Integrand size = 74, antiderivative size = 23 \[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=\log (3) \left (x-\left (e^x-\frac {\log (x)}{e x}\right )^2\right ) \] Output:
(x-(exp(x)-ln(x)/exp(1)/x)^2)*ln(3)
Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=-\frac {\log (3) \left (e^{2+2 x}-e^2 x-\frac {2 e^{1+x} \log (x)}{x}+\frac {\log ^2(x)}{x^2}\right )}{e^2} \] Input:
Integrate[(2*E^(1 + x)*x*Log[3] + E^2*x^3*Log[3] - 2*E^(2 + 2*x)*x^3*Log[3 ] + (-2*Log[3] + E^(1 + x)*(-2*x + 2*x^2)*Log[3])*Log[x] + 2*Log[3]*Log[x] ^2)/(E^2*x^3),x]
Output:
-((Log[3]*(E^(2 + 2*x) - E^2*x - (2*E^(1 + x)*Log[x])/x + Log[x]^2/x^2))/E ^2)
Leaf count is larger than twice the leaf count of optimal. \(52\) vs. \(2(23)=46\).
Time = 0.33 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.26, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {27, 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.
\(\displaystyle \int \frac {-2 e^{2 x+2} x^3 \log (3)+e^2 x^3 \log (3)+\left (e^{x+1} \left (2 x^2-2 x\right ) \log (3)-2 \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)+2 e^{x+1} x \log (3)}{e^2 x^3} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {-2 e^{2 x+2} \log (3) x^3+e^2 \log (3) x^3+2 e^{x+1} \log (3) x+2 \log (3) \log ^2(x)-\left (2 e^{x+1} \log (3) \left (x-x^2\right )+\log (9)\right ) \log (x)}{x^3}dx}{e^2}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {\int \left (\frac {2 e^{x+1} \log (3) (x \log (x)-\log (x)+1)}{x^2}+\frac {e^2 \log (3) x^3+\log (9) \log ^2(x)-\log (9) \log (x)}{x^3}-e^{2 x+2} \log (9)\right )dx}{e^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\log (9) \log ^2(x)}{2 x^2}+\frac {2 e^{x+1} \log (3) \log (x)}{x}-\frac {1}{2} e^{2 x+2} \log (9)+e^2 x \log (3)}{e^2}\) |
Input:
Int[(2*E^(1 + x)*x*Log[3] + E^2*x^3*Log[3] - 2*E^(2 + 2*x)*x^3*Log[3] + (- 2*Log[3] + E^(1 + x)*(-2*x + 2*x^2)*Log[3])*Log[x] + 2*Log[3]*Log[x]^2)/(E ^2*x^3),x]
Output:
(E^2*x*Log[3] - (E^(2 + 2*x)*Log[9])/2 + (2*E^(1 + x)*Log[3]*Log[x])/x - ( Log[9]*Log[x]^2)/(2*x^2))/E^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 = 2.14 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74
method | result | size |
risch | \(-\frac {{\mathrm e}^{-2} \ln \left (3\right ) \ln \left (x \right )^{2}}{x^{2}}+\frac {2 \ln \left (3\right ) \ln \left (x \right ) {\mathrm e}^{-1+x}}{x}-\ln \left (3\right ) {\mathrm e}^{2 x}+x \ln \left (3\right )\) | \(40\) |
parallelrisch | \(\frac {{\mathrm e}^{-2} \left (-x^{2} {\mathrm e}^{2} \ln \left (3\right ) {\mathrm e}^{2 x}+x^{3} {\mathrm e}^{2} \ln \left (3\right )+2 \ln \left (3\right ) {\mathrm e} x \ln \left (x \right ) {\mathrm e}^{x}-\ln \left (3\right ) \ln \left (x \right )^{2}\right )}{x^{2}}\) | \(54\) |
parts | \(\frac {2 \,{\mathrm e}^{-1} \ln \left (3\right ) {\mathrm e}^{x} \ln \left (x \right )}{x}-\ln \left (3\right ) {\mathrm e}^{2 x}-2 \ln \left (3\right ) {\mathrm e}^{-2} \left (-\frac {\ln \left (x \right )}{2 x^{2}}-\frac {1}{4 x^{2}}\right )+2 \ln \left (3\right ) {\mathrm e}^{-2} \left (-\frac {\ln \left (x \right )^{2}}{2 x^{2}}-\frac {\ln \left (x \right )}{2 x^{2}}-\frac {1}{4 x^{2}}\right )+x \ln \left (3\right )\) | \(80\) |
