\(\int \frac {e (3+3 x)}{25 e^3 x^2+10 e^4 x^2 \log (\frac {8 e^{\frac {1}{x}}}{x})+e^5 x^2 \log ^2(\frac {8 e^{\frac {1}{x}}}{x})} \, dx\) [3207]

Optimal result

Integrand size = 57, antiderivative size = 22 \[ \int \frac {e (3+3 x)}{25 e^3 x^2+10 e^4 x^2 \log \left (\frac {8 e^{\frac {1}{x}}}{x}\right )+e^5 x^2 \log ^2\left (\frac {8 e^{\frac {1}{x}}}{x}\right )} \, dx=\frac {3}{e^3 \left (5+e \log \left (\frac {8 e^{\frac {1}{x}}}{x}\right )\right )} \]

[Out]

3/(5+ln(8*exp(1/x)/x)*exp(1))/exp(3)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {12, 6820, 6818} \[ \int \frac {e (3+3 x)}{25 e^3 x^2+10 e^4 x^2 \log \left (\frac {8 e^{\frac {1}{x}}}{x}\right )+e^5 x^2 \log ^2\left (\frac {8 e^{\frac {1}{x}}}{x}\right )} \, dx=\frac {3}{e^3 \left (e \log \left (\frac {8 e^{\frac {1}{x}}}{x}\right )+5\right )} \]

[In]

Int[(E*(3 + 3*x))/(25*E^3*x^2 + 10*E^4*x^2*Log[(8*E^x^(-1))/x] + E^5*x^2*Log[(8*E^x^(-1))/x]^2),x]

[Out]

3/(E^3*(5 + E*Log[(8*E^x^(-1))/x]))

Rule 12

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

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = e \int \frac {3+3 x}{25 e^3 x^2+10 e^4 x^2 \log \left (\frac {8 e^{\frac {1}{x}}}{x}\right )+e^5 x^2 \log ^2\left (\frac {8 e^{\frac {1}{x}}}{x}\right )} \, dx \\ & = e \int \frac {3 (1+x)}{e^3 x^2 \left (5+e \log \left (\frac {8 e^{\frac {1}{x}}}{x}\right )\right )^2} \, dx \\ & = \frac {3 \int \frac {1+x}{x^2 \left (5+e \log \left (\frac {8 e^{\frac {1}{x}}}{x}\right )\right )^2} \, dx}{e^2} \\ & = \frac {3}{e^3 \left (5+e \log \left (\frac {8 e^{\frac {1}{x}}}{x}\right )\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e (3+3 x)}{25 e^3 x^2+10 e^4 x^2 \log \left (\frac {8 e^{\frac {1}{x}}}{x}\right )+e^5 x^2 \log ^2\left (\frac {8 e^{\frac {1}{x}}}{x}\right )} \, dx=\frac {3}{e^3 \left (5+e \log \left (\frac {8 e^{\frac {1}{x}}}{x}\right )\right )} \]

[In]

Integrate[(E*(3 + 3*x))/(25*E^3*x^2 + 10*E^4*x^2*Log[(8*E^x^(-1))/x] + E^5*x^2*Log[(8*E^x^(-1))/x]^2),x]

[Out]

3/(E^3*(5 + E*Log[(8*E^x^(-1))/x]))

Maple [A] (verified)

Time = 3.51 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64

method	result	size
norman	\(-\frac {3 \,{\mathrm e}^{-3} {\mathrm e} \ln \left (\frac {8 \,{\mathrm e}^{\frac {1}{x}}}{x}\right )}{5 \left (5+\ln \left (\frac {8 \,{\mathrm e}^{\frac {1}{x}}}{x}\right ) {\mathrm e}\right )}\)	\(36\)
parallelrisch	\(-\frac {3 \,{\mathrm e}^{-3} {\mathrm e} \ln \left (\frac {8 \,{\mathrm e}^{\frac {1}{x}}}{x}\right )}{5 \left (5+\ln \left (\frac {8 \,{\mathrm e}^{\frac {1}{x}}}{x}\right ) {\mathrm e}\right )}\)	\(36\)
default	\(-\frac {3 \,{\mathrm e}^{-3} x}{x \,{\mathrm e} \ln \left (x \right )-{\mathrm e} x \left (\ln \left (\frac {8 \,{\mathrm e}^{\frac {1}{x}}}{x}\right )-\frac {1}{x}+\ln \left (x \right )\right )-{\mathrm e}-5 x}\)	\(47\)
risch	\(-\frac {6 i {\mathrm e}^{-3}}{\pi \,{\mathrm e} \,\operatorname {csgn}\left (i {\mathrm e}^{\frac {1}{x}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {1}{x}}}{x}\right )^{2}-\pi \,{\mathrm e} \,\operatorname {csgn}\left (i {\mathrm e}^{\frac {1}{x}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {1}{x}}}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )-\pi \,{\mathrm e} \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {1}{x}}}{x}\right )^{3}+\pi \,{\mathrm e} \operatorname {csgn}\left (\frac {i {\mathrm e}^{\frac {1}{x}}}{x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right )-6 i {\mathrm e} \ln \left (2\right )+2 i {\mathrm e} \ln \left (x \right )-2 i {\mathrm e} \ln \left ({\mathrm e}^{\frac {1}{x}}\right )-10 i}\)	\(133\)

