3.47.12 \(\int \frac {1}{5} e^{-2 x} (-2+e^{2 x} (-5+10 x)) \, dx\)

Optimal. Leaf size=21 \[ 16+\frac {e^{-2 x}}{5}-x+x^2+\log ^2(2) \]

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Rubi [A]  time = 0.04, antiderivative size = 16, normalized size of antiderivative = 0.76, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {12, 6688, 2194} \begin {gather*} x^2-x+\frac {e^{-2 x}}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + E^(2*x)*(-5 + 10*x))/(5*E^(2*x)),x]

[Out]

1/(5*E^(2*x)) - x + x^2

Rule 12

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

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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int e^{-2 x} \left (-2+e^{2 x} (-5+10 x)\right ) \, dx\\ &=\frac {1}{5} \int \left (-5-2 e^{-2 x}+10 x\right ) \, dx\\ &=-x+x^2-\frac {2}{5} \int e^{-2 x} \, dx\\ &=\frac {e^{-2 x}}{5}-x+x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.76 \begin {gather*} \frac {e^{-2 x}}{5}-x+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + E^(2*x)*(-5 + 10*x))/(5*E^(2*x)),x]

[Out]

1/(5*E^(2*x)) - x + x^2

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fricas [A]  time = 0.45, size = 21, normalized size = 1.00 \begin {gather*} \frac {1}{5} \, {\left (5 \, {\left (x^{2} - x\right )} e^{\left (2 \, x\right )} + 1\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x-5)*exp(2*x)-2)/exp(2*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 13, normalized size = 0.62 \begin {gather*} x^{2} - x + \frac {1}{5} \, e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x-5)*exp(2*x)-2)/exp(2*x),x, algorithm="giac")

[Out]

x^2 - x + 1/5*e^(-2*x)

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maple [A]  time = 0.02, size = 14, normalized size = 0.67

method	result	size
risch	\(-x +x^{2}+\frac {{\mathrm e}^{-2 x}}{5}\)	\(14\)
derivativedivides	\(-x +x^{2}+\frac {{\mathrm e}^{-2 x}}{5}\)	\(16\)
default	\(-x +x^{2}+\frac {{\mathrm e}^{-2 x}}{5}\)	\(16\)
norman	\(\left (\frac {1}{5}+{\mathrm e}^{2 x} x^{2}-x \,{\mathrm e}^{2 x}\right ) {\mathrm e}^{-2 x}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((10*x-5)*exp(2*x)-2)/exp(2*x),x,method=_RETURNVERBOSE)

[Out]

-x+x^2+1/5*exp(-2*x)

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maxima [A]  time = 0.37, size = 13, normalized size = 0.62 \begin {gather*} x^{2} - x + \frac {1}{5} \, e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x-5)*exp(2*x)-2)/exp(2*x),x, algorithm="maxima")

[Out]

x^2 - x + 1/5*e^(-2*x)

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mupad [B]  time = 3.46, size = 13, normalized size = 0.62 \begin {gather*} \frac {{\mathrm {e}}^{-2\,x}}{5}-x+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-2*x)*((exp(2*x)*(10*x - 5))/5 - 2/5),x)

[Out]

exp(-2*x)/5 - x + x^2

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sympy [A]  time = 0.09, size = 12, normalized size = 0.57 \begin {gather*} x^{2} - x + \frac {e^{- 2 x}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((10*x-5)*exp(2*x)-2)/exp(2*x),x)

[Out]

x**2 - x + exp(-2*x)/5

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