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

Optimal. Leaf size=17 \[ \log \left (\frac {1}{16} e^{-4 e^x (-1+\log (2))} x\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 0.82, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {14, 2194} \begin {gather*} \log (x)+4 e^x (1-\log (2)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + E^x*(4*x - 4*x*Log[2]))/x,x]

[Out]

4*E^x*(1 - Log[2]) + Log[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} &=\int \left (\frac {1}{x}-4 e^x (-1+\log (2))\right ) \, dx\\ &=\log (x)-(4 (-1+\log (2))) \int e^x \, dx\\ &=4 e^x (1-\log (2))+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 12, normalized size = 0.71 \begin {gather*} -4 e^x (-1+\log (2))+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-4*E^x*(-1 + Log[2]) + Log[x]

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fricas [A]  time = 0.70, size = 11, normalized size = 0.65 \begin {gather*} -4 \, {\left (\log \relax (2) - 1\right )} e^{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(2)+4*x)*exp(x)+1)/x,x, algorithm="fricas")

[Out]

-4*(log(2) - 1)*e^x + log(x)

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giac [A]  time = 0.17, size = 13, normalized size = 0.76 \begin {gather*} -4 \, e^{x} \log \relax (2) + 4 \, e^{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(2)+4*x)*exp(x)+1)/x,x, algorithm="giac")

[Out]

-4*e^x*log(2) + 4*e^x + log(x)

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maple [A]  time = 0.02, size = 13, normalized size = 0.76

method	result	size
norman	\(\left (-4 \ln \relax (2)+4\right ) {\mathrm e}^{x}+\ln \relax (x )\)	\(13\)
default	\(\ln \relax (x )-4 \,{\mathrm e}^{x} \ln \relax (2)+4 \,{\mathrm e}^{x}\)	\(14\)
risch	\(\ln \relax (x )-4 \,{\mathrm e}^{x} \ln \relax (2)+4 \,{\mathrm e}^{x}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x*ln(2)+4*x)*exp(x)+1)/x,x,method=_RETURNVERBOSE)

[Out]

(-4*ln(2)+4)*exp(x)+ln(x)

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maxima [A]  time = 0.34, size = 13, normalized size = 0.76 \begin {gather*} -4 \, e^{x} \log \relax (2) + 4 \, e^{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(2)+4*x)*exp(x)+1)/x,x, algorithm="maxima")

[Out]

-4*e^x*log(2) + 4*e^x + log(x)

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mupad [B]  time = 0.05, size = 11, normalized size = 0.65 \begin {gather*} \ln \relax (x)-{\mathrm {e}}^x\,\left (\ln \left (16\right )-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(4*x - 4*x*log(2)) + 1)/x,x)

[Out]

log(x) - exp(x)*(log(16) - 4)

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sympy [A]  time = 0.10, size = 12, normalized size = 0.71 \begin {gather*} \left (4 - 4 \log {\relax (2 )}\right ) e^{x} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*ln(2)+4*x)*exp(x)+1)/x,x)

[Out]

(4 - 4*log(2))*exp(x) + log(x)

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