\(\int \frac {18+e^x (-8-4 x)-6 x-6 \log (\frac {3}{2})+e^x (36-4 x^2-(12+4 x) \log (\frac {3}{2})) \log (3-x-\log (\frac {3}{2}))}{9-3 x-3 \log (\frac {3}{2})} \, dx\) [898]

Optimal result

Integrand size = 68, antiderivative size = 26 \[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=2 x+\frac {4}{3} e^x (2+x) \log \left (3-x-\log \left (\frac {3}{2}\right )\right ) \] Output:

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

Mathematica [A] (verified)

Time = 4.73 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=\frac {2}{3} \left (3 x+2 e^x (2+x) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )\right ) \] Input:

Integrate[(18 + E^x*(-8 - 4*x) - 6*x - 6*Log[3/2] + E^x*(36 - 4*x^2 - (12 
+ 4*x)*Log[3/2])*Log[3 - x - Log[3/2]])/(9 - 3*x - 3*Log[3/2]),x]
 

Output:

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

Rubi [A] (verified)

Time = 1.21 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {7292, 7293, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {e^x \left (-4 x^2-(4 x+12) \log \left (\frac {3}{2}\right )+36\right ) \log \left (-x+3-\log \left (\frac {3}{2}\right )\right )+e^x (-4 x-8)-6 x+18-6 \log \left (\frac {3}{2}\right )}{-3 x+9-3 \log \left (\frac {3}{2}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^x \left (-4 x^2-(4 x+12) \log \left (\frac {3}{2}\right )+36\right ) \log \left (-x+3-\log \left (\frac {3}{2}\right )\right )+e^x (-4 x-8)-6 x+18 \left (1-\frac {1}{3} \log \left (\frac {3}{2}\right )\right )}{-3 x+9-3 \log \left (\frac {3}{2}\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {4 e^x \left (x^2 \left (-\log \left (-x+3-\log \left (\frac {3}{2}\right )\right )\right )-x-x \log \left (\frac {3}{2}\right ) \log \left (-x+3-\log \left (\frac {3}{2}\right )\right )+9 \left (1-\frac {1}{3} \log \left (\frac {3}{2}\right )\right ) \log \left (-x+3-\log \left (\frac {3}{2}\right )\right )-2\right )}{3 \left (-x+3-\log \left (\frac {3}{2}\right )\right )}+2\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 x-\frac {4}{3} e^x \log \left (-x+3-\log \left (\frac {3}{2}\right )\right )+\frac {4}{3} e^x (x+3) \log \left (-x+3-\log \left (\frac {3}{2}\right )\right )\)

Input:

Int[(18 + E^x*(-8 - 4*x) - 6*x - 6*Log[3/2] + E^x*(36 - 4*x^2 - (12 + 4*x) 
*Log[3/2])*Log[3 - x - Log[3/2]])/(9 - 3*x - 3*Log[3/2]),x]
 

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 8.92 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92

method	result	size
risch	\(\frac {4 \left (2+x \right ) {\mathrm e}^{x} \ln \left (\ln \left (2\right )-\ln \left (3\right )+3-x \right )}{3}+2 x\)	\(24\)
default	\(2 x +\frac {8 \,{\mathrm e}^{x} \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}+\frac {4 \,{\mathrm e}^{x} x \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}\)	\(30\)
norman	\(2 x +\frac {8 \,{\mathrm e}^{x} \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}+\frac {4 \,{\mathrm e}^{x} x \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}\)	\(30\)
parts	\(2 x +\frac {8 \,{\mathrm e}^{x} \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}+\frac {4 \,{\mathrm e}^{x} x \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}\)	\(30\)
parallelrisch	\(\frac {4 \,{\mathrm e}^{x} x \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}+12+\frac {8 \,{\mathrm e}^{x} \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}+4 \ln \left (\frac {2}{3}\right )+2 x\)	\(35\)

Input:

int((((4*x+12)*ln(2/3)-4*x^2+36)*exp(x)*ln(ln(2/3)+3-x)+(-4*x-8)*exp(x)+6* 
ln(2/3)-6*x+18)/(3*ln(2/3)-3*x+9),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=\frac {4}{3} \, {\left (x + 2\right )} e^{x} \log \left (-x + \log \left (\frac {2}{3}\right ) + 3\right ) + 2 \, x \] Input:

integrate((((4*x+12)*log(2/3)-4*x^2+36)*exp(x)*log(log(2/3)+3-x)+(-4*x-8)* 
exp(x)+6*log(2/3)-6*x+18)/(3*log(2/3)-3*x+9),x, algorithm="fricas")
 

Output:

4/3*(x + 2)*e^x*log(-x + log(2/3) + 3) + 2*x
 

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=2 x + \frac {\left (4 x \log {\left (- x + \log {\left (\frac {2}{3} \right )} + 3 \right )} + 8 \log {\left (- x + \log {\left (\frac {2}{3} \right )} + 3 \right )}\right ) e^{x}}{3} \] Input:

integrate((((4*x+12)*ln(2/3)-4*x**2+36)*exp(x)*ln(ln(2/3)+3-x)+(-4*x-8)*ex 
p(x)+6*ln(2/3)-6*x+18)/(3*ln(2/3)-3*x+9),x)
 

Output:

