Integrand size = 100, antiderivative size = 31 \[ \int \frac {e^{3+2 x} \left (-2 x+16 x^2\right )+e^{3+2 x} \left (2 x-6 x^2-8 x^3\right ) \log \left (x-4 x^2\right )+\left (-1+4 x+e^{3+2 x} \left (9-38 x+8 x^2\right )\right ) \log ^3\left (x-4 x^2\right )}{(-1+4 x) \log ^3\left (x-4 x^2\right )} \, dx=4+x-e^{3+2 x} \left (5-x+\frac {x^2}{\log ^2\left (x-4 x^2\right )}\right ) \]
Time = 0.47 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {e^{3+2 x} \left (-2 x+16 x^2\right )+e^{3+2 x} \left (2 x-6 x^2-8 x^3\right ) \log \left (x-4 x^2\right )+\left (-1+4 x+e^{3+2 x} \left (9-38 x+8 x^2\right )\right ) \log ^3\left (x-4 x^2\right )}{(-1+4 x) \log ^3\left (x-4 x^2\right )} \, dx=e^{3+2 x} (-5+x)+x-\frac {e^{3+2 x} x^2}{\log ^2\left (x-4 x^2\right )} \]
Integrate[(E^(3 + 2*x)*(-2*x + 16*x^2) + E^(3 + 2*x)*(2*x - 6*x^2 - 8*x^3) *Log[x - 4*x^2] + (-1 + 4*x + E^(3 + 2*x)*(9 - 38*x + 8*x^2))*Log[x - 4*x^ 2]^3)/((-1 + 4*x)*Log[x - 4*x^2]^3),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.
\(\displaystyle \int \frac {e^{2 x+3} \left (16 x^2-2 x\right )+\left (e^{2 x+3} \left (8 x^2-38 x+9\right )+4 x-1\right ) \log ^3\left (x-4 x^2\right )+e^{2 x+3} \left (-8 x^3-6 x^2+2 x\right ) \log \left (x-4 x^2\right )}{(4 x-1) \log ^3\left (x-4 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-e^{2 x+3} \left (16 x^2-2 x\right )-\left (\left (e^{2 x+3} \left (8 x^2-38 x+9\right )+4 x-1\right ) \log ^3\left (x-4 x^2\right )\right )-e^{2 x+3} \left (-8 x^3-6 x^2+2 x\right ) \log \left (x-4 x^2\right )}{(1-4 x) \log ^3((1-4 x) x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (1-\frac {e^{2 x+3} \left (-16 x^2-8 x^2 \log ^3\left (x-4 x^2\right )+38 x \log ^3\left (x-4 x^2\right )-9 \log ^3\left (x-4 x^2\right )+6 x^2 \log \left (x-4 x^2\right )-2 x \log \left (x-4 x^2\right )+8 x^3 \log \left (x-4 x^2\right )+2 x\right )}{(4 x-1) \log ^3((1-4 x) x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {e^{2 x+3} x^2}{\log ^2((1-4 x) x)}dx+\frac {1}{2} \int \frac {e^{2 x+3}}{\log ^3((1-4 x) x)}dx+4 \int \frac {e^{2 x+3} x}{\log ^3((1-4 x) x)}dx+\frac {1}{2} \int \frac {e^{2 x+3}}{(4 x-1) \log ^3((1-4 x) x)}dx-2 \int \frac {e^{2 x+3} x}{\log ^2((1-4 x) x)}dx+e^{2 x+3} x+x-5 e^{2 x+3}\) |
Int[(E^(3 + 2*x)*(-2*x + 16*x^2) + E^(3 + 2*x)*(2*x - 6*x^2 - 8*x^3)*Log[x - 4*x^2] + (-1 + 4*x + E^(3 + 2*x)*(9 - 38*x + 8*x^2))*Log[x - 4*x^2]^3)/ ((-1 + 4*x)*Log[x - 4*x^2]^3),x]
3.26.74.3.1 Defintions of rubi rules used
Leaf count of result is larger than twice the leaf count of optimal. \(84\) vs. \(2(30)=60\).