default | \({\mathrm e}^{-2} \left (\frac {2 \,{\mathrm e} \ln \left (3\right ) {\mathrm e}^{x} \ln \left (x \right )}{x}+x \,{\mathrm e}^{2} \ln \left (3\right )-{\mathrm e}^{2} \ln \left (3\right ) {\mathrm e}^{2 x}+\frac {\ln \left (3\right ) \ln \left (x \right )}{x^{2}}+\frac {\ln \left (3\right )}{2 x^{2}}+2 \ln \left (3\right ) \left (-\frac {\ln \left (x \right )^{2}}{2 x^{2}}-\frac {\ln \left (x \right )}{2 x^{2}}-\frac {1}{4 x^{2}}\right )\right )\) | \(81\) |
Input:
int((2*ln(3)*ln(x)^2+((2*x^2-2*x)*exp(1)*ln(3)*exp(x)-2*ln(3))*ln(x)-2*x^3 *exp(1)^2*ln(3)*exp(x)^2+2*x*exp(1)*ln(3)*exp(x)+x^3*exp(1)^2*ln(3))/x^3/e xp(1)^2,x,method=_RETURNVERBOSE)
Output:
-exp(-2)*ln(3)/x^2*ln(x)^2+2*ln(3)*ln(x)/x*exp(-1+x)-ln(3)*exp(2*x)+x*ln(3 )
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04 \[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=\frac {{\left (x^{3} e^{2} \log \left (3\right ) - x^{2} e^{\left (2 \, x + 2\right )} \log \left (3\right ) + 2 \, x e^{\left (x + 1\right )} \log \left (3\right ) \log \left (x\right ) - \log \left (3\right ) \log \left (x\right )^{2}\right )} e^{\left (-2\right )}}{x^{2}} \] Input:
integrate((2*log(3)*log(x)^2+((2*x^2-2*x)*exp(1)*log(3)*exp(x)-2*log(3))*l og(x)-2*x^3*exp(1)^2*log(3)*exp(x)^2+2*x*exp(1)*log(3)*exp(x)+x^3*exp(1)^2 *log(3))/x^3/exp(1)^2,x, algorithm="fricas")
Output:
(x^3*e^2*log(3) - x^2*e^(2*x + 2)*log(3) + 2*x*e^(x + 1)*log(3)*log(x) - l og(3)*log(x)^2)*e^(-2)/x^2
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (17) = 34\).
Time = 0.15 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.13 \[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=x \log {\left (3 \right )} + \frac {- e x e^{2 x} \log {\left (3 \right )} + 2 e^{x} \log {\left (3 \right )} \log {\left (x \right )}}{e x} - \frac {\log {\left (3 \right )} \log {\left (x \right )}^{2}}{x^{2} e^{2}} \] Input:
integrate((2*ln(3)*ln(x)**2+((2*x**2-2*x)*exp(1)*ln(3)*exp(x)-2*ln(3))*ln( x)-2*x**3*exp(1)**2*ln(3)*exp(x)**2+2*x*exp(1)*ln(3)*exp(x)+x**3*exp(1)**2 *ln(3))/x**3/exp(1)**2,x)
Output:
x*log(3) + (-E*x*exp(2*x)*log(3) + 2*exp(x)*log(3)*log(x))*exp(-1)/x - exp (-2)*log(3)*log(x)**2/x**2
\[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=\int { \frac {{\left (x^{3} e^{2} \log \left (3\right ) - 2 \, x^{3} e^{\left (2 \, x + 2\right )} \log \left (3\right ) + 2 \, x e^{\left (x + 1\right )} \log \left (3\right ) + 2 \, \log \left (3\right ) \log \left (x\right )^{2} + 2 \, {\left ({\left (x^{2} - x\right )} e^{\left (x + 1\right )} \log \left (3\right ) - \log \left (3\right )\right )} \log \left (x\right )\right )} e^{\left (-2\right )}}{x^{3}} \,d x } \] Input:
integrate((2*log(3)*log(x)^2+((2*x^2-2*x)*exp(1)*log(3)*exp(x)-2*log(3))*l og(x)-2*x^3*exp(1)^2*log(3)*exp(x)^2+2*x*exp(1)*log(3)*exp(x)+x^3*exp(1)^2 *log(3))/x^3/exp(1)^2,x, algorithm="maxima")
Output:
1/2*(2*x*e^2*log(3) + 4*e*gamma(-1, -x)*log(3) + (2*log(x)/x^2 + 1/x^2)*lo g(3) - 2*e^(2*x + 2)*log(3) - 4*integrate(e^(x + 1)/x^2, x)*log(3) + (4*x* e^(x + 1)*log(3)*log(x) - 2*log(3)*log(x)^2 - 2*log(3)*log(x) - log(3))/x^ 2)*e^(-2)
Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (23) = 46\).
