3.44.64 \(\int \frac {-2-8 e^4}{3-2 x-8 e^4 x+2 e^4 \log (5)} \, dx\)

Optimal. Leaf size=23 \[ \log \left (-2-4 x+4 \left (2-e^4 (4 x-\log (5))\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {6, 12, 31} \begin {gather*} \log \left (-2 \left (1+4 e^4\right ) x+3+e^4 \log (25)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 - 8*E^4)/(3 - 2*x - 8*E^4*x + 2*E^4*Log[5]),x]

[Out]

Log[3 - 2*(1 + 4*E^4)*x + E^4*Log[25]]

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2-8 e^4}{3+\left (-2-8 e^4\right ) x+2 e^4 \log (5)} \, dx\\ &=-\left (\left (2 \left (1+4 e^4\right )\right ) \int \frac {1}{3+\left (-2-8 e^4\right ) x+2 e^4 \log (5)} \, dx\right )\\ &=\log \left (3-2 \left (1+4 e^4\right ) x+e^4 \log (25)\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 19, normalized size = 0.83 \begin {gather*} \log \left (-3+\left (2+8 e^4\right ) x-2 e^4 \log (5)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 8*E^4)/(3 - 2*x - 8*E^4*x + 2*E^4*Log[5]),x]

[Out]

Log[-3 + (2 + 8*E^4)*x - 2*E^4*Log[5]]

________________________________________________________________________________________

fricas [A]  time = 0.71, size = 17, normalized size = 0.74 \begin {gather*} \log \left (8 \, x e^{4} - 2 \, e^{4} \log \relax (5) + 2 \, x - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(4)-2)/(2*exp(4)*log(5)-8*x*exp(4)+3-2*x),x, algorithm="fricas")

[Out]

log(8*x*e^4 - 2*e^4*log(5) + 2*x - 3)

________________________________________________________________________________________

giac [A]  time = 0.12, size = 18, normalized size = 0.78 \begin {gather*} \log \left ({\left | 8 \, x e^{4} - 2 \, e^{4} \log \relax (5) + 2 \, x - 3 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(4)-2)/(2*exp(4)*log(5)-8*x*exp(4)+3-2*x),x, algorithm="giac")

[Out]

log(abs(8*x*e^4 - 2*e^4*log(5) + 2*x - 3))

________________________________________________________________________________________

maple [A]  time = 0.16, size = 18, normalized size = 0.78

method	result	size
default	\(\ln \left (\left (-8 \,{\mathrm e}^{4}-2\right ) x +2 \,{\mathrm e}^{4} \ln \relax (5)+3\right )\)	\(18\)
norman	\(\ln \left (2 \,{\mathrm e}^{4} \ln \relax (5)-8 x \,{\mathrm e}^{4}+3-2 x \right )\)	\(18\)
risch	\(\frac {4 \ln \left (\left (-8 \,{\mathrm e}^{4}-2\right ) x +2 \,{\mathrm e}^{4} \ln \relax (5)+3\right ) {\mathrm e}^{4}}{4 \,{\mathrm e}^{4}+1}+\frac {\ln \left (\left (-8 \,{\mathrm e}^{4}-2\right ) x +2 \,{\mathrm e}^{4} \ln \relax (5)+3\right )}{4 \,{\mathrm e}^{4}+1}\)	\(57\)
meijerg	\(\frac {4 \,{\mathrm e}^{4} \ln \left (1-\frac {2 x \left (4 \,{\mathrm e}^{4}+1\right )}{2 \,{\mathrm e}^{4} \ln \relax (5)+3}\right )}{4 \,{\mathrm e}^{4}+1}+\frac {\ln \left (1-\frac {2 x \left (4 \,{\mathrm e}^{4}+1\right )}{2 \,{\mathrm e}^{4} \ln \relax (5)+3}\right )}{4 \,{\mathrm e}^{4}+1}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*exp(4)-2)/(2*exp(4)*ln(5)-8*x*exp(4)+3-2*x),x,method=_RETURNVERBOSE)

[Out]

ln((-8*exp(4)-2)*x+2*exp(4)*ln(5)+3)

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 17, normalized size = 0.74 \begin {gather*} \log \left (8 \, x e^{4} - 2 \, e^{4} \log \relax (5) + 2 \, x - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(4)-2)/(2*exp(4)*log(5)-8*x*exp(4)+3-2*x),x, algorithm="maxima")

[Out]

log(8*x*e^4 - 2*e^4*log(5) + 2*x - 3)

________________________________________________________________________________________

mupad [B]  time = 3.07, size = 17, normalized size = 0.74 \begin {gather*} \ln \left (x\,\left (8\,{\mathrm {e}}^4+2\right )-2\,{\mathrm {e}}^4\,\ln \relax (5)-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*exp(4) + 2)/(2*x - 2*exp(4)*log(5) + 8*x*exp(4) - 3),x)

[Out]

log(x*(8*exp(4) + 2) - 2*exp(4)*log(5) - 3)

________________________________________________________________________________________

sympy [A]  time = 0.08, size = 19, normalized size = 0.83 \begin {gather*} \log {\left (x \left (2 + 8 e^{4}\right ) - 2 e^{4} \log {\relax (5 )} - 3 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(4)-2)/(2*exp(4)*ln(5)-8*x*exp(4)+3-2*x),x)

[Out]

log(x*(2 + 8*exp(4)) - 2*exp(4)*log(5) - 3)

________________________________________________________________________________________