3.28.6 \(\int \frac {e^{\frac {100+e^3}{1+2 x+x^2}} (-999+e^3 (-10-2 x)-197 x+3 x^2+x^3)}{1+3 x+3 x^2+x^3} \, dx\)

Optimal. Leaf size=17 \[ e^{\frac {100+e^3}{(1+x)^2}} (5+x) \]

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Rubi [A]  time = 0.85, antiderivative size = 33, normalized size of antiderivative = 1.94, number of steps used = 10, number of rules used = 7, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6741, 6688, 6742, 2206, 2211, 2204, 2209} \begin {gather*} e^{\frac {100+e^3}{(x+1)^2}} (x+1)+4 e^{\frac {100+e^3}{(x+1)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((100 + E^3)/(1 + 2*x + x^2))*(-999 + E^3*(-10 - 2*x) - 197*x + 3*x^2 + x^3))/(1 + 3*x + 3*x^2 + x^3),x
]

[Out]

4*E^((100 + E^3)/(1 + x)^2) + E^((100 + E^3)/(1 + x)^2)*(1 + x)

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2206

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 2209

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 2211

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[1/(d*(m + 1))
, Subst[Int[F^(a + b*x^2), x], x, (c + d*x)^(m + 1)], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && EqQ[n, 2*(m + 1
)]

Rule 6688

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

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {100+e^3}{1+2 x+x^2}} \left (-999-10 e^3-\left (197+2 e^3\right ) x+3 x^2+x^3\right )}{1+3 x+3 x^2+x^3} \, dx\\ &=\int \frac {e^{\frac {100+e^3}{(1+x)^2}} \left (-999-10 e^3-\left (197+2 e^3\right ) x+3 x^2+x^3\right )}{(1+x)^3} \, dx\\ &=\int \left (e^{\frac {100+e^3}{(1+x)^2}}-\frac {8 e^{\frac {100+e^3}{(1+x)^2}} \left (100+e^3\right )}{(1+x)^3}-\frac {2 e^{\frac {100+e^3}{(1+x)^2}} \left (100+e^3\right )}{(1+x)^2}\right ) \, dx\\ &=-\left (\left (2 \left (100+e^3\right )\right ) \int \frac {e^{\frac {100+e^3}{(1+x)^2}}}{(1+x)^2} \, dx\right )-\left (8 \left (100+e^3\right )\right ) \int \frac {e^{\frac {100+e^3}{(1+x)^2}}}{(1+x)^3} \, dx+\int e^{\frac {100+e^3}{(1+x)^2}} \, dx\\ &=4 e^{\frac {100+e^3}{(1+x)^2}}+e^{\frac {100+e^3}{(1+x)^2}} (1+x)+\left (2 \left (100+e^3\right )\right ) \int \frac {e^{\frac {100+e^3}{(1+x)^2}}}{(1+x)^2} \, dx+\left (2 \left (100+e^3\right )\right ) \operatorname {Subst}\left (\int e^{\left (100+e^3\right ) x^2} \, dx,x,\frac {1}{1+x}\right )\\ &=4 e^{\frac {100+e^3}{(1+x)^2}}+e^{\frac {100+e^3}{(1+x)^2}} (1+x)+\sqrt {\left (100+e^3\right ) \pi } \text {erfi}\left (\frac {\sqrt {100+e^3}}{1+x}\right )-\left (2 \left (100+e^3\right )\right ) \operatorname {Subst}\left (\int e^{\left (100+e^3\right ) x^2} \, dx,x,\frac {1}{1+x}\right )\\ &=4 e^{\frac {100+e^3}{(1+x)^2}}+e^{\frac {100+e^3}{(1+x)^2}} (1+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 17, normalized size = 1.00 \begin {gather*} e^{\frac {100+e^3}{(1+x)^2}} (5+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((100 + E^3)/(1 + 2*x + x^2))*(-999 + E^3*(-10 - 2*x) - 197*x + 3*x^2 + x^3))/(1 + 3*x + 3*x^2 +
x^3),x]

