Integrand size = 46, antiderivative size = 28 \[ \int \frac {1}{25} \left (25+\left (900+1050 x+450 x^2+100 x^3+\left (60+60 x+30 x^2\right ) \log (2)+2 x \log ^2(2)\right ) \log ^2(9)\right ) \, dx=x+x^2 \left (x+\frac {3 (2+x)}{x}+\frac {\log (2)}{5}\right )^2 \log ^2(9) \] Output:
4*x^2*(x+1/5*ln(2)+3*(2+x)/x)^2*ln(3)^2+x
Leaf count is larger than twice the leaf count of optimal. \(59\) vs. \(2(28)=56\).
Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.11 \[ \int \frac {1}{25} \left (25+\left (900+1050 x+450 x^2+100 x^3+\left (60+60 x+30 x^2\right ) \log (2)+2 x \log ^2(2)\right ) \log ^2(9)\right ) \, dx=x+x^4 \log ^2(9)+\frac {12}{5} x (15+\log (2)) \log ^2(9)+\frac {2}{5} x^3 (15+\log (2)) \log ^2(9)+\frac {1}{25} x^2 \left (525+30 \log (2)+\log ^2(2)\right ) \log ^2(9) \] Input:
Integrate[(25 + (900 + 1050*x + 450*x^2 + 100*x^3 + (60 + 60*x + 30*x^2)*L og[2] + 2*x*Log[2]^2)*Log[9]^2)/25,x]
Output:
x + x^4*Log[9]^2 + (12*x*(15 + Log[2])*Log[9]^2)/5 + (2*x^3*(15 + Log[2])* Log[9]^2)/5 + (x^2*(525 + 30*Log[2] + Log[2]^2)*Log[9]^2)/25
Leaf count is larger than twice the leaf count of optimal. \(78\) vs. \(2(28)=56\).
Time = 0.40 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.79, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {27, 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.
\(\displaystyle \int \frac {1}{25} \left (\log ^2(9) \left (100 x^3+450 x^2+\left (30 x^2+60 x+60\right ) \log (2)+1050 x+2 x \log ^2(2)+900\right )+25\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{25} \int \left (2 \log ^2(9) \left (50 x^3+225 x^2+\log ^2(2) x+525 x+15 \left (x^2+2 x+2\right ) \log (2)+450\right )+25\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{25} \left (25 x^4 \log ^2(9)+10 x^3 \log (2) \log ^2(9)+150 x^3 \log ^2(9)+x^2 \left (525+\log ^2(2)\right ) \log ^2(9)+30 x^2 \log (2) \log ^2(9)+25 x+60 x \log (2) \log ^2(9)+900 x \log ^2(9)\right )\) |
Input:
Int[(25 + (900 + 1050*x + 450*x^2 + 100*x^3 + (60 + 60*x + 30*x^2)*Log[2] + 2*x*Log[2]^2)*Log[9]^2)/25,x]
Output:
(25*x + 900*x*Log[9]^2 + 150*x^3*Log[9]^2 + 25*x^4*Log[9]^2 + 60*x*Log[2]* Log[9]^2 + 30*x^2*Log[2]*Log[9]^2 + 10*x^3*Log[2]*Log[9]^2 + x^2*(525 + Lo g[2]^2)*Log[9]^2)/25
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.15 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.96
method | result | size |
parallelrisch | \(\frac {4 \ln \left (3\right )^{2} \left (x^{2} \ln \left (2\right )^{2}+10 x^{3} \ln \left (2\right )+25 x^{4}+30 x^{2} \ln \left (2\right )+150 x^{3}+60 x \ln \left (2\right )+525 x^{2}+900 x \right )}{25}+x\) | \(55\) |
gosper | \(\frac {x \left (4 \ln \left (3\right )^{2} \ln \left (2\right )^{2} x +40 \ln \left (3\right )^{2} \ln \left (2\right ) x^{2}+100 x^{3} \ln \left (3\right )^{2}+120 x \ln \left (2\right ) \ln \left (3\right )^{2}+600 x^{2} \ln \left (3\right )^{2}+240 \ln \left (3\right )^{2} \ln \left (2\right )+2100 x \ln \left (3\right )^{2}+3600 \ln \left (3\right )^{2}+25\right )}{25}\) | \(76\) |
