3.40.28 \(\int (e^{-5+x+\log ^2(2)}-\log (\frac {e^{32}}{5})) \, dx\)

Optimal. Leaf size=22 \[ 25+e^{-5+x+\log ^2(2)}-x \log \left (\frac {e^{32}}{5}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 0.86, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {2194} \begin {gather*} e^{x-5+\log ^2(2)}-x (32-\log (5)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(-5 + x + Log[2]^2) - Log[E^32/5],x]

[Out]

E^(-5 + x + Log[2]^2) - x*(32 - Log[5])

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} &=-x (32-\log (5))+\int e^{-5+x+\log ^2(2)} \, dx\\ &=e^{-5+x+\log ^2(2)}-x (32-\log (5))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 0.77 \begin {gather*} e^{-5+x+\log ^2(2)}-32 x+x \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(-5 + x + Log[2]^2) - Log[E^32/5],x]

[Out]

E^(-5 + x + Log[2]^2) - 32*x + x*Log[5]

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fricas [A]  time = 0.60, size = 16, normalized size = 0.73 \begin {gather*} x \log \relax (5) - 32 \, x + e^{\left (\log \relax (2)^{2} + x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(1/5*exp(16)^2)+exp(log(2)^2+x-5),x, algorithm="fricas")

[Out]

x*log(5) - 32*x + e^(log(2)^2 + x - 5)

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giac [A]  time = 0.20, size = 17, normalized size = 0.77 \begin {gather*} -x \log \left (\frac {1}{5} \, e^{32}\right ) + e^{\left (\log \relax (2)^{2} + x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(1/5*exp(16)^2)+exp(log(2)^2+x-5),x, algorithm="giac")

[Out]

-x*log(1/5*e^32) + e^(log(2)^2 + x - 5)

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maple [A]  time = 0.03, size = 16, normalized size = 0.73

method	result	size
norman	\(\left (-32+\ln \relax (5)\right ) x +{\mathrm e}^{\ln \relax (2)^{2}+x -5}\)	\(16\)
risch	\(x \ln \relax (5)+{\mathrm e}^{\ln \relax (2)^{2}+x -5}-32 x\)	\(17\)
default	\(-\ln \left (\frac {{\mathrm e}^{32}}{5}\right ) x +{\mathrm e}^{\ln \relax (2)^{2}+x -5}\)	\(20\)
derivativedivides	\({\mathrm e}^{\ln \relax (2)^{2}+x -5}-\ln \left (\frac {{\mathrm e}^{32}}{5}\right ) \ln \left ({\mathrm e}^{\ln \relax (2)^{2}+x -5}\right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-ln(1/5*exp(16)^2)+exp(ln(2)^2+x-5),x,method=_RETURNVERBOSE)

[Out]

(-32+ln(5))*x+exp(ln(2)^2+x-5)

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maxima [A]  time = 0.34, size = 17, normalized size = 0.77 \begin {gather*} -x \log \left (\frac {1}{5} \, e^{32}\right ) + e^{\left (\log \relax (2)^{2} + x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(1/5*exp(16)^2)+exp(log(2)^2+x-5),x, algorithm="maxima")

[Out]

-x*log(1/5*e^32) + e^(log(2)^2 + x - 5)

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mupad [B]  time = 2.20, size = 18, normalized size = 0.82 \begin {gather*} x\,\ln \relax (5)-32\,x+{\mathrm {e}}^{{\ln \relax (2)}^2}\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x + log(2)^2 - 5) - log(exp(32)/5),x)

[Out]

x*log(5) - 32*x + exp(log(2)^2)*exp(-5)*exp(x)

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sympy [A]  time = 0.08, size = 15, normalized size = 0.68 \begin {gather*} x \left (-32 + \log {\relax (5 )}\right ) + e^{x - 5 + \log {\relax (2 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-ln(1/5*exp(16)**2)+exp(ln(2)**2+x-5),x)

[Out]

x*(-32 + log(5)) + exp(x - 5 + log(2)**2)

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