3.51.96 \(\int \frac {-2+(16-15 e^x+12 x) \log (9+\log (5))}{1+(-5+5 e^x-6 x) \log (9+\log (5))} \, dx\) [5096]

Optimal. Leaf size=29 \[ 3-2 x-\log \left (1-\left (5+x+5 \left (-e^x+x\right )\right ) \log (9+\log (5))\right ) \]

[Out]

3-2*x-ln(1-(-5*exp(x)+6*x+5)*ln(ln(5)+9))

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Rubi [F]
time = 0.41, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-2+\left (16-15 e^x+12 x\right ) \log (9+\log (5))}{1+\left (-5+5 e^x-6 x\right ) \log (9+\log (5))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2 + (16 - 15*E^x + 12*x)*Log[9 + Log[5]])/(1 + (-5 + 5*E^x - 6*x)*Log[9 + Log[5]]),x]

[Out]

-3*x + (1 + Log[9 + Log[5]])*Defer[Int][(1 - 5*Log[9 + Log[5]] + 5*E^x*Log[9 + Log[5]] - 6*x*Log[9 + Log[5]])^
(-1), x] + 6*Log[9 + Log[5]]*Defer[Int][x/(-1 + 5*Log[9 + Log[5]] - 5*E^x*Log[9 + Log[5]] + 6*x*Log[9 + Log[5]
]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-3+\frac {-1-\log (9+\log (5))+6 x \log (9+\log (5))}{-1+5 \log (9+\log (5))-5 e^x \log (9+\log (5))+6 x \log (9+\log (5))}\right ) \, dx\\ &=-3 x+\int \frac {-1-\log (9+\log (5))+6 x \log (9+\log (5))}{-1+5 \log (9+\log (5))-5 e^x \log (9+\log (5))+6 x \log (9+\log (5))} \, dx\\ &=-3 x+\int \left (\frac {1+\log (9+\log (5))}{1-5 \log (9+\log (5))+5 e^x \log (9+\log (5))-6 x \log (9+\log (5))}+\frac {6 x \log (9+\log (5))}{-1+5 \log (9+\log (5))-5 e^x \log (9+\log (5))+6 x \log (9+\log (5))}\right ) \, dx\\ &=-3 x+(6 \log (9+\log (5))) \int \frac {x}{-1+5 \log (9+\log (5))-5 e^x \log (9+\log (5))+6 x \log (9+\log (5))} \, dx+(1+\log (9+\log (5))) \int \frac {1}{1-5 \log (9+\log (5))+5 e^x \log (9+\log (5))-6 x \log (9+\log (5))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.08, size = 34, normalized size = 1.17 \begin {gather*} -2 x-\log \left (1-5 \log (9+\log (5))+5 e^x \log (9+\log (5))-6 x \log (9+\log (5))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + (16 - 15*E^x + 12*x)*Log[9 + Log[5]])/(1 + (-5 + 5*E^x - 6*x)*Log[9 + Log[5]]),x]

[Out]

-2*x - Log[1 - 5*Log[9 + Log[5]] + 5*E^x*Log[9 + Log[5]] - 6*x*Log[9 + Log[5]]]

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Maple [A]
time = 0.51, size = 34, normalized size = 1.17

method	result	size
norman	\(-2 x -\ln \left (6 \ln \left (\ln \left (5\right )+9\right ) x -5 \ln \left (\ln \left (5\right )+9\right ) {\mathrm e}^{x}+5 \ln \left (\ln \left (5\right )+9\right )-1\right )\)	\(34\)
risch	\(-2 x -\ln \left ({\mathrm e}^{x}-\frac {6 \ln \left (\ln \left (5\right )+9\right ) x +5 \ln \left (\ln \left (5\right )+9\right )-1}{5 \ln \left (\ln \left (5\right )+9\right )}\right )\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-15*exp(x)+12*x+16)*ln(ln(5)+9)-2)/((5*exp(x)-5-6*x)*ln(ln(5)+9)+1),x,method=_RETURNVERBOSE)

[Out]

-2*x-ln(6*ln(ln(5)+9)*x-5*ln(ln(5)+9)*exp(x)+5*ln(ln(5)+9)-1)

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Maxima [A]
time = 0.51, size = 42, normalized size = 1.45 \begin {gather*} -2 \, x - \log \left (-\frac {6 \, x \log \left (\log \left (5\right ) + 9\right ) - 5 \, e^{x} \log \left (\log \left (5\right ) + 9\right ) + 5 \, \log \left (\log \left (5\right ) + 9\right ) - 1}{5 \, \log \left (\log \left (5\right ) + 9\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*exp(x)+12*x+16)*log(log(5)+9)-2)/((5*exp(x)-5-6*x)*log(log(5)+9)+1),x, algorithm="maxima")

[Out]

-2*x - log(-1/5*(6*x*log(log(5) + 9) - 5*e^x*log(log(5) + 9) + 5*log(log(5) + 9) - 1)/log(log(5) + 9))

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Fricas [A]
time = 0.39, size = 25, normalized size = 0.86 \begin {gather*} -2 \, x - \log \left (-{\left (6 \, x - 5 \, e^{x} + 5\right )} \log \left (\log \left (5\right ) + 9\right ) + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*exp(x)+12*x+16)*log(log(5)+9)-2)/((5*exp(x)-5-6*x)*log(log(5)+9)+1),x, algorithm="fricas")

[Out]

-2*x - log(-(6*x - 5*e^x + 5)*log(log(5) + 9) + 1)

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Sympy [A]
time = 0.09, size = 37, normalized size = 1.28 \begin {gather*} - 2 x - \log {\left (\frac {- 6 x \log {\left (\log {\left (5 \right )} + 9 \right )} - 5 \log {\left (\log {\left (5 \right )} + 9 \right )} + 1}{5 \log {\left (\log {\left (5 \right )} + 9 \right )}} + e^{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*exp(x)+12*x+16)*ln(ln(5)+9)-2)/((5*exp(x)-5-6*x)*ln(ln(5)+9)+1),x)

[Out]

-2*x - log((-6*x*log(log(5) + 9) - 5*log(log(5) + 9) + 1)/(5*log(log(5) + 9)) + exp(x))

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Giac [A]
time = 0.42, size = 33, normalized size = 1.14 \begin {gather*} -2 \, x - \log \left (-6 \, x \log \left (\log \left (5\right ) + 9\right ) + 5 \, e^{x} \log \left (\log \left (5\right ) + 9\right ) - 5 \, \log \left (\log \left (5\right ) + 9\right ) + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*exp(x)+12*x+16)*log(log(5)+9)-2)/((5*exp(x)-5-6*x)*log(log(5)+9)+1),x, algorithm="giac")

[Out]

-2*x - log(-6*x*log(log(5) + 9) + 5*e^x*log(log(5) + 9) - 5*log(log(5) + 9) + 1)

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Mupad [B]
time = 0.40, size = 33, normalized size = 1.14 \begin {gather*} -2\,x-\ln \left (5\,\ln \left (\ln \left (5\right )+9\right )+6\,x\,\ln \left (\ln \left (5\right )+9\right )-5\,\ln \left (\ln \left (5\right )+9\right )\,{\mathrm {e}}^x-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(5) + 9)*(12*x - 15*exp(x) + 16) - 2)/(log(log(5) + 9)*(6*x - 5*exp(x) + 5) - 1),x)

[Out]

- 2*x - log(5*log(log(5) + 9) + 6*x*log(log(5) + 9) - 5*log(log(5) + 9)*exp(x) - 1)

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