\(\int \frac {-144-289 x+e (1+2 x)}{-144 x+e x} \, dx\) [1807]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 22, antiderivative size = 21 \[ \int \frac {-144-289 x+e (1+2 x)}{-144 x+e x} \, dx=2 x+\frac {1+x}{144-e}+\log \left (\frac {x}{4}\right ) \]

[Out]

2*x+(1+x)/(144-exp(1))+ln(1/4*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6, 12, 192, 45} \[ \int \frac {-144-289 x+e (1+2 x)}{-144 x+e x} \, dx=\frac {(289-2 e) x}{144-e}+\log (x) \]

[In]

Int[(-144 - 289*x + E*(1 + 2*x))/(-144*x + E*x),x]

[Out]

((289 - 2*E)*x)/(144 - E) + Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 {-144-289 x+e (1+2 x)}{(-144+e) x} \, dx \\ & = \frac {\int \frac {-144-289 x+e (1+2 x)}{x} \, dx}{-144+e} \\ & = \frac {\int \frac {-144+e-(289-2 e) x}{x} \, dx}{-144+e} \\ & = \frac {\int \left (-289+2 e+\frac {-144+e}{x}\right ) \, dx}{-144+e} \\ & = \frac {(289-2 e) x}{144-e}+\log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {-144-289 x+e (1+2 x)}{-144 x+e x} \, dx=\frac {(-289+2 e) x+(-144+e) \log (x)}{-144+e} \]

[In]

Integrate[(-144 - 289*x + E*(1 + 2*x))/(-144*x + E*x),x]

[Out]

((-289 + 2*E)*x + (-144 + E)*Log[x])/(-144 + E)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86

method result size
norman \(\frac {\left (2 \,{\mathrm e}-289\right ) x}{{\mathrm e}-144}+\ln \left (x \right )\) \(18\)
default \(\frac {2 x \,{\mathrm e}-289 x +\left ({\mathrm e}-144\right ) \ln \left (x \right )}{{\mathrm e}-144}\) \(24\)
risch \(\frac {2 x \,{\mathrm e}}{{\mathrm e}-144}-\frac {289 x}{{\mathrm e}-144}+\ln \left (x \right )\) \(24\)
parallelrisch \(\frac {{\mathrm e} \ln \left (x \right )+2 x \,{\mathrm e}-144 \ln \left (x \right )-289 x}{{\mathrm e}-144}\) \(26\)

[In]

int(((1+2*x)*exp(1)-289*x-144)/(x*exp(1)-144*x),x,method=_RETURNVERBOSE)

[Out]

(2*exp(1)-289)/(exp(1)-144)*x+ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {-144-289 x+e (1+2 x)}{-144 x+e x} \, dx=\frac {2 \, x e + {\left (e - 144\right )} \log \left (x\right ) - 289 \, x}{e - 144} \]

[In]

integrate(((1+2*x)*exp(1)-289*x-144)/(x*exp(1)-144*x),x, algorithm="fricas")

[Out]

(2*x*e + (e - 144)*log(x) - 289*x)/(e - 144)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {-144-289 x+e (1+2 x)}{-144 x+e x} \, dx=\frac {- x \left (289 - 2 e\right ) - \left (144 - e\right ) \log {\left (x \right )}}{-144 + e} \]

[In]

integrate(((1+2*x)*exp(1)-289*x-144)/(x*exp(1)-144*x),x)

[Out]

(-x*(289 - 2*E) - (144 - E)*log(x))/(-144 + E)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-144-289 x+e (1+2 x)}{-144 x+e x} \, dx=\frac {x {\left (2 \, e - 289\right )}}{e - 144} + \log \left (x\right ) \]

[In]

integrate(((1+2*x)*exp(1)-289*x-144)/(x*exp(1)-144*x),x, algorithm="maxima")

[Out]

x*(2*e - 289)/(e - 144) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {-144-289 x+e (1+2 x)}{-144 x+e x} \, dx=\frac {2 \, x e - 289 \, x}{e - 144} + \log \left ({\left | x \right |}\right ) \]

[In]

integrate(((1+2*x)*exp(1)-289*x-144)/(x*exp(1)-144*x),x, algorithm="giac")

[Out]

(2*x*e - 289*x)/(e - 144) + log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-144-289 x+e (1+2 x)}{-144 x+e x} \, dx=\ln \left (x\right )+\frac {x\,\left (2\,\mathrm {e}-289\right )}{\mathrm {e}-144} \]

[In]

int((289*x - exp(1)*(2*x + 1) + 144)/(144*x - x*exp(1)),x)

[Out]

log(x) + (x*(2*exp(1) - 289))/(exp(1) - 144)