Integrand size = 103, antiderivative size = 20 \[ \int \frac {90 x^2+90 x \log (3)+(x+\log (3))^{\frac {2 x^2}{5}} \left (2 x^2+\left (4 x^2+4 x \log (3)\right ) \log (x+\log (3))\right )+(x+\log (3))^{\frac {x^2}{5}} \left (-30 x-6 x^3-30 \log (3)+\left (-12 x^3-12 x^2 \log (3)\right ) \log (x+\log (3))\right )}{5 x+5 \log (3)} \, dx=\left (3 x-(x+\log (3))^{\frac {x^2}{5}}\right )^2 \]
\[ \int \frac {90 x^2+90 x \log (3)+(x+\log (3))^{\frac {2 x^2}{5}} \left (2 x^2+\left (4 x^2+4 x \log (3)\right ) \log (x+\log (3))\right )+(x+\log (3))^{\frac {x^2}{5}} \left (-30 x-6 x^3-30 \log (3)+\left (-12 x^3-12 x^2 \log (3)\right ) \log (x+\log (3))\right )}{5 x+5 \log (3)} \, dx=\int \frac {90 x^2+90 x \log (3)+(x+\log (3))^{\frac {2 x^2}{5}} \left (2 x^2+\left (4 x^2+4 x \log (3)\right ) \log (x+\log (3))\right )+(x+\log (3))^{\frac {x^2}{5}} \left (-30 x-6 x^3-30 \log (3)+\left (-12 x^3-12 x^2 \log (3)\right ) \log (x+\log (3))\right )}{5 x+5 \log (3)} \, dx \]
Integrate[(90*x^2 + 90*x*Log[3] + (x + Log[3])^((2*x^2)/5)*(2*x^2 + (4*x^2 + 4*x*Log[3])*Log[x + Log[3]]) + (x + Log[3])^(x^2/5)*(-30*x - 6*x^3 - 30 *Log[3] + (-12*x^3 - 12*x^2*Log[3])*Log[x + Log[3]]))/(5*x + 5*Log[3]),x]
Integrate[(90*x^2 + 90*x*Log[3] + (x + Log[3])^((2*x^2)/5)*(2*x^2 + (4*x^2 + 4*x*Log[3])*Log[x + Log[3]]) + (x + Log[3])^(x^2/5)*(-30*x - 6*x^3 - 30 *Log[3] + (-12*x^3 - 12*x^2*Log[3])*Log[x + Log[3]]))/(5*x + 5*Log[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.
| \(\displaystyle \int \frac {90 x^2+\left (2 x^2+\left (4 x^2+4 x \log (3)\right ) \log (x+\log (3))\right ) (x+\log (3))^{\frac {2 x^2}{5}}+\left (-6 x^3+\left (-12 x^3-12 x^2 \log (3)\right ) \log (x+\log (3))-30 x-30 \log (3)\right ) (x+\log (3))^{\frac {x^2}{5}}+90 x \log (3)}{5 x+5 \log (3)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {2 \left (3 x-(x+\log (3))^{\frac {x^2}{5}}\right ) \left (-x^2 (x+\log (3))^{\frac {x^2}{5}}-2 x^2 \log (x+\log (3)) (x+\log (3))^{\frac {x^2}{5}}-x \log (9) \log (x+\log (3)) (x+\log (3))^{\frac {x^2}{5}}+15 x+15 \log (3)\right )}{5 x+5 \log (3)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \int \frac {\left (3 x-(x+\log (3))^{\frac {x^2}{5}}\right ) \left (-x^2 (x+\log (3))^{\frac {x^2}{5}}-2 x^2 \log (x+\log (3)) (x+\log (3))^{\frac {x^2}{5}}-x \log (9) \log (x+\log (3)) (x+\log (3))^{\frac {x^2}{5}}+15 x+15 \log (3)\right )}{5 x+\log (243)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 2 \int \left (\frac {3}{5} \left (-2 \log (x+\log (3)) x^3-x^3-\log (9) \log (x+\log (3)) x^2-5 x-\log (243)\right ) (x+\log (3))^{\frac {x^2}{5}-1}+\frac {1}{5} x (2 \log (x+\log (3)) x+x+\log (9) \log (x+\log (3))) (x+\log (3))^{\frac {2 x^2}{5}-1}+9 x\right )dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 2 \int \left (-\frac {3}{5} \left (2 \log (x+\log (3)) x^3+x^3+\log (9) \log (x+\log (3)) x^2+5 x+\log (243)\right ) (x+\log (3))^{\frac {x^2}{5}-1}+\frac {1}{5} x (2 \log (x+\log (3)) x+x+\log (9) \log (x+\log (3))) (x+\log (3))^{\frac {2 x^2}{5}-1}+9 x\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle 2 \int \left (-\frac {3}{5} \left (2 \log (x+\log (3)) x^3+x^3+\log (9) \log (x+\log (3)) x^2+5 x+\log (243)\right ) (x+\log (3))^{\frac {x^2}{5}-1}+\frac {1}{5} x (2 \log (x+\log (3)) x+x+\log (9) \log (x+\log (3))) (x+\log (3))^{\frac {2 x^2}{5}-1}+9 x\right )dx\) |
