(x)-12*x-9)*log(3-log(x))+2*x+3)/(2*log(x)-6),x, algorithm="fricas")

[Out]

1/2*(2*x^2 + 3*x)*log(-log(x) + 3)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {3+2 x+(-9-12 x+(3+4 x) \log (x)) \log (3-\log (x))}{-6+2 \log (x)} \, dx=\left (x^{2} + \frac {3 x}{2}\right ) \log {\left (3 - \log {\left (x \right )} \right )} \]

[In]

integrate((((3+4*x)*ln(x)-12*x-9)*ln(3-ln(x))+2*x+3)/(2*ln(x)-6),x)

[Out]

(x**2 + 3*x/2)*log(3 - log(x))

Maxima [F]

\[ \int \frac {3+2 x+(-9-12 x+(3+4 x) \log (x)) \log (3-\log (x))}{-6+2 \log (x)} \, dx=\int { \frac {{\left ({\left (4 \, x + 3\right )} \log \left (x\right ) - 12 \, x - 9\right )} \log \left (-\log \left (x\right ) + 3\right ) + 2 \, x + 3}{2 \, {\left (\log \left (x\right ) - 3\right )}} \,d x } \]

[In]

integrate((((3+4*x)*log(x)-12*x-9)*log(3-log(x))+2*x+3)/(2*log(x)-6),x, algorithm="maxima")

[Out]

-3/2*e^3*exp_integral_e(1, -log(x) + 3) - e^6*exp_integral_e(1, -2*log(x) + 6) + 1/2*(2*x^2 + 3*x)*log(-log(x)
+ 3) - 1/2*integrate((2*x + 3)/(log(x) - 3), x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.57 \[ \int \frac {3+2 x+(-9-12 x+(3+4 x) \log (x)) \log (3-\log (x))}{-6+2 \log (x)} \, dx=x^{2} \log \left (-\log \left (x\right ) + 3\right ) + \frac {3}{2} \, x \log \left (-\log \left (x\right ) + 3\right ) \]

[In]

integrate((((3+4*x)*log(x)-12*x-9)*log(3-log(x))+2*x+3)/(2*log(x)-6),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.75 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {3+2 x+(-9-12 x+(3+4 x) \log (x)) \log (3-\log (x))}{-6+2 \log (x)} \, dx=\frac {x\,\ln \left (3-\ln \left (x\right )\right )\,\left (2\,x+3\right )}{2} \]

[In]

int((2*x - log(3 - log(x))*(12*x - log(x)*(4*x + 3) + 9) + 3)/(2*log(x) - 6),x)

[Out]

(x*log(3 - log(x))*(2*x + 3))/2