Integrand size = 45, antiderivative size = 28 \[ \int \frac {-8 x^4-2 x^5+\left (x+x^2\right ) \log (1+x)+\left (-2-3 x-x^2\right ) \log ^2(1+x)}{2 x^5} \, dx=-x+\log \left (\frac {4}{x^4}\right )+\frac {(1+x)^2 \log ^2(1+x)}{4 x^4} \] Output:
1/4*(1+x)^2/x^4*ln(1+x)^2+ln(4/x^4)-x
Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-8 x^4-2 x^5+\left (x+x^2\right ) \log (1+x)+\left (-2-3 x-x^2\right ) \log ^2(1+x)}{2 x^5} \, dx=\frac {-4 x^5-16 x^4 \log (x)+(1+x)^2 \log ^2(1+x)}{4 x^4} \] Input:
Integrate[(-8*x^4 - 2*x^5 + (x + x^2)*Log[1 + x] + (-2 - 3*x - x^2)*Log[1 + x]^2)/(2*x^5),x]
Output:
(-4*x^5 - 16*x^4*Log[x] + (1 + x)^2*Log[1 + x]^2)/(4*x^4)
Time = 0.57 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {27, 25, 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.
\(\displaystyle \int \frac {-2 x^5-8 x^4+\left (-x^2-3 x-2\right ) \log ^2(x+1)+\left (x^2+x\right ) \log (x+1)}{2 x^5} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int -\frac {2 x^5+8 x^4+\left (x^2+3 x+2\right ) \log ^2(x+1)-\left (x^2+x\right ) \log (x+1)}{x^5}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {2 x^5+8 x^4+\left (x^2+3 x+2\right ) \log ^2(x+1)-\left (x^2+x\right ) \log (x+1)}{x^5}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {(x+1) (x+2) \log ^2(x+1)}{x^5}-\frac {(x+1) \log (x+1)}{x^4}+\frac {2 (x+4)}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {\log ^2(x+1)}{2 x^4}+\frac {\log ^2(x+1)}{x^3}+\frac {\log ^2(x+1)}{2 x^2}-2 x-8 \log (x)\right )\) |
Input:
Int[(-8*x^4 - 2*x^5 + (x + x^2)*Log[1 + x] + (-2 - 3*x - x^2)*Log[1 + x]^2 )/(2*x^5),x]
Output:
(-2*x - 8*Log[x] + Log[1 + x]^2/(2*x^4) + Log[1 + x]^2/x^3 + Log[1 + x]^2/ (2*x^2))/2
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 = 0.72 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {\left (x^{2}+2 x +1\right ) \ln \left (1+x \right )^{2}}{4 x^{4}}-x -4 \ln \left (x \right )\) | \(28\) |
norman | \(\frac {-x^{5}+\frac {\ln \left (1+x \right )^{2}}{4}+\frac {\ln \left (1+x \right )^{2} x}{2}+\frac {\ln \left (1+x \right )^{2} x^{2}}{4}}{x^{4}}-4 \ln \left (x \right )\) | \(44\) |
parallelrisch | \(-\frac {16 x^{4} \ln \left (x \right )+4 x^{5}-4 x^{4}-\ln \left (1+x \right )^{2} x^{2}-2 \ln \left (1+x \right )^{2} x -\ln \left (1+x \right )^{2}}{4 x^{4}}\) | \(52\) |
parts | \(-x -4 \ln \left (x \right )-\frac {1}{12 x^{2}}-\frac {1}{12 x}-\frac {\ln \left (1+x \right ) \left (1+x \right ) \left (\left (1+x \right )^{2}-3 x \right )}{6 x^{3}}+\frac {\ln \left (1+x \right ) \left (1+x \right ) \left (-1+x \right )}{4 x^{2}}+\frac {\frac {x^{2}}{12}+\frac {x^{3}}{12}+\frac {\ln \left (1+x \right )^{2}}{4}+\frac {\ln \left (1+x \right ) x}{6}+\frac {\ln \left (1+x \right )^{2} x}{2}+\frac {\ln \left (1+x \right ) x^{2}}{4}+\frac {\ln \left (1+x \right )^{2} x^{2}}{4}-\frac {x^{4} \ln \left (1+x \right )}{12}}{x^{4}}\) | \(123\) |
