\(\int \frac {-3 e^{e^4+\frac {3}{2+2 x}}+e^{\frac {3}{2+2 x}} (5-99 x-123 x^2-100 x^3)}{2+4 x+2 x^2} \, dx\) [6220]

Optimal result

Integrand size = 58, antiderivative size = 27 \[ \int \frac {-3 e^{e^4+\frac {3}{2+2 x}}+e^{\frac {3}{2+2 x}} \left (5-99 x-123 x^2-100 x^3\right )}{2+4 x+2 x^2} \, dx=4+e^{\frac {3}{2+2 x}} \left (-1+e^{e^4}+x-25 x^2\right ) \]

[Out]

4+(exp(exp(4))-1-25*x^2+x)*exp(3/(2+2*x))

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(27)=54\).

Time = 0.35 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.11, number of steps used = 16, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.155, Rules used = {27, 12, 6873, 6874, 2237, 2241, 2258, 2245, 2240} \[ \int \frac {-3 e^{e^4+\frac {3}{2+2 x}}+e^{\frac {3}{2+2 x}} \left (5-99 x-123 x^2-100 x^3\right )}{2+4 x+2 x^2} \, dx=-25 e^{\frac {3}{2 (x+1)}} (x+1)^2+51 e^{\frac {3}{2 (x+1)}} (x+1)-\left (27-e^{e^4}\right ) e^{\frac {3}{2 (x+1)}} \]

[In]

Int[(-3*E^(E^4 + 3/(2 + 2*x)) + E^(3/(2 + 2*x))*(5 - 99*x - 123*x^2 - 100*x^3))/(2 + 4*x + 2*x^2),x]

[Out]

-(E^(3/(2*(1 + x)))*(27 - E^E^4)) + 51*E^(3/(2*(1 + x)))*(1 + x) - 25*E^(3/(2*(1 + x)))*(1 + x)^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2237

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[(c + d*x)*(F^(a + b*(c + d*x)^n)/d), x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[n, 0]

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2241

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[F^a*(ExpIntegralEi[
b*(c + d*x)^n*Log[F]]/(f*n)), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2245

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))), x] - Dist[b*n*(Log[F]/(m + 1)), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rule 2258

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rule 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-3 e^{e^4+\frac {3}{2+2 x}}+e^{\frac {3}{2+2 x}} \left (5-99 x-123 x^2-100 x^3\right )}{2 (1+x)^2} \, dx \\ & = \frac {1}{2} \int \frac {-3 e^{e^4+\frac {3}{2+2 x}}+e^{\frac {3}{2+2 x}} \left (5-99 x-123 x^2-100 x^3\right )}{(1+x)^2} \, dx \\ & = \frac {1}{2} \int \frac {e^{\frac {3}{2+2 x}} \left (5-3 e^{e^4}-99 x-123 x^2-100 x^3\right )}{(1+x)^2} \, dx \\ & = \frac {1}{2} \int \left (77 e^{\frac {3}{2+2 x}}-100 e^{\frac {3}{2+2 x}} x-\frac {3 e^{\frac {3}{2+2 x}} \left (-27+e^{e^4}\right )}{(1+x)^2}-\frac {153 e^{\frac {3}{2+2 x}}}{1+x}\right ) \, dx \\ & = \frac {77}{2} \int e^{\frac {3}{2+2 x}} \, dx-50 \int e^{\frac {3}{2+2 x}} x \, dx-\frac {153}{2} \int \frac {e^{\frac {3}{2+2 x}}}{1+x} \, dx+\frac {1}{2} \left (3 \left (27-e^{e^4}\right )\right ) \int \frac {e^{\frac {3}{2+2 x}}}{(1+x)^2} \, dx \\ & = -e^{\frac {3}{2 (1+x)}} \left (27-e^{e^4}\right )+\frac {77}{2} e^{\frac {3}{2 (1+x)}} (1+x)+\frac {153}{2} \text {Ei}\left (\frac {3}{2 (1+x)}\right )-50 \int \left (-e^{\frac {3}{2+2 x}}+\frac {1}{2} e^{\frac {3}{2+2 x}} (2+2 x)\right ) \, dx+\frac {231}{2} \int \frac {e^{\frac {3}{2+2 x}}}{2+2 x} \, dx \\ & = -e^{\frac {3}{2 (1+x)}} \left (27-e^{e^4}\right )+\frac {77}{2} e^{\frac {3}{2 (1+x)}} (1+x)+\frac {75}{4} \text {Ei}\left (\frac {3}{2 (1+x)}\right )-25 \int e^{\frac {3}{2+2 x}} (2+2 x) \, dx+50 \int e^{\frac {3}{2+2 x}} \, dx \\ & = -e^{\frac {3}{2 (1+x)}} \left (27-e^{e^4}\right )+\frac {177}{2} e^{\frac {3}{2 (1+x)}} (1+x)-25 e^{\frac {3}{2 (1+x)}} (1+x)^2+\frac {75}{4} \text {Ei}\left (\frac {3}{2 (1+x)}\right )-\frac {75}{2} \int e^{\frac {3}{2+2 x}} \, dx+150 \int \frac {e^{\frac {3}{2+2 x}}}{2+2 x} \, dx \\ & = -e^{\frac {3}{2 (1+x)}} \left (27-e^{e^4}\right )+51 e^{\frac {3}{2 (1+x)}} (1+x)-25 e^{\frac {3}{2 (1+x)}} (1+x)^2-\frac {225}{4} \text {Ei}\left (\frac {3}{2 (1+x)}\right )-\frac {225}{2} \int \frac {e^{\frac {3}{2+2 x}}}{2+2 x} \, dx \\ & = -e^{\frac {3}{2 (1+x)}} \left (27-e^{e^4}\right )+51 e^{\frac {3}{2 (1+x)}} (1+x)-25 e^{\frac {3}{2 (1+x)}} (1+x)^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {-3 e^{e^4+\frac {3}{2+2 x}}+e^{\frac {3}{2+2 x}} \left (5-99 x-123 x^2-100 x^3\right )}{2+4 x+2 x^2} \, dx=e^{\frac {3}{2 (1+x)}} \left (-1+e^{e^4}+x-25 x^2\right ) \]

