3.43.49 \(\int \frac {100-50 x+e^{-e^{-3+x}+x} (-19-e^{-3+x}+10 x)}{-25+5 e^{-e^{-3+x}+x}} \, dx\)

Optimal. Leaf size=30 \[ (-4+x) x-\log (4)+\frac {1}{5} \log \left (5-e^{-e^{-3+x}+x}\right ) \]

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Rubi [F]  time = 0.87, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {100-50 x+e^{-e^{-3+x}+x} \left (-19-e^{-3+x}+10 x\right )}{-25+5 e^{-e^{-3+x}+x}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(100 - 50*x + E^(-E^(-3 + x) + x)*(-19 - E^(-3 + x) + 10*x))/(-25 + 5*E^(-E^(-3 + x) + x)),x]

[Out]

-4*x + x^2 - Defer[Subst][Defer[Int][(5*E^(x/E^3) - x)^(-1), x], x, E^x]/5 + Defer[Subst][Defer[Int][x/(5*E^(x
/E^3) - x), x], x, E^x]/(5*E^3)

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {1}{5} e^{-3+x}-\frac {e^{-3+e^{-3+x}} \left (-e^3+5 e^{e^{-3+x}}\right )}{-5 e^{e^{-3+x}}+e^x}+\frac {-19 e^3-5 e^{e^{-3+x}}+10 e^3 x}{5 e^3}\right ) \, dx\\ &=-\left (\frac {1}{5} \int e^{-3+x} \, dx\right )+\frac {\int \left (-19 e^3-5 e^{e^{-3+x}}+10 e^3 x\right ) \, dx}{5 e^3}-\int \frac {e^{-3+e^{-3+x}} \left (-e^3+5 e^{e^{-3+x}}\right )}{-5 e^{e^{-3+x}}+e^x} \, dx\\ &=-\frac {1}{5} e^{-3+x}-\frac {19 x}{5}+x^2-\frac {\int e^{e^{-3+x}} \, dx}{e^3}-\operatorname {Subst}\left (\int \frac {e^3-5 e^{\frac {x}{e^3}}}{e^3 x \left (5-e^{-\frac {x}{e^3}} x\right )} \, dx,x,e^x\right )\\ &=-\frac {1}{5} e^{-3+x}-\frac {19 x}{5}+x^2-\frac {\operatorname {Subst}\left (\int \frac {e^x}{x} \, dx,x,e^{-3+x}\right )}{e^3}-\frac {\operatorname {Subst}\left (\int \frac {e^3-5 e^{\frac {x}{e^3}}}{x \left (5-e^{-\frac {x}{e^3}} x\right )} \, dx,x,e^x\right )}{e^3}\\ &=-\frac {1}{5} e^{-3+x}-\frac {19 x}{5}+x^2-\frac {\text {Ei}\left (e^{-3+x}\right )}{e^3}-\frac {\operatorname {Subst}\left (\int \left (-\frac {e^{\frac {x}{e^3}}}{x}+\frac {e^3-x}{5 x}-\frac {-e^3+x}{5 \left (5 e^{\frac {x}{e^3}}-x\right )}\right ) \, dx,x,e^x\right )}{e^3}\\ &=-\frac {1}{5} e^{-3+x}-\frac {19 x}{5}+x^2-\frac {\text {Ei}\left (e^{-3+x}\right )}{e^3}-\frac {\operatorname {Subst}\left (\int \frac {e^3-x}{x} \, dx,x,e^x\right )}{5 e^3}+\frac {\operatorname {Subst}\left (\int \frac {-e^3+x}{5 e^{\frac {x}{e^3}}-x} \, dx,x,e^x\right )}{5 e^3}+\frac {\operatorname {Subst}\left (\int \frac {e^{\frac {x}{e^3}}}{x} \, dx,x,e^x\right )}{e^3}\\ &=-\frac {1}{5} e^{-3+x}-\frac {19 x}{5}+x^2-\frac {\operatorname {Subst}\left (\int \left (-1+\frac {e^3}{x}\right ) \, dx,x,e^x\right )}{5 e^3}+\frac {\operatorname {Subst}\left (\int \left (-\frac {e^3}{5 e^{\frac {x}{e^3}}-x}+\frac {x}{5 e^{\frac {x}{e^3}}-x}\right ) \, dx,x,e^x\right )}{5 e^3}\\ &=-4 x+x^2-\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{5 e^{\frac {x}{e^3}}-x} \, dx,x,e^x\right )+\frac {\operatorname {Subst}\left (\int \frac {x}{5 e^{\frac {x}{e^3}}-x} \, dx,x,e^x\right )}{5 e^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.46, size = 32, normalized size = 1.07 \begin {gather*} \frac {1}{5} \left (-e^{-3+x}+5 (-4+x) x+\log \left (-5 e^{e^{-3+x}}+e^x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(100 - 50*x + E^(-E^(-3 + x) + x)*(-19 - E^(-3 + x) + 10*x))/(-25 + 5*E^(-E^(-3 + x) + x)),x]

