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

Optimal. Leaf size=28 \[ e^{e^{9-x}}-\log \left (2+\frac {x^2}{5 x+x^2}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 22, normalized size of antiderivative = 0.79, number of steps used = 7, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6728, 2282, 2194, 616, 31} \begin {gather*} e^{e^{9-x}}+\log (x+5)-\log (3 x+10) \end {gather*}

Antiderivative was successfully verified.

[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 616

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 2194

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 2282

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 6728

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

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 22, normalized size = 0.79 \begin {gather*} e^{e^{9-x}}+\log (5+x)-\log (10+3 x) \end {gather*}

Antiderivative was successfully verified.

[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]

________________________________________________________________________________________

fricas [A]  time = 0.58, size = 46, normalized size = 1.64 \begin {gather*} -{\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [A]  time = 0.23, size = 46, normalized size = 1.64 \begin {gather*} -{\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [A]  time = 0.12, size = 21, normalized size = 0.75

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

Verification of antiderivative is not currently implemented for this CAS.

[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(3*x+10)+ln(5+x)

________________________________________________________________________________________

maxima [A]  time = 0.61, size = 20, normalized size = 0.71 \begin {gather*} e^{\left (e^{\left (-x + 9\right )}\right )} - \log \left (3 \, x + 10\right ) + \log \left (x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 3.23, size = 17, normalized size = 0.61 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{-x}\,{\mathrm {e}}^9}+2\,\mathrm {atanh}\left (\frac {6\,x}{5}+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [A]  time = 0.19, size = 17, normalized size = 0.61 \begin {gather*} e^{e^{9 - x}} - \log {\left (x + \frac {10}{3} \right )} + \log {\left (x + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________