\(\int \frac {4 e^{-8 x} (-2+\frac {1}{4} e^{8 x} \log (5))}{\log (5)} \, dx\) [7433]

Optimal result

Integrand size = 24, antiderivative size = 12 \[ \int \frac {4 e^{-8 x} \left (-2+\frac {1}{4} e^{8 x} \log (5)\right )}{\log (5)} \, dx=x+\frac {e^{-8 x}}{\log (5)} \]

[Out]

x+1/4/exp(-ln(2)+4*x)^2/ln(5)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 2280, 45} \[ \int \frac {4 e^{-8 x} \left (-2+\frac {1}{4} e^{8 x} \log (5)\right )}{\log (5)} \, dx=x+\frac {e^{-8 x}}{\log (5)} \]

[In]

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

[Out]

x + 1/(E^(8*x)*Log[5])

Rule 12

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2280

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[g*h*(Log[G]/(d*e*Log[F]))]}, Dist[Denominator[m]*(G^(f*h - c*g*(h/d))/(d*e*Log[F])), Subst
[Int[x^(Numerator[m] - 1)*(a + b*x^Denominator[m])^p, x], x, F^(e*((c + d*x)/Denominator[m]))], x] /; LeQ[m, -
1] || GeQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {4 \int e^{-8 x} \left (-2+\frac {1}{4} e^{8 x} \log (5)\right ) \, dx}{\log (5)} \\ & = \frac {\text {Subst}\left (\int \frac {-2+\frac {1}{4} x \log (5)}{x^2} \, dx,x,e^{8 x}\right )}{2 \log (5)} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {2}{x^2}+\frac {\log (5)}{4 x}\right ) \, dx,x,e^{8 x}\right )}{2 \log (5)} \\ & = x+\frac {e^{-8 x}}{\log (5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {4 e^{-8 x} \left (-2+\frac {1}{4} e^{8 x} \log (5)\right )}{\log (5)} \, dx=x+\frac {e^{-8 x}}{\log (5)} \]

[In]

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

[Out]

x + 1/(E^(8*x)*Log[5])

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00

method	result	size
risch	\(x +\frac {{\mathrm e}^{-8 x}}{\ln \left (5\right )}\)	\(12\)
parts	\(x +\frac {{\mathrm e}^{-8 x}}{\ln \left (5\right )}\)	\(20\)
derivativedivides	\(\frac {4 \,{\mathrm e}^{-8 x}+\ln \left (5\right ) \ln \left ({\mathrm e}^{-\ln \left (2\right )+4 x}\right )}{4 \ln \left (5\right )}\)	\(32\)
norman	\(4 \left (\frac {x \,{\mathrm e}^{8 x}}{4}+\frac {1}{4 \ln \left (5\right )}\right ) {\mathrm e}^{-8 x}\)	\(33\)
default	\(\frac {{\mathrm e}^{-8 x}+\frac {\ln \left (5\right ) \ln \left ({\mathrm e}^{-\ln \left (2\right )+4 x}\right )}{4}}{\ln \left (5\right )}\)	\(34\)
parallelrisch	\(\frac {\left ({\mathrm e}^{8 x} x \ln \left (5\right )+1\right ) {\mathrm e}^{-8 x}}{\ln \left (5\right )}\)	\(36\)

[In]

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

[Out]

x+exp(-8*x)/ln(5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.75 \[ \int \frac {4 e^{-8 x} \left (-2+\frac {1}{4} e^{8 x} \log (5)\right )}{\log (5)} \, dx=\frac {4 \, x \log \left (5\right ) + e^{\left (-8 \, x + 2 \, \log \left (2\right )\right )}}{4 \, \log \left (5\right )} \]

[In]

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

[Out]

1/4*(4*x*log(5) + e^(-8*x + 2*log(2)))/log(5)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {4 e^{-8 x} \left (-2+\frac {1}{4} e^{8 x} \log (5)\right )}{\log (5)} \, dx=x + \frac {e^{- 8 x}}{\log {\left (5 \right )}} \]

[In]

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

[Out]

x + exp(-8*x)/log(5)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {4 e^{-8 x} \left (-2+\frac {1}{4} e^{8 x} \log (5)\right )}{\log (5)} \, dx=\frac {x \log \left (5\right ) + e^{\left (-8 \, x\right )}}{\log \left (5\right )} \]

[In]

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

[Out]

(x*log(5) + e^(-8*x))/log(5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 2.17 \[ \int \frac {4 e^{-8 x} \left (-2+\frac {1}{4} e^{8 x} \log (5)\right )}{\log (5)} \, dx=-\frac {{\left (e^{\left (8 \, x\right )} \log \left (5\right ) - 8\right )} e^{\left (-8 \, x\right )} - 8 \, x \log \left (5\right )}{8 \, \log \left (5\right )} \]

[In]

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

[Out]

-1/8*((e^(8*x)*log(5) - 8)*e^(-8*x) - 8*x*log(5))/log(5)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {4 e^{-8 x} \left (-2+\frac {1}{4} e^{8 x} \log (5)\right )}{\log (5)} \, dx=x+\frac {{\mathrm {e}}^{-8\,x}}{\ln \left (5\right )} \]

[In]

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

[Out]

x + exp(-8*x)/log(5)