Integrand size = 81, antiderivative size = 26 \[ \int \frac {e^{15} (250-1000 x)+e^{20} \left (-x+4 x^2\right )+\left (e^{10} (127500-10000 x)-500 e^{15} x\right ) \log \left (x^2\right )+\left (1875000 e^5-2500 e^{10} x\right ) \log ^2\left (x^2\right )+6250000 \log ^3\left (x^2\right )}{32 e^{20} x} \, dx=\frac {1}{16} \left (6+x-\left (\frac {5}{2}+\frac {25 \log \left (x^2\right )}{e^5}\right )^2\right )^2 \]
Leaf count is larger than twice the leaf count of optimal. \(81\) vs. \(2(26)=52\).
Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 3.12 \[ \int \frac {e^{15} (250-1000 x)+e^{20} \left (-x+4 x^2\right )+\left (e^{10} (127500-10000 x)-500 e^{15} x\right ) \log \left (x^2\right )+\left (1875000 e^5-2500 e^{10} x\right ) \log ^2\left (x^2\right )+6250000 \log ^3\left (x^2\right )}{32 e^{20} x} \, dx=\frac {-e^{20} x+2 e^{20} x^2+250 e^{15} \log (x)-500 e^{15} x \log \left (x^2\right )+31875 e^{10} \log ^2\left (x^2\right )-2500 e^{10} x \log ^2\left (x^2\right )+312500 e^5 \log ^3\left (x^2\right )+781250 \log ^4\left (x^2\right )}{32 e^{20}} \]
Integrate[(E^15*(250 - 1000*x) + E^20*(-x + 4*x^2) + (E^10*(127500 - 10000 *x) - 500*E^15*x)*Log[x^2] + (1875000*E^5 - 2500*E^10*x)*Log[x^2]^2 + 6250 000*Log[x^2]^3)/(32*E^20*x),x]
(-(E^20*x) + 2*E^20*x^2 + 250*E^15*Log[x] - 500*E^15*x*Log[x^2] + 31875*E^ 10*Log[x^2]^2 - 2500*E^10*x*Log[x^2]^2 + 312500*E^5*Log[x^2]^3 + 781250*Lo g[x^2]^4)/(32*E^20)
Leaf count is larger than twice the leaf count of optimal. \(118\) vs. \(2(26)=52\).
Time = 0.35 (sec) , antiderivative size = 118, normalized size of antiderivative = 4.54, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {27, 2010, 2009}
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.
\(\displaystyle \int \frac {e^{20} \left (4 x^2-x\right )+6250000 \log ^3\left (x^2\right )+\left (1875000 e^5-2500 e^{10} x\right ) \log ^2\left (x^2\right )+\left (e^{10} (127500-10000 x)-500 e^{15} x\right ) \log \left (x^2\right )+e^{15} (250-1000 x)}{32 e^{20} x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {6250000 \log ^3\left (x^2\right )+2500 e^5 \left (750-e^5 x\right ) \log ^2\left (x^2\right )+500 \left (5 e^{10} (51-4 x)-e^{15} x\right ) \log \left (x^2\right )+250 e^{15} (1-4 x)-e^{20} \left (x-4 x^2\right )}{x}dx}{32 e^{20}}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {\int \left (\frac {6250000 \log ^3\left (x^2\right )}{x}-\frac {2500 e^5 \left (e^5 x-750\right ) \log ^2\left (x^2\right )}{x}+\frac {500 e^{10} \left (255-\left (20+e^5\right ) x\right ) \log \left (x^2\right )}{x}+\frac {e^{15} (4 x-1) \left (e^5 x-250\right )}{x}\right )dx}{32 e^{20}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 e^{20} x^2+781250 \log ^4\left (x^2\right )+312500 e^5 \log ^3\left (x^2\right )-2500 e^{10} x \log ^2\left (x^2\right )+31875 e^{10} \log ^2\left (x^2\right )-500 e^{10} \left (20+e^5\right ) x \log \left (x^2\right )+10000 e^{10} x \log \left (x^2\right )-e^{15} \left (1000+e^5\right ) x+1000 e^{10} \left (20+e^5\right ) x-20000 e^{10} x+250 e^{15} \log (x)}{32 e^{20}}\) |
Int[(E^15*(250 - 1000*x) + E^20*(-x + 4*x^2) + (E^10*(127500 - 10000*x) - 500*E^15*x)*Log[x^2] + (1875000*E^5 - 2500*E^10*x)*Log[x^2]^2 + 6250000*Lo g[x^2]^3)/(32*E^20*x),x]
