Integrand size = 80, antiderivative size = 25 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=-5+e^{\left (\frac {1}{3}+x\right )^{5-x}}+12 x \left (x+(3+x)^2\right ) \]
Time = 3.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=e^{\left (\frac {1}{3}+x\right )^{5-x}}+12 x \left (9+7 x+x^2\right ) \]
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]
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 |
\(\displaystyle 5 \int 3^{x-4} e^{\left (x+\frac {1}{3}\right )^{5-x}} (3 x+1)^{4-x}dx-\int 3^{x-4} e^{\left (x+\frac {1}{3}\right )^{5-x}} x (3 x+1)^{4-x}dx+3 \int \frac {\int e^{\left (x+\frac {1}{3}\right )^{5-x}} x \left (x+\frac {1}{3}\right )^{4-x}dx}{3 x+1}dx+\int \frac {\int 3^{x-5} e^{\left (x+\frac {1}{3}\right )^{5-x}} (3 x+1)^{4-x}dx}{x+\frac {1}{3}}dx-\log \left (x+\frac {1}{3}\right ) \int 3^{x-5} e^{\left (x+\frac {1}{3}\right )^{5-x}} (3 x+1)^{4-x}dx-\log \left (x+\frac {1}{3}\right ) \int 3^{x-4} e^{\left (x+\frac {1}{3}\right )^{5-x}} x (3 x+1)^{4-x}dx+12 x^3+84 x^2+108 x\) |
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]
3.11.53.3.1 Defintions of rubi rules used
Time = 0.54 (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 | \(12 x^{3}+84 x^{2}+108 x +{\mathrm e}^{{\mathrm e}^{\left (5-x \right ) \ln \left (x +\frac {1}{3}\right )}}-18\) | \(28\) |
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)
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=12 \, x^{3} + 84 \, x^{2} + 108 \, x + e^{\left ({\left (x + \frac {1}{3}\right )}^{-x + 5}\right )} \]
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=\
Time = 0.49 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=12 x^{3} + 84 x^{2} + 108 x + e^{e^{\left (5 - x\right ) \log {\left (x + \frac {1}{3} \right )}}} \]
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)
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (22) = 44\).
Time = 0.51 (sec) , antiderivative size = 130, normalized size of antiderivative = 5.20 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=12 \, x^{3} + 84 \, x^{2} + 108 \, x + e^{\left (x^{5} e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )} + \frac {5}{3} \, x^{4} e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )} + \frac {10}{9} \, x^{3} e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )} + \frac {10}{27} \, x^{2} e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )} + \frac {5}{81} \, x e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )} + \frac {1}{243} \, e^{\left (x \log \left (3\right ) - x \log \left (3 \, x + 1\right )\right )}\right )} \]
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=\
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)))
\[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=\int { \frac {108 \, x^{3} - {\left ({\left (3 \, x + 1\right )} \log \left (x + \frac {1}{3}\right ) + 3 \, x - 15\right )} {\left (x + \frac {1}{3}\right )}^{-x + 5} e^{\left ({\left (x + \frac {1}{3}\right )}^{-x + 5}\right )} + 540 \, x^{2} + 492 \, x + 108}{3 \, x + 1} \,d x } \]
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=\
Time = 12.75 (sec) , antiderivative size = 82, normalized size of antiderivative = 3.28 \[ \int \frac {108+492 x+540 x^2+108 x^3+3^{-5+x} e^{3^{-5+x} (1+3 x)^{5-x}} (1+3 x)^{5-x} \left (15-3 x+(-1-3 x) \log \left (\frac {1}{3} (1+3 x)\right )\right )}{1+3 x} \, dx=108\,x+{\mathrm {e}}^{\frac {1}{243\,{\left (x+\frac {1}{3}\right )}^x}+\frac {10\,x^2}{27\,{\left (x+\frac {1}{3}\right )}^x}+\frac {10\,x^3}{9\,{\left (x+\frac {1}{3}\right )}^x}+\frac {5\,x^4}{3\,{\left (x+\frac {1}{3}\right )}^x}+\frac {x^5}{{\left (x+\frac {1}{3}\right )}^x}+\frac {5\,x}{81\,{\left (x+\frac {1}{3}\right )}^x}}+84\,x^2+12\,x^3 \]
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)