[Out]

E^((100 + E^3)/(1 + x)^2)*(5 + x)

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fricas [A]  time = 0.59, size = 20, normalized size = 1.18 \begin {gather*} {\left (x + 5\right )} e^{\left (\frac {e^{3} + 100}{x^{2} + 2 \, x + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-10)*exp(3)+x^3+3*x^2-197*x-999)*exp((exp(3)+100)/(x^2+2*x+1))/(x^3+3*x^2+3*x+1),x, algorithm=
"fricas")

[Out]

(x + 5)*e^((e^3 + 100)/(x^2 + 2*x + 1))

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giac [B]  time = 0.22, size = 79, normalized size = 4.65 \begin {gather*} x e^{\left (-\frac {x^{2} e^{3} + 100 \, x^{2} + 2 \, x e^{3} + 200 \, x}{x^{2} + 2 \, x + 1} + e^{3} + 100\right )} + 5 \, e^{\left (-\frac {x^{2} e^{3} + 100 \, x^{2} + 2 \, x e^{3} + 200 \, x}{x^{2} + 2 \, x + 1} + e^{3} + 100\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-10)*exp(3)+x^3+3*x^2-197*x-999)*exp((exp(3)+100)/(x^2+2*x+1))/(x^3+3*x^2+3*x+1),x, algorithm=
"giac")

[Out]

x*e^(-(x^2*e^3 + 100*x^2 + 2*x*e^3 + 200*x)/(x^2 + 2*x + 1) + e^3 + 100) + 5*e^(-(x^2*e^3 + 100*x^2 + 2*x*e^3
+ 200*x)/(x^2 + 2*x + 1) + e^3 + 100)

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maple [A]  time = 0.12, size = 16, normalized size = 0.94

method	result	size
risch	\({\mathrm e}^{\frac {{\mathrm e}^{3}+100}{\left (x +1\right )^{2}}} \left (5+x \right )\)	\(16\)
gosper	\(\left (5+x \right ) {\mathrm e}^{\frac {{\mathrm e}^{3}+100}{x^{2}+2 x +1}}\)	\(21\)
norman	\(\frac {x^{3} {\mathrm e}^{\frac {{\mathrm e}^{3}+100}{x^{2}+2 x +1}}+11 x \,{\mathrm e}^{\frac {{\mathrm e}^{3}+100}{x^{2}+2 x +1}}+7 x^{2} {\mathrm e}^{\frac {{\mathrm e}^{3}+100}{x^{2}+2 x +1}}+5 \,{\mathrm e}^{\frac {{\mathrm e}^{3}+100}{x^{2}+2 x +1}}}{\left (x +1\right )^{2}}\)	\(86\)
derivativedivides	\(\left (x +1\right ) {\mathrm e}^{\frac {{\mathrm e}^{3}+100}{\left (x +1\right )^{2}}}+i \sqrt {{\mathrm e}^{3}+100}\, \sqrt {\pi }\, \erf \left (\frac {i \sqrt {{\mathrm e}^{3}+100}}{x +1}\right )-\frac {100 i \sqrt {\pi }\, \erf \left (\frac {i \sqrt {{\mathrm e}^{3}+100}}{x +1}\right )}{\sqrt {{\mathrm e}^{3}+100}}+\frac {400 \,{\mathrm e}^{\frac {{\mathrm e}^{3}+100}{\left (x +1\right )^{2}}}}{{\mathrm e}^{3}+100}-\frac {i \sqrt {\pi }\, {\mathrm e}^{3} \erf \left (\frac {i \sqrt {{\mathrm e}^{3}+100}}{x +1}\right )}{\sqrt {{\mathrm e}^{3}+100}}+\frac {4 \,{\mathrm e}^{3+\frac {{\mathrm e}^{3}+100}{\left (x +1\right )^{2}}}}{{\mathrm e}^{3}+100}\)	\(140\)
default	\(\left (x +1\right ) {\mathrm e}^{\frac {{\mathrm e}^{3}+100}{\left (x +1\right )^{2}}}+i \sqrt {{\mathrm e}^{3}+100}\, \sqrt {\pi }\, \erf \left (\frac {i \sqrt {{\mathrm e}^{3}+100}}{x +1}\right )-\frac {100 i \sqrt {\pi }\, \erf \left (\frac {i \sqrt {{\mathrm e}^{3}+100}}{x +1}\right )}{\sqrt {{\mathrm e}^{3}+100}}+\frac {400 \,{\mathrm e}^{\frac {{\mathrm e}^{3}+100}{\left (x +1\right )^{2}}}}{{\mathrm e}^{3}+100}-\frac {i \sqrt {\pi }\, {\mathrm e}^{3} \erf \left (\frac {i \sqrt {{\mathrm e}^{3}+100}}{x +1}\right )}{\sqrt {{\mathrm e}^{3}+100}}+\frac {4 \,{\mathrm e}^{3+\frac {{\mathrm e}^{3}+100}{\left (x +1\right )^{2}}}}{{\mathrm e}^{3}+100}\)	\(140\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x-10)*exp(3)+x^3+3*x^2-197*x-999)*exp((exp(3)+100)/(x^2+2*x+1))/(x^3+3*x^2+3*x+1),x,method=_RETURNVER
BOSE)