Time = 0.12 (sec) , antiderivative size = 114, normalized size of antiderivative = 4.96 \[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=\frac {{\left ({\left (x + 1\right )}^{3} e^{2} \log \left (3\right ) - 2 \, {\left (x + 1\right )}^{2} e^{2} \log \left (3\right ) - {\left (x + 1\right )}^{2} e^{\left (2 \, x + 2\right )} \log \left (3\right ) + 2 \, {\left (x + 1\right )} e^{\left (x + 1\right )} \log \left (3\right ) \log \left (x\right ) + {\left (x + 1\right )} e^{2} \log \left (3\right ) + 2 \, {\left (x + 1\right )} e^{\left (2 \, x + 2\right )} \log \left (3\right ) - 2 \, e^{\left (x + 1\right )} \log \left (3\right ) \log \left (x\right ) - \log \left (3\right ) \log \left (x\right )^{2} - e^{\left (2 \, x + 2\right )} \log \left (3\right )\right )} e^{\left (-2\right )}}{{\left (x + 1\right )}^{2} - 2 \, x - 1} \] Input:
integrate((2*log(3)*log(x)^2+((2*x^2-2*x)*exp(1)*log(3)*exp(x)-2*log(3))*l og(x)-2*x^3*exp(1)^2*log(3)*exp(x)^2+2*x*exp(1)*log(3)*exp(x)+x^3*exp(1)^2 *log(3))/x^3/exp(1)^2,x, algorithm="giac")
Output:
((x + 1)^3*e^2*log(3) - 2*(x + 1)^2*e^2*log(3) - (x + 1)^2*e^(2*x + 2)*log (3) + 2*(x + 1)*e^(x + 1)*log(3)*log(x) + (x + 1)*e^2*log(3) + 2*(x + 1)*e ^(2*x + 2)*log(3) - 2*e^(x + 1)*log(3)*log(x) - log(3)*log(x)^2 - e^(2*x + 2)*log(3))*e^(-2)/((x + 1)^2 - 2*x - 1)
Time = 3.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74 \[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=-\frac {{\mathrm {e}}^{-2}\,\ln \left (3\right )\,\left ({\ln \left (x\right )}^2-x^3\,{\mathrm {e}}^2+x^2\,{\mathrm {e}}^{2\,x+2}-2\,x\,{\mathrm {e}}^{x+1}\,\ln \left (x\right )\right )}{x^2} \] Input:
int((exp(-2)*(2*log(3)*log(x)^2 - log(x)*(2*log(3) + exp(1)*exp(x)*log(3)* (2*x - 2*x^2)) + x^3*exp(2)*log(3) - 2*x^3*exp(2*x)*exp(2)*log(3) + 2*x*ex p(1)*exp(x)*log(3)))/x^3,x)
Output:
-(exp(-2)*log(3)*(log(x)^2 - x^3*exp(2) + x^2*exp(2*x + 2) - 2*x*exp(x + 1 )*log(x)))/x^2
Time = 0.17 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.96 \[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=\frac {\mathrm {log}\left (3\right ) \left (-e^{2 x} e^{2} x^{2}+2 e^{x} \mathrm {log}\left (x \right ) e x -\mathrm {log}\left (x \right )^{2}+e^{2} x^{3}\right )}{e^{2} x^{2}} \] Input:
int((2*log(3)*log(x)^2+((2*x^2-2*x)*exp(1)*log(3)*exp(x)-2*log(3))*log(x)- 2*x^3*exp(1)^2*log(3)*exp(x)^2+2*x*exp(1)*log(3)*exp(x)+x^3*exp(1)^2*log(3 ))/x^3/exp(1)^2,x)
Output:
(log(3)*( - e**(2*x)*e**2*x**2 + 2*e**x*log(x)*e*x - log(x)**2 + e**2*x**3 ))/(e**2*x**2)