Integrand size = 97, antiderivative size = 24 \[ \int \frac {5^{\frac {9}{20 x^2-2 x^3+(-40+4 x) \log (5)}} \left (\left (-180 x+27 x^2\right ) \log (5)-18 \log ^2(5)\right )}{200 x^4-40 x^5+2 x^6+\left (-800 x^2+160 x^3-8 x^4\right ) \log (5)+\left (800-160 x+8 x^2\right ) \log ^2(5)} \, dx=e^{\frac {9}{2 (-10+x) \left (2-\frac {x^2}{\log (5)}\right )}} \]
Time = 0.34 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {5^{\frac {9}{20 x^2-2 x^3+(-40+4 x) \log (5)}} \left (\left (-180 x+27 x^2\right ) \log (5)-18 \log ^2(5)\right )}{200 x^4-40 x^5+2 x^6+\left (-800 x^2+160 x^3-8 x^4\right ) \log (5)+\left (800-160 x+8 x^2\right ) \log ^2(5)} \, dx=5^{-\frac {9}{2 (-10+x) \left (x^2-2 \log (5)\right )}} \]
Integrate[(5^(9/(20*x^2 - 2*x^3 + (-40 + 4*x)*Log[5]))*((-180*x + 27*x^2)* Log[5] - 18*Log[5]^2))/(200*x^4 - 40*x^5 + 2*x^6 + (-800*x^2 + 160*x^3 - 8 *x^4)*Log[5] + (800 - 160*x + 8*x^2)*Log[5]^2),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.
| \(\displaystyle \int \frac {5^{\frac {9}{-2 x^3+20 x^2+(4 x-40) \log (5)}} \left (\left (27 x^2-180 x\right ) \log (5)-18 \log ^2(5)\right )}{2 x^6-40 x^5+200 x^4+\left (8 x^2-160 x+800\right ) \log ^2(5)+\left (-8 x^4+160 x^3-800 x^2\right ) \log (5)} \, dx\) |
| \(\Big \downarrow \) 2463 |
| \(\displaystyle \int \left (\frac {5^{\frac {9}{-2 x^3+20 x^2+(4 x-40) \log (5)}} (10 x+50+\log (5)) \left (\left (27 x^2-180 x\right ) \log (5)-18 \log ^2(5)\right )}{4 (\log (5)-50)^2 \left (x^2-2 \log (5)\right )^2}+\frac {5^{\frac {9}{-2 x^3+20 x^2+(4 x-40) \log (5)}} \left (\left (27 x^2-180 x\right ) \log (5)-18 \log ^2(5)\right )}{8 (x-10)^2 (\log (5)-50)^2}+\frac {5^{\frac {9}{-2 x^3+20 x^2+(4 x-40) \log (5)}} (-20 x-150-\log (5)) \left (\left (27 x^2-180 x\right ) \log (5)-18 \log ^2(5)\right )}{8 (\log (5)-50)^3 \left (x^2-2 \log (5)\right )}+\frac {5^{\frac {9}{-2 x^3+20 x^2+(4 x-40) \log (5)}+1} \left (\left (27 x^2-180 x\right ) \log (5)-18 \log ^2(5)\right )}{2 (x-10) (\log (5)-50)^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {9 \log (5) (50+\log (125)) \int 5^{\frac {9}{-2 x^3+20 x^2+4 \log (5) x-40 \log (5)}}dx}{8 (50-\log (5))^3}+\frac {27 \log (5) \int 5^{\frac {9}{-2 x^3+20 x^2+4 \log (5) x-40 \log (5)}}dx}{8 (50-\log (5))^2}-\frac {9 \log (5) \int 5^{2+\frac {9}{-2 x^3+20 x^2+(4 x-40) \log (5)}}dx}{(50-\log (5))^3}+\frac {9 \log (5) \int \frac {5^{\frac {9}{-2 x^3+20 x^2+4 \log (5) x-40 \log (5)}}}{(x-10)^2}dx}{4 (50-\log (5))}+\frac {9 \log (5) \int \frac {5^{1+\frac {9}{-2 x^3+20 x^2+4 \log (5) x-40 \log (5)}}}{x-10}dx}{(50-\log (5))^2}-\frac {9 \log (5) \int \frac {5^{1+\frac {9}{-2 x^3+20 x^2+(4 x-40) \log (5)}}}{x-10}dx}{(50-\log (5))^2}+\frac {27 \log (5) \int 5^{1+\frac {9}{-2 x^3+20 x^2+4 \log (5) x-40 \log (5)}} xdx}{2 (50-\log (5))^3}-\frac {27 \log (5) \int 5^{1+\frac {9}{-2 x^3+20 x^2+(4 x-40) \log (5)}} xdx}{2 (50-\log (5))^3}-\frac {9 \log (5) \int \frac {5^{1+\frac {9}{-2 x^3+20 x^2+4 \log (5) x-40 \log (5)}} x}{\left (x^2-2 \log (5)\right )^2}dx}{50-\log (5)}-\frac {9 \log ^2(5) \int \frac {5^{\frac {9}{-2 x^3+20 x^2+4 \log (5) x-40 \log (5)}}}{\left (\sqrt {\log (25)}-x\right )^2}dx}{4 (50-\log (5)) \log (25)}+\frac {9 \sqrt {\log (25)} \left (50-30 \sqrt {\log (25)}-\log (125)\right ) \int \frac {5^{\frac {9}{-2 x^3+20 x^2+4 \log (5) x-40 \log (5)}}}{\sqrt {\log (25)}-x}dx}{16 (50-\log (5))^2}+\frac {9 \left (\log (5)+15 \sqrt {\log (25)}\right ) \sqrt {\log (25)} \int \frac {5^{\frac {9}{-2 x^3+20 x^2+4 \log (5) x-40 \log (5)}}}{\sqrt {\log (25)}-x}dx}{8 (50-\log (5))^2}-\frac {9 \log ^2(5) \int \frac {5^{\frac {9}{-2 x^3+20 x^2+4 \log (5) x-40 \log (5)}}}{\sqrt {\log (25)}-x}dx}{4 (50-\log (5)) \log ^{\frac {3}{2}}(25)}-\frac {9 \log ^2(5) \int \frac {5^{\frac {9}{-2 x^3+20 x^2+4 \log (5) x-40 \log (5)}}}{\left (x+\sqrt {\log (25)}\right )^2}dx}{4 (50-\log (5)) \log (25)}+\frac {9 \sqrt {\log (25)} \left (50+30 \sqrt {\log (25)}-\log (125)\right ) \int \frac {5^{\frac {9}{-2 x^3+20 x^2+4 \log (5) x-40 \log (5)}}}{x+\sqrt {\log (25)}}dx}{16 (50-\log (5))^2}+\frac {9 \left (\log (5)-15 \sqrt {\log (25)}\right ) \sqrt {\log (25)} \int \frac {5^{\frac {9}{-2 x^3+20 x^2+4 \log (5) x-40 \log (5)}}}{x+\sqrt {\log (25)}}dx}{8 (50-\log (5))^2}-\frac {9 \log ^2(5) \int \frac {5^{\frac {9}{-2 x^3+20 x^2+4 \log (5) x-40 \log (5)}}}{x+\sqrt {\log (25)}}dx}{4 (50-\log (5)) \log ^{\frac {3}{2}}(25)}\) |
Int[(5^(9/(20*x^2 - 2*x^3 + (-40 + 4*x)*Log[5]))*((-180*x + 27*x^2)*Log[5] - 18*Log[5]^2))/(200*x^4 - 40*x^5 + 2*x^6 + (-800*x^2 + 160*x^3 - 8*x^4)* Log[5] + (800 - 160*x + 8*x^2)*Log[5]^2),x]
3.13.10.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 0.77 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
| method | result | size |
| risch | \(5^{\frac {9}{2 \left (x -10\right ) \left (-x^{2}+2 \ln \left (5\right )\right )}}\) | \(22\) |
