\(\int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} (15-3 x+(-1-3 x) \log (\frac {1}{3} (1+3 x)))}{1+3 x} \, dx\) [1053]

Optimal result

Integrand size = 80, antiderivative size = 25 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=-5+e^{\left (\frac {1}{3}+x\right )^{5-x}}+12 x \left (x+(3+x)^2\right ) \] Output:

12*x*((3+x)^2+x)+exp(exp((5-x)*ln(x+1/3)))-5
 

Mathematica [A] (verified)

Time = 1.98 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=e^{\left (\frac {1}{3}+x\right )^{5-x}}+12 x \left (9+7 x+x^2\right ) \] Input:

Integrate[(108 + 492*x + 540*x^2 + 108*x^3 + 3^(-5 + x)*E^(3^(-5 + x)*(1 + 
 3*x)^(5 - x))*(1 + 3*x)^(5 - x)*(15 - 3*x + (-1 - 3*x)*Log[(1 + 3*x)/3])) 
/(1 + 3*x),x]
 

Output:

E^(1/3 + x)^(5 - x) + 12*x*(9 + 7*x + x^2)
 

Rubi [F]

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.

\(\displaystyle \int \frac {108 x^3+540 x^2+492 x+3^{x-5} e^{3^{x-5} (3 x+1)^{5-x}} (3 x+1)^{5-x} \left (-3 x+(-3 x-1) \log \left (\frac {1}{3} (3 x+1)\right )+15\right )+108}{3 x+1} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (12 \left (3 x^2+14 x+9\right )-3^{x-5} e^{\left (x+\frac {1}{3}\right )^{5-x}} (3 x+1)^{4-x} \left (3 x+3 x \log \left (x+\frac {1}{3}\right )+\log \left (x+\frac {1}{3}\right )-15\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 5 \int 3^{x-4} e^{\left (x+\frac {1}{3}\right )^{5-x}} (3 x+1)^{4-x}dx-\int 3^{x-4} e^{\left (x+\frac {1}{3}\right )^{5-x}} x (3 x+1)^{4-x}dx+3 \int \frac {\int e^{\left (x+\frac {1}{3}\right )^{5-x}} x \left (x+\frac {1}{3}\right )^{4-x}dx}{3 x+1}dx+\int \frac {\int 3^{x-5} e^{\left (x+\frac {1}{3}\right )^{5-x}} (3 x+1)^{4-x}dx}{x+\frac {1}{3}}dx-\log \left (x+\frac {1}{3}\right ) \int 3^{x-5} e^{\left (x+\frac {1}{3}\right )^{5-x}} (3 x+1)^{4-x}dx-\log \left (x+\frac {1}{3}\right ) \int 3^{x-4} e^{\left (x+\frac {1}{3}\right )^{5-x}} x (3 x+1)^{4-x}dx+12 x^3+84 x^2+108 x\)

Input:

Int[(108 + 492*x + 540*x^2 + 108*x^3 + 3^(-5 + x)*E^(3^(-5 + x)*(1 + 3*x)^ 
(5 - x))*(1 + 3*x)^(5 - x)*(15 - 3*x + (-1 - 3*x)*Log[(1 + 3*x)/3]))/(1 + 
3*x),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 2.85 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00

method	result	size
risch	\(108 x +{\mathrm e}^{\left (x +\frac {1}{3}\right )^{5-x}}+84 x^{2}+12 x^{3}\)	\(25\)
default	\(108 x +{\mathrm e}^{{\mathrm e}^{\left (5-x \right ) \ln \left (x +\frac {1}{3}\right )}}+84 x^{2}+12 x^{3}\)	\(27\)
parts	\(108 x +{\mathrm e}^{{\mathrm e}^{\left (5-x \right ) \ln \left (x +\frac {1}{3}\right )}}+84 x^{2}+12 x^{3}\)	\(27\)
parallelrisch	\(12 x^{3}+84 x^{2}+108 x +{\mathrm e}^{{\mathrm e}^{\left (5-x \right ) \ln \left (x +\frac {1}{3}\right )}}-18\)	\(28\)