[In]

Integrate[(-3*E^(E^4 + 3/(2 + 2*x)) + E^(3/(2 + 2*x))*(5 - 99*x - 123*x^2 - 100*x^3))/(2 + 4*x + 2*x^2),x]

[Out]

E^(3/(2*(1 + x)))*(-1 + E^E^4 + x - 25*x^2)

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78

method	result	size
gosper	\(\left ({\mathrm e}^{{\mathrm e}^{4}}-1-25 x^{2}+x \right ) {\mathrm e}^{\frac {3}{2 \left (1+x \right )}}\)	\(21\)
risch	\(\left ({\mathrm e}^{{\mathrm e}^{4}}-1-25 x^{2}+x \right ) {\mathrm e}^{\frac {3}{2 \left (1+x \right )}}\)	\(21\)
parallelrisch	\(-25 \,{\mathrm e}^{\frac {3}{2 \left (1+x \right )}} x^{2}+{\mathrm e}^{\frac {3}{2 \left (1+x \right )}} {\mathrm e}^{{\mathrm e}^{4}}+x \,{\mathrm e}^{\frac {3}{2 \left (1+x \right )}}-{\mathrm e}^{\frac {3}{2 \left (1+x \right )}}\)	\(47\)
derivativedivides	\(-25 \,{\mathrm e}^{\frac {3}{2 \left (1+x \right )}} \left (1+x \right )^{2}+51 \,{\mathrm e}^{\frac {3}{2 \left (1+x \right )}} \left (1+x \right )+{\mathrm e}^{\frac {3}{2 \left (1+x \right )}} {\mathrm e}^{{\mathrm e}^{4}}-27 \,{\mathrm e}^{\frac {3}{2 \left (1+x \right )}}\)	\(52\)
default	\(-25 \,{\mathrm e}^{\frac {3}{2 \left (1+x \right )}} \left (1+x \right )^{2}+51 \,{\mathrm e}^{\frac {3}{2 \left (1+x \right )}} \left (1+x \right )+{\mathrm e}^{\frac {3}{2 \left (1+x \right )}} {\mathrm e}^{{\mathrm e}^{4}}-27 \,{\mathrm e}^{\frac {3}{2 \left (1+x \right )}}\)	\(52\)
norman	\(\frac {\left ({\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{\frac {3}{2+2 x}}+x \,{\mathrm e}^{{\mathrm e}^{4}} {\mathrm e}^{\frac {3}{2+2 x}}-24 x^{2} {\mathrm e}^{\frac {3}{2+2 x}}-25 x^{3} {\mathrm e}^{\frac {3}{2+2 x}}}{1+x}\)	\(69\)

[In]

int((-3*exp(3/(2+2*x))*exp(exp(4))+(-100*x^3-123*x^2-99*x+5)*exp(3/(2+2*x)))/(2*x^2+4*x+2),x,method=_RETURNVER
BOSE)

[Out]

(exp(exp(4))-1-25*x^2+x)*exp(3/2/(1+x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {-3 e^{e^4+\frac {3}{2+2 x}}+e^{\frac {3}{2+2 x}} \left (5-99 x-123 x^2-100 x^3\right )}{2+4 x+2 x^2} \, dx=-{\left (25 \, x^{2} - x - e^{\left (e^{4}\right )} + 1\right )} e^{\left (\frac {2 \, {\left (x + 1\right )} e^{4} + 3}{2 \, {\left (x + 1\right )}} - e^{4}\right )} \]

