Integrand size = 78, antiderivative size = 25 \[ \int \frac {e^{-4 x} \left (288 x-576 x^2+\left (-2880 x+5760 x^2-1152 x^3+1152 x^4\right ) \log (5)+\left (7200 x-14400 x^2+5760 x^3-5760 x^4+864 x^5-576 x^6\right ) \log ^2(5)\right )}{25 \log ^2(5)} \, dx=\frac {144}{25} e^{-4 x} x^2 \left (5+x^2-\frac {1}{\log (5)}\right )^2 \]
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {e^{-4 x} \left (288 x-576 x^2+\left (-2880 x+5760 x^2-1152 x^3+1152 x^4\right ) \log (5)+\left (7200 x-14400 x^2+5760 x^3-5760 x^4+864 x^5-576 x^6\right ) \log ^2(5)\right )}{25 \log ^2(5)} \, dx=\frac {144 e^{-4 x} x^2 \left (-1+5 \log (5)+x^2 \log (5)\right )^2}{25 \log ^2(5)} \]
Integrate[(288*x - 576*x^2 + (-2880*x + 5760*x^2 - 1152*x^3 + 1152*x^4)*Lo g[5] + (7200*x - 14400*x^2 + 5760*x^3 - 5760*x^4 + 864*x^5 - 576*x^6)*Log[ 5]^2)/(25*E^(4*x)*Log[5]^2),x]
Leaf count is larger than twice the leaf count of optimal. \(91\) vs. \(2(25)=50\).
Time = 0.62 (sec) , antiderivative size = 91, normalized size of antiderivative = 3.64, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {27, 27, 2626, 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.
\(\displaystyle \int \frac {e^{-4 x} \left (-576 x^2+\left (1152 x^4-1152 x^3+5760 x^2-2880 x\right ) \log (5)+\left (-576 x^6+864 x^5-5760 x^4+5760 x^3-14400 x^2+7200 x\right ) \log ^2(5)+288 x\right )}{25 \log ^2(5)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int 288 e^{-4 x} \left (-2 x^2+x+\left (-2 x^6+3 x^5-20 x^4+20 x^3-50 x^2+25 x\right ) \log ^2(5)-2 \left (-2 x^4+2 x^3-10 x^2+5 x\right ) \log (5)\right )dx}{25 \log ^2(5)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {288 \int e^{-4 x} \left (-2 x^2+x+\left (-2 x^6+3 x^5-20 x^4+20 x^3-50 x^2+25 x\right ) \log ^2(5)-2 \left (-2 x^4+2 x^3-10 x^2+5 x\right ) \log (5)\right )dx}{25 \log ^2(5)}\) |
\(\Big \downarrow \) 2626 |
\(\displaystyle \frac {288 \int \left (-2 e^{-4 x} x^2+e^{-4 x} x-e^{-4 x} \left (2 x^5-3 x^4+20 x^3-20 x^2+50 x-25\right ) \log ^2(5) x+2 e^{-4 x} \left (2 x^3-2 x^2+10 x-5\right ) \log (5) x\right )dx}{25 \log ^2(5)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {288 \left (\frac {1}{2} e^{-4 x} x^6 \log ^2(5)+5 e^{-4 x} x^4 \log ^2(5)-e^{-4 x} x^4 \log (5)+\frac {1}{2} e^{-4 x} x^2+\frac {25}{2} e^{-4 x} x^2 \log ^2(5)-5 e^{-4 x} x^2 \log (5)\right )}{25 \log ^2(5)}\) |
Int[(288*x - 576*x^2 + (-2880*x + 5760*x^2 - 1152*x^3 + 1152*x^4)*Log[5] + (7200*x - 14400*x^2 + 5760*x^3 - 5760*x^4 + 864*x^5 - 576*x^6)*Log[5]^2)/ (25*E^(4*x)*Log[5]^2),x]
(288*(x^2/(2*E^(4*x)) - (5*x^2*Log[5])/E^(4*x) - (x^4*Log[5])/E^(4*x) + (2 5*x^2*Log[5]^2)/(2*E^(4*x)) + (5*x^4*Log[5]^2)/E^(4*x) + (x^6*Log[5]^2)/(2 *E^(4*x))))/(25*Log[5]^2)
