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

Optimal. Leaf size=12 \[ x+\frac {e^{-8 x}}{\log (5)} \]

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Rubi [A]  time = 0.02, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 2248, 43} \begin {gather*} x+\frac {e^{-8 x}}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[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 43

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 2248

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

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Mathematica [A]  time = 0.01, size = 12, normalized size = 1.00 \begin {gather*} x+\frac {e^{-8 x}}{\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + 1/(E^(8*x)*Log[5])

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fricas [A]  time = 0.62, size = 21, normalized size = 1.75 \begin {gather*} \frac {4 \, x \log \relax (5) + e^{\left (-8 \, x + 2 \, \log \relax (2)\right )}}{4 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.22, size = 26, normalized size = 2.17 \begin {gather*} -\frac {{\left (e^{\left (8 \, x\right )} \log \relax (5) - 8\right )} e^{\left (-8 \, x\right )} - 8 \, x \log \relax (5)}{8 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.03, size = 12, normalized size = 1.00

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.37, size = 14, normalized size = 1.17 \begin {gather*} \frac {x \log \relax (5) + e^{\left (-8 \, x\right )}}{\log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.10, size = 11, normalized size = 0.92 \begin {gather*} x+\frac {{\mathrm {e}}^{-8\,x}}{\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.11, size = 10, normalized size = 0.83 \begin {gather*} x + \frac {e^{- 8 x}}{\log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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