3.23.96 \(\int e^{-x} (-3+e^{\frac {1}{3} (59-42 x+15 x^2)} (-15+10 x)) \, dx\)

Optimal. Leaf size=27 \[ e^{-x} \left (3+e^{\frac {11}{3}+4 (-2+x)^2+2 x+x^2}\right ) \]

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Rubi [A]  time = 0.15, antiderivative size = 22, normalized size of antiderivative = 0.81, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6742, 2194, 2236} \begin {gather*} e^{5 x^2-15 x+\frac {59}{3}}+3 e^{-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 + E^((59 - 42*x + 15*x^2)/3)*(-15 + 10*x))/E^x,x]

[Out]

3/E^x + E^(59/3 - 15*x + 5*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 2236

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-3 e^{-x}+5 e^{\frac {59}{3}-15 x+5 x^2} (-3+2 x)\right ) \, dx\\ &=-\left (3 \int e^{-x} \, dx\right )+5 \int e^{\frac {59}{3}-15 x+5 x^2} (-3+2 x) \, dx\\ &=3 e^{-x}+e^{\frac {59}{3}-15 x+5 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 22, normalized size = 0.81 \begin {gather*} 3 e^{-x}+e^{\frac {59}{3}-15 x+5 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 + E^((59 - 42*x + 15*x^2)/3)*(-15 + 10*x))/E^x,x]

[Out]

3/E^x + E^(59/3 - 15*x + 5*x^2)

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fricas [A]  time = 0.94, size = 18, normalized size = 0.67 \begin {gather*} e^{\left (5 \, x^{2} - 15 \, x + \frac {59}{3}\right )} + 3 \, e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-15)*exp(5*x^2-14*x+59/3)-3)/exp(x),x, algorithm="fricas")

[Out]

e^(5*x^2 - 15*x + 59/3) + 3*e^(-x)

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giac [A]  time = 0.24, size = 18, normalized size = 0.67 \begin {gather*} e^{\left (5 \, x^{2} - 15 \, x + \frac {59}{3}\right )} + 3 \, e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-15)*exp(5*x^2-14*x+59/3)-3)/exp(x),x, algorithm="giac")

[Out]

e^(5*x^2 - 15*x + 59/3) + 3*e^(-x)

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maple [A]  time = 0.06, size = 19, normalized size = 0.70

method	result	size
default	\(3 \,{\mathrm e}^{-x}+{\mathrm e}^{-15 x +5 x^{2}+\frac {59}{3}}\)	\(19\)
norman	\(\left (3+{\mathrm e}^{5 x^{2}-14 x +\frac {59}{3}}\right ) {\mathrm e}^{-x}\)	\(19\)
risch	\(3 \,{\mathrm e}^{-x}+{\mathrm e}^{-15 x +5 x^{2}+\frac {59}{3}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x-15)*exp(5*x^2-14*x+59/3)-3)/exp(x),x,method=_RETURNVERBOSE)

[Out]

3/exp(x)+exp(-15*x+5*x^2+59/3)

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maxima [C]  time = 0.75, size = 93, normalized size = 3.44 \begin {gather*} \frac {3}{2} i \, \sqrt {5} \sqrt {\pi } \operatorname {erf}\left (i \, \sqrt {5} x - \frac {3}{2} i \, \sqrt {5}\right ) e^{\frac {101}{12}} + \frac {1}{10} \, \sqrt {5} {\left (\frac {15 \, \sqrt {\pi } {\left (2 \, x - 3\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {5} \sqrt {-{\left (2 \, x - 3\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x - 3\right )}^{2}}} + 2 \, \sqrt {5} e^{\left (\frac {5}{4} \, {\left (2 \, x - 3\right )}^{2}\right )}\right )} e^{\frac {101}{12}} + 3 \, e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-15)*exp(5*x^2-14*x+59/3)-3)/exp(x),x, algorithm="maxima")

[Out]

3/2*I*sqrt(5)*sqrt(pi)*erf(I*sqrt(5)*x - 3/2*I*sqrt(5))*e^(101/12) + 1/10*sqrt(5)*(15*sqrt(pi)*(2*x - 3)*(erf(
1/2*sqrt(5)*sqrt(-(2*x - 3)^2)) - 1)/sqrt(-(2*x - 3)^2) + 2*sqrt(5)*e^(5/4*(2*x - 3)^2))*e^(101/12) + 3*e^(-x)

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mupad [B]  time = 0.09, size = 20, normalized size = 0.74 \begin {gather*} 3\,{\mathrm {e}}^{-x}+{\mathrm {e}}^{-15\,x}\,{\mathrm {e}}^{59/3}\,{\mathrm {e}}^{5\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-x)*(exp(5*x^2 - 14*x + 59/3)*(10*x - 15) - 3),x)

[Out]

3*exp(-x) + exp(-15*x)*exp(59/3)*exp(5*x^2)

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sympy [A]  time = 0.17, size = 20, normalized size = 0.74 \begin {gather*} e^{- x} e^{5 x^{2} - 14 x + \frac {59}{3}} + 3 e^{- x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-15)*exp(5*x**2-14*x+59/3)-3)/exp(x),x)

[Out]

exp(-x)*exp(5*x**2 - 14*x + 59/3) + 3*exp(-x)

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