3.89.51 \(\int \frac {-x-e^{5+e^5 x} x}{x} \, dx\)

Optimal. Leaf size=23 \[ 3-e^{e^5 x}-\left (4+e^2\right )^2-x \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.57, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {14, 2194} \begin {gather*} -x-e^{e^5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x - E^(5 + E^5*x)*x)/x,x]

[Out]

-E^(E^5*x) - 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 (-1-e^{5+e^5 x}\right ) \, dx\\ &=-x-\int e^{5+e^5 x} \, dx\\ &=-e^{e^5 x}-x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 0.57 \begin {gather*} -e^{e^5 x}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x - E^(5 + E^5*x)*x)/x,x]

[Out]

-E^(E^5*x) - x

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fricas [A]  time = 0.62, size = 23, normalized size = 1.00 \begin {gather*} -\frac {{\left (x^{2} e^{5} + e^{\left (x e^{5} + \log \relax (x) + 5\right )}\right )} e^{\left (-5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x^2*e^5 + e^(x*e^5 + log(x) + 5))*e^(-5)/x

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giac [A]  time = 0.12, size = 11, normalized size = 0.48 \begin {gather*} -x - e^{\left (x e^{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - e^(x*e^5)

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maple [A]  time = 0.04, size = 12, normalized size = 0.52

method	result	size
norman	\(-x -{\mathrm e}^{x \,{\mathrm e}^{5}}\)	\(12\)
risch	\(-x -{\mathrm e}^{x \,{\mathrm e}^{5}}\)	\(12\)
default	\(-x -{\mathrm e}^{{\mathrm e}^{5+\ln \relax (x )}}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x-exp(x*exp(5))

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maxima [A]  time = 0.36, size = 11, normalized size = 0.48 \begin {gather*} -x - e^{\left (x e^{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - e^(x*e^5)

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mupad [B]  time = 5.13, size = 11, normalized size = 0.48 \begin {gather*} -x-{\mathrm {e}}^{x\,{\mathrm {e}}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + exp(exp(log(x) + 5))*exp(log(x) + 5))/x,x)

[Out]

- x - exp(x*exp(5))

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sympy [A]  time = 0.09, size = 8, normalized size = 0.35 \begin {gather*} - x - e^{x e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - exp(x*exp(5))

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