Integrand size = 65, antiderivative size = 19 \[ \int \frac {4+6 x+2 x^2+(2+2 x) \log ^2(2)+\left (4+2 x+2 \log ^2(2)\right ) \log \left (\frac {25 x^2+50 x \log ^2(2)+25 \log ^4(2)}{e^{10}}\right )}{x+\log ^2(2)} \, dx=\left (1+x+\log \left (\frac {25 \left (x+\log ^2(2)\right )^2}{e^{10}}\right )\right )^2 \]
Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {4+6 x+2 x^2+(2+2 x) \log ^2(2)+\left (4+2 x+2 \log ^2(2)\right ) \log \left (\frac {25 x^2+50 x \log ^2(2)+25 \log ^4(2)}{e^{10}}\right )}{x+\log ^2(2)} \, dx=\left (-9+x+\log (25)+\log \left (\left (x+\log ^2(2)\right )^2\right )\right )^2 \]
Integrate[(4 + 6*x + 2*x^2 + (2 + 2*x)*Log[2]^2 + (4 + 2*x + 2*Log[2]^2)*L og[(25*x^2 + 50*x*Log[2]^2 + 25*Log[2]^4)/E^10])/(x + Log[2]^2),x]
Time = 0.35 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {7239, 27, 25, 7237}
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.
\(\displaystyle \int \frac {2 x^2+\left (2 x+4+2 \log ^2(2)\right ) \log \left (\frac {25 x^2+50 x \log ^2(2)+25 \log ^4(2)}{e^{10}}\right )+6 x+(2 x+2) \log ^2(2)+4}{x+\log ^2(2)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 \left (x+2+\log ^2(2)\right ) \left (x+\log \left (\left (x+\log ^2(2)\right )^2\right )-9 \left (1-\frac {2 \log (5)}{9}\right )\right )}{x+\log ^2(2)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int -\frac {\left (x+\log ^2(2)+2\right ) \left (-x-\log \left (\left (x+\log ^2(2)\right )^2\right )-\log (25)+9\right )}{x+\log ^2(2)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {\left (x+\log ^2(2)+2\right ) \left (-x-\log \left (\left (x+\log ^2(2)\right )^2\right )-\log (25)+9\right )}{x+\log ^2(2)}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \left (-x-\log \left (\left (x+\log ^2(2)\right )^2\right )+9-\log (25)\right )^2\) |
Int[(4 + 6*x + 2*x^2 + (2 + 2*x)*Log[2]^2 + (4 + 2*x + 2*Log[2]^2)*Log[(25 *x^2 + 50*x*Log[2]^2 + 25*Log[2]^4)/E^10])/(x + Log[2]^2),x]
3.30.10.3.1 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_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(89\) vs. \(2(23)=46\).
Time = 0.52 (sec) , antiderivative size = 90, normalized size of antiderivative = 4.74
method | result | size |
norman | \(x^{2}+{\ln \left (\left (25 \ln \left (2\right )^{4}+50 x \ln \left (2\right )^{2}+25 x^{2}\right ) {\mathrm e}^{-10}\right )}^{2}+2 \ln \left (\left (25 \ln \left (2\right )^{4}+50 x \ln \left (2\right )^{2}+25 x^{2}\right ) {\mathrm e}^{-10}\right )+2 x +2 x \ln \left (\left (25 \ln \left (2\right )^{4}+50 x \ln \left (2\right )^{2}+25 x^{2}\right ) {\mathrm e}^{-10}\right )\) | \(90\) |
parallelrisch | \(-\ln \left (2\right )^{4}-4 \ln \left (2\right )^{2}+x^{2}+2 \ln \left (25 \left (\ln \left (2\right )^{4}+2 x \ln \left (2\right )^{2}+x^{2}\right ) {\mathrm e}^{-10}\right ) x +{\ln \left (25 \left (\ln \left (2\right )^{4}+2 x \ln \left (2\right )^{2}+x^{2}\right ) {\mathrm e}^{-10}\right )}^{2}+2 x +2 \ln \left (25 \left (\ln \left (2\right )^{4}+2 x \ln \left (2\right )^{2}+x^{2}\right ) {\mathrm e}^{-10}\right )\) | \(93\) |
