3.2.94 \(\int e^{-0.1 x} x \, dx\) [194]

Optimal. Leaf size=16 \[ -100. e^{-0.1 x}-10. e^{-0.1 x} x \]

[Out]

-100.*exp(-.1*x)-10.*exp(-.1*x)*x

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Rubi [A]
time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2207, 2225} \begin {gather*} -10. e^{-0.1 x} x-100. e^{-0.1 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/E^(0.1*x),x]

[Out]

-100./E^(0.1*x) - (10.*x)/E^(0.1*x)

Rule 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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]

Rubi steps

\begin {align*} \int e^{-0.1 x} x \, dx &=-10. e^{-0.1 x} x+10. \int e^{-0.1 x} \, dx\\ &=-100. e^{-0.1 x}-10. e^{-0.1 x} x\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 11, normalized size = 0.69 \begin {gather*} e^{-0.1 x} (-100.-10. x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/E^(0.1*x),x]

[Out]

(-99.99999999999999 - 10.*x)/E^(0.1*x)

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Maple [A]
time = 0.01, size = 15, normalized size = 0.94

method	result	size
gosper	\(- 10.0 \left (x + 10.0\right ) {\mathrm e}^{- 0.1000000000 x}\)	\(10\)
risch	\(\left (- 100.0- 10.0 x \right ) {\mathrm e}^{- 0.1000000000 x}\)	\(11\)
meijerg	\( 100.0- 50.0 \left ( 2.0+ 0.2000000000 x \right ) {\mathrm e}^{- 0.1000000000 x}\)	\(14\)
derivativedivides	\(- 10.0 \,{\mathrm e}^{- 0.1000000000 x} x - 100.0 \,{\mathrm e}^{- 0.1000000000 x}\)	\(15\)
default	\(- 10.0 \,{\mathrm e}^{- 0.1000000000 x} x - 100.0 \,{\mathrm e}^{- 0.1000000000 x}\)	\(15\)
norman	\(- 10.0 \,{\mathrm e}^{- 0.1000000000 x} x - 100.0 \,{\mathrm e}^{- 0.1000000000 x}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-.1*x)*x,x,method=_RETURNVERBOSE)

[Out]

-10.*exp(-.1000000000*x)*x-100.*exp(-.1000000000*x)

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Maxima [A]
time = 0.28, size = 9, normalized size = 0.56 \begin {gather*} -10 \, {\left (x + 10\right )} e^{\left (-\frac {1}{10} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(-.1*x)*x,x, algorithm="maxima")

[Out]

-10*(x + 10)*e^(-1/10*x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(-.1*x)*x,x, algorithm="fricas")

[Out]

Exception raised: TypeError >> An error occurred when FriCAS evaluated '(x)*(exp((x)*(((-0.10000000000000001):
:EXPR INT))))':    Cannot convert the value from type Float to Expression(Integer) .

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Sympy [A]
time = 0.03, size = 10, normalized size = 0.62 \begin {gather*} 1.0 \left (- 10.0 x - 100.0\right ) e^{- 0.1 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(-.1*x)*x,x)

[Out]

1.0*(-10.0*x - 100.0)*exp(-0.1*x)

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Giac [A]
time = 3.03, size = 10, normalized size = 0.62 \begin {gather*} {\left (-10.0000000000000 \, x - 100.000000000000\right )} e^{\left (-0.100000000000000 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(-.1*x)*x,x, algorithm="giac")

[Out]

(-10.0000000000000*x - 100.000000000000)*e^(-0.100000000000000*x)

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Mupad [B]
time = 0.03, size = 9, normalized size = 0.56 \begin {gather*} -10\,{\mathrm {e}}^{-0.1\,x}\,\left (x+10.0\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*exp(-0.1*x),x)

[Out]

-10*exp(-0.1*x)*(x + 10.0)

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