\(\int \frac {-5+e^{9+e^{9-x}-x} (-50-25 x-3 x^2)}{50+25 x+3 x^2} \, dx\) [4598]

Optimal result

Integrand size = 40, antiderivative size = 28 \[ \int \frac {-5+e^{9+e^{9-x}-x} \left (-50-25 x-3 x^2\right )}{50+25 x+3 x^2} \, dx=e^{e^{9-x}}-\log \left (2+\frac {x^2}{5 x+x^2}\right ) \]

[Out]

exp(exp(9-x))-ln(x^2/(x^2+5*x)+2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6860, 2320, 2225, 630, 31} \[ \int \frac {-5+e^{9+e^{9-x}-x} \left (-50-25 x-3 x^2\right )}{50+25 x+3 x^2} \, dx=e^{e^{9-x}}+\log (x+5)-\log (3 x+10) \]

[In]

Int[(-5 + E^(9 + E^(9 - x) - x)*(-50 - 25*x - 3*x^2))/(50 + 25*x + 3*x^2),x]

[Out]

E^E^(9 - x) + Log[5 + x] - Log[10 + 3*x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 630

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (-e^{9+e^{9-x}-x}-\frac {5}{50+25 x+3 x^2}\right ) \, dx \\ & = -\left (5 \int \frac {1}{50+25 x+3 x^2} \, dx\right )-\int e^{9+e^{9-x}-x} \, dx \\ & = -\left (3 \int \frac {1}{10+3 x} \, dx\right )+3 \int \frac {1}{15+3 x} \, dx+\text {Subst}\left (\int e^{9+e^9 x} \, dx,x,e^{-x}\right ) \\ & = e^{e^{9-x}}+\log (5+x)-\log (10+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-5+e^{9+e^{9-x}-x} \left (-50-25 x-3 x^2\right )}{50+25 x+3 x^2} \, dx=e^{e^{9-x}}+\log (5+x)-\log (10+3 x) \]

[In]

Integrate[(-5 + E^(9 + E^(9 - x) - x)*(-50 - 25*x - 3*x^2))/(50 + 25*x + 3*x^2),x]

[Out]

E^E^(9 - x) + Log[5 + x] - Log[10 + 3*x]

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68

method	result	size
parallelrisch	\({\mathrm e}^{{\mathrm e}^{9-x}}+\ln \left (5+x \right )-\ln \left (x +\frac {10}{3}\right )\)	\(19\)
norman	\({\mathrm e}^{{\mathrm e}^{9-x}}+\ln \left (5+x \right )-\ln \left (3 x +10\right )\)	\(21\)
risch	\({\mathrm e}^{{\mathrm e}^{9-x}}+\ln \left (5+x \right )-\ln \left (3 x +10\right )\)	\(21\)
parts	\({\mathrm e}^{{\mathrm e}^{9-x}}+\ln \left (5+x \right )-\ln \left (3 x +10\right )\)	\(21\)

[In]

int(((-3*x^2-25*x-50)*exp(9-x)*exp(exp(9-x))-5)/(3*x^2+25*x+50),x,method=_RETURNVERBOSE)

[Out]

exp(exp(9-x))+ln(5+x)-ln(x+10/3)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {-5+e^{9+e^{9-x}-x} \left (-50-25 x-3 x^2\right )}{50+25 x+3 x^2} \, dx=-{\left (e^{\left (-x + 9\right )} \log \left (3 \, x + 10\right ) - e^{\left (-x + 9\right )} \log \left (x + 5\right ) - e^{\left (-x + e^{\left (-x + 9\right )} + 9\right )}\right )} e^{\left (x - 9\right )} \]

[In]

integrate(((-3*x^2-25*x-50)*exp(9-x)*exp(exp(9-x))-5)/(3*x^2+25*x+50),x, algorithm="fricas")

[Out]

-(e^(-x + 9)*log(3*x + 10) - e^(-x + 9)*log(x + 5) - e^(-x + e^(-x + 9) + 9))*e^(x - 9)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {-5+e^{9+e^{9-x}-x} \left (-50-25 x-3 x^2\right )}{50+25 x+3 x^2} \, dx=e^{e^{9 - x}} - \log {\left (x + \frac {10}{3} \right )} + \log {\left (x + 5 \right )} \]

[In]

integrate(((-3*x**2-25*x-50)*exp(9-x)*exp(exp(9-x))-5)/(3*x**2+25*x+50),x)

[Out]

exp(exp(9 - x)) - log(x + 10/3) + log(x + 5)

Maxima [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {-5+e^{9+e^{9-x}-x} \left (-50-25 x-3 x^2\right )}{50+25 x+3 x^2} \, dx=e^{\left (e^{\left (-x + 9\right )}\right )} - \log \left (3 \, x + 10\right ) + \log \left (x + 5\right ) \]

[In]

integrate(((-3*x^2-25*x-50)*exp(9-x)*exp(exp(9-x))-5)/(3*x^2+25*x+50),x, algorithm="maxima")

[Out]

e^(e^(-x + 9)) - log(3*x + 10) + log(x + 5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {-5+e^{9+e^{9-x}-x} \left (-50-25 x-3 x^2\right )}{50+25 x+3 x^2} \, dx=-{\left (e^{\left (-x + 9\right )} \log \left (3 \, x + 10\right ) - e^{\left (-x + 9\right )} \log \left (x + 5\right ) - e^{\left (-x + e^{\left (-x + 9\right )} + 9\right )}\right )} e^{\left (x - 9\right )} \]

[In]

integrate(((-3*x^2-25*x-50)*exp(9-x)*exp(exp(9-x))-5)/(3*x^2+25*x+50),x, algorithm="giac")

[Out]

-(e^(-x + 9)*log(3*x + 10) - e^(-x + 9)*log(x + 5) - e^(-x + e^(-x + 9) + 9))*e^(x - 9)

Mupad [B] (verification not implemented)

Time = 10.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {-5+e^{9+e^{9-x}-x} \left (-50-25 x-3 x^2\right )}{50+25 x+3 x^2} \, dx={\mathrm {e}}^{{\mathrm {e}}^{-x}\,{\mathrm {e}}^9}+2\,\mathrm {atanh}\left (\frac {6\,x}{5}+5\right ) \]

[In]

int(-(exp(exp(9 - x))*exp(9 - x)*(25*x + 3*x^2 + 50) + 5)/(25*x + 3*x^2 + 50),x)

[Out]

exp(exp(-x)*exp(9)) + 2*atanh((6*x)/5 + 5)