3.52.34 \(\int \frac {e^{-2 x} (-544 x-e^{3 x} x+66 x^2-2 x^3+e^{2 x} (1+x))}{x} \, dx\)

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

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Rubi [A]  time = 0.31, antiderivative size = 33, normalized size of antiderivative = 1.57, number of steps used = 11, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6688, 2194, 2196, 2176} \begin {gather*} e^{-2 x} x^2-32 e^{-2 x} x+x+256 e^{-2 x}-e^x+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-544*x - E^(3*x)*x + 66*x^2 - 2*x^3 + E^(2*x)*(1 + x))/(E^(2*x)*x),x]

[Out]

256/E^(2*x) - E^x + x - (32*x)/E^(2*x) + x^2/E^(2*x) + Log[x]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

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} &=\int \left (1-e^x+\frac {1}{x}-2 e^{-2 x} \left (272-33 x+x^2\right )\right ) \, dx\\ &=x+\log (x)-2 \int e^{-2 x} \left (272-33 x+x^2\right ) \, dx-\int e^x \, dx\\ &=-e^x+x+\log (x)-2 \int \left (272 e^{-2 x}-33 e^{-2 x} x+e^{-2 x} x^2\right ) \, dx\\ &=-e^x+x+\log (x)-2 \int e^{-2 x} x^2 \, dx+66 \int e^{-2 x} x \, dx-544 \int e^{-2 x} \, dx\\ &=272 e^{-2 x}-e^x+x-33 e^{-2 x} x+e^{-2 x} x^2+\log (x)-2 \int e^{-2 x} x \, dx+33 \int e^{-2 x} \, dx\\ &=\frac {511 e^{-2 x}}{2}-e^x+x-32 e^{-2 x} x+e^{-2 x} x^2+\log (x)-\int e^{-2 x} \, dx\\ &=256 e^{-2 x}-e^x+x-32 e^{-2 x} x+e^{-2 x} x^2+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 28, normalized size = 1.33 \begin {gather*} -e^x+x-2 e^{-2 x} \left (-128+16 x-\frac {x^2}{2}\right )+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-544*x - E^(3*x)*x + 66*x^2 - 2*x^3 + E^(2*x)*(1 + x))/(E^(2*x)*x),x]

[Out]

-E^x + x - (2*(-128 + 16*x - x^2/2))/E^(2*x) + Log[x]

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fricas [A]  time = 0.54, size = 32, normalized size = 1.52 \begin {gather*} {\left (x^{2} + x e^{\left (2 \, x\right )} + e^{\left (2 \, x\right )} \log \relax (x) - 32 \, x - e^{\left (3 \, x\right )} + 256\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(x)^3+(x+1)*exp(x)^2-2*x^3+66*x^2-544*x)/x/exp(x)^2,x, algorithm="fricas")

[Out]

(x^2 + x*e^(2*x) + e^(2*x)*log(x) - 32*x - e^(3*x) + 256)*e^(-2*x)

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giac [A]  time = 0.12, size = 29, normalized size = 1.38 \begin {gather*} x^{2} e^{\left (-2 \, x\right )} - 32 \, x e^{\left (-2 \, x\right )} + x + 256 \, e^{\left (-2 \, x\right )} - e^{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(x)^3+(x+1)*exp(x)^2-2*x^3+66*x^2-544*x)/x/exp(x)^2,x, algorithm="giac")

[Out]

x^2*e^(-2*x) - 32*x*e^(-2*x) + x + 256*e^(-2*x) - e^x + log(x)

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maple [A]  time = 0.03, size = 22, normalized size = 1.05

method	result	size
risch	\(x +\ln \relax (x )-{\mathrm e}^{x}+\left (x^{2}-32 x +256\right ) {\mathrm e}^{-2 x}\)	\(22\)
norman	\(\left (256+x^{2}+x \,{\mathrm e}^{2 x}-32 x -{\mathrm e}^{3 x}\right ) {\mathrm e}^{-2 x}+\ln \relax (x )\)	\(29\)
default	\(x +\ln \relax (x )+256 \,{\mathrm e}^{-2 x}-32 x \,{\mathrm e}^{-2 x}+x^{2} {\mathrm e}^{-2 x}-{\mathrm e}^{x}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*exp(x)^3+(x+1)*exp(x)^2-2*x^3+66*x^2-544*x)/x/exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

x+ln(x)-exp(x)+(x^2-32*x+256)*exp(-2*x)

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maxima [B]  time = 0.36, size = 41, normalized size = 1.95 \begin {gather*} \frac {1}{2} \, {\left (2 \, x^{2} + 2 \, x + 1\right )} e^{\left (-2 \, x\right )} - \frac {33}{2} \, {\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )} + x + 272 \, e^{\left (-2 \, x\right )} - e^{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(x)^3+(x+1)*exp(x)^2-2*x^3+66*x^2-544*x)/x/exp(x)^2,x, algorithm="maxima")

[Out]

1/2*(2*x^2 + 2*x + 1)*e^(-2*x) - 33/2*(2*x + 1)*e^(-2*x) + x + 272*e^(-2*x) - e^x + log(x)

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mupad [B]  time = 0.10, size = 21, normalized size = 1.00 \begin {gather*} x-{\mathrm {e}}^x+\ln \relax (x)+{\mathrm {e}}^{-2\,x}\,\left (x^2-32\,x+256\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2*x)*(544*x + x*exp(3*x) - exp(2*x)*(x + 1) - 66*x^2 + 2*x^3))/x,x)

[Out]

x - exp(x) + log(x) + exp(-2*x)*(x^2 - 32*x + 256)

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sympy [A]  time = 0.13, size = 20, normalized size = 0.95 \begin {gather*} x + \left (x^{2} - 32 x + 256\right ) e^{- 2 x} - e^{x} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(x)**3+(x+1)*exp(x)**2-2*x**3+66*x**2-544*x)/x/exp(x)**2,x)

[Out]

x + (x**2 - 32*x + 256)*exp(-2*x) - exp(x) + log(x)

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