3.79.28 \(\int \frac {-e^{5+x} x-2 \log (x)}{x} \, dx\)

Optimal. Leaf size=15 \[ -159-e^{5+x}-\log ^2(x) \]

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {14, 2194, 2301} \begin {gather*} -e^{x+5}-\log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(E^(5 + x)*x) - 2*Log[x])/x,x]

[Out]

-E^(5 + x) - Log[x]^2

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]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-e^{5+x}-\frac {2 \log (x)}{x}\right ) \, dx\\ &=-\left (2 \int \frac {\log (x)}{x} \, dx\right )-\int e^{5+x} \, dx\\ &=-e^{5+x}-\log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 0.93 \begin {gather*} -e^{5+x}-\log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^(5 + x)*x) - 2*Log[x])/x,x]

[Out]

-E^(5 + x) - Log[x]^2

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fricas [A]  time = 0.52, size = 13, normalized size = 0.87 \begin {gather*} -\log \relax (x)^{2} - e^{\left (x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(x)-x*exp(5+x))/x,x, algorithm="fricas")

[Out]

-log(x)^2 - e^(x + 5)

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giac [A]  time = 0.21, size = 13, normalized size = 0.87 \begin {gather*} -\log \relax (x)^{2} - e^{\left (x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(x)-x*exp(5+x))/x,x, algorithm="giac")

[Out]

-log(x)^2 - e^(x + 5)

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maple [A]  time = 0.02, size = 14, normalized size = 0.93

method	result	size
default	\(-\ln \relax (x )^{2}-{\mathrm e}^{5+x}\)	\(14\)
norman	\(-\ln \relax (x )^{2}-{\mathrm e}^{5+x}\)	\(14\)
risch	\(-\ln \relax (x )^{2}-{\mathrm e}^{5+x}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*ln(x)-x*exp(5+x))/x,x,method=_RETURNVERBOSE)

[Out]

-ln(x)^2-exp(5+x)

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maxima [A]  time = 0.36, size = 13, normalized size = 0.87 \begin {gather*} -\log \relax (x)^{2} - e^{\left (x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(x)-x*exp(5+x))/x,x, algorithm="maxima")

[Out]

-log(x)^2 - e^(x + 5)

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mupad [B]  time = 4.98, size = 13, normalized size = 0.87 \begin {gather*} -{\ln \relax (x)}^2-{\mathrm {e}}^{x+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(x) + x*exp(x + 5))/x,x)

[Out]

- exp(x + 5) - log(x)^2

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sympy [A]  time = 0.25, size = 10, normalized size = 0.67 \begin {gather*} - e^{x + 5} - \log {\relax (x )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*ln(x)-x*exp(5+x))/x,x)

[Out]

-exp(x + 5) - log(x)**2

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