Time = 0.58 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.74
method | result | size |
parallelrisch | \(\frac {8 \ln \left (-4 x^{2}+x \right )^{2} x \,{\mathrm e}^{3+2 x}-8 x^{2} {\mathrm e}^{3+2 x}+8 \ln \left (-4 x^{2}+x \right )^{2} x -40 \,{\mathrm e}^{3+2 x} \ln \left (-4 x^{2}+x \right )^{2}+\ln \left (-4 x^{2}+x \right )^{2}}{8 \ln \left (-4 x^{2}+x \right )^{2}}\) | \(85\) |
risch | \({\mathrm e}^{3+2 x} x +x -5 \,{\mathrm e}^{3+2 x}+\frac {4 x^{2} {\mathrm e}^{3+2 x}}{\left (\pi \,\operatorname {csgn}\left (i \left (x -\frac {1}{4}\right )\right ) \operatorname {csgn}\left (i x \left (x -\frac {1}{4}\right )\right )^{2}-\pi \,\operatorname {csgn}\left (i \left (x -\frac {1}{4}\right )\right ) \operatorname {csgn}\left (i x \left (x -\frac {1}{4}\right )\right ) \operatorname {csgn}\left (i x \right )+\pi \operatorname {csgn}\left (i x \left (x -\frac {1}{4}\right )\right )^{3}-2 \pi \operatorname {csgn}\left (i x \left (x -\frac {1}{4}\right )\right )^{2}+\pi \operatorname {csgn}\left (i x \left (x -\frac {1}{4}\right )\right )^{2} \operatorname {csgn}\left (i x \right )+2 \pi -2 i \ln \left (x \right )-2 i \ln \left (x -\frac {1}{4}\right )\right )^{2}}\) | \(132\) |
int((((8*x^2-38*x+9)*exp(3+2*x)+4*x-1)*ln(-4*x^2+x)^3+(-8*x^3-6*x^2+2*x)*e xp(3+2*x)*ln(-4*x^2+x)+(16*x^2-2*x)*exp(3+2*x))/(-1+4*x)/ln(-4*x^2+x)^3,x, method=_RETURNVERBOSE)
1/8*(8*ln(-4*x^2+x)^2*x*exp(3+2*x)-8*x^2*exp(3+2*x)+8*ln(-4*x^2+x)^2*x-40* exp(3+2*x)*ln(-4*x^2+x)^2+ln(-4*x^2+x)^2)/ln(-4*x^2+x)^2
Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \frac {e^{3+2 x} \left (-2 x+16 x^2\right )+e^{3+2 x} \left (2 x-6 x^2-8 x^3\right ) \log \left (x-4 x^2\right )+\left (-1+4 x+e^{3+2 x} \left (9-38 x+8 x^2\right )\right ) \log ^3\left (x-4 x^2\right )}{(-1+4 x) \log ^3\left (x-4 x^2\right )} \, dx=-\frac {x^{2} e^{\left (2 \, x + 3\right )} - {\left ({\left (x - 5\right )} e^{\left (2 \, x + 3\right )} + x\right )} \log \left (-4 \, x^{2} + x\right )^{2}}{\log \left (-4 \, x^{2} + x\right )^{2}} \]
integrate((((8*x^2-38*x+9)*exp(3+2*x)+4*x-1)*log(-4*x^2+x)^3+(-8*x^3-6*x^2 +2*x)*exp(3+2*x)*log(-4*x^2+x)+(16*x^2-2*x)*exp(3+2*x))/(-1+4*x)/log(-4*x^ 2+x)^3,x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {e^{3+2 x} \left (-2 x+16 x^2\right )+e^{3+2 x} \left (2 x-6 x^2-8 x^3\right ) \log \left (x-4 x^2\right )+\left (-1+4 x+e^{3+2 x} \left (9-38 x+8 x^2\right )\right ) \log ^3\left (x-4 x^2\right )}{(-1+4 x) \log ^3\left (x-4 x^2\right )} \, dx=x + \frac {\left (- x^{2} + x \log {\left (- 4 x^{2} + x \right )}^{2} - 5 \log {\left (- 4 x^{2} + x \right )}^{2}\right ) e^{2 x + 3}}{\log {\left (- 4 x^{2} + x \right )}^{2}} \]
integrate((((8*x**2-38*x+9)*exp(3+2*x)+4*x-1)*ln(-4*x**2+x)**3+(-8*x**3-6* x**2+2*x)*exp(3+2*x)*ln(-4*x**2+x)+(16*x**2-2*x)*exp(3+2*x))/(-1+4*x)/ln(- 4*x**2+x)**3,x)
x + (-x**2 + x*log(-4*x**2 + x)**2 - 5*log(-4*x**2 + x)**2)*exp(2*x + 3)/l og(-4*x**2 + x)**2
Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (28) = 56\).
