3.16.44 \(\int \frac {-3+e^x (1-x) \log (8+4 \log (4))}{x^2} \, dx\)

Optimal. Leaf size=19 \[ \frac {3+x-e^x \log (4 (2+\log (4)))}{x} \]

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Rubi [A]  time = 0.04, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {14, 2197} \begin {gather*} \frac {3}{x}-\frac {e^x \log (8+\log (256))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 + E^x*(1 - x)*Log[8 + 4*Log[4]])/x^2,x]

[Out]

3/x - (E^x*Log[8 + Log[256]])/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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {3}{x^2}-\frac {e^x (-1+x) \log (8+\log (256))}{x^2}\right ) \, dx\\ &=\frac {3}{x}-\log (8+\log (256)) \int \frac {e^x (-1+x)}{x^2} \, dx\\ &=\frac {3}{x}-\frac {e^x \log (8+\log (256))}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 16, normalized size = 0.84 \begin {gather*} -\frac {-3+e^x \log (8+\log (256))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 + E^x*(1 - x)*Log[8 + 4*Log[4]])/x^2,x]

[Out]

-((-3 + E^x*Log[8 + Log[256]])/x)

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fricas [A]  time = 0.73, size = 17, normalized size = 0.89 \begin {gather*} -\frac {e^{x} \log \left (8 \, \log \relax (2) + 8\right ) - 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(x)*log(8*log(2)+8)-3)/x^2,x, algorithm="fricas")

[Out]

-(e^x*log(8*log(2) + 8) - 3)/x

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giac [A]  time = 0.17, size = 21, normalized size = 1.11 \begin {gather*} -\frac {3 \, e^{x} \log \relax (2) + e^{x} \log \left (\log \relax (2) + 1\right ) - 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(x)*log(8*log(2)+8)-3)/x^2,x, algorithm="giac")

[Out]

-(3*e^x*log(2) + e^x*log(log(2) + 1) - 3)/x

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maple [A]  time = 0.03, size = 22, normalized size = 1.16

method	result	size
norman	\(\frac {3+\left (-3 \ln \relax (2)-\ln \left (1+\ln \relax (2)\right )\right ) {\mathrm e}^{x}}{x}\)	\(22\)
risch	\(\frac {3}{x}-\frac {\left (3 \ln \relax (2)+\ln \left (1+\ln \relax (2)\right )\right ) {\mathrm e}^{x}}{x}\)	\(24\)
default	\(-\frac {{\mathrm e}^{x} \ln \left (1+\ln \relax (2)\right )}{x}+\frac {3}{x}-\frac {3 \ln \relax (2) {\mathrm e}^{x}}{x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1-x)*exp(x)*ln(8*ln(2)+8)-3)/x^2,x,method=_RETURNVERBOSE)

[Out]

(3+(-3*ln(2)-ln(1+ln(2)))*exp(x))/x

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maxima [C]  time = 0.47, size = 30, normalized size = 1.58 \begin {gather*} -{\rm Ei}\relax (x) \log \left (8 \, \log \relax (2) + 8\right ) + \Gamma \left (-1, -x\right ) \log \left (8 \, \log \relax (2) + 8\right ) + \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(x)*log(8*log(2)+8)-3)/x^2,x, algorithm="maxima")

[Out]

-Ei(x)*log(8*log(2) + 8) + gamma(-1, -x)*log(8*log(2) + 8) + 3/x

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mupad [B]  time = 1.01, size = 17, normalized size = 0.89 \begin {gather*} -\frac {{\mathrm {e}}^x\,\ln \left (8\,\ln \relax (2)+8\right )-3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*log(8*log(2) + 8)*(x - 1) + 3)/x^2,x)

[Out]

-(exp(x)*log(8*log(2) + 8) - 3)/x

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sympy [A]  time = 0.12, size = 20, normalized size = 1.05 \begin {gather*} \frac {\left (- 3 \log {\relax (2 )} - \log {\left (\log {\relax (2 )} + 1 \right )}\right ) e^{x}}{x} + \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(x)*ln(8*ln(2)+8)-3)/x**2,x)

[Out]

(-3*log(2) - log(log(2) + 1))*exp(x)/x + 3/x

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