3.60.20 \(\int \frac {3 \log ^2(4)+e^{\frac {13-25 x}{25 \log ^2(4)}} (-x+\log ^2(4))}{\log ^2(4)} \, dx\)

Optimal. Leaf size=18 \[ \left (3+e^{\frac {\frac {13}{25}-x}{\log ^2(4)}}\right ) x \]

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Rubi [B]  time = 0.04, antiderivative size = 48, normalized size of antiderivative = 2.67, number of steps used = 4, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 2176, 2194} \begin {gather*} 3 x+e^{\frac {13-25 x}{25 \log ^2(4)}} \left (x-\log ^2(4)\right )+\log ^2(4) e^{\frac {13-25 x}{25 \log ^2(4)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3*Log[4]^2 + E^((13 - 25*x)/(25*Log[4]^2))*(-x + Log[4]^2))/Log[4]^2,x]

[Out]

3*x + E^((13 - 25*x)/(25*Log[4]^2))*Log[4]^2 + E^((13 - 25*x)/(25*Log[4]^2))*(x - Log[4]^2)

Rule 12

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (3 \log ^2(4)+e^{\frac {13-25 x}{25 \log ^2(4)}} \left (-x+\log ^2(4)\right )\right ) \, dx}{\log ^2(4)}\\ &=3 x+\frac {\int e^{\frac {13-25 x}{25 \log ^2(4)}} \left (-x+\log ^2(4)\right ) \, dx}{\log ^2(4)}\\ &=3 x+e^{\frac {13-25 x}{25 \log ^2(4)}} \left (x-\log ^2(4)\right )-\int e^{\frac {13-25 x}{25 \log ^2(4)}} \, dx\\ &=3 x+e^{\frac {13-25 x}{25 \log ^2(4)}} \log ^2(4)+e^{\frac {13-25 x}{25 \log ^2(4)}} \left (x-\log ^2(4)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 24, normalized size = 1.33 \begin {gather*} 3 x+e^{\frac {13}{25 \log ^2(4)}-\frac {x}{\log ^2(4)}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*Log[4]^2 + E^((13 - 25*x)/(25*Log[4]^2))*(-x + Log[4]^2))/Log[4]^2,x]

[Out]

3*x + E^(13/(25*Log[4]^2) - x/Log[4]^2)*x

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fricas [A]  time = 0.58, size = 18, normalized size = 1.00 \begin {gather*} x e^{\left (-\frac {25 \, x - 13}{100 \, \log \relax (2)^{2}}\right )} + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((4*log(2)^2-x)*exp(1/100*(-25*x+13)/log(2)^2)+12*log(2)^2)/log(2)^2,x, algorithm="fricas")

[Out]

x*e^(-1/100*(25*x - 13)/log(2)^2) + 3*x

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giac [A]  time = 0.15, size = 31, normalized size = 1.72 \begin {gather*} \frac {x e^{\left (-\frac {25 \, x - 13}{100 \, \log \relax (2)^{2}}\right )} \log \relax (2)^{2} + 3 \, x \log \relax (2)^{2}}{\log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((4*log(2)^2-x)*exp(1/100*(-25*x+13)/log(2)^2)+12*log(2)^2)/log(2)^2,x, algorithm="giac")

[Out]

(x*e^(-1/100*(25*x - 13)/log(2)^2)*log(2)^2 + 3*x*log(2)^2)/log(2)^2

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maple [A]  time = 0.09, size = 19, normalized size = 1.06

method	result	size
risch	\({\mathrm e}^{-\frac {25 x -13}{100 \ln \relax (2)^{2}}} x +3 x\)	\(19\)
norman	\(\frac {x \ln \relax (2) {\mathrm e}^{\frac {-25 x +13}{100 \ln \relax (2)^{2}}}+3 x \ln \relax (2)}{\ln \relax (2)}\)	\(28\)
default	\(\frac {4 \ln \relax (2)^{2} {\mathrm e}^{-\frac {x}{4 \ln \relax (2)^{2}}+\frac {13}{100 \ln \relax (2)^{2}}} x +12 x \ln \relax (2)^{2}}{4 \ln \relax (2)^{2}}\)	\(37\)
derivativedivides	\(-12 \ln \relax (2)^{2} \left (-\frac {x}{4 \ln \relax (2)^{2}}+\frac {13}{100 \ln \relax (2)^{2}}\right )-4 \,{\mathrm e}^{-\frac {x}{4 \ln \relax (2)^{2}}+\frac {13}{100 \ln \relax (2)^{2}}} \ln \relax (2)^{2} \left (-\frac {x}{4 \ln \relax (2)^{2}}+\frac {13}{100 \ln \relax (2)^{2}}\right )+\frac {13 \,{\mathrm e}^{-\frac {x}{4 \ln \relax (2)^{2}}+\frac {13}{100 \ln \relax (2)^{2}}}}{25}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((4*ln(2)^2-x)*exp(1/100*(-25*x+13)/ln(2)^2)+12*ln(2)^2)/ln(2)^2,x,method=_RETURNVERBOSE)

[Out]

exp(-1/100*(25*x-13)/ln(2)^2)*x+3*x

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maxima [B]  time = 0.45, size = 72, normalized size = 4.00 \begin {gather*} -\frac {4 \, e^{\left (-\frac {x}{4 \, \log \relax (2)^{2}} + \frac {13}{100 \, \log \relax (2)^{2}}\right )} \log \relax (2)^{4} - 3 \, x \log \relax (2)^{2} - {\left (4 \, e^{\left (\frac {13}{100 \, \log \relax (2)^{2}}\right )} \log \relax (2)^{4} + x e^{\left (\frac {13}{100 \, \log \relax (2)^{2}}\right )} \log \relax (2)^{2}\right )} e^{\left (-\frac {x}{4 \, \log \relax (2)^{2}}\right )}}{\log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((4*log(2)^2-x)*exp(1/100*(-25*x+13)/log(2)^2)+12*log(2)^2)/log(2)^2,x, algorithm="maxima")

[Out]

-(4*e^(-1/4*x/log(2)^2 + 13/100/log(2)^2)*log(2)^4 - 3*x*log(2)^2 - (4*e^(13/100/log(2)^2)*log(2)^4 + x*e^(13/
100/log(2)^2)*log(2)^2)*e^(-1/4*x/log(2)^2))/log(2)^2

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mupad [B]  time = 4.32, size = 18, normalized size = 1.00 \begin {gather*} 3\,x+x\,{\mathrm {e}}^{-\frac {25\,x-13}{100\,{\ln \relax (2)}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(-(x/4 - 13/100)/log(2)^2)*(x - 4*log(2)^2))/4 - 3*log(2)^2)/log(2)^2,x)

[Out]

3*x + x*exp(-(25*x - 13)/(100*log(2)^2))

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sympy [A]  time = 0.11, size = 17, normalized size = 0.94 \begin {gather*} x e^{\frac {\frac {13}{100} - \frac {x}{4}}{\log {\relax (2 )}^{2}}} + 3 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((4*ln(2)**2-x)*exp(1/100*(-25*x+13)/ln(2)**2)+12*ln(2)**2)/ln(2)**2,x)

[Out]

x*exp((13/100 - x/4)/log(2)**2) + 3*x

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