3.18.65.3.1 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[(F_)^(v_)*(Px_), x_Symbol] :> Int[ExpandIntegrand[F^v, Px, x], x] /; Fr eeQ[F, x] && PolynomialQ[Px, x] && LinearQ[v, x] && !TrueQ[$UseGamma]
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12
method | result | size |
gosper | \(\frac {144 x^{2} \left (x^{2} \ln \left (5\right )+5 \ln \left (5\right )-1\right )^{2} {\mathrm e}^{-4 x}}{25 \ln \left (5\right )^{2}}\) | \(28\) |
risch | \(\frac {\left (144 x^{6} \ln \left (5\right )^{2}+1440 x^{4} \ln \left (5\right )^{2}-288 x^{4} \ln \left (5\right )+3600 x^{2} \ln \left (5\right )^{2}-1440 x^{2} \ln \left (5\right )+144 x^{2}\right ) {\mathrm e}^{-4 x}}{25 \ln \left (5\right )^{2}}\) | \(58\) |
parallelrisch | \(\frac {\left (144 x^{6} \ln \left (5\right )^{2}+1440 x^{4} \ln \left (5\right )^{2}-288 x^{4} \ln \left (5\right )+3600 x^{2} \ln \left (5\right )^{2}-1440 x^{2} \ln \left (5\right )+144 x^{2}\right ) {\mathrm e}^{-4 x}}{25 \ln \left (5\right )^{2}}\) | \(58\) |
meijerg | \(\frac {9 \left (28672 x^{6}+43008 x^{5}+53760 x^{4}+53760 x^{3}+40320 x^{2}+20160 x +5040\right ) {\mathrm e}^{-4 x}}{44800}-\frac {9 \left (6144 x^{5}+7680 x^{4}+7680 x^{3}+5760 x^{2}+2880 x +720\right ) {\mathrm e}^{-4 x}}{6400}+\frac {\left (-\frac {1152 \ln \left (5\right )^{2}}{5}+\frac {1152 \ln \left (5\right )}{25}\right ) \left (24-\frac {\left (1280 x^{4}+1280 x^{3}+960 x^{2}+480 x +120\right ) {\mathrm e}^{-4 x}}{5}\right )}{1024 \ln \left (5\right )^{2}}+\frac {\left (\frac {1152 \ln \left (5\right )^{2}}{5}-\frac {1152 \ln \left (5\right )}{25}\right ) \left (6-\frac {\left (256 x^{3}+192 x^{2}+96 x +24\right ) {\mathrm e}^{-4 x}}{4}\right )}{256 \ln \left (5\right )^{2}}+\frac {\left (-576 \ln \left (5\right )^{2}+\frac {1152 \ln \left (5\right )}{5}-\frac {576}{25}\right ) \left (2-\frac {\left (48 x^{2}+24 x +6\right ) {\mathrm e}^{-4 x}}{3}\right )}{64 \ln \left (5\right )^{2}}+\frac {\left (288 \ln \left (5\right )^{2}-\frac {576 \ln \left (5\right )}{5}+\frac {288}{25}\right ) \left (1-\frac {\left (8 x +2\right ) {\mathrm e}^{-4 x}}{2}\right )}{16 \ln \left (5\right )^{2}}\) | \(221\) |