2*x + (4*x*log(-x + log(2/3) + 3) + 8*log(-x + log(2/3) + 3))*exp(x)/3
 

Maxima [F]

\[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=\int { \frac {2 \, {\left (2 \, {\left (x^{2} - {\left (x + 3\right )} \log \left (\frac {2}{3}\right ) - 9\right )} e^{x} \log \left (-x + \log \left (\frac {2}{3}\right ) + 3\right ) + 2 \, {\left (x + 2\right )} e^{x} + 3 \, x - 3 \, \log \left (\frac {2}{3}\right ) - 9\right )}}{3 \, {\left (x - \log \left (\frac {2}{3}\right ) - 3\right )}} \,d x } \] Input:

integrate((((4*x+12)*log(2/3)-4*x^2+36)*exp(x)*log(log(2/3)+3-x)+(-4*x-8)* 
exp(x)+6*log(2/3)-6*x+18)/(3*log(2/3)-3*x+9),x, algorithm="maxima")
 

Output:

4/3*(x + 2)*e^x*log(-x - log(3) + log(2) + 3) - 16/9*e^3*exp_integral_e(1, 
 -x + log(2/3) + 3) + 2*(log(2/3) + 3)*log(x - log(2/3) - 3) - 2*log(2/3)* 
log(x - log(2/3) - 3) + 2*x - 8/3*integrate(e^x/(x + log(3) - log(2) - 3), 
 x) - 6*log(x - log(2/3) - 3)
 

Giac [C] (verification not implemented)

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

Time = 0.12 (sec) , antiderivative size = 128, normalized size of antiderivative = 4.92 \[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=-\frac {4}{3} \, {\rm Ei}\left (x + \log \left (3\right ) - \log \left (2\right ) - 3\right ) e^{\left (-\log \left (3\right ) + \log \left (2\right ) + 3\right )} \log \left (3\right ) + \frac {4}{3} \, {\rm Ei}\left (x + \log \left (3\right ) - \log \left (2\right ) - 3\right ) e^{\left (-\log \left (3\right ) + \log \left (2\right ) + 3\right )} \log \left (2\right ) - \frac {4}{3} \, {\rm Ei}\left (x - \log \left (\frac {2}{3}\right ) - 3\right ) e^{\left (\log \left (\frac {2}{3}\right ) + 3\right )} \log \left (\frac {2}{3}\right ) + \frac {4}{3} \, x e^{x} \log \left (-x + \log \left (\frac {2}{3}\right ) + 3\right ) + \frac {20}{3} \, {\rm Ei}\left (x + \log \left (3\right ) - \log \left (2\right ) - 3\right ) e^{\left (-\log \left (3\right ) + \log \left (2\right ) + 3\right )} - \frac {20}{3} \, {\rm Ei}\left (x - \log \left (\frac {2}{3}\right ) - 3\right ) e^{\left (\log \left (\frac {2}{3}\right ) + 3\right )} + \frac {8}{3} \, e^{x} \log \left (-x + \log \left (\frac {2}{3}\right ) + 3\right ) + 2 \, x \] Input:

integrate((((4*x+12)*log(2/3)-4*x^2+36)*exp(x)*log(log(2/3)+3-x)+(-4*x-8)* 
exp(x)+6*log(2/3)-6*x+18)/(3*log(2/3)-3*x+9),x, algorithm="giac")
 

Output:

-4/3*Ei(x + log(3) - log(2) - 3)*e^(-log(3) + log(2) + 3)*log(3) + 4/3*Ei( 
x + log(3) - log(2) - 3)*e^(-log(3) + log(2) + 3)*log(2) - 4/3*Ei(x - log( 
2/3) - 3)*e^(log(2/3) + 3)*log(2/3) + 4/3*x*e^x*log(-x + log(2/3) + 3) + 2 
0/3*Ei(x + log(3) - log(2) - 3)*e^(-log(3) + log(2) + 3) - 20/3*Ei(x - log 
(2/3) - 3)*e^(log(2/3) + 3) + 8/3*e^x*log(-x + log(2/3) + 3) + 2*x
 

Mupad [F(-1)]

Timed out. \[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=\int \frac {6\,\ln \left (\frac {2}{3}\right )-6\,x-{\mathrm {e}}^x\,\left (4\,x+8\right )+\ln \left (\ln \left (\frac {2}{3}\right )-x+3\right )\,{\mathrm {e}}^x\,\left (\ln \left (\frac {2}{3}\right )\,\left (4\,x+12\right )-4\,x^2+36\right )+18}{3\,\ln \left (\frac {2}{3}\right )-3\,x+9} \,d x \] Input:

int((6*log(2/3) - 6*x - exp(x)*(4*x + 8) + log(log(2/3) - x + 3)*exp(x)*(l 
og(2/3)*(4*x + 12) - 4*x^2 + 36) + 18)/(3*log(2/3) - 3*x + 9),x)
 

Output:

int((6*log(2/3) - 6*x - exp(x)*(4*x + 8) + log(log(2/3) - x + 3)*exp(x)*(l 
og(2/3)*(4*x + 12) - 4*x^2 + 36) + 18)/(3*log(2/3) - 3*x + 9), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=\frac {4 e^{x} \mathrm {log}\left (\mathrm {log}\left (\frac {2}{3}\right )-x +3\right ) x}{3}+\frac {8 e^{x} \mathrm {log}\left (\mathrm {log}\left (\frac {2}{3}\right )-x +3\right )}{3}+2 x \] Input:

int((((4*x+12)*log(2/3)-4*x^2+36)*exp(x)*log(log(2/3)+3-x)+(-4*x-8)*exp(x) 
+6*log(2/3)-6*x+18)/(3*log(2/3)-3*x+9),x)
 

Output:

(2*(2*e**x*log(log(2/3) - x + 3)*x + 4*e**x*log(log(2/3) - x + 3) + 3*x))/ 
3