Input:

int((((-3*x-1)*ln(x+1/3)+15-3*x)*exp((5-x)*ln(x+1/3))*exp(exp((5-x)*ln(x+1 
/3)))+108*x^3+540*x^2+492*x+108)/(1+3*x),x,method=_RETURNVERBOSE)
 

Output:

108*x+exp((x+1/3)^(5-x))+84*x^2+12*x^3
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=12 \, x^{3} + 84 \, x^{2} + 108 \, x + e^{\left ({\left (x + \frac {1}{3}\right )}^{-x + 5}\right )} \] Input:

integrate((((-3*x-1)*log(x+1/3)+15-3*x)*exp((5-x)*log(x+1/3))*exp(exp((5-x 
)*log(x+1/3)))+108*x^3+540*x^2+492*x+108)/(1+3*x),x, algorithm="fricas")
 

Output:

12*x^3 + 84*x^2 + 108*x + e^((x + 1/3)^(-x + 5))
 

Sympy [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=12 x^{3} + 84 x^{2} + 108 x + e^{e^{\left (5 - x\right ) \log {\left (x + \frac {1}{3} \right )}}} \] Input:

integrate((((-3*x-1)*ln(x+1/3)+15-3*x)*exp((5-x)*ln(x+1/3))*exp(exp((5-x)* 
ln(x+1/3)))+108*x**3+540*x**2+492*x+108)/(1+3*x),x)
 

Output:

12*x**3 + 84*x**2 + 108*x + exp(exp((5 - x)*log(x + 1/3)))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (22) = 44\).

Time = 0.33 (sec) , antiderivative size = 130, normalized size of antiderivative = 5.20 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=12 \, x^{3} + 84 \, x^{2} + 108 \, x + e^{\left (x^{5} e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )} + \frac {5}{3} \, x^{4} e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )} + \frac {10}{9} \, x^{3} e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )} + \frac {10}{27} \, x^{2} e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )} + \frac {5}{81} \, x e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )} + \frac {1}{243} \, e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )}\right )} \] Input:

integrate((((-3*x-1)*log(x+1/3)+15-3*x)*exp((5-x)*log(x+1/3))*exp(exp((5-x 
)*log(x+1/3)))+108*x^3+540*x^2+492*x+108)/(1+3*x),x, algorithm="maxima")
 

Output:

12*x^3 + 84*x^2 + 108*x + e^(x^5*e^(x*log(3) - x*log(3*x + 1)) + 5/3*x^4*e 
^(x*log(3) - x*log(3*x + 1)) + 10/9*x^3*e^(x*log(3) - x*log(3*x + 1)) + 10 
/27*x^2*e^(x*log(3) - x*log(3*x + 1)) + 5/81*x*e^(x*log(3) - x*log(3*x + 1 
)) + 1/243*e^(x*log(3) - x*log(3*x + 1)))
 

Giac [F]

\[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=\int { \frac {108 \, x^{3} - {\left ({\left (3 \, x + 1\right )} \log \left (x + \frac {1}{3}\right ) + 3 \, x - 15\right )} {\left (x + \frac {1}{3}\right )}^{-x + 5} e^{\left ({\left (x + \frac {1}{3}\right )}^{-x + 5}\right )} + 540 \, x^{2} + 492 \, x + 108}{3 \, x + 1} \,d x } \] Input:

integrate((((-3*x-1)*log(x+1/3)+15-3*x)*exp((5-x)*log(x+1/3))*exp(exp((5-x 
)*log(x+1/3)))+108*x^3+540*x^2+492*x+108)/(1+3*x),x, algorithm="giac")
 

Output:

sage0*x
 

Mupad [B] (verification not implemented)