| gosper | \({\mathrm e}^{\frac {9 \ln \left (5\right )}{2 \left (-x^{3}+2 x \ln \left (5\right )+10 x^{2}-20 \ln \left (5\right )\right )}}\) | \(28\) |
| parallelrisch | \({\mathrm e}^{\frac {9 \ln \left (5\right )}{2 \left (-x^{3}+2 x \ln \left (5\right )+10 x^{2}-20 \ln \left (5\right )\right )}}\) | \(28\) |
| norman | \(\frac {10 x^{2} {\mathrm e}^{\frac {9 \ln \left (5\right )}{\left (4 x -40\right ) \ln \left (5\right )-2 x^{3}+20 x^{2}}}-x^{3} {\mathrm e}^{\frac {9 \ln \left (5\right )}{\left (4 x -40\right ) \ln \left (5\right )-2 x^{3}+20 x^{2}}}-20 \ln \left (5\right ) {\mathrm e}^{\frac {9 \ln \left (5\right )}{\left (4 x -40\right ) \ln \left (5\right )-2 x^{3}+20 x^{2}}}+2 x \ln \left (5\right ) {\mathrm e}^{\frac {9 \ln \left (5\right )}{\left (4 x -40\right ) \ln \left (5\right )-2 x^{3}+20 x^{2}}}}{\left (x -10\right ) \left (-x^{2}+2 \ln \left (5\right )\right )}\) | \(143\) |
int((-18*ln(5)^2+(27*x^2-180*x)*ln(5))*exp(9*ln(5)/((4*x-40)*ln(5)-2*x^3+2 0*x^2))/((8*x^2-160*x+800)*ln(5)^2+(-8*x^4+160*x^3-800*x^2)*ln(5)+2*x^6-40 *x^5+200*x^4),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {5^{\frac {9}{20 x^2-2 x^3+(-40+4 x) \log (5)}} \left (\left (-180 x+27 x^2\right ) \log (5)-18 \log ^2(5)\right )}{200 x^4-40 x^5+2 x^6+\left (-800 x^2+160 x^3-8 x^4\right ) \log (5)+\left (800-160 x+8 x^2\right ) \log ^2(5)} \, dx=\frac {1}{5^{\frac {9}{2 \, {\left (x^{3} - 10 \, x^{2} - 2 \, {\left (x - 10\right )} \log \left (5\right )\right )}}}} \]
integrate((-18*log(5)^2+(27*x^2-180*x)*log(5))*exp(9*log(5)/((4*x-40)*log( 5)-2*x^3+20*x^2))/((8*x^2-160*x+800)*log(5)^2+(-8*x^4+160*x^3-800*x^2)*log (5)+2*x^6-40*x^5+200*x^4),x, algorithm=\
Time = 0.58 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {5^{\frac {9}{20 x^2-2 x^3+(-40+4 x) \log (5)}} \left (\left (-180 x+27 x^2\right ) \log (5)-18 \log ^2(5)\right )}{200 x^4-40 x^5+2 x^6+\left (-800 x^2+160 x^3-8 x^4\right ) \log (5)+\left (800-160 x+8 x^2\right ) \log ^2(5)} \, dx=e^{\frac {9 \log {\left (5 \right )}}{- 2 x^{3} + 20 x^{2} + \left (4 x - 40\right ) \log {\left (5 \right )}}} \]
integrate((-18*ln(5)**2+(27*x**2-180*x)*ln(5))*exp(9*ln(5)/((4*x-40)*ln(5) -2*x**3+20*x**2))/((8*x**2-160*x+800)*ln(5)**2+(-8*x**4+160*x**3-800*x**2) *ln(5)+2*x**6-40*x**5+200*x**4),x)
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (20) = 40\).
