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

Optimal. Leaf size=28 \[ \frac {3}{8}-2 e^{\frac {5+x}{x}}-\log (x)+\frac {\log (2+x)}{\log (3)} \]

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Rubi [A]  time = 0.41, antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {12, 1593, 6688, 2209} \begin {gather*} -2 e^{\frac {5}{x}+1}-\log (x)+\frac {\log (x+2)}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[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 1593

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 2209

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 29, normalized size = 1.04 \begin {gather*} -2 e^{1+\frac {5}{x}}-\log (x)+\frac {\log (x \log (3)+\log (9))}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.64, size = 30, normalized size = 1.07 \begin {gather*} -\frac {2 \, e^{\left (\frac {x + 5}{x}\right )} \log \relax (3) + \log \relax (3) \log \relax (x) - \log \left (x + 2\right )}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.18, size = 56, normalized size = 2.00 \begin {gather*} -\frac {2 \, e^{\left (\frac {x + 5}{x}\right )} \log \relax (3) - \log \relax (3) \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 \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.09, size = 25, normalized size = 0.89

method	result	size
norman	\(-2 \,{\mathrm e}^{\frac {5+x}{x}}+\frac {\ln \left (2+x \right )}{\ln \relax (3)}-\ln \relax (x )\)	\(25\)
risch	\(\frac {\ln \left (-x -2\right )}{\ln \relax (3)}-\ln \relax (x )-2 \,{\mathrm e}^{\frac {5+x}{x}}\)	\(27\)
default	\(\frac {-\ln \left (\frac {5}{x}\right )+\ln \left (5+\frac {10}{x}\right )+\ln \relax (3) \ln \left (\frac {5}{x}\right )-2 \ln \relax (3) {\mathrm e}^{1+\frac {5}{x}}}{\ln \relax (3)}\)	\(44\)
derivativedivides	\(-\frac {\ln \left (\frac {5}{x}\right )-\ln \left (5+\frac {10}{x}\right )-\ln \relax (3) \ln \left (\frac {5}{x}\right )+2 \ln \relax (3) {\mathrm e}^{1+\frac {5}{x}}}{\ln \relax (3)}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.50, size = 42, normalized size = 1.50 \begin {gather*} \frac {{\left (\log \left (x + 2\right ) - \log \relax (x)\right )} \log \relax (3) - 2 \, e^{\left (\frac {5}{x} + 1\right )} \log \relax (3) - \log \relax (3) \log \left (x + 2\right ) + \log \left (x + 2\right )}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 4.20, size = 24, normalized size = 0.86 \begin {gather*} \frac {\ln \left (x+2\right )}{\ln \relax (3)}-2\,\mathrm {e}\,{\mathrm {e}}^{5/x}-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.54, size = 29, normalized size = 1.04 \begin {gather*} - 2 e^{\frac {x + 5}{x}} - \log {\relax (x )} + \frac {\log {\left (x + \frac {2 + 2 \log {\relax (3 )}}{1 + \log {\relax (3 )}} \right )}}{\log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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