3.31.88 \(\int \frac {-52-16 e^4+240 e^{5 x}}{12 e^{5 x}-13 x-4 e^4 x-\log (5)} \, dx\)

Optimal. Leaf size=25 \[ 4 \log \left (x-4 \left (3 e^{5 x}-\left (3+e^4\right ) x\right )+\log (5)\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {6, 6684} \begin {gather*} 4 \log \left (-\left (\left (13+4 e^4\right ) x\right )+12 e^{5 x}-\log (5)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-52 - 16*E^4 + 240*E^(5*x))/(12*E^(5*x) - 13*x - 4*E^4*x - Log[5]),x]

[Out]

4*Log[12*E^(5*x) - (13 + 4*E^4)*x - Log[5]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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} &=\int \frac {-52-16 e^4+240 e^{5 x}}{12 e^{5 x}+\left (-13-4 e^4\right ) x-\log (5)} \, dx\\ &=4 \log \left (12 e^{5 x}-\left (13+4 e^4\right ) x-\log (5)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 24, normalized size = 0.96 \begin {gather*} 4 \log \left (12 e^{5 x}-13 x-4 e^4 x-\log (5)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-52 - 16*E^4 + 240*E^(5*x))/(12*E^(5*x) - 13*x - 4*E^4*x - Log[5]),x]

[Out]

4*Log[12*E^(5*x) - 13*x - 4*E^4*x - Log[5]]

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fricas [A]  time = 0.46, size = 22, normalized size = 0.88 \begin {gather*} 4 \, \log \left (-4 \, x e^{4} - 13 \, x + 12 \, e^{\left (5 \, x\right )} - \log \relax (5)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((240*exp(5*x)-16*exp(2)^2-52)/(12*exp(5*x)-log(5)-4*x*exp(2)^2-13*x),x, algorithm="fricas")

[Out]

4*log(-4*x*e^4 - 13*x + 12*e^(5*x) - log(5))

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giac [A]  time = 0.25, size = 22, normalized size = 0.88 \begin {gather*} 4 \, \log \left (-4 \, x e^{4} - 13 \, x + 12 \, e^{\left (5 \, x\right )} - \log \relax (5)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((240*exp(5*x)-16*exp(2)^2-52)/(12*exp(5*x)-log(5)-4*x*exp(2)^2-13*x),x, algorithm="giac")

[Out]

4*log(-4*x*e^4 - 13*x + 12*e^(5*x) - log(5))

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maple [A]  time = 0.11, size = 21, normalized size = 0.84

method	result	size
risch	\(4 \ln \left (-\frac {x \,{\mathrm e}^{4}}{3}-\frac {\ln \relax (5)}{12}-\frac {13 x}{12}+{\mathrm e}^{5 x}\right )\)	\(21\)
norman	\(4 \ln \left (4 x \,{\mathrm e}^{4}+\ln \relax (5)-12 \,{\mathrm e}^{5 x}+13 x \right )\)	\(23\)
derivativedivides	\(4 \ln \left (60 \,{\mathrm e}^{5 x}-5 \ln \relax (5)-20 x \,{\mathrm e}^{4}-65 x \right )\)	\(25\)
default	\(4 \ln \left (12 \,{\mathrm e}^{5 x}-\ln \relax (5)-4 x \,{\mathrm e}^{4}-13 x \right )\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((240*exp(5*x)-16*exp(2)^2-52)/(12*exp(5*x)-ln(5)-4*x*exp(2)^2-13*x),x,method=_RETURNVERBOSE)

[Out]

4*ln(-1/3*x*exp(4)-1/12*ln(5)-13/12*x+exp(5*x))

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maxima [A]  time = 0.37, size = 20, normalized size = 0.80 \begin {gather*} 4 \, \log \left (4 \, x e^{4} + 13 \, x - 12 \, e^{\left (5 \, x\right )} + \log \relax (5)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((240*exp(5*x)-16*exp(2)^2-52)/(12*exp(5*x)-log(5)-4*x*exp(2)^2-13*x),x, algorithm="maxima")

[Out]

4*log(4*x*e^4 + 13*x - 12*e^(5*x) + log(5))

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mupad [B]  time = 0.16, size = 20, normalized size = 0.80 \begin {gather*} 4\,\ln \left (13\,x-12\,{\mathrm {e}}^{5\,x}+\ln \relax (5)+4\,x\,{\mathrm {e}}^4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*exp(4) - 240*exp(5*x) + 52)/(13*x - 12*exp(5*x) + log(5) + 4*x*exp(4)),x)

[Out]

4*log(13*x - 12*exp(5*x) + log(5) + 4*x*exp(4))

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sympy [A]  time = 0.16, size = 29, normalized size = 1.16 \begin {gather*} \frac {55 x}{3} + \frac {\log {\left (- \frac {x e^{4}}{3} - \frac {13 x}{12} + e^{5 x} - \frac {\log {\relax (5 )}}{12} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((240*exp(5*x)-16*exp(2)**2-52)/(12*exp(5*x)-ln(5)-4*x*exp(2)**2-13*x),x)

[Out]

55*x/3 + log(-x*exp(4)/3 - 13*x/12 + exp(5*x) - log(5)/12)/3

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