norman | \(\left (\frac {8 \ln \left (3\right )^{2} \ln \left (2\right )}{5}+24 \ln \left (3\right )^{2}\right ) x^{3}+\left (\frac {48 \ln \left (3\right )^{2} \ln \left (2\right )}{5}+144 \ln \left (3\right )^{2}+1\right ) x +\left (\frac {4 \ln \left (3\right )^{2} \ln \left (2\right )^{2}}{25}+\frac {24 \ln \left (3\right )^{2} \ln \left (2\right )}{5}+84 \ln \left (3\right )^{2}\right ) x^{2}+4 x^{4} \ln \left (3\right )^{2}\) | \(77\) |
default | \(\frac {4 x^{2} \ln \left (3\right )^{2} \ln \left (2\right )^{2}}{25}+\frac {8 \ln \left (3\right )^{2} \ln \left (2\right ) x^{3}}{5}+4 x^{4} \ln \left (3\right )^{2}+\frac {24 \ln \left (3\right )^{2} \ln \left (2\right ) x^{2}}{5}+24 x^{3} \ln \left (3\right )^{2}+\frac {48 x \ln \left (2\right ) \ln \left (3\right )^{2}}{5}+84 x^{2} \ln \left (3\right )^{2}+144 x \ln \left (3\right )^{2}+x\) | \(81\) |
parts | \(\frac {4 x^{2} \ln \left (3\right )^{2} \ln \left (2\right )^{2}}{25}+\frac {8 \ln \left (3\right )^{2} \ln \left (2\right ) x^{3}}{5}+4 x^{4} \ln \left (3\right )^{2}+\frac {24 \ln \left (3\right )^{2} \ln \left (2\right ) x^{2}}{5}+24 x^{3} \ln \left (3\right )^{2}+\frac {48 x \ln \left (2\right ) \ln \left (3\right )^{2}}{5}+84 x^{2} \ln \left (3\right )^{2}+144 x \ln \left (3\right )^{2}+x\) | \(81\) |
risch | \(\frac {4 x^{2} \ln \left (3\right )^{2} \ln \left (2\right )^{2}}{25}+\frac {8 \ln \left (3\right )^{2} \ln \left (2\right ) x^{3}}{5}+4 x^{4} \ln \left (3\right )^{2}+\frac {24 \ln \left (3\right )^{2} \ln \left (2\right ) x^{2}}{5}+24 x^{3} \ln \left (3\right )^{2}+\frac {48 x \ln \left (2\right ) \ln \left (3\right )^{2}}{5}+84 x^{2} \ln \left (3\right )^{2}+144 x \ln \left (3\right )^{2}+144 \ln \left (3\right )^{2}+x\) | \(87\) |
Input:
int(4/25*(2*x*ln(2)^2+(30*x^2+60*x+60)*ln(2)+100*x^3+450*x^2+1050*x+900)*l n(3)^2+1,x,method=_RETURNVERBOSE)
Output:
4/25*ln(3)^2*(x^2*ln(2)^2+10*x^3*ln(2)+25*x^4+30*x^2*ln(2)+150*x^3+60*x*ln (2)+525*x^2+900*x)+x
Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82 \[ \int \frac {1}{25} \left (25+\left (900+1050 x+450 x^2+100 x^3+\left (60+60 x+30 x^2\right ) \log (2)+2 x \log ^2(2)\right ) \log ^2(9)\right ) \, dx=\frac {4}{25} \, {\left (25 \, x^{4} + x^{2} \log \left (2\right )^{2} + 150 \, x^{3} + 525 \, x^{2} + 10 \, {\left (x^{3} + 3 \, x^{2} + 6 \, x\right )} \log \left (2\right ) + 900 \, x\right )} \log \left (3\right )^{2} + x \] Input:
integrate(4/25*(2*x*log(2)^2+(30*x^2+60*x+60)*log(2)+100*x^3+450*x^2+1050* x+900)*log(3)^2+1,x, algorithm="fricas")
Output:
4/25*(25*x^4 + x^2*log(2)^2 + 150*x^3 + 525*x^2 + 10*(x^3 + 3*x^2 + 6*x)*l og(2) + 900*x)*log(3)^2 + x
Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (26) = 52\).
