3.68.75 \(\int \frac {-\frac {1}{x^2}-8 e^x x}{8 x} \, dx\)

Optimal. Leaf size=14 \[ 36-e^x+\frac {1}{16 x^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {12, 14, 2194} \begin {gather*} \frac {1}{16 x^2}-e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x^(-2) - 8*E^x*x)/(8*x),x]

[Out]

-E^x + 1/(16*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{8} \int \frac {-\frac {1}{x^2}-8 e^x x}{x} \, dx\\ &=\frac {1}{8} \int \left (-8 e^x-\frac {1}{x^3}\right ) \, dx\\ &=\frac {1}{16 x^2}-\int e^x \, dx\\ &=-e^x+\frac {1}{16 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 0.93 \begin {gather*} -e^x+\frac {1}{16 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x^(-2) - 8*E^x*x)/(8*x),x]

[Out]

-E^x + 1/(16*x^2)

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fricas [A]  time = 0.62, size = 14, normalized size = 1.00 \begin {gather*} -\frac {16 \, x^{2} e^{x} - 1}{16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(-1/x^2-8*exp(x)*x)/x,x, algorithm="fricas")

[Out]

-1/16*(16*x^2*e^x - 1)/x^2

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giac [A]  time = 0.13, size = 14, normalized size = 1.00 \begin {gather*} -\frac {16 \, x^{2} e^{x} - 1}{16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(-1/x^2-8*exp(x)*x)/x,x, algorithm="giac")

[Out]

-1/16*(16*x^2*e^x - 1)/x^2

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maple [A]  time = 0.02, size = 11, normalized size = 0.79

method	result	size
default	\(\frac {1}{16 x^{2}}-{\mathrm e}^{x}\)	\(11\)
risch	\(\frac {1}{16 x^{2}}-{\mathrm e}^{x}\)	\(11\)
norman	\(\frac {\frac {1}{16}-{\mathrm e}^{x} x^{2}}{x^{2}}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*(-1/x^2-8*exp(x)*x)/x,x,method=_RETURNVERBOSE)

[Out]

1/16/x^2-exp(x)

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maxima [A]  time = 0.37, size = 10, normalized size = 0.71 \begin {gather*} \frac {1}{16 \, x^{2}} - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(-1/x^2-8*exp(x)*x)/x,x, algorithm="maxima")

[Out]

1/16/x^2 - e^x

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mupad [B]  time = 4.10, size = 10, normalized size = 0.71 \begin {gather*} \frac {1}{16\,x^2}-{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*exp(x) + 1/(8*x^2))/x,x)

[Out]

1/(16*x^2) - exp(x)

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sympy [A]  time = 0.08, size = 8, normalized size = 0.57 \begin {gather*} - e^{x} + \frac {1}{16 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(-1/x**2-8*exp(x)*x)/x,x)

[Out]

-exp(x) + 1/(16*x**2)

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