risch | \(8 \ln \left (5\right ) \ln \left (\ln \left (2\right )^{2}+x \right )+4 x \ln \left (5\right )-4 \ln \left (\ln \left (2\right )^{2}+x \right )^{2}+4 \ln \left (\ln \left (2\right )^{2}+x \right ) \ln \left (\left (\ln \left (2\right )^{4}+2 x \ln \left (2\right )^{2}+x^{2}\right ) {\mathrm e}^{-10}\right )+2 \ln \left (\left (\ln \left (2\right )^{4}+2 x \ln \left (2\right )^{2}+x^{2}\right ) {\mathrm e}^{-10}\right ) x +x^{2}+4 \ln \left (\ln \left (2\right )^{2}+x \right )+2 x\) | \(94\) |
default | \(x^{2}-18 x +2 \left (-2 \ln \left (2\right )^{2}-18\right ) \ln \left (\ln \left (2\right )^{2}+x \right )+4 \ln \left (5\right ) \left (x +2 \ln \left (\ln \left (2\right )^{2}+x \right )\right )+2 x \ln \left (\ln \left (2\right )^{4}+2 x \ln \left (2\right )^{2}+x^{2}\right )+4 \ln \left (2\right )^{2} \ln \left (\ln \left (2\right )^{2}+x \right )+4 \ln \left (\ln \left (2\right )^{2}+x \right ) \ln \left (\ln \left (2\right )^{4}+2 x \ln \left (2\right )^{2}+x^{2}\right )-4 \ln \left (\ln \left (2\right )^{2}+x \right )^{2}\) | \(108\) |
parts | \(-18 x +x^{2}+2 \left (-2 \ln \left (2\right )^{2}+2\right ) \ln \left (\ln \left (2\right )^{2}+x \right )+2 x \ln \left (\ln \left (2\right )^{4}+2 x \ln \left (2\right )^{2}+x^{2}\right )+4 \ln \left (2\right )^{2} \ln \left (\ln \left (2\right )^{2}+x \right )+4 \ln \left (\ln \left (2\right )^{2}+x \right ) \ln \left (\ln \left (2\right )^{4}+2 x \ln \left (2\right )^{2}+x^{2}\right )-4 \ln \left (\ln \left (2\right )^{2}+x \right )^{2}-40 \ln \left (\ln \left (2\right )^{2}+x \right )+4 \ln \left (5\right ) \left (x +2 \ln \left (\ln \left (2\right )^{2}+x \right )\right )\) | \(117\) |
int(((2*ln(2)^2+2*x+4)*ln((25*ln(2)^4+50*x*ln(2)^2+25*x^2)/exp(5)^2)+(2+2* x)*ln(2)^2+2*x^2+6*x+4)/(ln(2)^2+x),x,method=_RETURNVERBOSE)
x^2+ln((25*ln(2)^4+50*x*ln(2)^2+25*x^2)/exp(5)^2)^2+2*ln((25*ln(2)^4+50*x* ln(2)^2+25*x^2)/exp(5)^2)+2*x+2*x*ln((25*ln(2)^4+50*x*ln(2)^2+25*x^2)/exp( 5)^2)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (18) = 36\).
Time = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.84 \[ \int \frac {4+6 x+2 x^2+(2+2 x) \log ^2(2)+\left (4+2 x+2 \log ^2(2)\right ) \log \left (\frac {25 x^2+50 x \log ^2(2)+25 \log ^4(2)}{e^{10}}\right )}{x+\log ^2(2)} \, dx=x^{2} + 2 \, {\left (x + 1\right )} \log \left (25 \, {\left (\log \left (2\right )^{4} + 2 \, x \log \left (2\right )^{2} + x^{2}\right )} e^{\left (-10\right )}\right ) + \log \left (25 \, {\left (\log \left (2\right )^{4} + 2 \, x \log \left (2\right )^{2} + x^{2}\right )} e^{\left (-10\right )}\right )^{2} + 2 \, x \]
integrate(((2*log(2)^2+2*x+4)*log((25*log(2)^4+50*x*log(2)^2+25*x^2)/exp(5 )^2)+(2+2*x)*log(2)^2+2*x^2+6*x+4)/(log(2)^2+x),x, algorithm=\
x^2 + 2*(x + 1)*log(25*(log(2)^4 + 2*x*log(2)^2 + x^2)*e^(-10)) + log(25*( log(2)^4 + 2*x*log(2)^2 + x^2)*e^(-10))^2 + 2*x
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (22) = 44\).
Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 3.74 \[ \int \frac {4+6 x+2 x^2+(2+2 x) \log ^2(2)+\left (4+2 x+2 \log ^2(2)\right ) \log \left (\frac {25 x^2+50 x \log ^2(2)+25 \log ^4(2)}{e^{10}}\right )}{x+\log ^2(2)} \, dx=x^{2} + 2 x \log {\left (\frac {25 x^{2} + 50 x \log {\left (2 \right )}^{2} + 25 \log {\left (2 \right )}^{4}}{e^{10}} \right )} + 2 x + \log {\left (\frac {25 x^{2} + 50 x \log {\left (2 \right )}^{2} + 25 \log {\left (2 \right )}^{4}}{e^{10}} \right )}^{2} + 4 \log {\left (x + \log {\left (2 \right )}^{2} \right )} \]
integrate(((2*ln(2)**2+2*x+4)*ln((25*ln(2)**4+50*x*ln(2)**2+25*x**2)/exp(5 )**2)+(2+2*x)*ln(2)**2+2*x**2+6*x+4)/(ln(2)**2+x),x)
x**2 + 2*x*log((25*x**2 + 50*x*log(2)**2 + 25*log(2)**4)*exp(-10)) + 2*x + log((25*x**2 + 50*x*log(2)**2 + 25*log(2)**4)*exp(-10))**2 + 4*log(x + lo g(2)**2)
Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (18) = 36\).
Time = 0.30 (sec) , antiderivative size = 244, normalized size of antiderivative = 12.84 \[ \int \frac {4+6 x+2 x^2+(2+2 x) \log ^2(2)+\left (4+2 x+2 \log ^2(2)\right ) \log \left (\frac {25 x^2+50 x \log ^2(2)+25 \log ^4(2)}{e^{10}}\right )}{x+\log ^2(2)} \, dx=2 \, \log \left (2\right )^{4} \log \left (\log \left (2\right )^{2} + x\right ) + 2 \, \log \left (2\right )^{2} \log \left (25 \, e^{\left (-10\right )} \log \left (2\right )^{4} + 50 \, x e^{\left (-10\right )} \log \left (2\right )^{2} + 25 \, x^{2} e^{\left (-10\right )}\right ) \log \left (\log \left (2\right )^{2} + x\right ) + 2 \, \log \left (2\right )^{2} \log \left (\log \left (2\right )^{2} + x\right )^{2} - 2 \, {\left (\log \left (2\right )^{2} \log \left (\log \left (2\right )^{2} + x\right ) - x\right )} \log \left (2\right )^{2} + 2 \, {\left (2 \, {\left (\log \left (5\right ) - 5\right )} \log \left (\log \left (2\right )^{2} + x\right ) - \log \left (25 \, e^{\left (-10\right )} \log \left (2\right )^{4} + 50 \, x e^{\left (-10\right )} \log \left (2\right )^{2} + 25 \, x^{2} e^{\left (-10\right )}\right ) \log \left (\log \left (2\right )^{2} + x\right ) + \log \left (\log \left (2\right )^{2} + x\right )^{2}\right )} \log \left (2\right )^{2} - 2 \, x \log \left (2\right )^{2} + x^{2} - 2 \, {\left (\log \left (2\right )^{2} \log \left (\log \left (2\right )^{2} + x\right ) - x\right )} \log \left (25 \, e^{\left (-10\right )} \log \left (2\right )^{4} + 50 \, x e^{\left (-10\right )} \log \left (2\right )^{2} + 25 \, x^{2} e^{\left (-10\right )}\right ) + 8 \, {\left (\log \left (5\right ) - 5\right )} \log \left (\log \left (2\right )^{2} + x\right ) + 4 \, \log \left (\log \left (2\right )^{2} + x\right )^{2} + 2 \, x + 4 \, \log \left (\log \left (2\right )^{2} + x\right ) \]
