3.14.89 \(\int \frac {e^x (25+10 x+21 x^2+4 x^3+4 x^4)+e^{\frac {e^3+5 x+x^2+2 x^3}{5+x+2 x^2}} (25+e^3 (-1-4 x)+10 x+21 x^2+4 x^3+4 x^4)}{25+10 x+21 x^2+4 x^3+4 x^4} \, dx\)

Optimal. Leaf size=28 \[ -1-e^5+e^x+e^{x+\frac {e^3}{5+x+2 x^2}} \]

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Rubi [A]  time = 0.90, antiderivative size = 22, normalized size of antiderivative = 0.79, number of steps used = 4, number of rules used = 3, integrand size = 106, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {6688, 2194, 6706} \begin {gather*} e^{\frac {e^3}{2 x^2+x+5}+x}+e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(25 + 10*x + 21*x^2 + 4*x^3 + 4*x^4) + E^((E^3 + 5*x + x^2 + 2*x^3)/(5 + x + 2*x^2))*(25 + E^3*(-1 -
4*x) + 10*x + 21*x^2 + 4*x^3 + 4*x^4))/(25 + 10*x + 21*x^2 + 4*x^3 + 4*x^4),x]

[Out]

E^x + E^(x + E^3/(5 + x + 2*x^2))

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 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^x+\frac {e^{x+\frac {e^3}{5+x+2 x^2}} \left (25-e^3+2 \left (5-2 e^3\right ) x+21 x^2+4 x^3+4 x^4\right )}{\left (5+x+2 x^2\right )^2}\right ) \, dx\\ &=\int e^x \, dx+\int \frac {e^{x+\frac {e^3}{5+x+2 x^2}} \left (25-e^3+2 \left (5-2 e^3\right ) x+21 x^2+4 x^3+4 x^4\right )}{\left (5+x+2 x^2\right )^2} \, dx\\ &=e^x+e^{x+\frac {e^3}{5+x+2 x^2}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.81, size = 22, normalized size = 0.79 \begin {gather*} e^x+e^{x+\frac {e^3}{5+x+2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(25 + 10*x + 21*x^2 + 4*x^3 + 4*x^4) + E^((E^3 + 5*x + x^2 + 2*x^3)/(5 + x + 2*x^2))*(25 + E^3*
(-1 - 4*x) + 10*x + 21*x^2 + 4*x^3 + 4*x^4))/(25 + 10*x + 21*x^2 + 4*x^3 + 4*x^4),x]

[Out]

E^x + E^(x + E^3/(5 + x + 2*x^2))

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fricas [A]  time = 0.54, size = 29, normalized size = 1.04 \begin {gather*} e^{x} + e^{\left (\frac {2 \, x^{3} + x^{2} + 5 \, x + e^{3}}{2 \, x^{2} + x + 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-1)*exp(3)+4*x^4+4*x^3+21*x^2+10*x+25)*exp((exp(3)+2*x^3+x^2+5*x)/(2*x^2+x+5))+(4*x^4+4*x^3+2
1*x^2+10*x+25)*exp(x))/(4*x^4+4*x^3+21*x^2+10*x+25),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.53, size = 53, normalized size = 1.89 \begin {gather*} {\left (e^{\left (x + 3\right )} + e^{\left (\frac {10 \, x^{3} - 2 \, x^{2} e^{3} + 5 \, x^{2} - x e^{3} + 25 \, x}{5 \, {\left (2 \, x^{2} + x + 5\right )}} + \frac {1}{5} \, e^{3} + 3\right )}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-1)*exp(3)+4*x^4+4*x^3+21*x^2+10*x+25)*exp((exp(3)+2*x^3+x^2+5*x)/(2*x^2+x+5))+(4*x^4+4*x^3+2
1*x^2+10*x+25)*exp(x))/(4*x^4+4*x^3+21*x^2+10*x+25),x, algorithm="giac")

[Out]

(e^(x + 3) + e^(1/5*(10*x^3 - 2*x^2*e^3 + 5*x^2 - x*e^3 + 25*x)/(2*x^2 + x + 5) + 1/5*e^3 + 3))*e^(-3)

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maple [A]  time = 0.22, size = 30, normalized size = 1.07

method	result	size
risch	\({\mathrm e}^{x}+{\mathrm e}^{\frac {{\mathrm e}^{3}+2 x^{3}+x^{2}+5 x}{2 x^{2}+x +5}}\)	\(30\)
norman	\(\frac {{\mathrm e}^{x} x +{\mathrm e}^{\frac {{\mathrm e}^{3}+2 x^{3}+x^{2}+5 x}{2 x^{2}+x +5}} x +2 \,{\mathrm e}^{x} x^{2}+2 \,{\mathrm e}^{\frac {{\mathrm e}^{3}+2 x^{3}+x^{2}+5 x}{2 x^{2}+x +5}} x^{2}+5 \,{\mathrm e}^{x}+5 \,{\mathrm e}^{\frac {{\mathrm e}^{3}+2 x^{3}+x^{2}+5 x}{2 x^{2}+x +5}}}{2 x^{2}+x +5}\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x-1)*exp(3)+4*x^4+4*x^3+21*x^2+10*x+25)*exp((exp(3)+2*x^3+x^2+5*x)/(2*x^2+x+5))+(4*x^4+4*x^3+21*x^2+
10*x+25)*exp(x))/(4*x^4+4*x^3+21*x^2+10*x+25),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.48, size = 19, normalized size = 0.68 \begin {gather*} e^{\left (x + \frac {e^{3}}{2 \, x^{2} + x + 5}\right )} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-1)*exp(3)+4*x^4+4*x^3+21*x^2+10*x+25)*exp((exp(3)+2*x^3+x^2+5*x)/(2*x^2+x+5))+(4*x^4+4*x^3+2
1*x^2+10*x+25)*exp(x))/(4*x^4+4*x^3+21*x^2+10*x+25),x, algorithm="maxima")

[Out]

e^(x + e^3/(2*x^2 + x + 5)) + e^x

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mupad [B]  time = 1.40, size = 63, normalized size = 2.25 \begin {gather*} {\mathrm {e}}^x+{\mathrm {e}}^{\frac {x^2}{2\,x^2+x+5}}\,{\mathrm {e}}^{\frac {2\,x^3}{2\,x^2+x+5}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^3}{2\,x^2+x+5}}\,{\mathrm {e}}^{\frac {5\,x}{2\,x^2+x+5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((5*x + exp(3) + x^2 + 2*x^3)/(x + 2*x^2 + 5))*(10*x + 21*x^2 + 4*x^3 + 4*x^4 - exp(3)*(4*x + 1) + 25)
+ exp(x)*(10*x + 21*x^2 + 4*x^3 + 4*x^4 + 25))/(10*x + 21*x^2 + 4*x^3 + 4*x^4 + 25),x)

[Out]

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

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sympy [A]  time = 0.58, size = 27, normalized size = 0.96 \begin {gather*} e^{x} + e^{\frac {2 x^{3} + x^{2} + 5 x + e^{3}}{2 x^{2} + x + 5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-1)*exp(3)+4*x**4+4*x**3+21*x**2+10*x+25)*exp((exp(3)+2*x**3+x**2+5*x)/(2*x**2+x+5))+(4*x**4+
4*x**3+21*x**2+10*x+25)*exp(x))/(4*x**4+4*x**3+21*x**2+10*x+25),x)

[Out]

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

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