3.15.98 \(\int \frac {-40 e^x+8 e^{2 x}+e^{\frac {x^2}{8}} x+e^{\frac {x^2}{16}} (e^x (-8-x)+5 x)}{100-40 e^x+4 e^{2 x}+4 e^{\frac {x^2}{8}}+e^{\frac {x^2}{16}} (40-8 e^x)+4 \log (4)} \, dx\)

Optimal. Leaf size=22 \[ \log \left (\left (5-e^x+e^{\frac {x^2}{16}}\right )^2+\log (4)\right ) \]

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Rubi [B]  time = 0.85, antiderivative size = 54, normalized size of antiderivative = 2.45, number of steps used = 4, number of rules used = 3, integrand size = 96, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {6741, 12, 6684} \begin {gather*} \log \left (-10 e^{\frac {x^2}{16}}-e^{\frac {x^2}{8}}+2 e^{\frac {x^2}{16}+x}+10 e^x-e^{2 x}-25-\log (4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-40*E^x + 8*E^(2*x) + E^(x^2/8)*x + E^(x^2/16)*(E^x*(-8 - x) + 5*x))/(100 - 40*E^x + 4*E^(2*x) + 4*E^(x^2
/8) + E^(x^2/16)*(40 - 8*E^x) + 4*Log[4]),x]

[Out]

Log[-25 + 10*E^x - E^(2*x) - 10*E^(x^2/16) - E^(x^2/8) + 2*E^(x + x^2/16) - Log[4]]

Rule 12

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

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (5-e^x+e^{\frac {x^2}{16}}\right ) \left (8 e^x-e^{\frac {x^2}{16}} x\right )}{40 e^x-4 e^{2 x}-4 e^{\frac {x^2}{8}}-e^{\frac {x^2}{16}} \left (40-8 e^x\right )-100 \left (1+\frac {2 \log (2)}{25}\right )} \, dx\\ &=\int \frac {\left (5-e^x+e^{\frac {x^2}{16}}\right ) \left (8 e^x-e^{\frac {x^2}{16}} x\right )}{4 \left (10 e^x-e^{2 x}-10 e^{\frac {x^2}{16}}-e^{\frac {x^2}{8}}+2 e^{x+\frac {x^2}{16}}-25 \left (1+\frac {2 \log (2)}{25}\right )\right )} \, dx\\ &=\frac {1}{4} \int \frac {\left (5-e^x+e^{\frac {x^2}{16}}\right ) \left (8 e^x-e^{\frac {x^2}{16}} x\right )}{10 e^x-e^{2 x}-10 e^{\frac {x^2}{16}}-e^{\frac {x^2}{8}}+2 e^{x+\frac {x^2}{16}}-25 \left (1+\frac {2 \log (2)}{25}\right )} \, dx\\ &=\log \left (-25+10 e^x-e^{2 x}-10 e^{\frac {x^2}{16}}-e^{\frac {x^2}{8}}+2 e^{x+\frac {x^2}{16}}-\log (4)\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.03, size = 48, normalized size = 2.18 \begin {gather*} \log \left (25-10 e^x+e^{2 x}+10 e^{\frac {x^2}{16}}+e^{\frac {x^2}{8}}-2 e^{x+\frac {x^2}{16}}+\log (4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-40*E^x + 8*E^(2*x) + E^(x^2/8)*x + E^(x^2/16)*(E^x*(-8 - x) + 5*x))/(100 - 40*E^x + 4*E^(2*x) + 4*
E^(x^2/8) + E^(x^2/16)*(40 - 8*E^x) + 4*Log[4]),x]

[Out]

Log[25 - 10*E^x + E^(2*x) + 10*E^(x^2/16) + E^(x^2/8) - 2*E^(x + x^2/16) + Log[4]]

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fricas [A]  time = 0.62, size = 33, normalized size = 1.50 \begin {gather*} \log \left (-2 \, {\left (e^{x} - 5\right )} e^{\left (\frac {1}{16} \, x^{2}\right )} + e^{\left (\frac {1}{8} \, x^{2}\right )} + e^{\left (2 \, x\right )} - 10 \, e^{x} + 2 \, \log \relax (2) + 25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(1/16*x^2)^2+((-x-8)*exp(x)+5*x)*exp(1/16*x^2)+8*exp(x)^2-40*exp(x))/(4*exp(1/16*x^2)^2+(-8*ex
p(x)+40)*exp(1/16*x^2)+4*exp(x)^2-40*exp(x)+8*log(2)+100),x, algorithm="fricas")

[Out]

log(-2*(e^x - 5)*e^(1/16*x^2) + e^(1/8*x^2) + e^(2*x) - 10*e^x + 2*log(2) + 25)

