3.26.6 \(\int \frac {1}{9} (9+e^{\frac {1}{9} (1314-198 e^5+9 e^{10}+x^2-6 x^3+9 x^4)} (9+2 x^2-18 x^3+36 x^4)) \, dx\)

Optimal. Leaf size=30 \[ x+e^{25+\left (11-e^5\right )^2+\left (\frac {x}{3}-x^2\right )^2} x \]

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Rubi [B]  time = 0.08, antiderivative size = 63, normalized size of antiderivative = 2.10, number of steps used = 3, number of rules used = 2, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {12, 2288} \begin {gather*} \frac {\left (18 x^4-9 x^3+x^2\right ) \exp \left (\frac {1}{9} \left (9 x^4-6 x^3+x^2+9 \left (146-22 e^5+e^{10}\right )\right )\right )}{18 x^3-9 x^2+x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9 + E^((1314 - 198*E^5 + 9*E^10 + x^2 - 6*x^3 + 9*x^4)/9)*(9 + 2*x^2 - 18*x^3 + 36*x^4))/9,x]

[Out]

x + (E^((9*(146 - 22*E^5 + E^10) + x^2 - 6*x^3 + 9*x^4)/9)*(x^2 - 9*x^3 + 18*x^4))/(x - 9*x^2 + 18*x^3)

Rule 12

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{9} \int \left (9+e^{\frac {1}{9} \left (1314-198 e^5+9 e^{10}+x^2-6 x^3+9 x^4\right )} \left (9+2 x^2-18 x^3+36 x^4\right )\right ) \, dx\\ &=x+\frac {1}{9} \int e^{\frac {1}{9} \left (1314-198 e^5+9 e^{10}+x^2-6 x^3+9 x^4\right )} \left (9+2 x^2-18 x^3+36 x^4\right ) \, dx\\ &=x+\frac {\exp \left (\frac {1}{9} \left (9 \left (146-22 e^5+e^{10}\right )+x^2-6 x^3+9 x^4\right )\right ) \left (x^2-9 x^3+18 x^4\right )}{x-9 x^2+18 x^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 33, normalized size = 1.10 \begin {gather*} \left (1+e^{146-22 e^5+e^{10}+\frac {x^2}{9}-\frac {2 x^3}{3}+x^4}\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9 + E^((1314 - 198*E^5 + 9*E^10 + x^2 - 6*x^3 + 9*x^4)/9)*(9 + 2*x^2 - 18*x^3 + 36*x^4))/9,x]

[Out]

(1 + E^(146 - 22*E^5 + E^10 + x^2/9 - (2*x^3)/3 + x^4))*x

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fricas [A]  time = 1.03, size = 26, normalized size = 0.87 \begin {gather*} x e^{\left (x^{4} - \frac {2}{3} \, x^{3} + \frac {1}{9} \, x^{2} + e^{10} - 22 \, e^{5} + 146\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(36*x^4-18*x^3+2*x^2+9)*exp(exp(5)^2-22*exp(5)+x^4-2/3*x^3+1/9*x^2+146)+1,x, algorithm="fricas")

[Out]

x*e^(x^4 - 2/3*x^3 + 1/9*x^2 + e^10 - 22*e^5 + 146) + x

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giac [A]  time = 0.18, size = 26, normalized size = 0.87 \begin {gather*} x e^{\left (x^{4} - \frac {2}{3} \, x^{3} + \frac {1}{9} \, x^{2} + e^{10} - 22 \, e^{5} + 146\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(36*x^4-18*x^3+2*x^2+9)*exp(exp(5)^2-22*exp(5)+x^4-2/3*x^3+1/9*x^2+146)+1,x, algorithm="giac")

[Out]

x*e^(x^4 - 2/3*x^3 + 1/9*x^2 + e^10 - 22*e^5 + 146) + x

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maple [A]  time = 0.08, size = 27, normalized size = 0.90

method	result	size
risch	\(x +x \,{\mathrm e}^{{\mathrm e}^{10}-22 \,{\mathrm e}^{5}+x^{4}-\frac {2 x^{3}}{3}+\frac {x^{2}}{9}+146}\)	\(27\)
default	\(x +x \,{\mathrm e}^{{\mathrm e}^{10}-22 \,{\mathrm e}^{5}+x^{4}-\frac {2 x^{3}}{3}+\frac {x^{2}}{9}+146}\)	\(29\)
norman	\(x +x \,{\mathrm e}^{{\mathrm e}^{10}-22 \,{\mathrm e}^{5}+x^{4}-\frac {2 x^{3}}{3}+\frac {x^{2}}{9}+146}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*(36*x^4-18*x^3+2*x^2+9)*exp(exp(5)^2-22*exp(5)+x^4-2/3*x^3+1/9*x^2+146)+1,x,method=_RETURNVERBOSE)

[Out]

x+x*exp(exp(10)-22*exp(5)+x^4-2/3*x^3+1/9*x^2+146)

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maxima [A]  time = 0.62, size = 26, normalized size = 0.87 \begin {gather*} x e^{\left (x^{4} - \frac {2}{3} \, x^{3} + \frac {1}{9} \, x^{2} + e^{10} - 22 \, e^{5} + 146\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(36*x^4-18*x^3+2*x^2+9)*exp(exp(5)^2-22*exp(5)+x^4-2/3*x^3+1/9*x^2+146)+1,x, algorithm="maxima")

[Out]

x*e^(x^4 - 2/3*x^3 + 1/9*x^2 + e^10 - 22*e^5 + 146) + x

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mupad [B]  time = 1.40, size = 30, normalized size = 1.00 \begin {gather*} x+\frac {x\,{\left ({\mathrm {e}}^{x^2}\right )}^{1/9}\,{\mathrm {e}}^{-22\,{\mathrm {e}}^5}\,{\mathrm {e}}^{x^4}\,{\mathrm {e}}^{146}\,{\mathrm {e}}^{{\mathrm {e}}^{10}}}{{\left ({\mathrm {e}}^{x^3}\right )}^{2/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(10) - 22*exp(5) + x^2/9 - (2*x^3)/3 + x^4 + 146)*(2*x^2 - 18*x^3 + 36*x^4 + 9))/9 + 1,x)

[Out]

x + (x*exp(x^2)^(1/9)*exp(-22*exp(5))*exp(x^4)*exp(146)*exp(exp(10)))/exp(x^3)^(2/3)

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sympy [A]  time = 0.14, size = 29, normalized size = 0.97 \begin {gather*} x e^{x^{4} - \frac {2 x^{3}}{3} + \frac {x^{2}}{9} - 22 e^{5} + 146 + e^{10}} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(36*x**4-18*x**3+2*x**2+9)*exp(exp(5)**2-22*exp(5)+x**4-2/3*x**3+1/9*x**2+146)+1,x)

[Out]

x*exp(x**4 - 2*x**3/3 + x**2/9 - 22*exp(5) + 146 + exp(10)) + x

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