Time = 0.41 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.96 \[ \int \frac {5^{\frac {9}{20 x^2-2 x^3+(-40+4 x) \log (5)}} \left (\left (-180 x+27 x^2\right ) \log (5)-18 \log ^2(5)\right )}{200 x^4-40 x^5+2 x^6+\left (-800 x^2+160 x^3-8 x^4\right ) \log (5)+\left (800-160 x+8 x^2\right ) \log ^2(5)} \, dx=e^{\left (-\frac {9 \, x \log \left (5\right )}{4 \, {\left (x^{2} {\left (\log \left (5\right ) - 50\right )} - 2 \, \log \left (5\right )^{2} + 100 \, \log \left (5\right )\right )}} - \frac {45 \, \log \left (5\right )}{2 \, {\left (x^{2} {\left (\log \left (5\right ) - 50\right )} - 2 \, \log \left (5\right )^{2} + 100 \, \log \left (5\right )\right )}} + \frac {9 \, \log \left (5\right )}{4 \, {\left (x {\left (\log \left (5\right ) - 50\right )} - 10 \, \log \left (5\right ) + 500\right )}}\right )} \]
integrate((-18*log(5)^2+(27*x^2-180*x)*log(5))*exp(9*log(5)/((4*x-40)*log( 5)-2*x^3+20*x^2))/((8*x^2-160*x+800)*log(5)^2+(-8*x^4+160*x^3-800*x^2)*log (5)+2*x^6-40*x^5+200*x^4),x, algorithm=\
e^(-9/4*x*log(5)/(x^2*(log(5) - 50) - 2*log(5)^2 + 100*log(5)) - 45/2*log( 5)/(x^2*(log(5) - 50) - 2*log(5)^2 + 100*log(5)) + 9/4*log(5)/(x*(log(5) - 50) - 10*log(5) + 500))
\[ \int \frac {5^{\frac {9}{20 x^2-2 x^3+(-40+4 x) \log (5)}} \left (\left (-180 x+27 x^2\right ) \log (5)-18 \log ^2(5)\right )}{200 x^4-40 x^5+2 x^6+\left (-800 x^2+160 x^3-8 x^4\right ) \log (5)+\left (800-160 x+8 x^2\right ) \log ^2(5)} \, dx=\int { \frac {9 \, {\left ({\left (3 \, x^{2} - 20 \, x\right )} \log \left (5\right ) - 2 \, \log \left (5\right )^{2}\right )}}{2 \, {\left (x^{6} - 20 \, x^{5} + 100 \, x^{4} + 4 \, {\left (x^{2} - 20 \, x + 100\right )} \log \left (5\right )^{2} - 4 \, {\left (x^{4} - 20 \, x^{3} + 100 \, x^{2}\right )} \log \left (5\right )\right )} 5^{\frac {9}{2 \, {\left (x^{3} - 10 \, x^{2} - 2 \, {\left (x - 10\right )} \log \left (5\right )\right )}}}} \,d x } \]
integrate((-18*log(5)^2+(27*x^2-180*x)*log(5))*exp(9*log(5)/((4*x-40)*log( 5)-2*x^3+20*x^2))/((8*x^2-160*x+800)*log(5)^2+(-8*x^4+160*x^3-800*x^2)*log (5)+2*x^6-40*x^5+200*x^4),x, algorithm=\
integrate(9/2*((3*x^2 - 20*x)*log(5) - 2*log(5)^2)/((x^6 - 20*x^5 + 100*x^ 4 + 4*(x^2 - 20*x + 100)*log(5)^2 - 4*(x^4 - 20*x^3 + 100*x^2)*log(5))*5^( 9/2/(x^3 - 10*x^2 - 2*(x - 10)*log(5)))), x)
Time = 13.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {5^{\frac {9}{20 x^2-2 x^3+(-40+4 x) \log (5)}} \left (\left (-180 x+27 x^2\right ) \log (5)-18 \log ^2(5)\right )}{200 x^4-40 x^5+2 x^6+\left (-800 x^2+160 x^3-8 x^4\right ) \log (5)+\left (800-160 x+8 x^2\right ) \log ^2(5)} \, dx=\frac {1}{5^{\frac {9}{2\,x^3-20\,x^2-4\,\ln \left (5\right )\,x+40\,\ln \left (5\right )}}} \]