Time = 0.28 (sec) , antiderivative size = 115, normalized size of antiderivative = 3.71 \[ \int \frac {e^{3+2 x} \left (-2 x+16 x^2\right )+e^{3+2 x} \left (2 x-6 x^2-8 x^3\right ) \log \left (x-4 x^2\right )+\left (-1+4 x+e^{3+2 x} \left (9-38 x+8 x^2\right )\right ) \log ^3\left (x-4 x^2\right )}{(-1+4 x) \log ^3\left (x-4 x^2\right )} \, dx=\frac {x \log \left (x\right )^{2} + {\left ({\left (x e^{3} - 5 \, e^{3}\right )} e^{\left (2 \, x\right )} + x\right )} \log \left (-4 \, x + 1\right )^{2} - {\left (x^{2} e^{3} - {\left (x e^{3} - 5 \, e^{3}\right )} \log \left (x\right )^{2}\right )} e^{\left (2 \, x\right )} + 2 \, {\left ({\left (x e^{3} - 5 \, e^{3}\right )} e^{\left (2 \, x\right )} \log \left (x\right ) + x \log \left (x\right )\right )} \log \left (-4 \, x + 1\right )}{\log \left (x\right )^{2} + 2 \, \log \left (x\right ) \log \left (-4 \, x + 1\right ) + \log \left (-4 \, x + 1\right )^{2}} \]
integrate((((8*x^2-38*x+9)*exp(3+2*x)+4*x-1)*log(-4*x^2+x)^3+(-8*x^3-6*x^2 +2*x)*exp(3+2*x)*log(-4*x^2+x)+(16*x^2-2*x)*exp(3+2*x))/(-1+4*x)/log(-4*x^ 2+x)^3,x, algorithm=\
(x*log(x)^2 + ((x*e^3 - 5*e^3)*e^(2*x) + x)*log(-4*x + 1)^2 - (x^2*e^3 - ( x*e^3 - 5*e^3)*log(x)^2)*e^(2*x) + 2*((x*e^3 - 5*e^3)*e^(2*x)*log(x) + x*l og(x))*log(-4*x + 1))/(log(x)^2 + 2*log(x)*log(-4*x + 1) + log(-4*x + 1)^2 )
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (28) = 56\).
