\(\int \frac {3 \log ^2(4)+e^{\frac {13-25 x}{25 \log ^2(4)}} (-x+\log ^2(4))}{\log ^2(4)} \, dx\) [5920]

Optimal result

Integrand size = 36, antiderivative size = 18 \[ \int \frac {3 \log ^2(4)+e^{\frac {13-25 x}{25 \log ^2(4)}} \left (-x+\log ^2(4)\right )}{\log ^2(4)} \, dx=\left (3+e^{\frac {\frac {13}{25}-x}{\log ^2(4)}}\right ) x \]

[Out]

(3+exp(1/4*(13/25-x)/ln(2)^2))*x

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(48\) vs. \(2(18)=36\).

Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.67, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 2207, 2225} \[ \int \frac {3 \log ^2(4)+e^{\frac {13-25 x}{25 \log ^2(4)}} \left (-x+\log ^2(4)\right )}{\log ^2(4)} \, dx=3 x+e^{\frac {13-25 x}{25 \log ^2(4)}} \left (x-\log ^2(4)\right )+\log ^2(4) e^{\frac {13-25 x}{25 \log ^2(4)}} \]

[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 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {3 \log ^2(4)+e^{\frac {13-25 x}{25 \log ^2(4)}} \left (-x+\log ^2(4)\right )}{\log ^2(4)} \, dx=3 x+e^{\frac {13}{25 \log ^2(4)}-\frac {x}{\log ^2(4)}} x \]

[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

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method	result	size
risch	\({\mathrm e}^{-\frac {25 x -13}{100 \ln \left (2\right )^{2}}} x +3 x\)	\(19\)
parts	\(3 x +{\mathrm e}^{-\frac {x}{4 \ln \left (2\right )^{2}}+\frac {13}{100 \ln \left (2\right )^{2}}} x\)	\(22\)
norman	\(\frac {x \ln \left (2\right ) {\mathrm e}^{\frac {-25 x +13}{100 \ln \left (2\right )^{2}}}+3 x \ln \left (2\right )}{\ln \left (2\right )}\)	\(28\)
parallelrisch	\(\frac {4 x \ln \left (2\right )^{2} {\mathrm e}^{-\frac {25 x -13}{100 \ln \left (2\right )^{2}}}+12 x \ln \left (2\right )^{2}}{4 \ln \left (2\right )^{2}}\)	\(34\)
default	\(\frac {4 \ln \left (2\right )^{2} {\mathrm e}^{-\frac {x}{4 \ln \left (2\right )^{2}}+\frac {13}{100 \ln \left (2\right )^{2}}} x +12 x \ln \left (2\right )^{2}}{4 \ln \left (2\right )^{2}}\)	\(37\)
derivativedivides	\(-12 \ln \left (2\right )^{2} \left (-\frac {x}{4 \ln \left (2\right )^{2}}+\frac {13}{100 \ln \left (2\right )^{2}}\right )-4 \,{\mathrm e}^{-\frac {x}{4 \ln \left (2\right )^{2}}+\frac {13}{100 \ln \left (2\right )^{2}}} \ln \left (2\right )^{2} \left (-\frac {x}{4 \ln \left (2\right )^{2}}+\frac {13}{100 \ln \left (2\right )^{2}}\right )+\frac {13 \,{\mathrm e}^{-\frac {x}{4 \ln \left (2\right )^{2}}+\frac {13}{100 \ln \left (2\right )^{2}}}}{25}\)	\(74\)

[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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {3 \log ^2(4)+e^{\frac {13-25 x}{25 \log ^2(4)}} \left (-x+\log ^2(4)\right )}{\log ^2(4)} \, dx=x e^{\left (-\frac {25 \, x - 13}{100 \, \log \left (2\right )^{2}}\right )} + 3 \, x \]

[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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {3 \log ^2(4)+e^{\frac {13-25 x}{25 \log ^2(4)}} \left (-x+\log ^2(4)\right )}{\log ^2(4)} \, dx=x e^{\frac {\frac {13}{100} - \frac {x}{4}}{\log {\left (2 \right )}^{2}}} + 3 x \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (16) = 32\).

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 4.00 \[ \int \frac {3 \log ^2(4)+e^{\frac {13-25 x}{25 \log ^2(4)}} \left (-x+\log ^2(4)\right )}{\log ^2(4)} \, dx=-\frac {4 \, e^{\left (-\frac {x}{4 \, \log \left (2\right )^{2}} + \frac {13}{100 \, \log \left (2\right )^{2}}\right )} \log \left (2\right )^{4} - 3 \, x \log \left (2\right )^{2} - {\left (4 \, e^{\left (\frac {13}{100 \, \log \left (2\right )^{2}}\right )} \log \left (2\right )^{4} + x e^{\left (\frac {13}{100 \, \log \left (2\right )^{2}}\right )} \log \left (2\right )^{2}\right )} e^{\left (-\frac {x}{4 \, \log \left (2\right )^{2}}\right )}}{\log \left (2\right )^{2}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.72 \[ \int \frac {3 \log ^2(4)+e^{\frac {13-25 x}{25 \log ^2(4)}} \left (-x+\log ^2(4)\right )}{\log ^2(4)} \, dx=\frac {x e^{\left (-\frac {25 \, x - 13}{100 \, \log \left (2\right )^{2}}\right )} \log \left (2\right )^{2} + 3 \, x \log \left (2\right )^{2}}{\log \left (2\right )^{2}} \]

[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

Mupad [B] (verification not implemented)

Time = 11.70 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {3 \log ^2(4)+e^{\frac {13-25 x}{25 \log ^2(4)}} \left (-x+\log ^2(4)\right )}{\log ^2(4)} \, dx=3\,x+x\,{\mathrm {e}}^{-\frac {25\,x-13}{100\,{\ln \left (2\right )}^2}} \]

[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))