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.14 \[ \int \frac {1}{25} \left (25+\left (900+1050 x+450 x^2+100 x^3+\left (60+60 x+30 x^2\right ) \log (2)+2 x \log ^2(2)\right ) \log ^2(9)\right ) \, dx=4 x^{4} \log {\left (3 \right )}^{2} + x^{3} \cdot \left (\frac {8 \log {\left (2 \right )} \log {\left (3 \right )}^{2}}{5} + 24 \log {\left (3 \right )}^{2}\right ) + x^{2} \cdot \left (\frac {4 \log {\left (2 \right )}^{2} \log {\left (3 \right )}^{2}}{25} + \frac {24 \log {\left (2 \right )} \log {\left (3 \right )}^{2}}{5} + 84 \log {\left (3 \right )}^{2}\right ) + x \left (1 + \frac {48 \log {\left (2 \right )} \log {\left (3 \right )}^{2}}{5} + 144 \log {\left (3 \right )}^{2}\right ) \] Input:
integrate(4/25*(2*x*ln(2)**2+(30*x**2+60*x+60)*ln(2)+100*x**3+450*x**2+105 0*x+900)*ln(3)**2+1,x)
Output:
4*x**4*log(3)**2 + x**3*(8*log(2)*log(3)**2/5 + 24*log(3)**2) + x**2*(4*lo g(2)**2*log(3)**2/25 + 24*log(2)*log(3)**2/5 + 84*log(3)**2) + x*(1 + 48*l og(2)*log(3)**2/5 + 144*log(3)**2)
Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82 \[ \int \frac {1}{25} \left (25+\left (900+1050 x+450 x^2+100 x^3+\left (60+60 x+30 x^2\right ) \log (2)+2 x \log ^2(2)\right ) \log ^2(9)\right ) \, dx=\frac {4}{25} \, {\left (25 \, x^{4} + x^{2} \log \left (2\right )^{2} + 150 \, x^{3} + 525 \, x^{2} + 10 \, {\left (x^{3} + 3 \, x^{2} + 6 \, x\right )} \log \left (2\right ) + 900 \, x\right )} \log \left (3\right )^{2} + x \] Input:
integrate(4/25*(2*x*log(2)^2+(30*x^2+60*x+60)*log(2)+100*x^3+450*x^2+1050* x+900)*log(3)^2+1,x, algorithm="maxima")
Output:
4/25*(25*x^4 + x^2*log(2)^2 + 150*x^3 + 525*x^2 + 10*(x^3 + 3*x^2 + 6*x)*l og(2) + 900*x)*log(3)^2 + x
Time = 0.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82 \[ \int \frac {1}{25} \left (25+\left (900+1050 x+450 x^2+100 x^3+\left (60+60 x+30 x^2\right ) \log (2)+2 x \log ^2(2)\right ) \log ^2(9)\right ) \, dx=\frac {4}{25} \, {\left (25 \, x^{4} + x^{2} \log \left (2\right )^{2} + 150 \, x^{3} + 525 \, x^{2} + 10 \, {\left (x^{3} + 3 \, x^{2} + 6 \, x\right )} \log \left (2\right ) + 900 \, x\right )} \log \left (3\right )^{2} + x \] Input:
integrate(4/25*(2*x*log(2)^2+(30*x^2+60*x+60)*log(2)+100*x^3+450*x^2+1050* x+900)*log(3)^2+1,x, algorithm="giac")
Output:
4/25*(25*x^4 + x^2*log(2)^2 + 150*x^3 + 525*x^2 + 10*(x^3 + 3*x^2 + 6*x)*l og(2) + 900*x)*log(3)^2 + x
Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.21 \[ \int \frac {1}{25} \left (25+\left (900+1050 x+450 x^2+100 x^3+\left (60+60 x+30 x^2\right ) \log (2)+2 x \log ^2(2)\right ) \log ^2(9)\right ) \, dx=4\,{\ln \left (3\right )}^2\,x^4+\frac {4\,{\ln \left (3\right )}^2\,\left (30\,\ln \left (2\right )+450\right )\,x^3}{75}+\frac {2\,{\ln \left (3\right )}^2\,\left (60\,\ln \left (2\right )+2\,{\ln \left (2\right )}^2+1050\right )\,x^2}{25}+\left (\frac {4\,{\ln \left (3\right )}^2\,\left (60\,\ln \left (2\right )+900\right )}{25}+1\right )\,x \] Input:
int((4*log(3)^2*(1050*x + log(2)*(60*x + 30*x^2 + 60) + 2*x*log(2)^2 + 450 *x^2 + 100*x^3 + 900))/25 + 1,x)
Output:
4*x^4*log(3)^2 + x*((4*log(3)^2*(60*log(2) + 900))/25 + 1) + (4*x^3*log(3) ^2*(30*log(2) + 450))/75 + (2*x^2*log(3)^2*(60*log(2) + 2*log(2)^2 + 1050) )/25
Time = 0.18 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.68 \[ \int \frac {1}{25} \left (25+\left (900+1050 x+450 x^2+100 x^3+\left (60+60 x+30 x^2\right ) \log (2)+2 x \log ^2(2)\right ) \log ^2(9)\right ) \, dx=\frac {x \left (4 \mathrm {log}\left (3\right )^{2} \mathrm {log}\left (2\right )^{2} x +40 \mathrm {log}\left (3\right )^{2} \mathrm {log}\left (2\right ) x^{2}+120 \mathrm {log}\left (3\right )^{2} \mathrm {log}\left (2\right ) x +240 \mathrm {log}\left (3\right )^{2} \mathrm {log}\left (2\right )+100 \mathrm {log}\left (3\right )^{2} x^{3}+600 \mathrm {log}\left (3\right )^{2} x^{2}+2100 \mathrm {log}\left (3\right )^{2} x +3600 \mathrm {log}\left (3\right )^{2}+25\right )}{25} \] Input:
int(4/25*(2*x*log(2)^2+(30*x^2+60*x+60)*log(2)+100*x^3+450*x^2+1050*x+900) *log(3)^2+1,x)
Output:
(x*(4*log(3)**2*log(2)**2*x + 40*log(3)**2*log(2)*x**2 + 120*log(3)**2*log (2)*x + 240*log(3)**2*log(2) + 100*log(3)**2*x**3 + 600*log(3)**2*x**2 + 2 100*log(3)**2*x + 3600*log(3)**2 + 25))/25