\(\int \frac {x^2+e^{\frac {5+x}{x}} (20+10 x) \log (3)+(-2 x-x^2) \log (3)}{(2 x^2+x^3) \log (3)} \, dx\) [6693]

Optimal result

Integrand size = 49, antiderivative size = 28 \[ \int \frac {x^2+e^{\frac {5+x}{x}} (20+10 x) \log (3)+\left (-2 x-x^2\right ) \log (3)}{\left (2 x^2+x^3\right ) \log (3)} \, dx=\frac {3}{8}-2 e^{\frac {5+x}{x}}-\log (x)+\frac {\log (2+x)}{\log (3)} \]

[Out]

ln(2+x)/ln(3)+3/8-ln(x)-2*exp(1/x*(5+x))

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {12, 1607, 6820, 2240} \[ \int \frac {x^2+e^{\frac {5+x}{x}} (20+10 x) \log (3)+\left (-2 x-x^2\right ) \log (3)}{\left (2 x^2+x^3\right ) \log (3)} \, dx=-2 e^{\frac {5}{x}+1}-\log (x)+\frac {\log (x+2)}{\log (3)} \]

[In]

Int[(x^2 + E^((5 + x)/x)*(20 + 10*x)*Log[3] + (-2*x - x^2)*Log[3])/((2*x^2 + x^3)*Log[3]),x]

[Out]

-2*E^(1 + 5/x) - Log[x] + Log[2 + x]/Log[3]

Rule 12

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

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {x^2+e^{\frac {5+x}{x}} (20+10 x) \log (3)+\left (-2 x-x^2\right ) \log (3)}{2 x^2+x^3} \, dx}{\log (3)} \\ & = \frac {\int \frac {x^2+e^{\frac {5+x}{x}} (20+10 x) \log (3)+\left (-2 x-x^2\right ) \log (3)}{x^2 (2+x)} \, dx}{\log (3)} \\ & = \frac {\int \left (\frac {1}{2+x}+\frac {10 e^{1+\frac {5}{x}} \log (3)}{x^2}-\frac {\log (3)}{x}\right ) \, dx}{\log (3)} \\ & = -\log (x)+\frac {\log (2+x)}{\log (3)}+10 \int \frac {e^{1+\frac {5}{x}}}{x^2} \, dx \\ & = -2 e^{1+\frac {5}{x}}-\log (x)+\frac {\log (2+x)}{\log (3)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {x^2+e^{\frac {5+x}{x}} (20+10 x) \log (3)+\left (-2 x-x^2\right ) \log (3)}{\left (2 x^2+x^3\right ) \log (3)} \, dx=-2 e^{1+\frac {5}{x}}-\log (x)+\frac {\log (x \log (3)+\log (9))}{\log (3)} \]

[In]

Integrate[(x^2 + E^((5 + x)/x)*(20 + 10*x)*Log[3] + (-2*x - x^2)*Log[3])/((2*x^2 + x^3)*Log[3]),x]

[Out]

-2*E^(1 + 5/x) - Log[x] + Log[x*Log[3] + Log[9]]/Log[3]

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89

method	result	size
norman	\(-2 \,{\mathrm e}^{\frac {5+x}{x}}+\frac {\ln \left (2+x \right )}{\ln \left (3\right )}-\ln \left (x \right )\)	\(25\)
parts	\(-2 \,{\mathrm e}^{\frac {5+x}{x}}+\frac {\ln \left (2+x \right )}{\ln \left (3\right )}-\ln \left (x \right )\)	\(25\)
risch	\(\frac {\ln \left (-2-x \right )}{\ln \left (3\right )}-\ln \left (x \right )-2 \,{\mathrm e}^{\frac {5+x}{x}}\)	\(27\)
parallelrisch	\(\frac {-\ln \left (3\right ) \ln \left (x \right )-2 \ln \left (3\right ) {\mathrm e}^{\frac {5+x}{x}}+\ln \left (2+x \right )}{\ln \left (3\right )}\)	\(29\)
default	\(\frac {\ln \left (5+\frac {10}{x}\right )-\ln \left (\frac {5}{x}\right )+125 \ln \left (3\right ) \left (-\frac {\ln \left (5+\frac {10}{x}\right )}{125}+\frac {\ln \left (\frac {5}{x}\right )}{125}\right )+\ln \left (3\right ) \ln \left (5+\frac {10}{x}\right )-2 \ln \left (3\right ) {\mathrm e}^{1+\frac {5}{x}}}{\ln \left (3\right )}\)	\(69\)
derivativedivides	\(-\frac {-\ln \left (5+\frac {10}{x}\right )+\ln \left (\frac {5}{x}\right )-125 \ln \left (3\right ) \left (-\frac {\ln \left (5+\frac {10}{x}\right )}{125}+\frac {\ln \left (\frac {5}{x}\right )}{125}\right )-\ln \left (3\right ) \ln \left (5+\frac {10}{x}\right )+2 \ln \left (3\right ) {\mathrm e}^{1+\frac {5}{x}}}{\ln \left (3\right )}\)	\(71\)

