3.52.22 \(\int \frac {-20 x-40 x^2+e^{\frac {e^x}{5}} (20+8 e^x x^2)+(-40 e^{\frac {e^x}{5}} x+40 x^2) \log (e^{\frac {e^x}{5}}-x)}{5 e^{\frac {e^x}{5}} x^3 \log (2)-5 x^4 \log (2)} \, dx\)

Optimal. Leaf size=34 \[ \frac {8 \left (-\frac {1}{4 x}-x+\log \left (e^{\frac {e^x}{5}}-x\right )\right )}{x \log (2)} \]

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Rubi [A]  time = 2.11, antiderivative size = 33, normalized size of antiderivative = 0.97, number of steps used = 12, number of rules used = 5, integrand size = 89, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6741, 12, 6742, 14, 2551} \begin {gather*} \frac {8 \log \left (e^{\frac {e^x}{5}}-x\right )}{x \log (2)}-\frac {2}{x^2 \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20*x - 40*x^2 + E^(E^x/5)*(20 + 8*E^x*x^2) + (-40*E^(E^x/5)*x + 40*x^2)*Log[E^(E^x/5) - x])/(5*E^(E^x/5)
*x^3*Log[2] - 5*x^4*Log[2]),x]

[Out]

-2/(x^2*Log[2]) + (8*Log[E^(E^x/5) - x])/(x*Log[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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-20 x-40 x^2+e^{\frac {e^x}{5}} \left (20+8 e^x x^2\right )+\left (-40 e^{\frac {e^x}{5}} x+40 x^2\right ) \log \left (e^{\frac {e^x}{5}}-x\right )}{5 \left (e^{\frac {e^x}{5}}-x\right ) x^3 \log (2)} \, dx\\ &=\frac {\int \frac {-20 x-40 x^2+e^{\frac {e^x}{5}} \left (20+8 e^x x^2\right )+\left (-40 e^{\frac {e^x}{5}} x+40 x^2\right ) \log \left (e^{\frac {e^x}{5}}-x\right )}{\left (e^{\frac {e^x}{5}}-x\right ) x^3} \, dx}{5 \log (2)}\\ &=\frac {\int \left (\frac {8 e^{\frac {1}{5} \left (e^x+5 x\right )}}{\left (e^{\frac {e^x}{5}}-x\right ) x}-\frac {20 \left (-e^{\frac {e^x}{5}}+x+2 x^2+2 e^{\frac {e^x}{5}} x \log \left (e^{\frac {e^x}{5}}-x\right )-2 x^2 \log \left (e^{\frac {e^x}{5}}-x\right )\right )}{\left (e^{\frac {e^x}{5}}-x\right ) x^3}\right ) \, dx}{5 \log (2)}\\ &=\frac {8 \int \frac {e^{\frac {1}{5} \left (e^x+5 x\right )}}{\left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{5 \log (2)}-\frac {4 \int \frac {-e^{\frac {e^x}{5}}+x+2 x^2+2 e^{\frac {e^x}{5}} x \log \left (e^{\frac {e^x}{5}}-x\right )-2 x^2 \log \left (e^{\frac {e^x}{5}}-x\right )}{\left (e^{\frac {e^x}{5}}-x\right ) x^3} \, dx}{\log (2)}\\ &=\frac {8 \int \frac {e^{\frac {1}{5} \left (e^x+5 x\right )}}{\left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{5 \log (2)}-\frac {4 \int \left (\frac {2}{\left (e^{\frac {e^x}{5}}-x\right ) x}+\frac {-1+2 x \log \left (e^{\frac {e^x}{5}}-x\right )}{x^3}\right ) \, dx}{\log (2)}\\ &=\frac {8 \int \frac {e^{\frac {1}{5} \left (e^x+5 x\right )}}{\left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{5 \log (2)}-\frac {4 \int \frac {-1+2 x \log \left (e^{\frac {e^x}{5}}-x\right )}{x^3} \, dx}{\log (2)}-\frac {8 \int \frac {1}{\left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{\log (2)}\\ &=\frac {8 \int \frac {e^{\frac {1}{5} \left (e^x+5 x\right )}}{\left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{5 \log (2)}-\frac {4 \int \left (-\frac {1}{x^3}+\frac {2 \log \left (e^{\frac {e^x}{5}}-x\right )}{x^2}\right ) \, dx}{\log (2)}-\frac {8 \int \frac {1}{\left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{\log (2)}\\ &=-\frac {2}{x^2 \log (2)}+\frac {8 \int \frac {e^{\frac {1}{5} \left (e^x+5 x\right )}}{\left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{5 \log (2)}-\frac {8 \int \frac {1}{\left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{\log (2)}-\frac {8 \int \frac {\log \left (e^{\frac {e^x}{5}}-x\right )}{x^2} \, dx}{\log (2)}\\ &=-\frac {2}{x^2 \log (2)}+\frac {8 \log \left (e^{\frac {e^x}{5}}-x\right )}{x \log (2)}+\frac {8 \int \frac {e^{\frac {1}{5} \left (e^x+5 x\right )}}{\left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{5 \log (2)}-\frac {8 \int \frac {1}{\left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{\log (2)}-\frac {8 \int \frac {-5+e^{\frac {e^x}{5}+x}}{5 \left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{\log (2)}\\ &=-\frac {2}{x^2 \log (2)}+\frac {8 \log \left (e^{\frac {e^x}{5}}-x\right )}{x \log (2)}+\frac {8 \int \frac {e^{\frac {1}{5} \left (e^x+5 x\right )}}{\left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{5 \log (2)}-\frac {8 \int \frac {-5+e^{\frac {e^x}{5}+x}}{\left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{5 \log (2)}-\frac {8 \int \frac {1}{\left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{\log (2)}\\ &=-\frac {2}{x^2 \log (2)}+\frac {8 \log \left (e^{\frac {e^x}{5}}-x\right )}{x \log (2)}-\frac {8 \int \left (-\frac {5}{\left (e^{\frac {e^x}{5}}-x\right ) x}+\frac {e^{\frac {1}{5} \left (e^x+5 x\right )}}{\left (e^{\frac {e^x}{5}}-x\right ) x}\right ) \, dx}{5 \log (2)}+\frac {8 \int \frac {e^{\frac {1}{5} \left (e^x+5 x\right )}}{\left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{5 \log (2)}-\frac {8 \int \frac {1}{\left (e^{\frac {e^x}{5}}-x\right ) x} \, dx}{\log (2)}\\ &=-\frac {2}{x^2 \log (2)}+\frac {8 \log \left (e^{\frac {e^x}{5}}-x\right )}{x \log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.32, size = 33, normalized size = 0.97 \begin {gather*} -\frac {\frac {10}{x^2}-\frac {40 \log \left (e^{\frac {e^x}{5}}-x\right )}{x}}{5 \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20*x - 40*x^2 + E^(E^x/5)*(20 + 8*E^x*x^2) + (-40*E^(E^x/5)*x + 40*x^2)*Log[E^(E^x/5) - x])/(5*E^(
E^x/5)*x^3*Log[2] - 5*x^4*Log[2]),x]