default | \(\frac {144 \,{\mathrm e}^{-4 x} x^{2}+720 \,{\mathrm e}^{-4 x} x \ln \left (5\right )+180 \ln \left (5\right ) {\mathrm e}^{-4 x}-1800 \,{\mathrm e}^{-4 x} x \ln \left (5\right )^{2}-450 \ln \left (5\right )^{2} {\mathrm e}^{-4 x}+5760 \ln \left (5\right ) \left (-\frac {{\mathrm e}^{-4 x} x^{2}}{4}-\frac {{\mathrm e}^{-4 x} x}{8}-\frac {{\mathrm e}^{-4 x}}{32}\right )-14400 \ln \left (5\right )^{2} \left (-\frac {{\mathrm e}^{-4 x} x^{2}}{4}-\frac {{\mathrm e}^{-4 x} x}{8}-\frac {{\mathrm e}^{-4 x}}{32}\right )-1152 \ln \left (5\right ) \left (-\frac {{\mathrm e}^{-4 x} x^{3}}{4}-\frac {3 \,{\mathrm e}^{-4 x} x^{2}}{16}-\frac {3 \,{\mathrm e}^{-4 x} x}{32}-\frac {3 \,{\mathrm e}^{-4 x}}{128}\right )+5760 \ln \left (5\right )^{2} \left (-\frac {{\mathrm e}^{-4 x} x^{3}}{4}-\frac {3 \,{\mathrm e}^{-4 x} x^{2}}{16}-\frac {3 \,{\mathrm e}^{-4 x} x}{32}-\frac {3 \,{\mathrm e}^{-4 x}}{128}\right )+1152 \ln \left (5\right ) \left (-\frac {{\mathrm e}^{-4 x} x^{4}}{4}-\frac {{\mathrm e}^{-4 x} x^{3}}{4}-\frac {3 \,{\mathrm e}^{-4 x} x^{2}}{16}-\frac {3 \,{\mathrm e}^{-4 x} x}{32}-\frac {3 \,{\mathrm e}^{-4 x}}{128}\right )-5760 \ln \left (5\right )^{2} \left (-\frac {{\mathrm e}^{-4 x} x^{4}}{4}-\frac {{\mathrm e}^{-4 x} x^{3}}{4}-\frac {3 \,{\mathrm e}^{-4 x} x^{2}}{16}-\frac {3 \,{\mathrm e}^{-4 x} x}{32}-\frac {3 \,{\mathrm e}^{-4 x}}{128}\right )+864 \ln \left (5\right )^{2} \left (-\frac {{\mathrm e}^{-4 x} x^{5}}{4}-\frac {5 \,{\mathrm e}^{-4 x} x^{4}}{16}-\frac {5 \,{\mathrm e}^{-4 x} x^{3}}{16}-\frac {15 \,{\mathrm e}^{-4 x} x^{2}}{64}-\frac {15 \,{\mathrm e}^{-4 x} x}{128}-\frac {15 \,{\mathrm e}^{-4 x}}{512}\right )-576 \ln \left (5\right )^{2} \left (-\frac {{\mathrm e}^{-4 x} x^{6}}{4}-\frac {3 \,{\mathrm e}^{-4 x} x^{5}}{8}-\frac {15 \,{\mathrm e}^{-4 x} x^{4}}{32}-\frac {15 \,{\mathrm e}^{-4 x} x^{3}}{32}-\frac {45 \,{\mathrm e}^{-4 x} x^{2}}{128}-\frac {45 \,{\mathrm e}^{-4 x} x}{256}-\frac {45 \,{\mathrm e}^{-4 x}}{1024}\right )}{25 \ln \left (5\right )^{2}}\) | \(398\) |
int(1/25*((-576*x^6+864*x^5-5760*x^4+5760*x^3-14400*x^2+7200*x)*ln(5)^2+(1 152*x^4-1152*x^3+5760*x^2-2880*x)*ln(5)-576*x^2+288*x)/ln(5)^2/exp(x)^4,x, method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).
Time = 0.23 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.84 \[ \int \frac {e^{-4 x} \left (288 x-576 x^2+\left (-2880 x+5760 x^2-1152 x^3+1152 x^4\right ) \log (5)+\left (7200 x-14400 x^2+5760 x^3-5760 x^4+864 x^5-576 x^6\right ) \log ^2(5)\right )}{25 \log ^2(5)} \, dx=\frac {144 \, {\left ({\left (x^{6} + 10 \, x^{4} + 25 \, x^{2}\right )} \log \left (5\right )^{2} + x^{2} - 2 \, {\left (x^{4} + 5 \, x^{2}\right )} \log \left (5\right )\right )} e^{\left (-4 \, x\right )}}{25 \, \log \left (5\right )^{2}} \]
integrate(1/25*((-576*x^6+864*x^5-5760*x^4+5760*x^3-14400*x^2+7200*x)*log( 5)^2+(1152*x^4-1152*x^3+5760*x^2-2880*x)*log(5)-576*x^2+288*x)/log(5)^2/ex p(x)^4,x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (22) = 44\).
Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.52 \[ \int \frac {e^{-4 x} \left (288 x-576 x^2+\left (-2880 x+5760 x^2-1152 x^3+1152 x^4\right ) \log (5)+\left (7200 x-14400 x^2+5760 x^3-5760 x^4+864 x^5-576 x^6\right ) \log ^2(5)\right )}{25 \log ^2(5)} \, dx=\frac {\left (144 x^{6} \log {\left (5 \right )}^{2} - 288 x^{4} \log {\left (5 \right )} + 1440 x^{4} \log {\left (5 \right )}^{2} - 1440 x^{2} \log {\left (5 \right )} + 144 x^{2} + 3600 x^{2} \log {\left (5 \right )}^{2}\right ) e^{- 4 x}}{25 \log {\left (5 \right )}^{2}} \]
integrate(1/25*((-576*x**6+864*x**5-5760*x**4+5760*x**3-14400*x**2+7200*x) *ln(5)**2+(1152*x**4-1152*x**3+5760*x**2-2880*x)*ln(5)-576*x**2+288*x)/ln( 5)**2/exp(x)**4,x)
(144*x**6*log(5)**2 - 288*x**4*log(5) + 1440*x**4*log(5)**2 - 1440*x**2*lo g(5) + 144*x**2 + 3600*x**2*log(5)**2)*exp(-4*x)/(25*log(5)**2)
Leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (22) = 44\).
Time = 0.21 (sec) , antiderivative size = 280, normalized size of antiderivative = 11.20 \[ \int \frac {e^{-4 x} \left (288 x-576 x^2+\left (-2880 x+5760 x^2-1152 x^3+1152 x^4\right ) \log (5)+\left (7200 x-14400 x^2+5760 x^3-5760 x^4+864 x^5-576 x^6\right ) \log ^2(5)\right )}{25 \log ^2(5)} \, dx=\frac {9 \, {\left ({\left (256 \, x^{6} + 384 \, x^{5} + 480 \, x^{4} + 480 \, x^{3} + 360 \, x^{2} + 180 \, x + 45\right )} e^{\left (-4 \, x\right )} \log \left (5\right )^{2} - 3 \, {\left (128 \, x^{5} + 160 \, x^{4} + 160 \, x^{3} + 120 \, x^{2} + 60 \, x + 15\right )} e^{\left (-4 \, x\right )} \log \left (5\right )^{2} + 80 \, {\left (32 \, x^{4} + 32 \, x^{3} + 24 \, x^{2} + 12 \, x + 3\right )} e^{\left (-4 \, x\right )} \log \left (5\right )^{2} - 80 \, {\left (32 \, x^{3} + 24 \, x^{2} + 12 \, x + 3\right )} e^{\left (-4 \, x\right )} \log \left (5\right )^{2} + 800 \, {\left (8 \, x^{2} + 4 \, x + 1\right )} e^{\left (-4 \, x\right )} \log \left (5\right )^{2} - 800 \, {\left (4 \, x + 1\right )} e^{\left (-4 \, x\right )} \log \left (5\right )^{2} - 16 \, {\left (32 \, x^{4} + 32 \, x^{3} + 24 \, x^{2} + 12 \, x + 3\right )} e^{\left (-4 \, x\right )} \log \left (5\right ) + 16 \, {\left (32 \, x^{3} + 24 \, x^{2} + 12 \, x + 3\right )} e^{\left (-4 \, x\right )} \log \left (5\right ) - 320 \, {\left (8 \, x^{2} + 4 \, x + 1\right )} e^{\left (-4 \, x\right )} \log \left (5\right ) + 320 \, {\left (4 \, x + 1\right )} e^{\left (-4 \, x\right )} \log \left (5\right ) + 32 \, {\left (8 \, x^{2} + 4 \, x + 1\right )} e^{\left (-4 \, x\right )} - 32 \, {\left (4 \, x + 1\right )} e^{\left (-4 \, x\right )}\right )}}{400 \, \log \left (5\right )^{2}} \]