Int[(90*x^2 + 90*x*Log[3] + (x + Log[3])^((2*x^2)/5)*(2*x^2 + (4*x^2 + 4*x *Log[3])*Log[x + Log[3]]) + (x + Log[3])^(x^2/5)*(-30*x - 6*x^3 - 30*Log[3 ] + (-12*x^3 - 12*x^2*Log[3])*Log[x + Log[3]]))/(5*x + 5*Log[3]),x]
3.5.46.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.42 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60
| method | result | size |
| risch | \(\left (\ln \left (3\right )+x \right )^{\frac {2 x^{2}}{5}}-6 x \left (\ln \left (3\right )+x \right )^{\frac {x^{2}}{5}}+9 x^{2}\) | \(32\) |
| default | \({\mathrm e}^{x^{2} \ln \left (\left (\ln \left (3\right )+x \right )^{\frac {2}{5}}\right )}-6 x \,{\mathrm e}^{\frac {x^{2} \ln \left (\ln \left (3\right )+x \right )}{5}}+9 x^{2}\) | \(34\) |
| parts | \({\mathrm e}^{x^{2} \ln \left (\left (\ln \left (3\right )+x \right )^{\frac {2}{5}}\right )}-6 x \,{\mathrm e}^{\frac {x^{2} \ln \left (\ln \left (3\right )+x \right )}{5}}+9 x^{2}\) | \(34\) |
| parallelrisch | \(-9 \ln \left (3\right )^{2}+9 x^{2}-6 x \,{\mathrm e}^{\frac {x^{2} \ln \left (\ln \left (3\right )+x \right )}{5}}+{\mathrm e}^{x^{2} \ln \left (\left (\ln \left (3\right )+x \right )^{\frac {2}{5}}\right )}\) | \(40\) |
int((((4*x*ln(3)+4*x^2)*ln(ln(3)+x)+2*x^2)*exp(1/5*x^2*ln(ln(3)+x))^2+((-1 2*x^2*ln(3)-12*x^3)*ln(ln(3)+x)-30*ln(3)-6*x^3-30*x)*exp(1/5*x^2*ln(ln(3)+ x))+90*x*ln(3)+90*x^2)/(5*ln(3)+5*x),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {90 x^2+90 x \log (3)+(x+\log (3))^{\frac {2 x^2}{5}} \left (2 x^2+\left (4 x^2+4 x \log (3)\right ) \log (x+\log (3))\right )+(x+\log (3))^{\frac {x^2}{5}} \left (-30 x-6 x^3-30 \log (3)+\left (-12 x^3-12 x^2 \log (3)\right ) \log (x+\log (3))\right )}{5 x+5 \log (3)} \, dx=-6 \, {\left (x + \log \left (3\right )\right )}^{\frac {1}{5} \, x^{2}} x + 9 \, x^{2} + {\left (x + \log \left (3\right )\right )}^{\frac {2}{5} \, x^{2}} \]
integrate((((4*x*log(3)+4*x^2)*log(log(3)+x)+2*x^2)*exp(1/5*x^2*log(log(3) +x))^2+((-12*x^2*log(3)-12*x^3)*log(log(3)+x)-30*log(3)-6*x^3-30*x)*exp(1/ 5*x^2*log(log(3)+x))+90*x*log(3)+90*x^2)/(5*log(3)+5*x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (17) = 34\).
Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80 \[ \int \frac {90 x^2+90 x \log (3)+(x+\log (3))^{\frac {2 x^2}{5}} \left (2 x^2+\left (4 x^2+4 x \log (3)\right ) \log (x+\log (3))\right )+(x+\log (3))^{\frac {x^2}{5}} \left (-30 x-6 x^3-30 \log (3)+\left (-12 x^3-12 x^2 \log (3)\right ) \log (x+\log (3))\right )}{5 x+5 \log (3)} \, dx=9 x^{2} - 6 x e^{\frac {x^{2} \log {\left (x + \log {\left (3 \right )} \right )}}{5}} + e^{\frac {2 x^{2} \log {\left (x + \log {\left (3 \right )} \right )}}{5}} \]
integrate((((4*x*ln(3)+4*x**2)*ln(ln(3)+x)+2*x**2)*exp(1/5*x**2*ln(ln(3)+x ))**2+((-12*x**2*ln(3)-12*x**3)*ln(ln(3)+x)-30*ln(3)-6*x**3-30*x)*exp(1/5* x**2*ln(ln(3)+x))+90*x*ln(3)+90*x**2)/(5*ln(3)+5*x),x)
Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (16) = 32\).
Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 3.05 \[ \int \frac {90 x^2+90 x \log (3)+(x+\log (3))^{\frac {2 x^2}{5}} \left (2 x^2+\left (4 x^2+4 x \log (3)\right ) \log (x+\log (3))\right )+(x+\log (3))^{\frac {x^2}{5}} \left (-30 x-6 x^3-30 \log (3)+\left (-12 x^3-12 x^2 \log (3)\right ) \log (x+\log (3))\right )}{5 x+5 \log (3)} \, dx=18 \, \log \left (3\right )^{2} \log \left (x + \log \left (3\right )\right ) - 6 \, {\left (x + \log \left (3\right )\right )}^{\frac {1}{5} \, x^{2}} x + 9 \, x^{2} - 18 \, {\left (\log \left (3\right ) \log \left (x + \log \left (3\right )\right ) - x\right )} \log \left (3\right ) - 18 \, x \log \left (3\right ) + {\left (x + \log \left (3\right )\right )}^{\frac {2}{5} \, x^{2}} \]
integrate((((4*x*log(3)+4*x^2)*log(log(3)+x)+2*x^2)*exp(1/5*x^2*log(log(3) +x))^2+((-12*x^2*log(3)-12*x^3)*log(log(3)+x)-30*log(3)-6*x^3-30*x)*exp(1/ 5*x^2*log(log(3)+x))+90*x*log(3)+90*x^2)/(5*log(3)+5*x),x, algorithm=\
18*log(3)^2*log(x + log(3)) - 6*(x + log(3))^(1/5*x^2)*x + 9*x^2 - 18*(log (3)*log(x + log(3)) - x)*log(3) - 18*x*log(3) + (x + log(3))^(2/5*x^2)
\[ \int \frac {90 x^2+90 x \log (3)+(x+\log (3))^{\frac {2 x^2}{5}} \left (2 x^2+\left (4 x^2+4 x \log (3)\right ) \log (x+\log (3))\right )+(x+\log (3))^{\frac {x^2}{5}} \left (-30 x-6 x^3-30 \log (3)+\left (-12 x^3-12 x^2 \log (3)\right ) \log (x+\log (3))\right )}{5 x+5 \log (3)} \, dx=\int { \frac {2 \, {\left ({\left (x^{2} + 2 \, {\left (x^{2} + x \log \left (3\right )\right )} \log \left (x + \log \left (3\right )\right )\right )} {\left (x + \log \left (3\right )\right )}^{\frac {2}{5} \, x^{2}} - 3 \, {\left (x^{3} + 2 \, {\left (x^{3} + x^{2} \log \left (3\right )\right )} \log \left (x + \log \left (3\right )\right ) + 5 \, x + 5 \, \log \left (3\right )\right )} {\left (x + \log \left (3\right )\right )}^{\frac {1}{5} \, x^{2}} + 45 \, x^{2} + 45 \, x \log \left (3\right )\right )}}{5 \, {\left (x + \log \left (3\right )\right )}} \,d x } \]
integrate((((4*x*log(3)+4*x^2)*log(log(3)+x)+2*x^2)*exp(1/5*x^2*log(log(3) +x))^2+((-12*x^2*log(3)-12*x^3)*log(log(3)+x)-30*log(3)-6*x^3-30*x)*exp(1/ 5*x^2*log(log(3)+x))+90*x*log(3)+90*x^2)/(5*log(3)+5*x),x, algorithm=\
integrate(2/5*((x^2 + 2*(x^2 + x*log(3))*log(x + log(3)))*(x + log(3))^(2/ 5*x^2) - 3*(x^3 + 2*(x^3 + x^2*log(3))*log(x + log(3)) + 5*x + 5*log(3))*( x + log(3))^(1/5*x^2) + 45*x^2 + 45*x*log(3))/(x + log(3)), x)
Timed out. \[ \int \frac {90 x^2+90 x \log (3)+(x+\log (3))^{\frac {2 x^2}{5}} \left (2 x^2+\left (4 x^2+4 x \log (3)\right ) \log (x+\log (3))\right )+(x+\log (3))^{\frac {x^2}{5}} \left (-30 x-6 x^3-30 \log (3)+\left (-12 x^3-12 x^2 \log (3)\right ) \log (x+\log (3))\right )}{5 x+5 \log (3)} \, dx=\int \frac {{\mathrm {e}}^{\frac {2\,x^2\,\ln \left (x+\ln \left (3\right )\right )}{5}}\,\left (2\,x^2+\ln \left (x+\ln \left (3\right )\right )\,\left (4\,x^2+4\,\ln \left (3\right )\,x\right )\right )-{\mathrm {e}}^{\frac {x^2\,\ln \left (x+\ln \left (3\right )\right )}{5}}\,\left (30\,x+30\,\ln \left (3\right )+\ln \left (x+\ln \left (3\right )\right )\,\left (12\,x^3+12\,\ln \left (3\right )\,x^2\right )+6\,x^3\right )+90\,x\,\ln \left (3\right )+90\,x^2}{5\,x+5\,\ln \left (3\right )} \,d x \]
int((exp((2*x^2*log(x + log(3)))/5)*(2*x^2 + log(x + log(3))*(4*x*log(3) + 4*x^2)) - exp((x^2*log(x + log(3)))/5)*(30*x + 30*log(3) + log(x + log(3) )*(12*x^2*log(3) + 12*x^3) + 6*x^3) + 90*x*log(3) + 90*x^2)/(5*x + 5*log(3 )),x)