[Out]

-1/5*(10/x^2 - (40*Log[E^(E^x/5) - x])/x)/Log[2]

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fricas [A]  time = 0.53, size = 24, normalized size = 0.71 \begin {gather*} \frac {2 \, {\left (4 \, x \log \left (-x + e^{\left (\frac {1}{5} \, e^{x}\right )}\right ) - 1\right )}}{x^{2} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x*exp(1/5*exp(x))+40*x^2)*log(exp(1/5*exp(x))-x)+(8*exp(x)*x^2+20)*exp(1/5*exp(x))-40*x^2-20*x
)/(5*x^3*log(2)*exp(1/5*exp(x))-5*x^4*log(2)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 35, normalized size = 1.03 \begin {gather*} \frac {2 \, {\left (4 \, x \log \left (-{\left (x e^{x} - e^{\left (x + \frac {1}{5} \, e^{x}\right )}\right )} e^{\left (-x\right )}\right ) - 1\right )}}{x^{2} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x*exp(1/5*exp(x))+40*x^2)*log(exp(1/5*exp(x))-x)+(8*exp(x)*x^2+20)*exp(1/5*exp(x))-40*x^2-20*x
)/(5*x^3*log(2)*exp(1/5*exp(x))-5*x^4*log(2)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.10, size = 30, normalized size = 0.88

method	result	size
risch	\(\frac {8 \ln \left ({\mathrm e}^{\frac {{\mathrm e}^{x}}{5}}-x \right )}{x \ln \relax (2)}-\frac {2}{\ln \relax (2) x^{2}}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-40*x*exp(1/5*exp(x))+40*x^2)*ln(exp(1/5*exp(x))-x)+(8*exp(x)*x^2+20)*exp(1/5*exp(x))-40*x^2-20*x)/(5*x^
3*ln(2)*exp(1/5*exp(x))-5*x^4*ln(2)),x,method=_RETURNVERBOSE)

[Out]

8/x/ln(2)*ln(exp(1/5*exp(x))-x)-2/ln(2)/x^2

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maxima [A]  time = 0.48, size = 24, normalized size = 0.71 \begin {gather*} \frac {2 \, {\left (4 \, x \log \left (-x + e^{\left (\frac {1}{5} \, e^{x}\right )}\right ) - 1\right )}}{x^{2} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x*exp(1/5*exp(x))+40*x^2)*log(exp(1/5*exp(x))-x)+(8*exp(x)*x^2+20)*exp(1/5*exp(x))-40*x^2-20*x
)/(5*x^3*log(2)*exp(1/5*exp(x))-5*x^4*log(2)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.38, size = 23, normalized size = 0.68 \begin {gather*} \frac {8\,x\,\ln \left ({\mathrm {e}}^{\frac {{\mathrm {e}}^x}{5}}-x\right )-2}{x^2\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*x - exp(exp(x)/5)*(8*x^2*exp(x) + 20) + log(exp(exp(x)/5) - x)*(40*x*exp(exp(x)/5) - 40*x^2) + 40*x^2)
/(5*x^4*log(2) - 5*x^3*exp(exp(x)/5)*log(2)),x)

[Out]

(8*x*log(exp(exp(x)/5) - x) - 2)/(x^2*log(2))

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sympy [A]  time = 0.50, size = 24, normalized size = 0.71 \begin {gather*} \frac {8 \log {\left (- x + e^{\frac {e^{x}}{5}} \right )}}{x \log {\relax (2 )}} - \frac {2}{x^{2} \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x*exp(1/5*exp(x))+40*x**2)*ln(exp(1/5*exp(x))-x)+(8*exp(x)*x**2+20)*exp(1/5*exp(x))-40*x**2-20
*x)/(5*x**3*ln(2)*exp(1/5*exp(x))-5*x**4*ln(2)),x)

[Out]

8*log(-x + exp(exp(x)/5))/(x*log(2)) - 2/(x**2*log(2))

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