3.75.68 \(\int (-125+125 x+125 \log (\frac {e^{10+x}}{x})) \, dx\)

Optimal. Leaf size=15 \[ 5+125 x \log \left (\frac {e^{10+x}}{x}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.87, number of steps used = 3, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2548} \begin {gather*} 125 x \log \left (\frac {e^{x+10}}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-125 + 125*x + 125*Log[E^(10 + x)/x],x]

[Out]

125*x*Log[E^(10 + x)/x]

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-125 x+\frac {125 x^2}{2}+125 \int \log \left (\frac {e^{10+x}}{x}\right ) \, dx\\ &=-125 x+\frac {125 x^2}{2}+125 x \log \left (\frac {e^{10+x}}{x}\right )-125 \int (-1+x) \, dx\\ &=125 x \log \left (\frac {e^{10+x}}{x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 0.87 \begin {gather*} 125 x \log \left (\frac {e^{10+x}}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-125 + 125*x + 125*Log[E^(10 + x)/x],x]

[Out]

125*x*Log[E^(10 + x)/x]

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fricas [A]  time = 0.56, size = 12, normalized size = 0.80 \begin {gather*} 125 \, x \log \left (\frac {e^{\left (x + 10\right )}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(125*log(exp(2*x+9)/x/exp(x-1))+125*x-125,x, algorithm="fricas")

[Out]

125*x*log(e^(x + 10)/x)

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giac [A]  time = 0.22, size = 14, normalized size = 0.93 \begin {gather*} 125 \, x^{2} - 125 \, x \log \relax (x) + 1250 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(125*log(exp(2*x+9)/x/exp(x-1))+125*x-125,x, algorithm="giac")

[Out]

125*x^2 - 125*x*log(x) + 1250*x

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maple [A]  time = 0.08, size = 15, normalized size = 1.00

method	result	size
norman	\(125 x \ln \left (\frac {{\mathrm e}^{11} {\mathrm e}^{x -1}}{x}\right )\)	\(15\)
default	\(125 \ln \left (\frac {{\mathrm e}^{2 x +9} {\mathrm e}^{1-x}}{x}\right ) x\)	\(21\)
risch	\(125 x \ln \left ({\mathrm e}^{x -1}\right )-125 x \ln \relax (x )+\frac {125 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x -1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x -1}}{x}\right )^{2} x}{2}-\frac {125 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x -1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{x -1}}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right ) x}{2}+\frac {125 i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{x -1}}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right ) x}{2}-\frac {125 i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{x -1}}{x}\right )^{3} x}{2}+1375 x\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(125*ln(exp(2*x+9)/x/exp(x-1))+125*x-125,x,method=_RETURNVERBOSE)

[Out]

125*x*ln(exp(11)*exp(x-1)/x)

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maxima [A]  time = 0.36, size = 12, normalized size = 0.80 \begin {gather*} 125 \, x \log \left (\frac {e^{\left (x + 10\right )}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(125*log(exp(2*x+9)/x/exp(x-1))+125*x-125,x, algorithm="maxima")

[Out]

125*x*log(e^(x + 10)/x)

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mupad [B]  time = 5.29, size = 10, normalized size = 0.67 \begin {gather*} 125\,x\,\left (x+\ln \left (\frac {1}{x}\right )+10\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(125*x + 125*log((exp(1 - x)*exp(2*x + 9))/x) - 125,x)

[Out]

125*x*(x + log(1/x) + 10)

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sympy [A]  time = 0.17, size = 14, normalized size = 0.93 \begin {gather*} 125 x \log {\left (\frac {e^{11} e^{x - 1}}{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(125*ln(exp(2*x+9)/x/exp(x-1))+125*x-125,x)

[Out]

125*x*log(exp(11)*exp(x - 1)/x)

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