3.96.29 \(\int \frac {e^{\frac {15+3 x-3 \log (\frac {-1+x+\log (x)}{4 x+\log (x)})}{x}} (-3+39 x-60 x^2+(15-66 x) \log (x)-15 \log ^2(x)+(-12 x+12 x^2+(-3+15 x) \log (x)+3 \log ^2(x)) \log (\frac {-1+x+\log (x)}{4 x+\log (x)}))}{-4 x^3+4 x^4+(-x^2+5 x^3) \log (x)+x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ -3+e^{\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}} \]

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Rubi [F]  time = 22.41, 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 {\exp \left (\frac {15+3 x-3 \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )}{x}\right ) \left (-3+39 x-60 x^2+(15-66 x) \log (x)-15 \log ^2(x)+\left (-12 x+12 x^2+(-3+15 x) \log (x)+3 \log ^2(x)\right ) \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{-4 x^3+4 x^4+\left (-x^2+5 x^3\right ) \log (x)+x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((15 + 3*x - 3*Log[(-1 + x + Log[x])/(4*x + Log[x])])/x)*(-3 + 39*x - 60*x^2 + (15 - 66*x)*Log[x] - 15*
Log[x]^2 + (-12*x + 12*x^2 + (-3 + 15*x)*Log[x] + 3*Log[x]^2)*Log[(-1 + x + Log[x])/(4*x + Log[x])]))/(-4*x^3
+ 4*x^4 + (-x^2 + 5*x^3)*Log[x] + x^2*Log[x]^2),x]

[Out]

-15*Defer[Int][E^((3*(5 + x - Log[(-1 + x + Log[x])/(4*x + Log[x])]))/x)/x^2, x] - 3*Defer[Int][E^((3*(5 + x -
Log[(-1 + x + Log[x])/(4*x + Log[x])]))/x)/(x^2*(-1 + x + Log[x])), x] - 3*Defer[Int][E^((3*(5 + x - Log[(-1
+ x + Log[x])/(4*x + Log[x])]))/x)/(x*(-1 + x + Log[x])), x] + 3*Defer[Int][E^((3*(5 + x - Log[(-1 + x + Log[x
])/(4*x + Log[x])]))/x)/(x^2*(4*x + Log[x])), x] + 12*Defer[Int][E^((3*(5 + x - Log[(-1 + x + Log[x])/(4*x + L
og[x])]))/x)/(x*(4*x + Log[x])), x] + 3*Defer[Int][(E^((3*(5 + x - Log[(-1 + x + Log[x])/(4*x + Log[x])]))/x)*
Log[(-1 + x + Log[x])/(4*x + Log[x])])/x^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right ) \left (3-39 x+60 x^2-(15-66 x) \log (x)+15 \log ^2(x)-\left (-12 x+12 x^2+(-3+15 x) \log (x)+3 \log ^2(x)\right ) \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{4 x^3-4 x^4-\left (-x^2+5 x^3\right ) \log (x)-x^2 \log ^2(x)} \, dx\\ &=\int \left (-\frac {60 \exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{(-1+x+\log (x)) (4 x+\log (x))}-\frac {3 \exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{x^2 (-1+x+\log (x)) (4 x+\log (x))}+\frac {39 \exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{x (-1+x+\log (x)) (4 x+\log (x))}-\frac {3 \exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right ) (-5+22 x) \log (x)}{x^2 (-1+x+\log (x)) (4 x+\log (x))}-\frac {15 \exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right ) \log ^2(x)}{x^2 (-1+x+\log (x)) (4 x+\log (x))}+\frac {3 \exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right ) \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )}{x^2}\right ) \, dx\\ &=-\left (3 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{x^2 (-1+x+\log (x)) (4 x+\log (x))} \, dx\right )-3 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right ) (-5+22 x) \log (x)}{x^2 (-1+x+\log (x)) (4 x+\log (x))} \, dx+3 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right ) \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )}{x^2} \, dx-15 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right ) \log ^2(x)}{x^2 (-1+x+\log (x)) (4 x+\log (x))} \, dx+39 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{x (-1+x+\log (x)) (4 x+\log (x))} \, dx-60 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{(-1+x+\log (x)) (4 x+\log (x))} \, dx\\ &=-\left (3 \int \left (\frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{x^2 (1+3 x) (-1+x+\log (x))}-\frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{x^2 (1+3 x) (4 x+\log (x))}\right ) \, dx\right )-3 \int \left (-\frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right ) (-1+x) (-5+22 x)}{x^2 (1+3 x) (-1+x+\log (x))}+\frac {4 \exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right ) (-5+22 x)}{x (1+3 x) (4 x+\log (x))}\right ) \, dx+3 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right ) \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )}{x^2} \, dx-15 \int \left (\frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{x^2}+\frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right ) (-1+x)^2}{x^2 (1+3 x) (-1+x+\log (x))}-\frac {16 \exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{(1+3 x) (4 x+\log (x))}\right ) \, dx+39 \int \left (\frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{x (1+3 x) (-1+x+\log (x))}-\frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{x (1+3 x) (4 x+\log (x))}\right ) \, dx-60 \int \left (\frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{(1+3 x) (-1+x+\log (x))}-\frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{(1+3 x) (4 x+\log (x))}\right ) \, dx\\ &=-\left (3 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{x^2 (1+3 x) (-1+x+\log (x))} \, dx\right )+3 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right ) (-1+x) (-5+22 x)}{x^2 (1+3 x) (-1+x+\log (x))} \, dx+3 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{x^2 (1+3 x) (4 x+\log (x))} \, dx+3 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right ) \log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )}{x^2} \, dx-12 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right ) (-5+22 x)}{x (1+3 x) (4 x+\log (x))} \, dx-15 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{x^2} \, dx-15 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right ) (-1+x)^2}{x^2 (1+3 x) (-1+x+\log (x))} \, dx+39 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{x (1+3 x) (-1+x+\log (x))} \, dx-39 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{x (1+3 x) (4 x+\log (x))} \, dx-60 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{(1+3 x) (-1+x+\log (x))} \, dx+60 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{(1+3 x) (4 x+\log (x))} \, dx+240 \int \frac {\exp \left (\frac {3 \left (5+x-\log \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )\right )}{x}\right )}{(1+3 x) (4 x+\log (x))} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 30, normalized size = 1.03 \begin {gather*} e^{3+\frac {15}{x}} \left (\frac {-1+x+\log (x)}{4 x+\log (x)}\right )^{-3/x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((15 + 3*x - 3*Log[(-1 + x + Log[x])/(4*x + Log[x])])/x)*(-3 + 39*x - 60*x^2 + (15 - 66*x)*Log[x]
- 15*Log[x]^2 + (-12*x + 12*x^2 + (-3 + 15*x)*Log[x] + 3*Log[x]^2)*Log[(-1 + x + Log[x])/(4*x + Log[x])]))/(-
4*x^3 + 4*x^4 + (-x^2 + 5*x^3)*Log[x] + x^2*Log[x]^2),x]