(-20000*E^10*x + 1000*E^10*(20 + E^5)*x - E^15*(1000 + E^5)*x + 2*E^20*x^2 + 250*E^15*Log[x] + 10000*E^10*x*Log[x^2] - 500*E^10*(20 + E^5)*x*Log[x^2 ] + 31875*E^10*Log[x^2]^2 - 2500*E^10*x*Log[x^2]^2 + 312500*E^5*Log[x^2]^3 + 781250*Log[x^2]^4)/(32*E^20)
3.9.27.3.1 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]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.85
method | result | size |
risch | \(\frac {x^{2}}{16}-\frac {x}{32}+\frac {125 \ln \left (x \right ) {\mathrm e}^{-5}}{16}+\frac {390625 \,{\mathrm e}^{-20} \ln \left (x^{2}\right )^{4}}{16}+\frac {78125 \,{\mathrm e}^{-15} \ln \left (x^{2}\right )^{3}}{8}-\frac {625 \,{\mathrm e}^{-10} \ln \left (x^{2}\right )^{2} x}{8}-\frac {125 x \ln \left (x^{2}\right ) {\mathrm e}^{-5}}{8}-\frac {31875 \ln \left (x \right )^{2} {\mathrm e}^{-10}}{8}+\frac {31875 \ln \left (x \right ) \ln \left (x^{2}\right ) {\mathrm e}^{-10}}{8}\) | \(74\) |
norman | \(\left (\frac {125 \,{\mathrm e}^{10} \ln \left (x^{2}\right )}{32}+\frac {78125 \ln \left (x^{2}\right )^{3}}{8}-\frac {x \,{\mathrm e}^{15}}{32}+\frac {x^{2} {\mathrm e}^{15}}{16}+\frac {390625 \,{\mathrm e}^{-5} \ln \left (x^{2}\right )^{4}}{16}+\frac {31875 \,{\mathrm e}^{5} \ln \left (x^{2}\right )^{2}}{32}-\frac {625 x \,{\mathrm e}^{5} \ln \left (x^{2}\right )^{2}}{8}-\frac {125 \,{\mathrm e}^{10} \ln \left (x^{2}\right ) x}{8}\right ) {\mathrm e}^{-15}\) | \(85\) |
parts | \(\frac {390625 \,{\mathrm e}^{-20} \ln \left (x^{2}\right )^{4}}{16}+\frac {x^{2}}{16}-\frac {x}{32}+\frac {125 \ln \left (x \right ) {\mathrm e}^{-5}}{16}-\frac {125 \,{\mathrm e}^{-5} x}{4}+\left (-625 x +\frac {625 x \ln \left (x^{2}\right )}{2}+\frac {78125 \,{\mathrm e}^{-5} \ln \left (x^{2}\right )^{3}}{8}-\frac {625 x \ln \left (x^{2}\right )^{2}}{8}\right ) {\mathrm e}^{-10}-\frac {125 \,{\mathrm e}^{-10} \left (x \,{\mathrm e}^{5} \ln \left (x^{2}\right )-2 x \,{\mathrm e}^{5}+20 x \ln \left (x^{2}\right )-40 x -255 \ln \left (x \right ) \ln \left (x^{2}\right )+255 \ln \left (x \right )^{2}\right )}{8}\) | \(118\) |
default | \(\frac {{\mathrm e}^{-20} \left ({\mathrm e}^{15} \left (2 x^{2} {\mathrm e}^{5}-x \,{\mathrm e}^{5}-1000 x +250 \ln \left (x \right )\right )+781250 \ln \left (x^{2}\right )^{4}+312500 \,{\mathrm e}^{5} \ln \left (x^{2}\right )^{3}-20000 x \,{\mathrm e}^{10}+10000 \,{\mathrm e}^{10} \ln \left (x^{2}\right ) x -2500 \,{\mathrm e}^{10} \ln \left (x^{2}\right )^{2} x -500 \,{\mathrm e}^{10} \left (x \,{\mathrm e}^{5} \ln \left (x^{2}\right )-2 x \,{\mathrm e}^{5}+20 x \ln \left (x^{2}\right )-40 x -255 \ln \left (x \right ) \ln \left (x^{2}\right )+255 \ln \left (x \right )^{2}\right )\right )}{32}\) | \(126\) |
int(1/32*(6250000*ln(x^2)^3+(-2500*x*exp(5)^2+1875000*exp(5))*ln(x^2)^2+(- 500*x*exp(5)^3+(-10000*x+127500)*exp(5)^2)*ln(x^2)+(4*x^2-x)*exp(5)^4+(-10 00*x+250)*exp(5)^3)/x/exp(5)^4,x,method=_RETURNVERBOSE)
1/16*x^2-1/32*x+125/16*ln(x)*exp(-5)+390625/16*exp(-20)*ln(x^2)^4+78125/8* exp(-15)*ln(x^2)^3-625/8*exp(-10)*ln(x^2)^2*x-125/8*x*ln(x^2)*exp(-5)-3187 5/8*ln(x)^2*exp(-10)+31875/8*ln(x)*ln(x^2)*exp(-10)
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (23) = 46\).