[In]

int((3*x+3)*exp(1)/(x^2*exp(1)^2*exp(3)*ln(8*exp(1/x)/x)^2+10*x^2*exp(1)*exp(3)*ln(8*exp(1/x)/x)+25*x^2*exp(3)
),x,method=_RETURNVERBOSE)

[Out]

-3/5/exp(3)*exp(1)*ln(8*exp(1/x)/x)/(5+ln(8*exp(1/x)/x)*exp(1))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e (3+3 x)}{25 e^3 x^2+10 e^4 x^2 \log \left (\frac {8 e^{\frac {1}{x}}}{x}\right )+e^5 x^2 \log ^2\left (\frac {8 e^{\frac {1}{x}}}{x}\right )} \, dx=\frac {3}{e^{4} \log \left (\frac {8 \, e^{\frac {1}{x}}}{x}\right ) + 5 \, e^{3}} \]

[In]

integrate((3*x+3)*exp(1)/(x^2*exp(1)^2*exp(3)*log(8*exp(1/x)/x)^2+10*x^2*exp(1)*exp(3)*log(8*exp(1/x)/x)+25*x^
2*exp(3)),x, algorithm="fricas")

[Out]

3/(e^4*log(8*e^(1/x)/x) + 5*e^3)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e (3+3 x)}{25 e^3 x^2+10 e^4 x^2 \log \left (\frac {8 e^{\frac {1}{x}}}{x}\right )+e^5 x^2 \log ^2\left (\frac {8 e^{\frac {1}{x}}}{x}\right )} \, dx=\frac {3}{e^{4} \log {\left (\frac {8 e^{\frac {1}{x}}}{x} \right )} + 5 e^{3}} \]

[In]

integrate((3*x+3)*exp(1)/(x**2*exp(1)**2*exp(3)*ln(8*exp(1/x)/x)**2+10*x**2*exp(1)*exp(3)*ln(8*exp(1/x)/x)+25*
x**2*exp(3)),x)

[Out]

3/(exp(4)*log(8*exp(1/x)/x) + 5*exp(3))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {e (3+3 x)}{25 e^3 x^2+10 e^4 x^2 \log \left (\frac {8 e^{\frac {1}{x}}}{x}\right )+e^5 x^2 \log ^2\left (\frac {8 e^{\frac {1}{x}}}{x}\right )} \, dx=-\frac {3 \, x e}{x e^{5} \log \left (x\right ) - {\left (3 \, e^{5} \log \left (2\right ) + 5 \, e^{4}\right )} x - e^{5}} \]

[In]

integrate((3*x+3)*exp(1)/(x^2*exp(1)^2*exp(3)*log(8*exp(1/x)/x)^2+10*x^2*exp(1)*exp(3)*log(8*exp(1/x)/x)+25*x^
2*exp(3)),x, algorithm="maxima")

[Out]

-3*x*e/(x*e^5*log(x) - (3*e^5*log(2) + 5*e^4)*x - e^5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {e (3+3 x)}{25 e^3 x^2+10 e^4 x^2 \log \left (\frac {8 e^{\frac {1}{x}}}{x}\right )+e^5 x^2 \log ^2\left (\frac {8 e^{\frac {1}{x}}}{x}\right )} \, dx=\frac {3 \, x e}{3 \, x e^{5} \log \left (2\right ) - x e^{5} \log \left (x\right ) + 5 \, x e^{4} + e^{5}} \]

[In]

integrate((3*x+3)*exp(1)/(x^2*exp(1)^2*exp(3)*log(8*exp(1/x)/x)^2+10*x^2*exp(1)*exp(3)*log(8*exp(1/x)/x)+25*x^
2*exp(3)),x, algorithm="giac")

[Out]

3*x*e/(3*x*e^5*log(2) - x*e^5*log(x) + 5*x*e^4 + e^5)

Mupad [B] (verification not implemented)

Time = 9.65 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e (3+3 x)}{25 e^3 x^2+10 e^4 x^2 \log \left (\frac {8 e^{\frac {1}{x}}}{x}\right )+e^5 x^2 \log ^2\left (\frac {8 e^{\frac {1}{x}}}{x}\right )} \, dx=\frac {3\,{\mathrm {e}}^{-4}}{5\,{\mathrm {e}}^{-1}+\ln \left (\frac {8}{x}\right )+\frac {1}{x}} \]

[In]

int((exp(1)*(3*x + 3))/(25*x^2*exp(3) + 10*x^2*exp(4)*log((8*exp(1/x))/x) + x^2*exp(5)*log((8*exp(1/x))/x)^2),
x)

[Out]

(3*exp(-4))/(5*exp(-1) + log(8/x) + 1/x)