3.16.33 \(\int \frac {-100-50 x+\frac {(-4-2 x) x^2}{e^{10}}+\frac {x (40+20 x)}{e^5}+e^{\frac {1}{-5+\frac {x}{e^5}}} (25-\frac {11 x}{e^5}+\frac {x^2}{e^{10}})}{-100 x-25 x^2+\frac {x^2 (-4 x-x^2)}{e^{10}}+\frac {x (40 x+10 x^2)}{e^5}+e^{\frac {1}{-5+\frac {x}{e^5}}} (25 x-\frac {10 x^2}{e^5}+\frac {x^3}{e^{10}})} \, dx\)

Optimal. Leaf size=19 \[ \log \left (\left (-4+e^{\frac {1}{-5+\frac {x}{e^5}}}-x\right ) x\right ) \]

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Rubi [A]  time = 0.82, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 127, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6741, 12, 6685} \begin {gather*} \log \left (x \left (x-e^{\frac {1}{\frac {x}{e^5}-5}}+4\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-100 - 50*x + ((-4 - 2*x)*x^2)/E^10 + (x*(40 + 20*x))/E^5 + E^(-5 + x/E^5)^(-1)*(25 - (11*x)/E^5 + x^2/E^
10))/(-100*x - 25*x^2 + (x^2*(-4*x - x^2))/E^10 + (x*(40*x + 10*x^2))/E^5 + E^(-5 + x/E^5)^(-1)*(25*x - (10*x^
2)/E^5 + x^3/E^10)),x]

[Out]

Log[x*(4 - E^(-5 + x/E^5)^(-1) + x)]

Rule 12

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

Rule 6685

Int[(u_)/((w_)*(y_)), x_Symbol] :> With[{q = DerivativeDivides[y*w, u, x]}, Simp[q*Log[RemoveContent[y*w, x]],
x] /;  !FalseQ[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 {e^{10} \left (100+50 x-\frac {(-4-2 x) x^2}{e^{10}}-\frac {x (40+20 x)}{e^5}-e^{\frac {1}{-5+\frac {x}{e^5}}} \left (25-\frac {11 x}{e^5}+\frac {x^2}{e^{10}}\right )\right )}{\left (5 e^5-x\right )^2 x \left (4-e^{\frac {1}{-5+\frac {x}{e^5}}}+x\right )} \, dx\\ &=e^{10} \int \frac {100+50 x-\frac {(-4-2 x) x^2}{e^{10}}-\frac {x (40+20 x)}{e^5}-e^{\frac {1}{-5+\frac {x}{e^5}}} \left (25-\frac {11 x}{e^5}+\frac {x^2}{e^{10}}\right )}{\left (5 e^5-x\right )^2 x \left (4-e^{\frac {1}{-5+\frac {x}{e^5}}}+x\right )} \, dx\\ &=\log \left (x \left (4-e^{\frac {1}{-5+\frac {x}{e^5}}}+x\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 24, normalized size = 1.26 \begin {gather*} \log (x)+\log \left (4-e^{\frac {e^5}{-5 e^5+x}}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-100 - 50*x + ((-4 - 2*x)*x^2)/E^10 + (x*(40 + 20*x))/E^5 + E^(-5 + x/E^5)^(-1)*(25 - (11*x)/E^5 +
x^2/E^10))/(-100*x - 25*x^2 + (x^2*(-4*x - x^2))/E^10 + (x*(40*x + 10*x^2))/E^5 + E^(-5 + x/E^5)^(-1)*(25*x -
(10*x^2)/E^5 + x^3/E^10)),x]

[Out]

Log[x] + Log[4 - E^(E^5/(-5*E^5 + x)) + x]

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fricas [A]  time = 0.79, size = 21, normalized size = 1.11 \begin {gather*} \log \relax (x) + \log \left (-x + e^{\left (\frac {e^{5}}{x - 5 \, e^{5}}\right )} - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(log(x)-5)^2-11*exp(log(x)-5)+25)*exp(1/(exp(log(x)-5)-5))+(-2*x-4)*exp(log(x)-5)^2+(20*x+40)*e
xp(log(x)-5)-50*x-100)/((x*exp(log(x)-5)^2-10*x*exp(log(x)-5)+25*x)*exp(1/(exp(log(x)-5)-5))+(-x^2-4*x)*exp(lo
g(x)-5)^2+(10*x^2+40*x)*exp(log(x)-5)-25*x^2-100*x),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.82, size = 93, normalized size = 4.89 \begin {gather*} {\left (e^{5} \log \left (\frac {5 \, e^{10}}{x - 5 \, e^{5}} + \frac {4 \, e^{5}}{x - 5 \, e^{5}} - \frac {e^{\left (\frac {e^{5}}{x - 5 \, e^{5}} + 5\right )}}{x - 5 \, e^{5}} + e^{5}\right ) + e^{5} \log \left (\frac {5 \, e^{5}}{x - 5 \, e^{5}} + 1\right ) - 2 \, e^{5} \log \left (\frac {e^{5}}{x - 5 \, e^{5}}\right )\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(log(x)-5)^2-11*exp(log(x)-5)+25)*exp(1/(exp(log(x)-5)-5))+(-2*x-4)*exp(log(x)-5)^2+(20*x+40)*e
xp(log(x)-5)-50*x-100)/((x*exp(log(x)-5)^2-10*x*exp(log(x)-5)+25*x)*exp(1/(exp(log(x)-5)-5))+(-x^2-4*x)*exp(lo
g(x)-5)^2+(10*x^2+40*x)*exp(log(x)-5)-25*x^2-100*x),x, algorithm="giac")