Time = 0.47 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.29 \[ \int \frac {e^{3+2 x} \left (-2 x+16 x^2\right )+e^{3+2 x} \left (2 x-6 x^2-8 x^3\right ) \log \left (x-4 x^2\right )+\left (-1+4 x+e^{3+2 x} \left (9-38 x+8 x^2\right )\right ) \log ^3\left (x-4 x^2\right )}{(-1+4 x) \log ^3\left (x-4 x^2\right )} \, dx=\frac {x e^{\left (2 \, x + 3\right )} \log \left (-4 \, x^{2} + x\right )^{2} - x^{2} e^{\left (2 \, x + 3\right )} + x \log \left (-4 \, x^{2} + x\right )^{2} - 5 \, e^{\left (2 \, x + 3\right )} \log \left (-4 \, x^{2} + x\right )^{2}}{\log \left (-4 \, x^{2} + x\right )^{2}} \]
integrate((((8*x^2-38*x+9)*exp(3+2*x)+4*x-1)*log(-4*x^2+x)^3+(-8*x^3-6*x^2 +2*x)*exp(3+2*x)*log(-4*x^2+x)+(16*x^2-2*x)*exp(3+2*x))/(-1+4*x)/log(-4*x^ 2+x)^3,x, algorithm=\
(x*e^(2*x + 3)*log(-4*x^2 + x)^2 - x^2*e^(2*x + 3) + x*log(-4*x^2 + x)^2 - 5*e^(2*x + 3)*log(-4*x^2 + x)^2)/log(-4*x^2 + x)^2
Time = 8.21 (sec) , antiderivative size = 216, normalized size of antiderivative = 6.97 \[ \int \frac {e^{3+2 x} \left (-2 x+16 x^2\right )+e^{3+2 x} \left (2 x-6 x^2-8 x^3\right ) \log \left (x-4 x^2\right )+\left (-1+4 x+e^{3+2 x} \left (9-38 x+8 x^2\right )\right ) \log ^3\left (x-4 x^2\right )}{(-1+4 x) \log ^3\left (x-4 x^2\right )} \, dx=x+\frac {\frac {\left (x-4\,x^2\right )\,\left (x\,{\mathrm {e}}^{2\,x+3}+x^2\,{\mathrm {e}}^{2\,x+3}\right )}{8\,x-1}-\frac {{\mathrm {e}}^{2\,x+3}\,\ln \left (x-4\,x^2\right )\,\left (x-4\,x^2\right )\,\left (64\,x^5+136\,x^4+10\,x^3-15\,x^2+2\,x\right )}{{\left (8\,x-1\right )}^3}}{\ln \left (x-4\,x^2\right )}-\frac {x^2\,{\mathrm {e}}^{2\,x+3}+\frac {x\,{\mathrm {e}}^{2\,x+3}\,\ln \left (x-4\,x^2\right )\,\left (x-4\,x^2\right )\,\left (x+1\right )}{8\,x-1}}{{\ln \left (x-4\,x^2\right )}^2}-\frac {{\mathrm {e}}^{2\,x+3}\,\left (\frac {x^7}{2}+\frac {15\,x^6}{16}-\frac {3\,x^5}{16}-\frac {291\,x^4}{256}+\frac {2775\,x^3}{512}-\frac {493\,x^2}{256}+\frac {121\,x}{512}-\frac {5}{512}\right )}{x^3-\frac {3\,x^2}{8}+\frac {3\,x}{64}-\frac {1}{512}} \]
int(-(exp(2*x + 3)*(2*x - 16*x^2) - log(x - 4*x^2)^3*(4*x + exp(2*x + 3)*( 8*x^2 - 38*x + 9) - 1) + exp(2*x + 3)*log(x - 4*x^2)*(6*x^2 - 2*x + 8*x^3) )/(log(x - 4*x^2)^3*(4*x - 1)),x)
x + (((x - 4*x^2)*(x*exp(2*x + 3) + x^2*exp(2*x + 3)))/(8*x - 1) - (exp(2* x + 3)*log(x - 4*x^2)*(x - 4*x^2)*(2*x - 15*x^2 + 10*x^3 + 136*x^4 + 64*x^ 5))/(8*x - 1)^3)/log(x - 4*x^2) - (x^2*exp(2*x + 3) + (x*exp(2*x + 3)*log( x - 4*x^2)*(x - 4*x^2)*(x + 1))/(8*x - 1))/log(x - 4*x^2)^2 - (exp(2*x + 3 )*((121*x)/512 - (493*x^2)/256 + (2775*x^3)/512 - (291*x^4)/256 - (3*x^5)/ 16 + (15*x^6)/16 + x^7/2 - 5/512))/((3*x)/64 - (3*x^2)/8 + x^3 - 1/512)