Time = 3.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 3.28 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=108\,x+{\mathrm {e}}^{\frac {1}{243\,{\left (x+\frac {1}{3}\right )}^x}+\frac {10\,x^2}{27\,{\left (x+\frac {1}{3}\right )}^x}+\frac {10\,x^3}{9\,{\left (x+\frac {1}{3}\right )}^x}+\frac {5\,x^4}{3\,{\left (x+\frac {1}{3}\right )}^x}+\frac {x^5}{{\left (x+\frac {1}{3}\right )}^x}+\frac {5\,x}{81\,{\left (x+\frac {1}{3}\right )}^x}}+84\,x^2+12\,x^3 \] Input:

int((492*x + 540*x^2 + 108*x^3 - exp(exp(-log(x + 1/3)*(x - 5)))*exp(-log( 
x + 1/3)*(x - 5))*(3*x + log(x + 1/3)*(3*x + 1) - 15) + 108)/(3*x + 1),x)
 

Output:

108*x + exp(1/(243*(x + 1/3)^x) + (10*x^2)/(27*(x + 1/3)^x) + (10*x^3)/(9* 
(x + 1/3)^x) + (5*x^4)/(3*(x + 1/3)^x) + x^5/(x + 1/3)^x + (5*x)/(81*(x + 
1/3)^x)) + 84*x^2 + 12*x^3
 

Reduce [F]

\[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx =\text {Too large to display} \] Input:

int((((-3*x-1)*log(x+1/3)+15-3*x)*exp((5-x)*log(x+1/3))*exp(exp((5-x)*log( 
x+1/3)))+108*x^3+540*x^2+492*x+108)/(1+3*x),x)
 

Output:

( - 243*int((e**((243*3**x*x**5 + 405*3**x*x**4 + 270*3**x*x**3 + 90*3**x* 
x**2 + 15*3**x*x + 3**x)/(243*(3*x + 1)**x))*3**x*x**5)/(3*x + 1)**x,x) + 
891*int((e**((243*3**x*x**5 + 405*3**x*x**4 + 270*3**x*x**3 + 90*3**x*x**2 
 + 15*3**x*x + 3**x)/(243*(3*x + 1)**x))*3**x*x**4)/(3*x + 1)**x,x) + 1458 
*int((e**((243*3**x*x**5 + 405*3**x*x**4 + 270*3**x*x**3 + 90*3**x*x**2 + 
15*3**x*x + 3**x)/(243*(3*x + 1)**x))*3**x*x**3)/(3*x + 1)**x,x) + 774*int 
((e**((243*3**x*x**5 + 405*3**x*x**4 + 270*3**x*x**3 + 90*3**x*x**2 + 15*3 
**x*x + 3**x)/(243*(3*x + 1)**x))*3**x*x**2)/(3*x + 1)**x,x) - 243*int((e* 
*((243*3**x*x**5 + 405*3**x*x**4 + 270*3**x*x**3 + 90*3**x*x**2 + 15*3**x* 
x + 3**x)/(243*(3*x + 1)**x))*3**x*log((3*x + 1)/3)*x**5)/(3*x + 1)**x,x) 
- 405*int((e**((243*3**x*x**5 + 405*3**x*x**4 + 270*3**x*x**3 + 90*3**x*x* 
*2 + 15*3**x*x + 3**x)/(243*(3*x + 1)**x))*3**x*log((3*x + 1)/3)*x**4)/(3* 
x + 1)**x,x) - 270*int((e**((243*3**x*x**5 + 405*3**x*x**4 + 270*3**x*x**3 
 + 90*3**x*x**2 + 15*3**x*x + 3**x)/(243*(3*x + 1)**x))*3**x*log((3*x + 1) 
/3)*x**3)/(3*x + 1)**x,x) - 90*int((e**((243*3**x*x**5 + 405*3**x*x**4 + 2 
70*3**x*x**3 + 90*3**x*x**2 + 15*3**x*x + 3**x)/(243*(3*x + 1)**x))*3**x*l 
og((3*x + 1)/3)*x**2)/(3*x + 1)**x,x) - 15*int((e**((243*3**x*x**5 + 405*3 
**x*x**4 + 270*3**x*x**3 + 90*3**x*x**2 + 15*3**x*x + 3**x)/(243*(3*x + 1) 
**x))*3**x*log((3*x + 1)/3)*x)/(3*x + 1)**x,x) - int((e**((243*3**x*x**5 + 
 405*3**x*x**4 + 270*3**x*x**3 + 90*3**x*x**2 + 15*3**x*x + 3**x)/(243*...