[Out]

(e^5*log(5*e^10/(x - 5*e^5) + 4*e^5/(x - 5*e^5) - e^(e^5/(x - 5*e^5) + 5)/(x - 5*e^5) + e^5) + e^5*log(5*e^5/(
x - 5*e^5) + 1) - 2*e^5*log(e^5/(x - 5*e^5)))*e^(-5)

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maple [A]  time = 1.80, size = 19, normalized size = 1.00

method	result	size
norman	\(\ln \relax (x )+\ln \left (x -{\mathrm e}^{\frac {1}{x \,{\mathrm e}^{-5}-5}}+4\right )\)	\(19\)
risch	\(\ln \left (-x \right )+\ln \left (-x +{\mathrm e}^{\frac {1}{x \,{\mathrm e}^{-5}-5}}-4\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(ln(x)-5)^2-11*exp(ln(x)-5)+25)*exp(1/(exp(ln(x)-5)-5))+(-2*x-4)*exp(ln(x)-5)^2+(20*x+40)*exp(ln(x)-5
)-50*x-100)/((x*exp(ln(x)-5)^2-10*x*exp(ln(x)-5)+25*x)*exp(1/(exp(ln(x)-5)-5))+(-x^2-4*x)*exp(ln(x)-5)^2+(10*x
^2+40*x)*exp(ln(x)-5)-25*x^2-100*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.48, size = 21, normalized size = 1.11 \begin {gather*} \log \relax (x) + \log \left (-x + e^{\left (\frac {e^{5}}{x - 5 \, e^{5}}\right )} - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(log(x)-5)^2-11*exp(log(x)-5)+25)*exp(1/(exp(log(x)-5)-5))+(-2*x-4)*exp(log(x)-5)^2+(20*x+40)*e
xp(log(x)-5)-50*x-100)/((x*exp(log(x)-5)^2-10*x*exp(log(x)-5)+25*x)*exp(1/(exp(log(x)-5)-5))+(-x^2-4*x)*exp(lo
g(x)-5)^2+(10*x^2+40*x)*exp(log(x)-5)-25*x^2-100*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.40, size = 18, normalized size = 0.95 \begin {gather*} \ln \left (x-{\mathrm {e}}^{\frac {1}{x\,{\mathrm {e}}^{-5}-5}}+4\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*x - exp(log(x) - 5)*(20*x + 40) - exp(1/(exp(log(x) - 5) - 5))*(exp(2*log(x) - 10) - 11*exp(log(x) - 5
) + 25) + exp(2*log(x) - 10)*(2*x + 4) + 100)/(100*x + exp(2*log(x) - 10)*(4*x + x^2) - exp(log(x) - 5)*(40*x
+ 10*x^2) - exp(1/(exp(log(x) - 5) - 5))*(25*x - 10*x*exp(log(x) - 5) + x*exp(2*log(x) - 10)) + 25*x^2),x)

[Out]

log(x - exp(1/(x*exp(-5) - 5)) + 4) + log(x)

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sympy [A]  time = 0.39, size = 17, normalized size = 0.89 \begin {gather*} \log {\relax (x )} + \log {\left (- x + e^{\frac {1}{\frac {x}{e^{5}} - 5}} - 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(ln(x)-5)**2-11*exp(ln(x)-5)+25)*exp(1/(exp(ln(x)-5)-5))+(-2*x-4)*exp(ln(x)-5)**2+(20*x+40)*exp
(ln(x)-5)-50*x-100)/((x*exp(ln(x)-5)**2-10*x*exp(ln(x)-5)+25*x)*exp(1/(exp(ln(x)-5)-5))+(-x**2-4*x)*exp(ln(x)-
5)**2+(10*x**2+40*x)*exp(ln(x)-5)-25*x**2-100*x),x)

[Out]

log(x) + log(-x + exp(1/(x*exp(-5) - 5)) - 4)

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