Time = 0.25 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.46 \[ \int \frac {e^{15} (250-1000 x)+e^{20} \left (-x+4 x^2\right )+\left (e^{10} (127500-10000 x)-500 e^{15} x\right ) \log \left (x^2\right )+\left (1875000 e^5-2500 e^{10} x\right ) \log ^2\left (x^2\right )+6250000 \log ^3\left (x^2\right )}{32 e^{20} x} \, dx=-\frac {1}{32} \, {\left (625 \, {\left (4 \, x - 51\right )} e^{10} \log \left (x^{2}\right )^{2} - 312500 \, e^{5} \log \left (x^{2}\right )^{3} - 781250 \, \log \left (x^{2}\right )^{4} + 125 \, {\left (4 \, x - 1\right )} e^{15} \log \left (x^{2}\right ) - {\left (2 \, x^{2} - x\right )} e^{20}\right )} e^{\left (-20\right )} \]
integrate(1/32*(6250000*log(x^2)^3+(-2500*x*exp(5)^2+1875000*exp(5))*log(x ^2)^2+(-500*x*exp(5)^3+(-10000*x+127500)*exp(5)^2)*log(x^2)+(4*x^2-x)*exp( 5)^4+(-1000*x+250)*exp(5)^3)/x/exp(5)^4,x, algorithm=\
-1/32*(625*(4*x - 51)*e^10*log(x^2)^2 - 312500*e^5*log(x^2)^3 - 781250*log (x^2)^4 + 125*(4*x - 1)*e^15*log(x^2) - (2*x^2 - x)*e^20)*e^(-20)
Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (24) = 48\).
Time = 0.15 (sec) , antiderivative size = 80, normalized size of antiderivative = 3.08 \[ \int \frac {e^{15} (250-1000 x)+e^{20} \left (-x+4 x^2\right )+\left (e^{10} (127500-10000 x)-500 e^{15} x\right ) \log \left (x^2\right )+\left (1875000 e^5-2500 e^{10} x\right ) \log ^2\left (x^2\right )+6250000 \log ^3\left (x^2\right )}{32 e^{20} x} \, dx=- \frac {125 x \log {\left (x^{2} \right )}}{8 e^{5}} + \frac {\left (31875 - 2500 x\right ) \log {\left (x^{2} \right )}^{2}}{32 e^{10}} + \frac {2 x^{2} e^{5} - x e^{5} + 250 \log {\left (x \right )}}{32 e^{5}} + \frac {390625 \log {\left (x^{2} \right )}^{4}}{16 e^{20}} + \frac {78125 \log {\left (x^{2} \right )}^{3}}{8 e^{15}} \]
integrate(1/32*(6250000*ln(x**2)**3+(-2500*x*exp(5)**2+1875000*exp(5))*ln( x**2)**2+(-500*x*exp(5)**3+(-10000*x+127500)*exp(5)**2)*ln(x**2)+(4*x**2-x )*exp(5)**4+(-1000*x+250)*exp(5)**3)/x/exp(5)**4,x)
-125*x*exp(-5)*log(x**2)/8 + (31875 - 2500*x)*exp(-10)*log(x**2)**2/32 + ( 2*x**2*exp(5) - x*exp(5) + 250*log(x))*exp(-5)/32 + 390625*exp(-20)*log(x* *2)**4/16 + 78125*exp(-15)*log(x**2)**3/8
Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (23) = 46\).
Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 3.08 \[ \int \frac {e^{15} (250-1000 x)+e^{20} \left (-x+4 x^2\right )+\left (e^{10} (127500-10000 x)-500 e^{15} x\right ) \log \left (x^2\right )+\left (1875000 e^5-2500 e^{10} x\right ) \log ^2\left (x^2\right )+6250000 \log ^3\left (x^2\right )}{32 e^{20} x} \, dx=-\frac {1}{32} \, {\left (2500 \, x e^{10} \log \left (x^{2}\right )^{2} - 312500 \, e^{5} \log \left (x^{2}\right )^{3} - 781250 \, \log \left (x^{2}\right )^{4} - 2 \, x^{2} e^{20} - 31875 \, e^{10} \log \left (x^{2}\right )^{2} + x e^{20} + 500 \, {\left (x \log \left (x^{2}\right ) - 2 \, x\right )} e^{15} + 1000 \, x e^{15} - 250 \, e^{15} \log \left (x\right )\right )} e^{\left (-20\right )} \]
integrate(1/32*(6250000*log(x^2)^3+(-2500*x*exp(5)^2+1875000*exp(5))*log(x ^2)^2+(-500*x*exp(5)^3+(-10000*x+127500)*exp(5)^2)*log(x^2)+(4*x^2-x)*exp( 5)^4+(-1000*x+250)*exp(5)^3)/x/exp(5)^4,x, algorithm=\
-1/32*(2500*x*e^10*log(x^2)^2 - 312500*e^5*log(x^2)^3 - 781250*log(x^2)^4 - 2*x^2*e^20 - 31875*e^10*log(x^2)^2 + x*e^20 + 500*(x*log(x^2) - 2*x)*e^1 5 + 1000*x*e^15 - 250*e^15*log(x))*e^(-20)
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (23) = 46\).
Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.69 \[ \int \frac {e^{15} (250-1000 x)+e^{20} \left (-x+4 x^2\right )+\left (e^{10} (127500-10000 x)-500 e^{15} x\right ) \log \left (x^2\right )+\left (1875000 e^5-2500 e^{10} x\right ) \log ^2\left (x^2\right )+6250000 \log ^3\left (x^2\right )}{32 e^{20} x} \, dx=-\frac {1}{32} \, {\left (2500 \, x e^{10} \log \left (x^{2}\right )^{2} - 312500 \, e^{5} \log \left (x^{2}\right )^{3} - 781250 \, \log \left (x^{2}\right )^{4} - 2 \, x^{2} e^{20} + 500 \, x e^{15} \log \left (x^{2}\right ) - 31875 \, e^{10} \log \left (x^{2}\right )^{2} + x e^{20} - 250 \, e^{15} \log \left (x\right )\right )} e^{\left (-20\right )} \]
integrate(1/32*(6250000*log(x^2)^3+(-2500*x*exp(5)^2+1875000*exp(5))*log(x ^2)^2+(-500*x*exp(5)^3+(-10000*x+127500)*exp(5)^2)*log(x^2)+(4*x^2-x)*exp( 5)^4+(-1000*x+250)*exp(5)^3)/x/exp(5)^4,x, algorithm=\
-1/32*(2500*x*e^10*log(x^2)^2 - 312500*e^5*log(x^2)^3 - 781250*log(x^2)^4 - 2*x^2*e^20 + 500*x*e^15*log(x^2) - 31875*e^10*log(x^2)^2 + x*e^20 - 250* e^15*log(x))*e^(-20)
Time = 9.39 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.58 \[ \int \frac {e^{15} (250-1000 x)+e^{20} \left (-x+4 x^2\right )+\left (e^{10} (127500-10000 x)-500 e^{15} x\right ) \log \left (x^2\right )+\left (1875000 e^5-2500 e^{10} x\right ) \log ^2\left (x^2\right )+6250000 \log ^3\left (x^2\right )}{32 e^{20} x} \, dx=\frac {x^2}{16}-\frac {625\,{\mathrm {e}}^{-10}\,x\,{\ln \left (x^2\right )}^2}{8}-\frac {125\,{\mathrm {e}}^{-5}\,x\,\ln \left (x^2\right )}{8}-\frac {x}{32}+\frac {390625\,{\mathrm {e}}^{-20}\,{\ln \left (x^2\right )}^4}{16}+\frac {78125\,{\mathrm {e}}^{-15}\,{\ln \left (x^2\right )}^3}{8}+\frac {31875\,{\mathrm {e}}^{-10}\,{\ln \left (x^2\right )}^2}{32}+\frac {125\,{\mathrm {e}}^{-5}\,\ln \left (x^2\right )}{32} \]
int(-(exp(-20)*((log(x^2)*(500*x*exp(15) + exp(10)*(10000*x - 127500)))/32 - (log(x^2)^2*(1875000*exp(5) - 2500*x*exp(10)))/32 - (390625*log(x^2)^3) /2 + (exp(20)*(x - 4*x^2))/32 + (exp(15)*(1000*x - 250))/32))/x,x)