\(\int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+(-2 \log (3)+e^{1+x} (-2 x+2 x^2) \log (3)) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx\) [7448]

Optimal result

Integrand size = 74, antiderivative size = 23 \[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=\log (3) \left (x-\left (e^x-\frac {\log (x)}{e x}\right )^2\right ) \]

[Out]

(x-(exp(x)-ln(x)/exp(1)/x)^2)*ln(3)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {12, 14, 2225, 2326, 2341, 2342} \[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=-\frac {\log (3) \log ^2(x)}{e^2 x^2}+\frac {2 e^{x-1} \log (3) \log (x)}{x}-\frac {1}{2} e^{2 x} \log (9)+x \log (3) \]

[In]

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]

[Out]

x*Log[3] - (E^(2*x)*Log[9])/2 + (2*E^(-1 + x)*Log[3]*Log[x])/x - (Log[3]*Log[x]^2)/(E^2*x^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

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]]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2342

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Lo
g[c*x^n])^p/(d*(m + 1))), x] - Dist[b*n*(p/(m + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\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)}{x^3} \, dx}{e^2} \\ & = \frac {\int \left (-e^{2+2 x} \log (9)+\frac {2 e^{1+x} \log (3) (1-\log (x)+x \log (x))}{x^2}+\frac {\log (3) \left (e^2 x^3-2 \log (x)+2 \log ^2(x)\right )}{x^3}\right ) \, dx}{e^2} \\ & = \frac {\log (3) \int \frac {e^2 x^3-2 \log (x)+2 \log ^2(x)}{x^3} \, dx}{e^2}+\frac {(2 \log (3)) \int \frac {e^{1+x} (1-\log (x)+x \log (x))}{x^2} \, dx}{e^2}-\frac {\log (9) \int e^{2+2 x} \, dx}{e^2} \\ & = -\frac {1}{2} e^{2 x} \log (9)+\frac {2 e^{-1+x} \log (3) \log (x)}{x}+\frac {\log (3) \int \left (e^2-\frac {2 \log (x)}{x^3}+\frac {2 \log ^2(x)}{x^3}\right ) \, dx}{e^2} \\ & = x \log (3)-\frac {1}{2} e^{2 x} \log (9)+\frac {2 e^{-1+x} \log (3) \log (x)}{x}-\frac {(2 \log (3)) \int \frac {\log (x)}{x^3} \, dx}{e^2}+\frac {(2 \log (3)) \int \frac {\log ^2(x)}{x^3} \, dx}{e^2} \\ & = \frac {\log (3)}{2 e^2 x^2}+x \log (3)-\frac {1}{2} e^{2 x} \log (9)+\frac {\log (3) \log (x)}{e^2 x^2}+\frac {2 e^{-1+x} \log (3) \log (x)}{x}-\frac {\log (3) \log ^2(x)}{e^2 x^2}+\frac {(2 \log (3)) \int \frac {\log (x)}{x^3} \, dx}{e^2} \\ & = x \log (3)-\frac {1}{2} e^{2 x} \log (9)+\frac {2 e^{-1+x} \log (3) \log (x)}{x}-\frac {\log (3) \log ^2(x)}{e^2 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=-\frac {\log (3) \left (e^{2+2 x}-e^2 x-\frac {2 e^{1+x} \log (x)}{x}+\frac {\log ^2(x)}{x^2}\right )}{e^2} \]

[In]

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]

[Out]

-((Log[3]*(E^(2 + 2*x) - E^2*x - (2*E^(1 + x)*Log[x])/x + Log[x]^2/x^2))/E^2)

Maple [A] (verified)

Time = 0.25 (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	\({\mathrm e}^{-2} \left (\frac {2 \,{\mathrm e} \ln \left (3\right ) {\mathrm e}^{x} \ln \left (x \right )}{x}+x \,{\mathrm e}^{2} \ln \left (3\right )-{\mathrm e}^{2} \ln \left (3\right ) {\mathrm e}^{2 x}+\frac {\ln \left (3\right ) \ln \left (x \right )}{x^{2}}+\frac {\ln \left (3\right )}{2 x^{2}}+2 \ln \left (3\right ) \left (-\frac {\ln \left (x \right )^{2}}{2 x^{2}}-\frac {\ln \left (x \right )}{2 x^{2}}-\frac {1}{4 x^{2}}\right )\right )\)	\(81\)

[In]

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/exp(1)^2,x,method=_RETURNVERBOSE)

[Out]

-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)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).

Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04 \[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=\frac {{\left (x^{3} e^{2} \log \left (3\right ) - x^{2} e^{\left (2 \, x + 2\right )} \log \left (3\right ) + 2 \, x e^{\left (x + 1\right )} \log \left (3\right ) \log \left (x\right ) - \log \left (3\right ) \log \left (x\right )^{2}\right )} e^{\left (-2\right )}}{x^{2}} \]

[In]

integrate((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, algorithm="fricas")

[Out]

(x^3*e^2*log(3) - x^2*e^(2*x + 2)*log(3) + 2*x*e^(x + 1)*log(3)*log(x) - log(3)*log(x)^2)*e^(-2)/x^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (17) = 34\).

Time = 0.14 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.13 \[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=x \log {\left (3 \right )} + \frac {- e x e^{2 x} \log {\left (3 \right )} + 2 e^{x} \log {\left (3 \right )} \log {\left (x \right )}}{e x} - \frac {\log {\left (3 \right )} \log {\left (x \right )}^{2}}{x^{2} e^{2}} \]

[In]

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)

[Out]

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

Maxima [F]

\[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=\int { \frac {{\left (x^{3} e^{2} \log \left (3\right ) - 2 \, x^{3} e^{\left (2 \, x + 2\right )} \log \left (3\right ) + 2 \, x e^{\left (x + 1\right )} \log \left (3\right ) + 2 \, \log \left (3\right ) \log \left (x\right )^{2} + 2 \, {\left ({\left (x^{2} - x\right )} e^{\left (x + 1\right )} \log \left (3\right ) - \log \left (3\right )\right )} \log \left (x\right )\right )} e^{\left (-2\right )}}{x^{3}} \,d x } \]

[In]

integrate((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, algorithm="maxima")

[Out]

1/2*(2*x*e^2*log(3) + 4*e*gamma(-1, -x)*log(3) + (2*log(x)/x^2 + 1/x^2)*log(3) - 2*e^(2*x + 2)*log(3) - 4*inte
grate(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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (23) = 46\).

Time = 0.26 (sec) , antiderivative size = 114, normalized size of antiderivative = 4.96 \[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=\frac {{\left ({\left (x + 1\right )}^{3} e^{2} \log \left (3\right ) - 2 \, {\left (x + 1\right )}^{2} e^{2} \log \left (3\right ) - {\left (x + 1\right )}^{2} e^{\left (2 \, x + 2\right )} \log \left (3\right ) + 2 \, {\left (x + 1\right )} e^{\left (x + 1\right )} \log \left (3\right ) \log \left (x\right ) + {\left (x + 1\right )} e^{2} \log \left (3\right ) + 2 \, {\left (x + 1\right )} e^{\left (2 \, x + 2\right )} \log \left (3\right ) - 2 \, e^{\left (x + 1\right )} \log \left (3\right ) \log \left (x\right ) - \log \left (3\right ) \log \left (x\right )^{2} - e^{\left (2 \, x + 2\right )} \log \left (3\right )\right )} e^{\left (-2\right )}}{{\left (x + 1\right )}^{2} - 2 \, x - 1} \]

[In]

integrate((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, algorithm="giac")

[Out]

((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)

Mupad [B] (verification not implemented)

Time = 13.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74 \[ \int \frac {2 e^{1+x} x \log (3)+e^2 x^3 \log (3)-2 e^{2+2 x} x^3 \log (3)+\left (-2 \log (3)+e^{1+x} \left (-2 x+2 x^2\right ) \log (3)\right ) \log (x)+2 \log (3) \log ^2(x)}{e^2 x^3} \, dx=-\frac {{\mathrm {e}}^{-2}\,\ln \left (3\right )\,\left ({\ln \left (x\right )}^2-x^3\,{\mathrm {e}}^2+x^2\,{\mathrm {e}}^{2\,x+2}-2\,x\,{\mathrm {e}}^{x+1}\,\ln \left (x\right )\right )}{x^2} \]

[In]

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*exp(1)*exp(x)*log(3)))/x^3,x)

[Out]

-(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