\(\int \frac {90-60 x+15 x^2+e^{e^x+x} (60-60 x+15 x^2)}{4-4 x+x^2} \, dx\) [8585]

Optimal result

Integrand size = 39, antiderivative size = 30 \[ \int \frac {90-60 x+15 x^2+e^{e^x+x} \left (60-60 x+15 x^2\right )}{4-4 x+x^2} \, dx=15 \left (-1-e^{e^{e^{e^5}}}+e^{e^x}+x+\frac {x}{2-x}\right ) \]

[Out]

15*exp(exp(x))+15*x-15*exp(exp(exp(exp(5))))-15+15*x/(2-x)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {27, 6874, 2320, 2225, 697} \[ \int \frac {90-60 x+15 x^2+e^{e^x+x} \left (60-60 x+15 x^2\right )}{4-4 x+x^2} \, dx=15 x+15 e^{e^x}+\frac {30}{2-x} \]

[In]

Int[(90 - 60*x + 15*x^2 + E^(E^x + x)*(60 - 60*x + 15*x^2))/(4 - 4*x + x^2),x]

[Out]

15*E^E^x + 30/(2 - x) + 15*x

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 697

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

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 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {90-60 x+15 x^2+e^{e^x+x} \left (60-60 x+15 x^2\right )}{(-2+x)^2} \, dx \\ & = \int \left (15 e^{e^x+x}+\frac {15 \left (6-4 x+x^2\right )}{(-2+x)^2}\right ) \, dx \\ & = 15 \int e^{e^x+x} \, dx+15 \int \frac {6-4 x+x^2}{(-2+x)^2} \, dx \\ & = 15 \int \left (1+\frac {2}{(-2+x)^2}\right ) \, dx+15 \text {Subst}\left (\int e^x \, dx,x,e^x\right ) \\ & = 15 e^{e^x}+\frac {30}{2-x}+15 x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.60 \[ \int \frac {90-60 x+15 x^2+e^{e^x+x} \left (60-60 x+15 x^2\right )}{4-4 x+x^2} \, dx=15 \left (e^{e^x}+\frac {2}{2-x}+x\right ) \]

[In]

Integrate[(90 - 60*x + 15*x^2 + E^(E^x + x)*(60 - 60*x + 15*x^2))/(4 - 4*x + x^2),x]

[Out]

15*(E^E^x + 2/(2 - x) + x)

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.57

method	result	size
risch	\(-\frac {30}{-2+x}+15 x +15 \,{\mathrm e}^{{\mathrm e}^{x}}\)	\(17\)
parts	\(-\frac {30}{-2+x}+15 x +15 \,{\mathrm e}^{{\mathrm e}^{x}}\)	\(17\)
norman	\(\frac {15 x^{2}+15 x \,{\mathrm e}^{{\mathrm e}^{x}}-30 \,{\mathrm e}^{{\mathrm e}^{x}}-90}{-2+x}\)	\(25\)
parallelrisch	\(\frac {15 x^{2}+15 x \,{\mathrm e}^{{\mathrm e}^{x}}-30 \,{\mathrm e}^{{\mathrm e}^{x}}-90}{-2+x}\)	\(25\)

[In]

int(((15*x^2-60*x+60)*exp(x)*exp(exp(x))+15*x^2-60*x+90)/(x^2-4*x+4),x,method=_RETURNVERBOSE)

[Out]

-30/(-2+x)+15*x+15*exp(exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {90-60 x+15 x^2+e^{e^x+x} \left (60-60 x+15 x^2\right )}{4-4 x+x^2} \, dx=\frac {15 \, {\left ({\left (x - 2\right )} e^{\left (x + e^{x}\right )} + {\left (x^{2} - 2 \, x - 2\right )} e^{x}\right )} e^{\left (-x\right )}}{x - 2} \]

[In]

integrate(((15*x^2-60*x+60)*exp(x)*exp(exp(x))+15*x^2-60*x+90)/(x^2-4*x+4),x, algorithm="fricas")

[Out]

15*((x - 2)*e^(x + e^x) + (x^2 - 2*x - 2)*e^x)*e^(-x)/(x - 2)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.47 \[ \int \frac {90-60 x+15 x^2+e^{e^x+x} \left (60-60 x+15 x^2\right )}{4-4 x+x^2} \, dx=15 x + 15 e^{e^{x}} - \frac {30}{x - 2} \]

[In]

integrate(((15*x**2-60*x+60)*exp(x)*exp(exp(x))+15*x**2-60*x+90)/(x**2-4*x+4),x)

[Out]

15*x + 15*exp(exp(x)) - 30/(x - 2)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.53 \[ \int \frac {90-60 x+15 x^2+e^{e^x+x} \left (60-60 x+15 x^2\right )}{4-4 x+x^2} \, dx=15 \, x - \frac {30}{x - 2} + 15 \, e^{\left (e^{x}\right )} \]

[In]

integrate(((15*x^2-60*x+60)*exp(x)*exp(exp(x))+15*x^2-60*x+90)/(x^2-4*x+4),x, algorithm="maxima")

[Out]

15*x - 30/(x - 2) + 15*e^(e^x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.43 \[ \int \frac {90-60 x+15 x^2+e^{e^x+x} \left (60-60 x+15 x^2\right )}{4-4 x+x^2} \, dx=\frac {15 \, {\left (x^{2} e^{x} + x e^{\left (x + e^{x}\right )} - 2 \, x e^{x} - 2 \, e^{\left (x + e^{x}\right )} - 2 \, e^{x}\right )}}{x e^{x} - 2 \, e^{x}} \]

[In]

integrate(((15*x^2-60*x+60)*exp(x)*exp(exp(x))+15*x^2-60*x+90)/(x^2-4*x+4),x, algorithm="giac")

[Out]

15*(x^2*e^x + x*e^(x + e^x) - 2*x*e^x - 2*e^(x + e^x) - 2*e^x)/(x*e^x - 2*e^x)

Mupad [B] (verification not implemented)

Time = 14.61 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.53 \[ \int \frac {90-60 x+15 x^2+e^{e^x+x} \left (60-60 x+15 x^2\right )}{4-4 x+x^2} \, dx=15\,x+15\,{\mathrm {e}}^{{\mathrm {e}}^x}-\frac {30}{x-2} \]

[In]

int((15*x^2 - 60*x + exp(exp(x))*exp(x)*(15*x^2 - 60*x + 60) + 90)/(x^2 - 4*x + 4),x)

[Out]

15*x + 15*exp(exp(x)) - 30/(x - 2)