Integrand size = 55, antiderivative size = 33 \[ \int \frac {-1+2 e^x x^2 \log (5)+\left (-2 x^2+e^2 \left (8 x^2+4 x^3\right )\right ) \log (5)-\log \left (\frac {4}{x}\right )}{2 x^2 \log (5)} \, dx=e^x-x+e^2 (2+x)^2+\frac {\log \left (\frac {4}{x}\right )}{2 x \log (5)} \] Output:
exp(x)-x+exp(1)^2*(2+x)^2+1/2*ln(4/x)/ln(5)/x
Time = 0.05 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {-1+2 e^x x^2 \log (5)+\left (-2 x^2+e^2 \left (8 x^2+4 x^3\right )\right ) \log (5)-\log \left (\frac {4}{x}\right )}{2 x^2 \log (5)} \, dx=\frac {-2 \left (1-4 e^2\right ) x \log (5)+2 e^2 x^2 \log (5)+e^x \log (25)+\frac {\log \left (\frac {4}{x}\right )}{x}}{\log (25)} \] Input:
Integrate[(-1 + 2*E^x*x^2*Log[5] + (-2*x^2 + E^2*(8*x^2 + 4*x^3))*Log[5] - Log[4/x])/(2*x^2*Log[5]),x]
Output:
(-2*(1 - 4*E^2)*x*Log[5] + 2*E^2*x^2*Log[5] + E^x*Log[25] + Log[4/x]/x)/Lo g[25]
Time = 0.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, 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.
\(\displaystyle \int \frac {2 e^x x^2 \log (5)+\left (e^2 \left (4 x^3+8 x^2\right )-2 x^2\right ) \log (5)-\log \left (\frac {4}{x}\right )-1}{2 x^2 \log (5)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {-2 e^x \log (5) x^2+\log \left (\frac {4}{x}\right )+2 \left (x^2-2 e^2 \left (x^3+2 x^2\right )\right ) \log (5)+1}{x^2}dx}{2 \log (5)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {-2 e^x \log (5) x^2+\log \left (\frac {4}{x}\right )+2 \left (x^2-2 e^2 \left (x^3+2 x^2\right )\right ) \log (5)+1}{x^2}dx}{2 \log (5)}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {\int \left (\frac {-4 e^2 \log (5) x^3-8 e^2 \log (5) \left (1-\frac {\log (25)}{8 e^2 \log (5)}\right ) x^2+\log \left (\frac {4}{x}\right )+1}{x^2}-2 e^x \log (5)\right )dx}{2 \log (5)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-2 e^2 x^2 \log (5)+2 \left (1-4 e^2\right ) x \log (5)-2 e^x \log (5)-\frac {\log \left (\frac {4}{x}\right )}{x}}{2 \log (5)}\) |
Input:
Int[(-1 + 2*E^x*x^2*Log[5] + (-2*x^2 + E^2*(8*x^2 + 4*x^3))*Log[5] - Log[4 /x])/(2*x^2*Log[5]),x]
Output:
-1/2*(-2*E^x*Log[5] + 2*(1 - 4*E^2)*x*Log[5] - 2*E^2*x^2*Log[5] - Log[4/x] /x)/Log[5]
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 = 0.42 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27
method | result | size |
norman | \(\frac {x^{3} {\mathrm e}^{2}+\left (-1+4 \,{\mathrm e}^{2}\right ) x^{2}+{\mathrm e}^{x} x +\frac {\ln \left (\frac {4}{x}\right )}{2 \ln \left (5\right )}}{x}\) | \(42\) |