derivativedivides | \(-\frac {1}{12 x}-4 \ln \left (x \right )+\frac {\ln \left (1+x \right ) \left (1+x \right ) \left (-1+x \right )}{4 x^{2}}-\frac {1}{12 x^{2}}-\frac {\ln \left (1+x \right ) \left (1+x \right ) \left (\left (1+x \right )^{2}-3 x \right )}{6 x^{3}}-1-x +\frac {-\frac {\left (1+x \right )^{2}}{3}+\frac {1}{6}+\frac {x}{6}+\frac {\left (1+x \right )^{3}}{6}+\frac {2 \left (1+x \right )^{3} \ln \left (1+x \right )}{3}-\frac {\left (1+x \right )^{4} \ln \left (1+x \right )}{6}-\frac {\ln \left (1+x \right ) \left (1+x \right )^{2}}{2}+\frac {\ln \left (1+x \right )^{2} \left (1+x \right )^{2}}{2}}{2 x^{4}}\) | \(126\) |
default | \(-\frac {1}{12 x}-4 \ln \left (x \right )+\frac {\ln \left (1+x \right ) \left (1+x \right ) \left (-1+x \right )}{4 x^{2}}-\frac {1}{12 x^{2}}-\frac {\ln \left (1+x \right ) \left (1+x \right ) \left (\left (1+x \right )^{2}-3 x \right )}{6 x^{3}}-1-x +\frac {-\frac {\left (1+x \right )^{2}}{3}+\frac {1}{6}+\frac {x}{6}+\frac {\left (1+x \right )^{3}}{6}+\frac {2 \left (1+x \right )^{3} \ln \left (1+x \right )}{3}-\frac {\left (1+x \right )^{4} \ln \left (1+x \right )}{6}-\frac {\ln \left (1+x \right ) \left (1+x \right )^{2}}{2}+\frac {\ln \left (1+x \right )^{2} \left (1+x \right )^{2}}{2}}{2 x^{4}}\) | \(126\) |
Input:
int(1/2*((-x^2-3*x-2)*ln(1+x)^2+(x^2+x)*ln(1+x)-2*x^5-8*x^4)/x^5,x,method= _RETURNVERBOSE)
Output:
1/4*(x^2+2*x+1)/x^4*ln(1+x)^2-x-4*ln(x)
Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {-8 x^4-2 x^5+\left (x+x^2\right ) \log (1+x)+\left (-2-3 x-x^2\right ) \log ^2(1+x)}{2 x^5} \, dx=-\frac {4 \, x^{5} + 16 \, x^{4} \log \left (x\right ) - {\left (x^{2} + 2 \, x + 1\right )} \log \left (x + 1\right )^{2}}{4 \, x^{4}} \] Input:
integrate(1/2*((-x^2-3*x-2)*log(1+x)^2+(x^2+x)*log(1+x)-2*x^5-8*x^4)/x^5,x , algorithm="fricas")
Output:
-1/4*(4*x^5 + 16*x^4*log(x) - (x^2 + 2*x + 1)*log(x + 1)^2)/x^4
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {-8 x^4-2 x^5+\left (x+x^2\right ) \log (1+x)+\left (-2-3 x-x^2\right ) \log ^2(1+x)}{2 x^5} \, dx=- x - 4 \log {\left (x \right )} + \frac {\left (x^{2} + 2 x + 1\right ) \log {\left (x + 1 \right )}^{2}}{4 x^{4}} \] Input:
integrate(1/2*((-x**2-3*x-2)*ln(1+x)**2+(x**2+x)*ln(1+x)-2*x**5-8*x**4)/x* *5,x)
Output:
-x - 4*log(x) + (x**2 + 2*x + 1)*log(x + 1)**2/(4*x**4)
Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (26) = 52\).
Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 3.32 \[ \int \frac {-8 x^4-2 x^5+\left (x+x^2\right ) \log (1+x)+\left (-2-3 x-x^2\right ) \log ^2(1+x)}{2 x^5} \, dx=-x + \frac {2 \, x - 1}{12 \, x^{2}} - \frac {1}{4 \, x} - \frac {\log \left (x + 1\right )}{4 \, x^{2}} - \frac {\log \left (x + 1\right )}{6 \, x^{3}} + \frac {x^{3} + 3 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (x + 1\right )^{2} + x^{2} - {\left (x^{4} - 3 \, x^{2} - 2 \, x\right )} \log \left (x + 1\right )}{12 \, x^{4}} + \frac {1}{12} \, \log \left (x + 1\right ) - 4 \, \log \left (x\right ) \] Input:
integrate(1/2*((-x^2-3*x-2)*log(1+x)^2+(x^2+x)*log(1+x)-2*x^5-8*x^4)/x^5,x , algorithm="maxima")
Output:
-x + 1/12*(2*x - 1)/x^2 - 1/4/x - 1/4*log(x + 1)/x^2 - 1/6*log(x + 1)/x^3 + 1/12*(x^3 + 3*(x^2 + 2*x + 1)*log(x + 1)^2 + x^2 - (x^4 - 3*x^2 - 2*x)*l og(x + 1))/x^4 + 1/12*log(x + 1) - 4*log(x)
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {-8 x^4-2 x^5+\left (x+x^2\right ) \log (1+x)+\left (-2-3 x-x^2\right ) \log ^2(1+x)}{2 x^5} \, dx=-x + \frac {{\left (x^{2} + 2 \, x + 1\right )} \log \left (x + 1\right )^{2}}{4 \, x^{4}} - 4 \, \log \left (x\right ) \] Input:
integrate(1/2*((-x^2-3*x-2)*log(1+x)^2+(x^2+x)*log(1+x)-2*x^5-8*x^4)/x^5,x , algorithm="giac")
Output:
-x + 1/4*(x^2 + 2*x + 1)*log(x + 1)^2/x^4 - 4*log(x)
Time = 7.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {-8 x^4-2 x^5+\left (x+x^2\right ) \log (1+x)+\left (-2-3 x-x^2\right ) \log ^2(1+x)}{2 x^5} \, dx=\frac {\frac {x^2\,{\ln \left (x+1\right )}^2}{4}+\frac {x\,{\ln \left (x+1\right )}^2}{2}+\frac {{\ln \left (x+1\right )}^2}{4}}{x^4}-4\,\ln \left (x\right )-x \] Input:
int(-((log(x + 1)^2*(3*x + x^2 + 2))/2 - (log(x + 1)*(x + x^2))/2 + 4*x^4 + x^5)/x^5,x)
Output:
((x*log(x + 1)^2)/2 + log(x + 1)^2/4 + (x^2*log(x + 1)^2)/4)/x^4 - 4*log(x ) - x
Time = 0.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {-8 x^4-2 x^5+\left (x+x^2\right ) \log (1+x)+\left (-2-3 x-x^2\right ) \log ^2(1+x)}{2 x^5} \, dx=\frac {\mathrm {log}\left (x +1\right )^{2} x^{2}+2 \mathrm {log}\left (x +1\right )^{2} x +\mathrm {log}\left (x +1\right )^{2}-16 \,\mathrm {log}\left (x \right ) x^{4}-4 x^{5}}{4 x^{4}} \] Input:
int(1/2*((-x^2-3*x-2)*log(1+x)^2+(x^2+x)*log(1+x)-2*x^5-8*x^4)/x^5,x)
Output:
(log(x + 1)**2*x**2 + 2*log(x + 1)**2*x + log(x + 1)**2 - 16*log(x)*x**4 - 4*x**5)/(4*x**4)