[Out]

exp((exp(3)+100)/(x+1)^2)*(5+x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {999 \, e^{\left (\frac {e^{3}}{x^{2} + 2 \, x + 1} + \frac {100}{x^{2} + 2 \, x + 1}\right )}}{2 \, {\left (e^{3} + 100\right )}} + \int \frac {{\left (x^{3} + 3 \, x^{2} - x {\left (2 \, e^{3} + 197\right )} - 10 \, e^{3}\right )} e^{\left (\frac {e^{3}}{x^{2} + 2 \, x + 1} + \frac {100}{x^{2} + 2 \, x + 1}\right )}}{x^{3} + 3 \, x^{2} + 3 \, x + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-10)*exp(3)+x^3+3*x^2-197*x-999)*exp((exp(3)+100)/(x^2+2*x+1))/(x^3+3*x^2+3*x+1),x, algorithm=
"maxima")

[Out]

999/2*e^(e^3/(x^2 + 2*x + 1) + 100/(x^2 + 2*x + 1))/(e^3 + 100) + integrate((x^3 + 3*x^2 - x*(2*e^3 + 197) - 1
0*e^3)*e^(e^3/(x^2 + 2*x + 1) + 100/(x^2 + 2*x + 1))/(x^3 + 3*x^2 + 3*x + 1), x)

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mupad [B]  time = 1.95, size = 31, normalized size = 1.82 \begin {gather*} {\mathrm {e}}^{\frac {100}{x^2+2\,x+1}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^3}{x^2+2\,x+1}}\,\left (x+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(3) + 100)/(2*x + x^2 + 1))*(197*x - 3*x^2 - x^3 + exp(3)*(2*x + 10) + 999))/(3*x + 3*x^2 + x^3
+ 1),x)

[Out]

exp(100/(2*x + x^2 + 1))*exp(exp(3)/(2*x + x^2 + 1))*(x + 5)

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sympy [A]  time = 0.31, size = 17, normalized size = 1.00 \begin {gather*} \left (x + 5\right ) e^{\frac {e^{3} + 100}{x^{2} + 2 x + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-10)*exp(3)+x**3+3*x**2-197*x-999)*exp((exp(3)+100)/(x**2+2*x+1))/(x**3+3*x**2+3*x+1),x)

[Out]

(x + 5)*exp((exp(3) + 100)/(x**2 + 2*x + 1))

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