integrate(((2*log(2)^2+2*x+4)*log((25*log(2)^4+50*x*log(2)^2+25*x^2)/exp(5 )^2)+(2+2*x)*log(2)^2+2*x^2+6*x+4)/(log(2)^2+x),x, algorithm=\
2*log(2)^4*log(log(2)^2 + x) + 2*log(2)^2*log(25*e^(-10)*log(2)^4 + 50*x*e ^(-10)*log(2)^2 + 25*x^2*e^(-10))*log(log(2)^2 + x) + 2*log(2)^2*log(log(2 )^2 + x)^2 - 2*(log(2)^2*log(log(2)^2 + x) - x)*log(2)^2 + 2*(2*(log(5) - 5)*log(log(2)^2 + x) - log(25*e^(-10)*log(2)^4 + 50*x*e^(-10)*log(2)^2 + 2 5*x^2*e^(-10))*log(log(2)^2 + x) + log(log(2)^2 + x)^2)*log(2)^2 - 2*x*log (2)^2 + x^2 - 2*(log(2)^2*log(log(2)^2 + x) - x)*log(25*e^(-10)*log(2)^4 + 50*x*e^(-10)*log(2)^2 + 25*x^2*e^(-10)) + 8*(log(5) - 5)*log(log(2)^2 + x ) + 4*log(log(2)^2 + x)^2 + 2*x + 4*log(log(2)^2 + x)
Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (18) = 36\).
Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 3.21 \[ \int \frac {4+6 x+2 x^2+(2+2 x) \log ^2(2)+\left (4+2 x+2 \log ^2(2)\right ) \log \left (\frac {25 x^2+50 x \log ^2(2)+25 \log ^4(2)}{e^{10}}\right )}{x+\log ^2(2)} \, dx=x^{2} + 2 \, x \log \left (25 \, \log \left (2\right )^{4} + 50 \, x \log \left (2\right )^{2} + 25 \, x^{2}\right ) + \log \left (25 \, \log \left (2\right )^{4} + 50 \, x \log \left (2\right )^{2} + 25 \, x^{2}\right )^{2} - 18 \, x - 36 \, \log \left (\log \left (2\right )^{2} + x\right ) \]
integrate(((2*log(2)^2+2*x+4)*log((25*log(2)^4+50*x*log(2)^2+25*x^2)/exp(5 )^2)+(2+2*x)*log(2)^2+2*x^2+6*x+4)/(log(2)^2+x),x, algorithm=\
x^2 + 2*x*log(25*log(2)^4 + 50*x*log(2)^2 + 25*x^2) + log(25*log(2)^4 + 50 *x*log(2)^2 + 25*x^2)^2 - 18*x - 36*log(log(2)^2 + x)
Time = 0.79 (sec) , antiderivative size = 63, normalized size of antiderivative = 3.32 \[ \int \frac {4+6 x+2 x^2+(2+2 x) \log ^2(2)+\left (4+2 x+2 \log ^2(2)\right ) \log \left (\frac {25 x^2+50 x \log ^2(2)+25 \log ^4(2)}{e^{10}}\right )}{x+\log ^2(2)} \, dx=2\,x+2\,\ln \left ({\left (x+{\ln \left (2\right )}^2\right )}^2\right )+{\ln \left (25\,{\mathrm {e}}^{-10}\,\left (x^2+2\,{\ln \left (2\right )}^2\,x+{\ln \left (2\right )}^4\right )\right )}^2+2\,x\,\ln \left (25\,{\mathrm {e}}^{-10}\,\left (x^2+2\,{\ln \left (2\right )}^2\,x+{\ln \left (2\right )}^4\right )\right )+x^2 \]
int((6*x + log(exp(-10)*(50*x*log(2)^2 + 25*log(2)^4 + 25*x^2))*(2*x + 2*l og(2)^2 + 4) + log(2)^2*(2*x + 2) + 2*x^2 + 4)/(x + log(2)^2),x)