[In]

int(((10*x+20)*ln(3)*exp(1/x*(5+x))+(-x^2-2*x)*ln(3)+x^2)/(x^3+2*x^2)/ln(3),x,method=_RETURNVERBOSE)

[Out]

-2*exp(1/x*(5+x))+ln(2+x)/ln(3)-ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {x^2+e^{\frac {5+x}{x}} (20+10 x) \log (3)+\left (-2 x-x^2\right ) \log (3)}{\left (2 x^2+x^3\right ) \log (3)} \, dx=-\frac {2 \, e^{\left (\frac {x + 5}{x}\right )} \log \left (3\right ) + \log \left (3\right ) \log \left (x\right ) - \log \left (x + 2\right )}{\log \left (3\right )} \]

[In]

integrate(((10*x+20)*log(3)*exp(1/x*(5+x))+(-x^2-2*x)*log(3)+x^2)/(x^3+2*x^2)/log(3),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {x^2+e^{\frac {5+x}{x}} (20+10 x) \log (3)+\left (-2 x-x^2\right ) \log (3)}{\left (2 x^2+x^3\right ) \log (3)} \, dx=- 2 e^{\frac {x + 5}{x}} - \log {\left (x \right )} + \frac {\log {\left (x + \frac {2 + 2 \log {\left (3 \right )}}{1 + \log {\left (3 \right )}} \right )}}{\log {\left (3 \right )}} \]

[In]

integrate(((10*x+20)*ln(3)*exp(1/x*(5+x))+(-x**2-2*x)*ln(3)+x**2)/(x**3+2*x**2)/ln(3),x)

[Out]

-2*exp((x + 5)/x) - log(x) + log(x + (2 + 2*log(3))/(1 + log(3)))/log(3)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {x^2+e^{\frac {5+x}{x}} (20+10 x) \log (3)+\left (-2 x-x^2\right ) \log (3)}{\left (2 x^2+x^3\right ) \log (3)} \, dx=\frac {{\left (\log \left (x + 2\right ) - \log \left (x\right )\right )} \log \left (3\right ) - 2 \, e^{\left (\frac {5}{x} + 1\right )} \log \left (3\right ) - \log \left (3\right ) \log \left (x + 2\right ) + \log \left (x + 2\right )}{\log \left (3\right )} \]

[In]

integrate(((10*x+20)*log(3)*exp(1/x*(5+x))+(-x^2-2*x)*log(3)+x^2)/(x^3+2*x^2)/log(3),x, algorithm="maxima")

[Out]

((log(x + 2) - log(x))*log(3) - 2*e^(5/x + 1)*log(3) - log(3)*log(x + 2) + log(x + 2))/log(3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (25) = 50\).

Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.00 \[ \int \frac {x^2+e^{\frac {5+x}{x}} (20+10 x) \log (3)+\left (-2 x-x^2\right ) \log (3)}{\left (2 x^2+x^3\right ) \log (3)} \, dx=-\frac {2 \, e^{\left (\frac {x + 5}{x}\right )} \log \left (3\right ) - \log \left (3\right ) \log \left (\frac {x + 5}{x} - 1\right ) - \log \left (\frac {2 \, {\left (x + 5\right )}}{x} + 3\right ) + \log \left (\frac {x + 5}{x} - 1\right )}{\log \left (3\right )} \]

[In]

integrate(((10*x+20)*log(3)*exp(1/x*(5+x))+(-x^2-2*x)*log(3)+x^2)/(x^3+2*x^2)/log(3),x, algorithm="giac")

[Out]

-(2*e^((x + 5)/x)*log(3) - log(3)*log((x + 5)/x - 1) - log(2*(x + 5)/x + 3) + log((x + 5)/x - 1))/log(3)

Mupad [B] (verification not implemented)

Time = 12.50 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {x^2+e^{\frac {5+x}{x}} (20+10 x) \log (3)+\left (-2 x-x^2\right ) \log (3)}{\left (2 x^2+x^3\right ) \log (3)} \, dx=\frac {\ln \left (x+2\right )}{\ln \left (3\right )}-2\,\mathrm {e}\,{\mathrm {e}}^{5/x}-\ln \left (x\right ) \]

[In]

int((x^2 - log(3)*(2*x + x^2) + exp((x + 5)/x)*log(3)*(10*x + 20))/(log(3)*(2*x^2 + x^3)),x)

[Out]

log(x + 2)/log(3) - 2*exp(1)*exp(5/x) - log(x)