3.75.97 \(\int \frac {-20-4 x+(5+x) \log (\frac {5+x}{x})+\log (x) (5-20 x-4 x^2+e^x (20 x+4 x^2)+(5 x+x^2+e^x (-5 x-x^2)) \log (\frac {5+x}{x}))}{\log (x) (-20 x-4 x^2+(5 x+x^2) \log (\frac {5+x}{x}))} \, dx\)

Optimal. Leaf size=25 \[ \log \left (\frac {e^{-e^x+x} \log (x)}{-4+\log \left (\frac {5+x}{x}\right )}\right ) \]

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Rubi [F]  time = 1.14, 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 {-20-4 x+(5+x) \log \left (\frac {5+x}{x}\right )+\log (x) \left (5-20 x-4 x^2+e^x \left (20 x+4 x^2\right )+\left (5 x+x^2+e^x \left (-5 x-x^2\right )\right ) \log \left (\frac {5+x}{x}\right )\right )}{\log (x) \left (-20 x-4 x^2+\left (5 x+x^2\right ) \log \left (\frac {5+x}{x}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 5.12, size = 25, normalized size = 1.00 \begin {gather*} -e^x+x+\log (\log (x))-\log \left (4-\log \left (\frac {5+x}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.08, size = 22, normalized size = 0.88 \begin {gather*} x - e^{x} - \log \left (\log \left (\frac {x + 5}{x}\right ) - 4\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 22, normalized size = 0.88 \begin {gather*} x - e^{x} - \log \left (\log \left (x + 5\right ) - \log \relax (x) - 4\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.11, size = 23, normalized size = 0.92

method	result	size
default	\(x -\ln \left (\ln \left (\frac {5+x}{x}\right )-4\right )+\ln \left (\ln \relax (x )\right )-{\mathrm e}^{x}\)	\(23\)
risch	\(x -{\mathrm e}^{x}+\ln \left (\ln \relax (x )\right )-\ln \left (\ln \left (5+x \right )-\frac {i \left (\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (5+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (5+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{3}-2 i \ln \relax (x )-8 i\right )}{2}\right )\)	\(113\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 22, normalized size = 0.88 \begin {gather*} x - e^{x} - \log \left (\log \left (x + 5\right ) - \log \relax (x) - 4\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.78, size = 22, normalized size = 0.88 \begin {gather*} x-\ln \left (\ln \left (\frac {x+5}{x}\right )-4\right )+\ln \left (\ln \relax (x)\right )-{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.60, size = 19, normalized size = 0.76 \begin {gather*} x - e^{x} - \log {\left (\log {\left (\frac {x + 5}{x} \right )} - 4 \right )} + \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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