\(\int \frac {-e^5+e^{\frac {4+e^5 (12-3 x)-13 x+3 x^2+(4-x) \log (3)}{e^5}} (-13-3 e^5+6 x-\log (3))}{e^5} \, dx\) [931]

Optimal result

Integrand size = 58, antiderivative size = 28 \[ \int \frac {-e^5+e^{\frac {4+e^5 (12-3 x)-13 x+3 x^2+(4-x) \log (3)}{e^5}} \left (-13-3 e^5+6 x-\log (3)\right )}{e^5} \, dx=e^{\frac {(-4+x) \left (-1+3 \left (-e^5+x\right )-\log (3)\right )}{e^5}}-x \]

[Out]

exp((x-4)/exp(5)*(-3*exp(5)+3*x-1-ln(3)))-x

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {12, 2276, 2268} \[ \int \frac {-e^5+e^{\frac {4+e^5 (12-3 x)-13 x+3 x^2+(4-x) \log (3)}{e^5}} \left (-13-3 e^5+6 x-\log (3)\right )}{e^5} \, dx=\exp \left (\frac {3 x^2}{e^5}-\frac {x \left (13+3 e^5+\log (3)\right )}{e^5}+\frac {4+12 e^5+\log (81)}{e^5}\right )-x \]

[In]

Int[(-E^5 + E^((4 + E^5*(12 - 3*x) - 13*x + 3*x^2 + (4 - x)*Log[3])/E^5)*(-13 - 3*E^5 + 6*x - Log[3]))/E^5,x]

[Out]

E^((3*x^2)/E^5 - (x*(13 + 3*E^5 + Log[3]))/E^5 + (4 + 12*E^5 + Log[81])/E^5) - x

Rule 12

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

Rule 2268

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

Rule 2276

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (-e^5+\exp \left (\frac {4+e^5 (12-3 x)-13 x+3 x^2+(4-x) \log (3)}{e^5}\right ) \left (-13-3 e^5+6 x-\log (3)\right )\right ) \, dx}{e^5} \\ & = -x+\frac {\int \exp \left (\frac {4+e^5 (12-3 x)-13 x+3 x^2+(4-x) \log (3)}{e^5}\right ) \left (-13-3 e^5+6 x-\log (3)\right ) \, dx}{e^5} \\ & = -x+\frac {\int \exp \left (\frac {3 x^2}{e^5}+\frac {4 \left (1+3 e^5+\log (3)\right )}{e^5}-\frac {x \left (13+3 e^5+\log (3)\right )}{e^5}\right ) \left (-13-3 e^5+6 x-\log (3)\right ) \, dx}{e^5} \\ & = \exp \left (\frac {3 x^2}{e^5}-\frac {x \left (13+3 e^5+\log (3)\right )}{e^5}+\frac {4+12 e^5+\log (81)}{e^5}\right )-x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {-e^5+e^{\frac {4+e^5 (12-3 x)-13 x+3 x^2+(4-x) \log (3)}{e^5}} \left (-13-3 e^5+6 x-\log (3)\right )}{e^5} \, dx=e^{5+\frac {3 x^2}{e^5}-\frac {x \left (13+3 e^5+\log (3)\right )}{e^5}+\frac {4+7 e^5+\log (81)}{e^5}}-x \]

[In]

Integrate[(-E^5 + E^((4 + E^5*(12 - 3*x) - 13*x + 3*x^2 + (4 - x)*Log[3])/E^5)*(-13 - 3*E^5 + 6*x - Log[3]))/E
^5,x]

[Out]

E^(5 + (3*x^2)/E^5 - (x*(13 + 3*E^5 + Log[3]))/E^5 + (4 + 7*E^5 + Log[81])/E^5) - x