[Out]

(-E^(-3 + x) + 5*(-4 + x)*x + Log[-5*E^E^(-3 + x) + E^x])/5

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fricas [A]  time = 0.65, size = 21, normalized size = 0.70 \begin {gather*} x^{2} - 4 \, x + \frac {1}{5} \, \log \left (e^{\left (x - e^{\left (x - 3\right )}\right )} - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x-3)+10*x-19)*exp(-exp(x-3)+x)-50*x+100)/(5*exp(-exp(x-3)+x)-25),x, algorithm="fricas")

[Out]

x^2 - 4*x + 1/5*log(e^(x - e^(x - 3)) - 5)

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giac [A]  time = 0.18, size = 28, normalized size = 0.93 \begin {gather*} x^{2} - 4 \, x - \frac {1}{5} \, e^{\left (x - 3\right )} + \frac {1}{5} \, \log \left (-e^{x} + 5 \, e^{\left (e^{\left (x - 3\right )}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x-3)+10*x-19)*exp(-exp(x-3)+x)-50*x+100)/(5*exp(-exp(x-3)+x)-25),x, algorithm="giac")

[Out]

x^2 - 4*x - 1/5*e^(x - 3) + 1/5*log(-e^x + 5*e^(e^(x - 3)))

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maple [A]  time = 0.05, size = 22, normalized size = 0.73

method	result	size
risch	\(x^{2}-4 x +\frac {\ln \left ({\mathrm e}^{-{\mathrm e}^{x -3}+x}-5\right )}{5}\)	\(22\)
norman	\(x^{2}-4 x +\frac {\ln \left (5 \,{\mathrm e}^{-{\mathrm e}^{x -3}+x}-25\right )}{5}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(x-3)+10*x-19)*exp(-exp(x-3)+x)-50*x+100)/(5*exp(-exp(x-3)+x)-25),x,method=_RETURNVERBOSE)

[Out]

x^2-4*x+1/5*ln(exp(-exp(x-3)+x)-5)

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maxima [A]  time = 0.39, size = 35, normalized size = 1.17 \begin {gather*} \frac {1}{5} \, {\left (5 \, x^{2} e^{3} - 20 \, x e^{3} - e^{x}\right )} e^{\left (-3\right )} + \frac {1}{5} \, \log \left (-\frac {1}{5} \, e^{x} + e^{\left (e^{\left (x - 3\right )}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x-3)+10*x-19)*exp(-exp(x-3)+x)-50*x+100)/(5*exp(-exp(x-3)+x)-25),x, algorithm="maxima")

[Out]

1/5*(5*x^2*e^3 - 20*x*e^3 - e^x)*e^(-3) + 1/5*log(-1/5*e^x + e^(e^(x - 3)))

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mupad [B]  time = 0.10, size = 22, normalized size = 0.73 \begin {gather*} \frac {\ln \left ({\mathrm {e}}^{-{\mathrm {e}}^{-3}\,{\mathrm {e}}^x}\,{\mathrm {e}}^x-5\right )}{5}-4\,x+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(50*x + exp(x - exp(x - 3))*(exp(x - 3) - 10*x + 19) - 100)/(5*exp(x - exp(x - 3)) - 25),x)

[Out]

log(exp(-exp(-3)*exp(x))*exp(x) - 5)/5 - 4*x + x^2

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sympy [A]  time = 0.15, size = 19, normalized size = 0.63 \begin {gather*} x^{2} - 4 x + \frac {\log {\left (e^{x - e^{x - 3}} - 5 \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x-3)+10*x-19)*exp(-exp(x-3)+x)-50*x+100)/(5*exp(-exp(x-3)+x)-25),x)

[Out]

x**2 - 4*x + log(exp(x - exp(x - 3)) - 5)/5

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