3.3.66 \(\int \frac {e^5-15 x-\log (5)+x \log (x)}{x} \, dx\)

Optimal. Leaf size=17 \[ 3+x+\left (e^5+x-\log (5)\right ) (-17+\log (x)) \]

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Rubi [A]  time = 0.02, antiderivative size = 19, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {14, 43, 2295} \begin {gather*} -16 x+x \log (x)+\left (e^5-\log (5)\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^5 - 15*x - Log[5] + x*Log[x])/x,x]

[Out]

-16*x + x*Log[x] + (E^5 - Log[5])*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {e^5-15 x-\log (5)}{x}+\log (x)\right ) \, dx\\ &=\int \frac {e^5-15 x-\log (5)}{x} \, dx+\int \log (x) \, dx\\ &=-x+x \log (x)+\int \left (-15+\frac {e^5-\log (5)}{x}\right ) \, dx\\ &=-16 x+x \log (x)+\left (e^5-\log (5)\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 20, normalized size = 1.18 \begin {gather*} -16 x+e^5 \log (x)+x \log (x)-\log (5) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5 - 15*x - Log[5] + x*Log[x])/x,x]

[Out]

-16*x + E^5*Log[x] + x*Log[x] - Log[5]*Log[x]

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fricas [A]  time = 1.30, size = 15, normalized size = 0.88 \begin {gather*} {\left (x + e^{5} - \log \relax (5)\right )} \log \relax (x) - 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)-log(5)+exp(5)-15*x)/x,x, algorithm="fricas")

[Out]

(x + e^5 - log(5))*log(x) - 16*x

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giac [A]  time = 0.45, size = 19, normalized size = 1.12 \begin {gather*} x \log \relax (x) + e^{5} \log \relax (x) - \log \relax (5) \log \relax (x) - 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)-log(5)+exp(5)-15*x)/x,x, algorithm="giac")

[Out]

x*log(x) + e^5*log(x) - log(5)*log(x) - 16*x

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maple [A]  time = 0.08, size = 19, normalized size = 1.12

method	result	size
norman	\(x \ln \relax (x )+\left (-\ln \relax (5)+{\mathrm e}^{5}\right ) \ln \relax (x )-16 x\)	\(19\)
default	\(x \ln \relax (x )-16 x -\ln \relax (5) \ln \relax (x )+{\mathrm e}^{5} \ln \relax (x )\)	\(20\)
risch	\(x \ln \relax (x )-16 x -\ln \relax (5) \ln \relax (x )+{\mathrm e}^{5} \ln \relax (x )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*ln(x)-ln(5)+exp(5)-15*x)/x,x,method=_RETURNVERBOSE)

[Out]

x*ln(x)+(-ln(5)+exp(5))*ln(x)-16*x

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maxima [A]  time = 0.43, size = 19, normalized size = 1.12 \begin {gather*} x \log \relax (x) + e^{5} \log \relax (x) - \log \relax (5) \log \relax (x) - 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(x)-log(5)+exp(5)-15*x)/x,x, algorithm="maxima")

[Out]

x*log(x) + e^5*log(x) - log(5)*log(x) - 16*x

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mupad [B]  time = 0.32, size = 19, normalized size = 1.12 \begin {gather*} {\mathrm {e}}^5\,\ln \relax (x)-16\,x-\ln \relax (5)\,\ln \relax (x)+x\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*x - exp(5) + log(5) - x*log(x))/x,x)

[Out]

exp(5)*log(x) - 16*x - log(5)*log(x) + x*log(x)

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sympy [A]  time = 0.14, size = 17, normalized size = 1.00 \begin {gather*} x \log {\relax (x )} - 16 x - \left (- e^{5} + \log {\relax (5 )}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*ln(x)-ln(5)+exp(5)-15*x)/x,x)

[Out]

x*log(x) - 16*x - (-exp(5) + log(5))*log(x)

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