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86

method	result	size
risch	\(-x +{\mathrm e}^{-\left (x -4\right ) \left (-3 x +\ln \left (3\right )+3 \,{\mathrm e}^{5}+1\right ) {\mathrm e}^{-5}}\)	\(24\)
norman	\(-x +{\mathrm e}^{\left (\left (-x +4\right ) \ln \left (3\right )+\left (-3 x +12\right ) {\mathrm e}^{5}+3 x^{2}-13 x +4\right ) {\mathrm e}^{-5}}\)	\(37\)
default	\({\mathrm e}^{-5} \left ({\mathrm e}^{\left (\left (-x +4\right ) \ln \left (3\right )+\left (-3 x +12\right ) {\mathrm e}^{5}+3 x^{2}-13 x +4\right ) {\mathrm e}^{-5}} {\mathrm e}^{5}-x \,{\mathrm e}^{5}\right )\)	\(47\)
parallelrisch	\({\mathrm e}^{-5} \left ({\mathrm e}^{\left (\left (-x +4\right ) \ln \left (3\right )+\left (-3 x +12\right ) {\mathrm e}^{5}+3 x^{2}-13 x +4\right ) {\mathrm e}^{-5}} {\mathrm e}^{5}-x \,{\mathrm e}^{5}\right )\)	\(47\)
parts	\(-x +{\mathrm e}^{3 \,{\mathrm e}^{-5} x^{2}+\left (-\ln \left (3\right )-3 \,{\mathrm e}^{5}-13\right ) {\mathrm e}^{-5} x +\left (4 \ln \left (3\right )+12 \,{\mathrm e}^{5}+4\right ) {\mathrm e}^{-5}}\)	\(47\)

[In]

int(((-ln(3)-3*exp(5)+6*x-13)*exp(((-x+4)*ln(3)+(-3*x+12)*exp(5)+3*x^2-13*x+4)/exp(5))-exp(5))/exp(5),x,method
=_RETURNVERBOSE)

[Out]

-x+exp(-(x-4)*(-3*x+ln(3)+3*exp(5)+1)*exp(-5))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-e^5+e^{\frac {4+e^5 (12-3 x)-13 x+3 x^2+(4-x) \log (3)}{e^5}} \left (-13-3 e^5+6 x-\log (3)\right )}{e^5} \, dx=-x + e^{\left ({\left (3 \, x^{2} - 3 \, {\left (x - 4\right )} e^{5} - {\left (x - 4\right )} \log \left (3\right ) - 13 \, x + 4\right )} e^{\left (-5\right )}\right )} \]

[In]

integrate(((-log(3)-3*exp(5)+6*x-13)*exp(((-x+4)*log(3)+(-3*x+12)*exp(5)+3*x^2-13*x+4)/exp(5))-exp(5))/exp(5),
x, algorithm="fricas")

[Out]

-x + e^((3*x^2 - 3*(x - 4)*e^5 - (x - 4)*log(3) - 13*x + 4)*e^(-5))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {-e^5+e^{\frac {4+e^5 (12-3 x)-13 x+3 x^2+(4-x) \log (3)}{e^5}} \left (-13-3 e^5+6 x-\log (3)\right )}{e^5} \, dx=- x + e^{\frac {3 x^{2} - 13 x + \left (4 - x\right ) \log {\left (3 \right )} + \left (12 - 3 x\right ) e^{5} + 4}{e^{5}}} \]

[In]

integrate(((-ln(3)-3*exp(5)+6*x-13)*exp(((-x+4)*ln(3)+(-3*x+12)*exp(5)+3*x**2-13*x+4)/exp(5))-exp(5))/exp(5),x
)

[Out]

-x + exp((3*x**2 - 13*x + (4 - x)*log(3) + (12 - 3*x)*exp(5) + 4)*exp(-5))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {-e^5+e^{\frac {4+e^5 (12-3 x)-13 x+3 x^2+(4-x) \log (3)}{e^5}} \left (-13-3 e^5+6 x-\log (3)\right )}{e^5} \, dx=-{\left (x e^{5} - e^{\left ({\left (3 \, x^{2} - 3 \, {\left (x - 4\right )} e^{5} - {\left (x - 4\right )} \log \left (3\right ) - 13 \, x + 4\right )} e^{\left (-5\right )} + 5\right )}\right )} e^{\left (-5\right )} \]

[In]

integrate(((-log(3)-3*exp(5)+6*x-13)*exp(((-x+4)*log(3)+(-3*x+12)*exp(5)+3*x^2-13*x+4)/exp(5))-exp(5))/exp(5),
x, algorithm="maxima")

[Out]

-(x*e^5 - e^((3*x^2 - 3*(x - 4)*e^5 - (x - 4)*log(3) - 13*x + 4)*e^(-5) + 5))*e^(-5)

