Integrand size = 31, antiderivative size = 19 \[ \int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx=-3+\log ^2\left (\frac {36}{25} e^{2 x} (4+x)^2\right ) \]
Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx=\log ^2\left (\frac {36}{25} e^{2 x} (4+x)^2\right ) \]
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.
\(\displaystyle \int \frac {(4 x+20) \log \left (\frac {1}{25} e^{2 x} \left (36 x^2+288 x+576\right )\right )}{x+4} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 \log \left (\frac {36}{25} e^{2 x} (x+4)^2\right )}{x+4}+4 \log \left (\frac {36}{25} e^{2 x} (x+4)^2\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \int \frac {\log \left (\frac {36}{25} e^{2 x} (x+4)^2\right )}{x+4}dx-4 x^2-8 x+4 x \log \left (\frac {36}{25} e^{2 x} (x+4)^2\right )+32 \log (x+4)\) |
3.30.48.3.1 Defintions of rubi rules used
Time = 1.87 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95
method | result | size |
parallelrisch | \({\ln \left (\frac {36 \left (x^{2}+8 x +16\right ) {\mathrm e}^{2 x}}{25}\right )}^{2}\) | \(18\) |
default | \({\ln \left (\frac {\left (36 x^{2}+288 x +576\right ) {\mathrm e}^{2 x}}{25}\right )}^{2}\) | \(20\) |
norman | \({\ln \left (\frac {\left (36 x^{2}+288 x +576\right ) {\mathrm e}^{2 x}}{25}\right )}^{2}\) | \(20\) |
parts | \(4 \ln \left (\frac {\left (36 x^{2}+288 x +576\right ) {\mathrm e}^{2 x}}{25}\right ) \ln \left (4+x \right )+4 \ln \left (\frac {\left (36 x^{2}+288 x +576\right ) {\mathrm e}^{2 x}}{25}\right ) x -4 x^{2}+32 \ln \left (4+x \right )-8 \left (4+x \right ) \ln \left (4+x \right )+32-4 \ln \left (4+x \right )^{2}\) | \(74\) |
risch | \(4 i \ln \left (4+x \right ) \pi \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )+4 i \pi x \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )-2 i \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2}+2 i \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{2}+2 i \ln \left (4+x \right ) \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{2}+2 i \ln \left (4+x \right ) \pi \,\operatorname {csgn}\left (i \left (4+x \right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{2}+2 i \pi x \,\operatorname {csgn}\left (i \left (4+x \right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{2}+4 \ln \left (4+x \right )^{2}+8 \ln \left (4+x \right ) \ln \left (2\right )+8 \ln \left (4+x \right ) \ln \left (3\right )-8 \ln \left (4+x \right ) \ln \left (5\right )+\left (8 x +8 \ln \left (4+x \right )\right ) \ln \left ({\mathrm e}^{x}\right )-8 x \ln \left (5\right )+8 x \ln \left (2\right )+8 x \ln \left (3\right )-2 i \ln \left (4+x \right ) \pi \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{3}-2 i \pi x \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-2 i \pi x \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{3}-2 i \ln \left (4+x \right ) \pi \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-2 i \ln \left (4+x \right ) \pi \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )^{3}-2 i \pi x \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )^{3}-4 x^{2}-2 i \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i \left (4+x \right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )-2 i \ln \left (4+x \right ) \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i \left (4+x \right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )-2 i \ln \left (4+x \right ) \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2}-2 i \ln \left (4+x \right ) \pi \operatorname {csgn}\left (i \left (4+x \right )\right )^{2} \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )+4 i \ln \left (4+x \right ) \pi \,\operatorname {csgn}\left (i \left (4+x \right )\right ) \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )^{2}-2 i \pi x \operatorname {csgn}\left (i \left (4+x \right )\right )^{2} \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )+4 i \pi x \,\operatorname {csgn}\left (i \left (4+x \right )\right ) \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )^{2}\) | \(560\) |
Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx=\log \left (\frac {36}{25} \, {\left (x^{2} + 8 \, x + 16\right )} e^{\left (2 \, x\right )}\right )^{2} \]
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx=\log {\left (\left (\frac {36 x^{2}}{25} + \frac {288 x}{25} + \frac {576}{25}\right ) e^{2 x} \right )}^{2} \]
Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (16) = 32\).
Time = 0.21 (sec) , antiderivative size = 184, normalized size of antiderivative = 9.68 \[ \int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx=4 \, {\left (x + \log \left (x + 4\right )\right )} \log \left (\frac {36}{25} \, {\left (x^{2} + 8 \, x + 16\right )} e^{\left (2 \, x\right )}\right ) - 8 \, {\left (\frac {x^{2} + 4 \, x - 16}{x + 4} - 8 \, \log \left (x + 4\right )\right )} \log \left (x + 4\right ) - 72 \, {\left (\frac {4}{x + 4} + \log \left (x + 4\right )\right )} \log \left (x + 4\right ) - \frac {4 \, {\left (x^{3} - 12 \, x^{2} - 64 \, x + 128\right )}}{x + 4} - \frac {8 \, {\left (4 \, {\left (x + 4\right )} \log \left (x + 4\right )^{2} - x^{2} + 4 \, {\left (x + 4\right )} \log \left (x + 4\right ) - 4 \, x - 16\right )}}{x + 4} + \frac {36 \, {\left ({\left (x + 4\right )} \log \left (x + 4\right )^{2} - 8\right )}}{x + 4} - \frac {72 \, {\left (x^{2} + 4 \, x - 16\right )}}{x + 4} + \frac {160 \, \log \left (x + 4\right )}{x + 4} - \frac {480}{x + 4} + 32 \, \log \left (x + 4\right ) \]
4*(x + log(x + 4))*log(36/25*(x^2 + 8*x + 16)*e^(2*x)) - 8*((x^2 + 4*x - 1 6)/(x + 4) - 8*log(x + 4))*log(x + 4) - 72*(4/(x + 4) + log(x + 4))*log(x + 4) - 4*(x^3 - 12*x^2 - 64*x + 128)/(x + 4) - 8*(4*(x + 4)*log(x + 4)^2 - x^2 + 4*(x + 4)*log(x + 4) - 4*x - 16)/(x + 4) + 36*((x + 4)*log(x + 4)^2 - 8)/(x + 4) - 72*(x^2 + 4*x - 16)/(x + 4) + 160*log(x + 4)/(x + 4) - 480 /(x + 4) + 32*log(x + 4)
Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (16) = 32\).
Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx=4 \, x^{2} + 4 \, x \log \left (\frac {36}{25} \, x^{2} + \frac {288}{25} \, x + \frac {576}{25}\right ) + \log \left (\frac {36}{25} \, x^{2} + \frac {288}{25} \, x + \frac {576}{25}\right )^{2} \]
Time = 0.67 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx={\left (2\,x+\ln \left (\frac {36\,x^2}{25}+\frac {288\,x}{25}+\frac {576}{25}\right )\right )}^2 \]