3.104.20 \(\int \frac {-e^{25-x} \log (3)+(8-2 x) \log (3)}{-13+e^{25-x}+8 x-x^2} \, dx\)

Optimal. Leaf size=33 \[ \log (3) \log \left (\frac {1}{2} \left (-4+\frac {1}{3} \left (e^{25-x}-(1-x)^2\right )\right )+x\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 21, normalized size of antiderivative = 0.64, number of steps used = 1, number of rules used = 1, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6684} \begin {gather*} \log (3) \log \left (x^2-8 x-e^{25-x}+13\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(E^(25 - x)*Log[3]) + (8 - 2*x)*Log[3])/(-13 + E^(25 - x) + 8*x - x^2),x]

[Out]

Log[3]*Log[13 - E^(25 - x) - 8*x + x^2]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (3) \log \left (13-e^{25-x}-8 x+x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.46, size = 32, normalized size = 0.97 \begin {gather*} \log (3) \left (-x+\log \left (-e^{25}+13 e^x-8 e^x x+e^x x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^(25 - x)*Log[3]) + (8 - 2*x)*Log[3])/(-13 + E^(25 - x) + 8*x - x^2),x]

[Out]

Log[3]*(-x + Log[-E^25 + 13*E^x - 8*E^x*x + E^x*x^2])

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fricas [A]  time = 0.62, size = 20, normalized size = 0.61 \begin {gather*} \log \relax (3) \log \left (-x^{2} + 8 \, x + e^{\left (-x + 25\right )} - 13\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(3)*exp(-x+25)+(-2*x+8)*log(3))/(exp(-x+25)-x^2+8*x-13),x, algorithm="fricas")

[Out]

log(3)*log(-x^2 + 8*x + e^(-x + 25) - 13)

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giac [A]  time = 0.16, size = 20, normalized size = 0.61 \begin {gather*} \log \relax (3) \log \left (-x^{2} + 8 \, x + e^{\left (-x + 25\right )} - 13\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(3)*exp(-x+25)+(-2*x+8)*log(3))/(exp(-x+25)-x^2+8*x-13),x, algorithm="giac")

[Out]

log(3)*log(-x^2 + 8*x + e^(-x + 25) - 13)

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maple [A]  time = 0.08, size = 21, normalized size = 0.64

method	result	size
norman	\(\ln \relax (3) \ln \left (x^{2}-8 x -{\mathrm e}^{-x +25}+13\right )\)	\(21\)
risch	\(-25 \ln \relax (3)+\ln \relax (3) \ln \left ({\mathrm e}^{-x +25}-x^{2}+8 x -13\right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(3)*exp(-x+25)+(-2*x+8)*ln(3))/(exp(-x+25)-x^2+8*x-13),x,method=_RETURNVERBOSE)

[Out]

ln(3)*ln(x^2-8*x-exp(-x+25)+13)

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maxima [B]  time = 0.47, size = 49, normalized size = 1.48 \begin {gather*} -x \log \relax (3) + \log \relax (3) \log \left (x^{2} - 8 \, x + 13\right ) + \log \relax (3) \log \left (\frac {{\left (x^{2} - 8 \, x + 13\right )} e^{x} - e^{25}}{x^{2} - 8 \, x + 13}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(3)*exp(-x+25)+(-2*x+8)*log(3))/(exp(-x+25)-x^2+8*x-13),x, algorithm="maxima")

[Out]

-x*log(3) + log(3)*log(x^2 - 8*x + 13) + log(3)*log(((x^2 - 8*x + 13)*e^x - e^25)/(x^2 - 8*x + 13))

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mupad [B]  time = 8.05, size = 20, normalized size = 0.61 \begin {gather*} \ln \relax (3)\,\ln \left (x^2-{\mathrm {e}}^{25-x}-8\,x+13\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(3)*(2*x - 8) + exp(25 - x)*log(3))/(8*x + exp(25 - x) - x^2 - 13),x)

[Out]

log(3)*log(x^2 - exp(25 - x) - 8*x + 13)

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sympy [A]  time = 0.15, size = 17, normalized size = 0.52 \begin {gather*} \log {\relax (3 )} \log {\left (- x^{2} + 8 x + e^{25 - x} - 13 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(3)*exp(-x+25)+(-2*x+8)*ln(3))/(exp(-x+25)-x**2+8*x-13),x)

[Out]

log(3)*log(-x**2 + 8*x + exp(25 - x) - 13)

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