integrate(1/25*((-576*x^6+864*x^5-5760*x^4+5760*x^3-14400*x^2+7200*x)*log( 5)^2+(1152*x^4-1152*x^3+5760*x^2-2880*x)*log(5)-576*x^2+288*x)/log(5)^2/ex p(x)^4,x, algorithm=\
9/400*((256*x^6 + 384*x^5 + 480*x^4 + 480*x^3 + 360*x^2 + 180*x + 45)*e^(- 4*x)*log(5)^2 - 3*(128*x^5 + 160*x^4 + 160*x^3 + 120*x^2 + 60*x + 15)*e^(- 4*x)*log(5)^2 + 80*(32*x^4 + 32*x^3 + 24*x^2 + 12*x + 3)*e^(-4*x)*log(5)^2 - 80*(32*x^3 + 24*x^2 + 12*x + 3)*e^(-4*x)*log(5)^2 + 800*(8*x^2 + 4*x + 1)*e^(-4*x)*log(5)^2 - 800*(4*x + 1)*e^(-4*x)*log(5)^2 - 16*(32*x^4 + 32*x ^3 + 24*x^2 + 12*x + 3)*e^(-4*x)*log(5) + 16*(32*x^3 + 24*x^2 + 12*x + 3)* e^(-4*x)*log(5) - 320*(8*x^2 + 4*x + 1)*e^(-4*x)*log(5) + 320*(4*x + 1)*e^ (-4*x)*log(5) + 32*(8*x^2 + 4*x + 1)*e^(-4*x) - 32*(4*x + 1)*e^(-4*x))/log (5)^2
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).
Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \[ \int \frac {e^{-4 x} \left (288 x-576 x^2+\left (-2880 x+5760 x^2-1152 x^3+1152 x^4\right ) \log (5)+\left (7200 x-14400 x^2+5760 x^3-5760 x^4+864 x^5-576 x^6\right ) \log ^2(5)\right )}{25 \log ^2(5)} \, dx=\frac {144 \, {\left (x^{6} \log \left (5\right )^{2} + 10 \, x^{4} \log \left (5\right )^{2} - 2 \, x^{4} \log \left (5\right ) + 25 \, x^{2} \log \left (5\right )^{2} - 10 \, x^{2} \log \left (5\right ) + x^{2}\right )} e^{\left (-4 \, x\right )}}{25 \, \log \left (5\right )^{2}} \]
integrate(1/25*((-576*x^6+864*x^5-5760*x^4+5760*x^3-14400*x^2+7200*x)*log( 5)^2+(1152*x^4-1152*x^3+5760*x^2-2880*x)*log(5)-576*x^2+288*x)/log(5)^2/ex p(x)^4,x, algorithm=\
144/25*(x^6*log(5)^2 + 10*x^4*log(5)^2 - 2*x^4*log(5) + 25*x^2*log(5)^2 - 10*x^2*log(5) + x^2)*e^(-4*x)/log(5)^2
Time = 10.97 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.00 \[ \int \frac {e^{-4 x} \left (288 x-576 x^2+\left (-2880 x+5760 x^2-1152 x^3+1152 x^4\right ) \log (5)+\left (7200 x-14400 x^2+5760 x^3-5760 x^4+864 x^5-576 x^6\right ) \log ^2(5)\right )}{25 \log ^2(5)} \, dx=\frac {144\,x^6\,{\mathrm {e}}^{-4\,x}}{25}+\frac {288\,x^4\,{\mathrm {e}}^{-4\,x}\,\left (5\,\ln \left (5\right )-1\right )}{25\,\ln \left (5\right )}+\frac {144\,x^2\,{\mathrm {e}}^{-4\,x}\,{\left (5\,\ln \left (5\right )-1\right )}^2}{25\,{\ln \left (5\right )}^2} \]
int((exp(-4*x)*((288*x)/25 + (log(5)^2*(7200*x - 14400*x^2 + 5760*x^3 - 57 60*x^4 + 864*x^5 - 576*x^6))/25 - (log(5)*(2880*x - 5760*x^2 + 1152*x^3 - 1152*x^4))/25 - (576*x^2)/25))/log(5)^2,x)