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.30 (sec) , antiderivative size = 166, normalized size of antiderivative = 5.93 \[ \int \frac {-e^5+e^{\frac {4+e^5 (12-3 x)-13 x+3 x^2+(4-x) \log (3)}{e^5}} \left (-13-3 e^5+6 x-\log (3)\right )}{e^5} \, dx=-\frac {1}{2} \, {\left (-i \, \sqrt {3} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{6} i \, \sqrt {3} {\left (6 \, x - 3 \, e^{5} - \log \left (3\right ) - 13\right )} e^{\left (-\frac {5}{2}\right )}\right ) e^{\left (-\frac {1}{12} \, {\left (6 \, e^{5} \log \left (3\right ) + \log \left (3\right )^{2} + 9 \, e^{10} - 66 \, e^{5} - 22 \, \log \left (3\right ) + 121\right )} e^{\left (-5\right )} + \frac {15}{2}\right )} + i \, \sqrt {3} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{6} i \, \sqrt {3} {\left (6 \, x - 3 \, e^{5} - \log \left (3\right ) - 13\right )} e^{\left (-\frac {5}{2}\right )}\right ) e^{\left (-\frac {1}{12} \, {\left (6 \, e^{5} \log \left (3\right ) + \log \left (3\right )^{2} + 9 \, e^{10} - 126 \, e^{5} - 22 \, \log \left (3\right ) + 121\right )} e^{\left (-5\right )} + \frac {5}{2}\right )} + 2 \, x e^{5} - 2 \, e^{\left ({\left (3 \, x^{2} - 3 \, x e^{5} - x \log \left (3\right ) - 13 \, x + 12 \, e^{5} + 4 \, \log \left (3\right ) + 4\right )} e^{\left (-5\right )} + 5\right )}\right )} e^{\left (-5\right )} \]

[In]

integrate(((-log(3)-3*exp(5)+6*x-13)*exp(((-x+4)*log(3)+(-3*x+12)*exp(5)+3*x^2-13*x+4)/exp(5))-exp(5))/exp(5),
x, algorithm="giac")

[Out]

-1/2*(-I*sqrt(3)*sqrt(pi)*erf(-1/6*I*sqrt(3)*(6*x - 3*e^5 - log(3) - 13)*e^(-5/2))*e^(-1/12*(6*e^5*log(3) + lo
g(3)^2 + 9*e^10 - 66*e^5 - 22*log(3) + 121)*e^(-5) + 15/2) + I*sqrt(3)*sqrt(pi)*erf(-1/6*I*sqrt(3)*(6*x - 3*e^
5 - log(3) - 13)*e^(-5/2))*e^(-1/12*(6*e^5*log(3) + log(3)^2 + 9*e^10 - 126*e^5 - 22*log(3) + 121)*e^(-5) + 5/
2) + 2*x*e^5 - 2*e^((3*x^2 - 3*x*e^5 - x*log(3) - 13*x + 12*e^5 + 4*log(3) + 4)*e^(-5) + 5))*e^(-5)

Mupad [B] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {-e^5+e^{\frac {4+e^5 (12-3 x)-13 x+3 x^2+(4-x) \log (3)}{e^5}} \left (-13-3 e^5+6 x-\log (3)\right )}{e^5} \, dx=\frac {3^{4\,{\mathrm {e}}^{-5}}\,{\mathrm {e}}^{3\,x^2\,{\mathrm {e}}^{-5}}\,{\mathrm {e}}^{4\,{\mathrm {e}}^{-5}}\,{\mathrm {e}}^{-3\,x}\,{\mathrm {e}}^{12}\,{\mathrm {e}}^{-13\,x\,{\mathrm {e}}^{-5}}}{3^{x\,{\mathrm {e}}^{-5}}}-x \]

[In]

int(-exp(-5)*(exp(5) + exp(-exp(-5)*(13*x + log(3)*(x - 4) - 3*x^2 + exp(5)*(3*x - 12) - 4))*(3*exp(5) - 6*x +
log(3) + 13)),x)

[Out]

(3^(4*exp(-5))*exp(3*x^2*exp(-5))*exp(4*exp(-5))*exp(-3*x)*exp(12)*exp(-13*x*exp(-5)))/3^(x*exp(-5)) - x