default | \(\frac {2 \ln \left (5\right ) {\mathrm e}^{2} x^{2}+8 \,{\mathrm e}^{2} \ln \left (5\right ) x -2 x \ln \left (5\right )+\frac {\ln \left (\frac {4}{x}\right )}{x}+2 \,{\mathrm e}^{x} \ln \left (5\right )}{2 \ln \left (5\right )}\) | \(45\) |
parts | \(-\frac {-\frac {\ln \left (\frac {4}{x}\right )}{x}+\frac {1}{x}}{2 \ln \left (5\right )}+x^{2} {\mathrm e}^{2}+4 \,{\mathrm e}^{2} x -x +\frac {1}{2 \ln \left (5\right ) x}+{\mathrm e}^{x}\) | \(48\) |
parallelrisch | \(\frac {2 \ln \left (5\right ) {\mathrm e}^{2} x^{3}+8 \ln \left (5\right ) {\mathrm e}^{2} x^{2}-2 x^{2} \ln \left (5\right )+2 x \,{\mathrm e}^{x} \ln \left (5\right )+\ln \left (\frac {4}{x}\right )}{2 \ln \left (5\right ) x}\) | \(53\) |
risch | \(-\frac {\ln \left (x \right )}{2 \ln \left (5\right ) x}+\frac {4 \ln \left (5\right ) {\mathrm e}^{2} x^{3}+16 \ln \left (5\right ) {\mathrm e}^{2} x^{2}-4 x^{2} \ln \left (5\right )+4 x \,{\mathrm e}^{x} \ln \left (5\right )+4 \ln \left (2\right )}{4 \ln \left (5\right ) x}\) | \(59\) |
Input:
int(1/2*(-ln(4/x)+2*x^2*ln(5)*exp(x)+((4*x^3+8*x^2)*exp(1)^2-2*x^2)*ln(5)- 1)/x^2/ln(5),x,method=_RETURNVERBOSE)
Output:
(x^3*exp(1)^2+(4*exp(1)^2-1)*x^2+exp(x)*x+1/2/ln(5)*ln(4/x))/x
Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {-1+2 e^x x^2 \log (5)+\left (-2 x^2+e^2 \left (8 x^2+4 x^3\right )\right ) \log (5)-\log \left (\frac {4}{x}\right )}{2 x^2 \log (5)} \, dx=\frac {2 \, x e^{x} \log \left (5\right ) - 2 \, {\left (x^{2} - {\left (x^{3} + 4 \, x^{2}\right )} e^{2}\right )} \log \left (5\right ) + \log \left (\frac {4}{x}\right )}{2 \, x \log \left (5\right )} \] Input:
integrate(1/2*(-log(4/x)+2*x^2*log(5)*exp(x)+((4*x^3+8*x^2)*exp(1)^2-2*x^2 )*log(5)-1)/x^2/log(5),x, algorithm="fricas")
Output:
1/2*(2*x*e^x*log(5) - 2*(x^2 - (x^3 + 4*x^2)*e^2)*log(5) + log(4/x))/(x*lo g(5))
Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {-1+2 e^x x^2 \log (5)+\left (-2 x^2+e^2 \left (8 x^2+4 x^3\right )\right ) \log (5)-\log \left (\frac {4}{x}\right )}{2 x^2 \log (5)} \, dx=x^{2} e^{2} + x \left (-1 + 4 e^{2}\right ) + e^{x} + \frac {\log {\left (\frac {4}{x} \right )}}{2 x \log {\left (5 \right )}} \] Input:
integrate(1/2*(-ln(4/x)+2*x**2*ln(5)*exp(x)+((4*x**3+8*x**2)*exp(1)**2-2*x **2)*ln(5)-1)/x**2/ln(5),x)
Output:
x**2*exp(2) + x*(-1 + 4*exp(2)) + exp(x) + log(4/x)/(2*x*log(5))
Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {-1+2 e^x x^2 \log (5)+\left (-2 x^2+e^2 \left (8 x^2+4 x^3\right )\right ) \log (5)-\log \left (\frac {4}{x}\right )}{2 x^2 \log (5)} \, dx=\frac {2 \, x^{2} e^{2} \log \left (5\right ) + 8 \, x e^{2} \log \left (5\right ) - 2 \, x \log \left (5\right ) + 2 \, e^{x} \log \left (5\right ) + \frac {\log \left (\frac {4}{x}\right )}{x}}{2 \, \log \left (5\right )} \] Input:
integrate(1/2*(-log(4/x)+2*x^2*log(5)*exp(x)+((4*x^3+8*x^2)*exp(1)^2-2*x^2 )*log(5)-1)/x^2/log(5),x, algorithm="maxima")
Output:
1/2*(2*x^2*e^2*log(5) + 8*x*e^2*log(5) - 2*x*log(5) + 2*e^x*log(5) + log(4 /x)/x)/log(5)
Time = 0.12 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.52 \[ \int \frac {-1+2 e^x x^2 \log (5)+\left (-2 x^2+e^2 \left (8 x^2+4 x^3\right )\right ) \log (5)-\log \left (\frac {4}{x}\right )}{2 x^2 \log (5)} \, dx=\frac {2 \, x^{3} e^{2} \log \left (5\right ) + 8 \, x^{2} e^{2} \log \left (5\right ) - 2 \, x^{2} \log \left (5\right ) + 2 \, x e^{x} \log \left (5\right ) + 2 \, \log \left (2\right ) - \log \left (x\right )}{2 \, x \log \left (5\right )} \] Input:
integrate(1/2*(-log(4/x)+2*x^2*log(5)*exp(x)+((4*x^3+8*x^2)*exp(1)^2-2*x^2 )*log(5)-1)/x^2/log(5),x, algorithm="giac")
Output:
1/2*(2*x^3*e^2*log(5) + 8*x^2*e^2*log(5) - 2*x^2*log(5) + 2*x*e^x*log(5) + 2*log(2) - log(x))/(x*log(5))
Time = 3.98 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {-1+2 e^x x^2 \log (5)+\left (-2 x^2+e^2 \left (8 x^2+4 x^3\right )\right ) \log (5)-\log \left (\frac {4}{x}\right )}{2 x^2 \log (5)} \, dx=x^2\,{\mathrm {e}}^2+x\,\left (4\,{\mathrm {e}}^2-1\right )+\frac {{\mathrm {e}}^x\,\ln \left (25\right )}{2\,\ln \left (5\right )}+\frac {\ln \left (\frac {4}{x}\right )}{2\,x\,\ln \left (5\right )} \] Input:
int(-(log(4/x)/2 - (log(5)*(exp(2)*(8*x^2 + 4*x^3) - 2*x^2))/2 - x^2*exp(x )*log(5) + 1/2)/(x^2*log(5)),x)
Output:
x^2*exp(2) + x*(4*exp(2) - 1) + (exp(x)*log(25))/(2*log(5)) + log(4/x)/(2* x*log(5))
Time = 0.18 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.55 \[ \int \frac {-1+2 e^x x^2 \log (5)+\left (-2 x^2+e^2 \left (8 x^2+4 x^3\right )\right ) \log (5)-\log \left (\frac {4}{x}\right )}{2 x^2 \log (5)} \, dx=\frac {2 e^{x} \mathrm {log}\left (5\right ) x +\mathrm {log}\left (\frac {4}{x}\right )+2 \,\mathrm {log}\left (5\right ) e^{2} x^{3}+8 \,\mathrm {log}\left (5\right ) e^{2} x^{2}-2 \,\mathrm {log}\left (5\right ) x^{2}}{2 \,\mathrm {log}\left (5\right ) x} \] Input:
int(1/2*(-log(4/x)+2*x^2*log(5)*exp(x)+((4*x^3+8*x^2)*exp(1)^2-2*x^2)*log( 5)-1)/x^2/log(5),x)
Output:
(2*e**x*log(5)*x + log(4/x) + 2*log(5)*e**2*x**3 + 8*log(5)*e**2*x**2 - 2* log(5)*x**2)/(2*log(5)*x)