[In]

integrate((-3*exp(3/(2+2*x))*exp(exp(4))+(-100*x^3-123*x^2-99*x+5)*exp(3/(2+2*x)))/(2*x^2+4*x+2),x, algorithm=
"fricas")

[Out]

-(25*x^2 - x - e^(e^4) + 1)*e^(1/2*(2*(x + 1)*e^4 + 3)/(x + 1) - e^4)

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-3 e^{e^4+\frac {3}{2+2 x}}+e^{\frac {3}{2+2 x}} \left (5-99 x-123 x^2-100 x^3\right )}{2+4 x+2 x^2} \, dx=\left (- 25 x^{2} + x - 1 + e^{e^{4}}\right ) e^{\frac {3}{2 x + 2}} \]

[In]

integrate((-3*exp(3/(2+2*x))*exp(exp(4))+(-100*x**3-123*x**2-99*x+5)*exp(3/(2+2*x)))/(2*x**2+4*x+2),x)

[Out]

(-25*x**2 + x - 1 + exp(exp(4)))*exp(3/(2*x + 2))

Maxima [F]

\[ \int \frac {-3 e^{e^4+\frac {3}{2+2 x}}+e^{\frac {3}{2+2 x}} \left (5-99 x-123 x^2-100 x^3\right )}{2+4 x+2 x^2} \, dx=\int { -\frac {{\left (100 \, x^{3} + 123 \, x^{2} + 99 \, x - 5\right )} e^{\left (\frac {3}{2 \, {\left (x + 1\right )}}\right )} + 3 \, e^{\left (\frac {3}{2 \, {\left (x + 1\right )}} + e^{4}\right )}}{2 \, {\left (x^{2} + 2 \, x + 1\right )}} \,d x } \]

[In]

integrate((-3*exp(3/(2+2*x))*exp(exp(4))+(-100*x^3-123*x^2-99*x+5)*exp(3/(2+2*x)))/(2*x^2+4*x+2),x, algorithm=
"maxima")

[Out]

-(25*x^2 - x)*e^(3/2/(x + 1)) - 1/2*(3*e^(e^4) + 2)*integrate(e^(3/2/(x + 1))/(x^2 + 2*x + 1), x) - 5/3*e^(3/2
/(x + 1))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (27) = 54\).

Time = 0.29 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.37 \[ \int \frac {-3 e^{e^4+\frac {3}{2+2 x}}+e^{\frac {3}{2+2 x}} \left (5-99 x-123 x^2-100 x^3\right )}{2+4 x+2 x^2} \, dx={\left (x + 1\right )}^{2} {\left (\frac {51 \, e^{\left (\frac {3}{2 \, {\left (x + 1\right )}}\right )}}{x + 1} - \frac {27 \, e^{\left (\frac {3}{2 \, {\left (x + 1\right )}}\right )}}{{\left (x + 1\right )}^{2}} + \frac {e^{\left (\frac {3}{2 \, {\left (x + 1\right )}} + e^{4}\right )}}{{\left (x + 1\right )}^{2}} - 25 \, e^{\left (\frac {3}{2 \, {\left (x + 1\right )}}\right )}\right )} \]

[In]

integrate((-3*exp(3/(2+2*x))*exp(exp(4))+(-100*x^3-123*x^2-99*x+5)*exp(3/(2+2*x)))/(2*x^2+4*x+2),x, algorithm=
"giac")

[Out]

(x + 1)^2*(51*e^(3/2/(x + 1))/(x + 1) - 27*e^(3/2/(x + 1))/(x + 1)^2 + e^(3/2/(x + 1) + e^4)/(x + 1)^2 - 25*e^
(3/2/(x + 1)))

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-3 e^{e^4+\frac {3}{2+2 x}}+e^{\frac {3}{2+2 x}} \left (5-99 x-123 x^2-100 x^3\right )}{2+4 x+2 x^2} \, dx={\mathrm {e}}^{\frac {3}{2\,\left (x+1\right )}}\,\left (-25\,x^2+x+{\mathrm {e}}^{{\mathrm {e}}^4}-1\right ) \]

[In]

int(-(exp(3/(2*x + 2))*(99*x + 123*x^2 + 100*x^3 - 5) + 3*exp(3/(2*x + 2))*exp(exp(4)))/(4*x + 2*x^2 + 2),x)

[Out]

exp(3/(2*(x + 1)))*(x + exp(exp(4)) - 25*x^2 - 1)