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giac [A]  time = 0.36, size = 39, normalized size = 1.77 \begin {gather*} \log \left (e^{\left (\frac {1}{8} \, x^{2}\right )} + 10 \, e^{\left (\frac {1}{16} \, x^{2}\right )} - 2 \, e^{\left (\frac {1}{16} \, x^{2} + x\right )} + e^{\left (2 \, x\right )} - 10 \, e^{x} + 2 \, \log \relax (2) + 25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(1/16*x^2)^2+((-x-8)*exp(x)+5*x)*exp(1/16*x^2)+8*exp(x)^2-40*exp(x))/(4*exp(1/16*x^2)^2+(-8*ex
p(x)+40)*exp(1/16*x^2)+4*exp(x)^2-40*exp(x)+8*log(2)+100),x, algorithm="giac")

[Out]

log(e^(1/8*x^2) + 10*e^(1/16*x^2) - 2*e^(1/16*x^2 + x) + e^(2*x) - 10*e^x + 2*log(2) + 25)

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maple [A]  time = 0.17, size = 35, normalized size = 1.59

method	result	size
risch	\(\ln \left ({\mathrm e}^{\frac {x^{2}}{8}}+\left (-2 \,{\mathrm e}^{x}+10\right ) {\mathrm e}^{\frac {x^{2}}{16}}+{\mathrm e}^{2 x}+2 \ln \relax (2)-10 \,{\mathrm e}^{x}+25\right )\)	\(35\)
norman	\(\ln \left (4 \,{\mathrm e}^{2 x}-8 \,{\mathrm e}^{x} {\mathrm e}^{\frac {x^{2}}{16}}+4 \,{\mathrm e}^{\frac {x^{2}}{8}}-40 \,{\mathrm e}^{x}+8 \ln \relax (2)+40 \,{\mathrm e}^{\frac {x^{2}}{16}}+100\right )\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(1/16*x^2)^2+((-x-8)*exp(x)+5*x)*exp(1/16*x^2)+8*exp(x)^2-40*exp(x))/(4*exp(1/16*x^2)^2+(-8*exp(x)+4
0)*exp(1/16*x^2)+4*exp(x)^2-40*exp(x)+8*ln(2)+100),x,method=_RETURNVERBOSE)

[Out]

ln(exp(1/8*x^2)+(-2*exp(x)+10)*exp(1/16*x^2)+exp(2*x)+2*ln(2)-10*exp(x)+25)

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maxima [A]  time = 0.73, size = 33, normalized size = 1.50 \begin {gather*} \log \left (-2 \, {\left (e^{x} - 5\right )} e^{\left (\frac {1}{16} \, x^{2}\right )} + e^{\left (\frac {1}{8} \, x^{2}\right )} + e^{\left (2 \, x\right )} - 10 \, e^{x} + 2 \, \log \relax (2) + 25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(1/16*x^2)^2+((-x-8)*exp(x)+5*x)*exp(1/16*x^2)+8*exp(x)^2-40*exp(x))/(4*exp(1/16*x^2)^2+(-8*ex
p(x)+40)*exp(1/16*x^2)+4*exp(x)^2-40*exp(x)+8*log(2)+100),x, algorithm="maxima")

[Out]

log(-2*(e^x - 5)*e^(1/16*x^2) + e^(1/8*x^2) + e^(2*x) - 10*e^x + 2*log(2) + 25)

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mupad [B]  time = 1.17, size = 37, normalized size = 1.68 \begin {gather*} \ln \left ({\mathrm {e}}^{2\,x}+\ln \relax (4)-2\,{\mathrm {e}}^{\frac {x^2}{16}+x}+{\mathrm {e}}^{\frac {x^2}{8}}+10\,{\mathrm {e}}^{\frac {x^2}{16}}-10\,{\mathrm {e}}^x+25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*exp(2*x) - 40*exp(x) + exp(x^2/16)*(5*x - exp(x)*(x + 8)) + x*exp(x^2/8))/(4*exp(2*x) + 8*log(2) + 4*ex
p(x^2/8) - 40*exp(x) - exp(x^2/16)*(8*exp(x) - 40) + 100),x)

[Out]

log(exp(2*x) + log(4) - 2*exp(x + x^2/16) + exp(x^2/8) + 10*exp(x^2/16) - 10*exp(x) + 25)

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sympy [B]  time = 0.32, size = 46, normalized size = 2.09 \begin {gather*} \frac {3 x^{2}}{32} + \frac {\log {\left (\left (10 - 2 e^{x}\right ) e^{\frac {x^{2}}{16}} + e^{2 x} - 10 e^{x} + e^{\frac {x^{2}}{8}} + 2 \log {\relax (2 )} + 25 \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(1/16*x**2)**2+((-x-8)*exp(x)+5*x)*exp(1/16*x**2)+8*exp(x)**2-40*exp(x))/(4*exp(1/16*x**2)**2+
(-8*exp(x)+40)*exp(1/16*x**2)+4*exp(x)**2-40*exp(x)+8*ln(2)+100),x)

[Out]

3*x**2/32 + log((10 - 2*exp(x))*exp(x**2/16) + exp(2*x) - 10*exp(x) + exp(x**2/8) + 2*log(2) + 25)/4

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