\(\int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx\) [2796]

Optimal result

Integrand size = 34, antiderivative size = 23 \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=1-\frac {2+x}{e}-\frac {x}{5 (-4+3 \log (3))} \]

[Out]

-exp(ln(2+x)-1)+1-1/5*x/(3*ln(3)-4)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {192, 21, 8} \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=-\frac {1}{5} x \left (\frac {5}{e}+\frac {1}{\log (27)-4}\right ) \]

[In]

Int[(-2 - x + ((2 + x)*(20 - 15*Log[3]))/E)/(-40 - 20*x + (30 + 15*x)*Log[3]),x]

[Out]

-1/5*(x*(5/E + (-4 + Log[27])^(-1)))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 192

Int[(u_)^(m_.)*(v_)^(n_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^n, x] /; FreeQ[{m, n}, x] &&
LinearQ[{u, v}, x] &&  !LinearMatchQ[{u, v}, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\frac {2 (20-e-5 \log (27))}{e}+\frac {x (20-e-5 \log (27))}{e}}{-10 (4-3 \log (3))-5 x (4-\log (27))} \, dx \\ & = -\left (\frac {1}{5} \left (\frac {5}{e}+\frac {1}{-4+\log (27)}\right ) \int 1 \, dx\right ) \\ & = -\frac {1}{5} x \left (\frac {5}{e}+\frac {1}{-4+\log (27)}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=x \left (-\frac {1}{e}-\frac {1}{5 (-4+\log (27))}\right ) \]

[In]

Integrate[(-2 - x + ((2 + x)*(20 - 15*Log[3]))/E)/(-40 - 20*x + (30 + 15*x)*Log[3]),x]

[Out]

x*(-E^(-1) - 1/(5*(-4 + Log[27])))

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96

method	result	size
default	\(-\frac {x}{5 \left (3 \ln \left (3\right )-4\right )}-{\mathrm e}^{\ln \left (2+x \right )-1}\)	\(22\)
parts	\(-\frac {x}{5 \left (3 \ln \left (3\right )-4\right )}-{\mathrm e}^{\ln \left (2+x \right )-1}\)	\(22\)
norman	\(-\frac {{\mathrm e}^{-1} \left ({\mathrm e}+15 \ln \left (3\right )-20\right ) x}{5 \left (3 \ln \left (3\right )-4\right )}\)	\(24\)
risch	\(-\frac {15 x \ln \left (3\right ) {\mathrm e}^{-1}}{15 \ln \left (3\right )-20}+\frac {20 x \,{\mathrm e}^{-1}}{15 \ln \left (3\right )-20}-\frac {x}{15 \ln \left (3\right )-20}\)	\(41\)
parallelrisch	\(\frac {16+12 x -15 \ln \left (3\right ) x^{2} {\mathrm e}^{\ln \left (2+x \right )-1}-80 \,{\mathrm e}^{\ln \left (2+x \right )-1}-x^{3}+20 x^{2} {\mathrm e}^{\ln \left (2+x \right )-1}+60 \,{\mathrm e}^{\ln \left (2+x \right )-1} \ln \left (3\right )}{5 \left (2+x \right )^{2} \left (3 \ln \left (3\right )-4\right )}\)	\(72\)

[In]

int(((-15*ln(3)+20)*exp(ln(2+x)-1)-x-2)/((15*x+30)*ln(3)-20*x-40),x,method=_RETURNVERBOSE)

[Out]

-1/5*x/(3*ln(3)-4)-exp(ln(2+x)-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=-\frac {x e + 15 \, x \log \left (3\right ) - 20 \, x}{5 \, {\left (3 \, e \log \left (3\right ) - 4 \, e\right )}} \]

[In]

integrate(((-15*log(3)+20)*exp(log(2+x)-1)-x-2)/((15*x+30)*log(3)-20*x-40),x, algorithm="fricas")

[Out]

-1/5*(x*e + 15*x*log(3) - 20*x)/(3*e*log(3) - 4*e)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=\frac {x \left (- 15 \log {\left (3 \right )} - e + 20\right )}{- 20 e + 15 e \log {\left (3 \right )}} \]

[In]

integrate(((-15*ln(3)+20)*exp(ln(2+x)-1)-x-2)/((15*x+30)*ln(3)-20*x-40),x)

[Out]

x*(-15*log(3) - E + 20)/(-20*E + 15*E*log(3))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=-\frac {x {\left (e + 15 \, \log \left (3\right ) - 20\right )}}{5 \, {\left (3 \, e \log \left (3\right ) - 4 \, e\right )}} \]

[In]

integrate(((-15*log(3)+20)*exp(log(2+x)-1)-x-2)/((15*x+30)*log(3)-20*x-40),x, algorithm="maxima")

[Out]

-1/5*x*(e + 15*log(3) - 20)/(3*e*log(3) - 4*e)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=-\frac {x e + 15 \, x \log \left (3\right ) - 20 \, x}{5 \, {\left (3 \, e \log \left (3\right ) - 4 \, e\right )}} \]

[In]

integrate(((-15*log(3)+20)*exp(log(2+x)-1)-x-2)/((15*x+30)*log(3)-20*x-40),x, algorithm="giac")

[Out]

-1/5*(x*e + 15*x*log(3) - 20*x)/(3*e*log(3) - 4*e)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-2-x+\frac {(2+x) (20-15 \log (3))}{e}}{-40-20 x+(30+15 x) \log (3)} \, dx=\frac {x\,\left (\mathrm {e}+15\,\ln \left (3\right )-20\right )}{20\,\mathrm {e}-15\,\mathrm {e}\,\ln \left (3\right )} \]

[In]

int((x + exp(log(x + 2) - 1)*(15*log(3) - 20) + 2)/(20*x - log(3)*(15*x + 30) + 40),x)

[Out]

(x*(exp(1) + 15*log(3) - 20))/(20*exp(1) - 15*exp(1)*log(3))