3.102.36 \(\int \frac {e^3 (-1-x) (e^{-1+x} (-3 e^{1-x}+x))^{e^3}}{3 e^{1-x}-x} \, dx\)

Optimal. Leaf size=13 \[ \left (-3+e^{-1+x} x\right )^{e^3} \]

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Rubi [A]  time = 0.27, antiderivative size = 21, normalized size of antiderivative = 1.62, number of steps used = 3, number of rules used = 3, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 6688, 6686} \begin {gather*} e^{-e^3} \left (e^x x-3 e\right )^{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^3*(-1 - x)*(E^(-1 + x)*(-3*E^(1 - x) + x))^E^3)/(3*E^(1 - x) - x),x]

[Out]

(-3*E + E^x*x)^E^3/E^E^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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^3 \int \frac {(-1-x) \left (e^{-1+x} \left (-3 e^{1-x}+x\right )\right )^{e^3}}{3 e^{1-x}-x} \, dx\\ &=e^3 \int e^{-e^3+x} (1+x) \left (-3 e+e^x x\right )^{-1+e^3} \, dx\\ &=e^{-e^3} \left (-3 e+e^x x\right )^{e^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 13, normalized size = 1.00 \begin {gather*} \left (-3+e^{-1+x} x\right )^{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3*(-1 - x)*(E^(-1 + x)*(-3*E^(1 - x) + x))^E^3)/(3*E^(1 - x) - x),x]

[Out]

(-3 + E^(-1 + x)*x)^E^3

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fricas [A]  time = 0.54, size = 11, normalized size = 0.85 \begin {gather*} {\left (x e^{\left (x - 1\right )} - 3\right )}^{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x-1)*exp(3)*exp(exp(3)*log((-3*exp(-x+1)+x)/exp(-x+1)))/(3*exp(-x+1)-x),x, algorithm="fricas")

[Out]

(x*e^(x - 1) - 3)^e^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x-1)*exp(3)*exp(exp(3)*log((-3*exp(-x+1)+x)/exp(-x+1)))/(3*exp(-x+1)-x),x, algorithm="giac")

[Out]

integrate(((x - 3*e^(-x + 1))*e^(x - 1))^e^3*(x + 1)*e^3/(x - 3*e^(-x + 1)), x)

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maple [A]  time = 0.23, size = 25, normalized size = 1.92

method	result	size
norman	\({\mathrm e}^{{\mathrm e}^{3} \ln \left (\left (-3 \,{\mathrm e}^{1-x}+x \right ) {\mathrm e}^{x -1}\right )}\)	\(25\)
risch	\({\mathrm e}^{-\frac {{\mathrm e}^{3} \left (i \pi \mathrm {csgn}\left (i \left (-3 \,{\mathrm e}^{1-x}+x \right ) {\mathrm e}^{x -1}\right )^{3}-i \pi \mathrm {csgn}\left (i \left (-3 \,{\mathrm e}^{1-x}+x \right ) {\mathrm e}^{x -1}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{x -1}\right )-i \pi \mathrm {csgn}\left (i \left (-3 \,{\mathrm e}^{1-x}+x \right ) {\mathrm e}^{x -1}\right )^{2} \mathrm {csgn}\left (i \left (-3 \,{\mathrm e}^{1-x}+x \right )\right )+i \pi \,\mathrm {csgn}\left (i \left (-3 \,{\mathrm e}^{1-x}+x \right ) {\mathrm e}^{x -1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x -1}\right ) \mathrm {csgn}\left (i \left (-3 \,{\mathrm e}^{1-x}+x \right )\right )+2 \ln \left ({\mathrm e}^{1-x}\right )-2 \ln \left (-3 \,{\mathrm e}^{1-x}+x \right )\right )}{2}}\)	\(167\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x-1)*exp(3)*exp(exp(3)*ln((-3*exp(1-x)+x)/exp(1-x)))/(3*exp(1-x)-x),x,method=_RETURNVERBOSE)

[Out]

exp(exp(3)*ln((-3*exp(1-x)+x)/exp(1-x)))

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maxima [A]  time = 0.39, size = 19, normalized size = 1.46 \begin {gather*} e^{\left (e^{3} \log \left (x e^{x} - 3 \, e\right ) - e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x-1)*exp(3)*exp(exp(3)*log((-3*exp(-x+1)+x)/exp(-x+1)))/(3*exp(-x+1)-x),x, algorithm="maxima")

[Out]

e^(e^3*log(x*e^x - 3*e) - e^3)

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mupad [B]  time = 7.33, size = 11, normalized size = 0.85 \begin {gather*} {\left (x\,{\mathrm {e}}^{x-1}-3\right )}^{{\mathrm {e}}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3)*(exp(x - 1)*(x - 3*exp(1 - x)))^exp(3)*(x + 1))/(x - 3*exp(1 - x)),x)

[Out]

(x*exp(x - 1) - 3)^exp(3)

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sympy [A]  time = 15.87, size = 15, normalized size = 1.15 \begin {gather*} \left (\left (x - 3 e^{1 - x}\right ) e^{x - 1}\right )^{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x-1)*exp(3)*exp(exp(3)*ln((-3*exp(-x+1)+x)/exp(-x+1)))/(3*exp(-x+1)-x),x)

[Out]

((x - 3*exp(1 - x))*exp(x - 1))**exp(3)

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