[Out]

E^(3 + 15/x)/((-1 + x + Log[x])/(4*x + Log[x]))^(3/x)

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fricas [A]  time = 0.61, size = 26, normalized size = 0.90 \begin {gather*} e^{\left (\frac {3 \, {\left (x - \log \left (\frac {x + \log \relax (x) - 1}{4 \, x + \log \relax (x)}\right ) + 5\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*log(x)^2+(15*x-3)*log(x)+12*x^2-12*x)*log((-1+log(x)+x)/(4*x+log(x)))-15*log(x)^2+(-66*x+15)*log
(x)-60*x^2+39*x-3)*exp((-3*log((-1+log(x)+x)/(4*x+log(x)))+15+3*x)/x)/(x^2*log(x)^2+(5*x^3-x^2)*log(x)+4*x^4-4
*x^3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 3.96, size = 46, normalized size = 1.59 \begin {gather*} e^{\left (-\frac {3 \, \log \left (\frac {x}{4 \, x + \log \relax (x)} + \frac {\log \relax (x)}{4 \, x + \log \relax (x)} - \frac {1}{4 \, x + \log \relax (x)}\right )}{x} + \frac {15}{x} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*log(x)^2+(15*x-3)*log(x)+12*x^2-12*x)*log((-1+log(x)+x)/(4*x+log(x)))-15*log(x)^2+(-66*x+15)*log
(x)-60*x^2+39*x-3)*exp((-3*log((-1+log(x)+x)/(4*x+log(x)))+15+3*x)/x)/(x^2*log(x)^2+(5*x^3-x^2)*log(x)+4*x^4-4
*x^3),x, algorithm="giac")

[Out]

e^(-3*log(x/(4*x + log(x)) + log(x)/(4*x + log(x)) - 1/(4*x + log(x)))/x + 15/x + 3)

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maple [C]  time = 0.12, size = 165, normalized size = 5.69

method	result	size
risch	\({\mathrm e}^{\frac {\frac {3 i \pi \mathrm {csgn}\left (\frac {i \left (-1+\ln \relax (x )+x \right )}{x +\frac {\ln \relax (x )}{4}}\right )^{3}}{2}-\frac {3 i \pi \mathrm {csgn}\left (\frac {i \left (-1+\ln \relax (x )+x \right )}{x +\frac {\ln \relax (x )}{4}}\right )^{2} \mathrm {csgn}\left (\frac {i}{x +\frac {\ln \relax (x )}{4}}\right )}{2}-\frac {3 i \pi \mathrm {csgn}\left (\frac {i \left (-1+\ln \relax (x )+x \right )}{x +\frac {\ln \relax (x )}{4}}\right )^{2} \mathrm {csgn}\left (i \left (-1+\ln \relax (x )+x \right )\right )}{2}+\frac {3 i \pi \,\mathrm {csgn}\left (\frac {i \left (-1+\ln \relax (x )+x \right )}{x +\frac {\ln \relax (x )}{4}}\right ) \mathrm {csgn}\left (\frac {i}{x +\frac {\ln \relax (x )}{4}}\right ) \mathrm {csgn}\left (i \left (-1+\ln \relax (x )+x \right )\right )}{2}+6 \ln \relax (2)+3 \ln \left (x +\frac {\ln \relax (x )}{4}\right )-3 \ln \left (-1+\ln \relax (x )+x \right )+3 x +15}{x}}\)	\(165\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*ln(x)^2+(15*x-3)*ln(x)+12*x^2-12*x)*ln((-1+ln(x)+x)/(4*x+ln(x)))-15*ln(x)^2+(-66*x+15)*ln(x)-60*x^2+39
*x-3)*exp((-3*ln((-1+ln(x)+x)/(4*x+ln(x)))+15+3*x)/x)/(x^2*ln(x)^2+(5*x^3-x^2)*ln(x)+4*x^4-4*x^3),x,method=_RE
TURNVERBOSE)

[Out]

exp(3/2*(I*Pi*csgn(I/(x+1/4*ln(x))*(-1+ln(x)+x))^3-I*Pi*csgn(I/(x+1/4*ln(x))*(-1+ln(x)+x))^2*csgn(I/(x+1/4*ln(
x)))-I*Pi*csgn(I/(x+1/4*ln(x))*(-1+ln(x)+x))^2*csgn(I*(-1+ln(x)+x))+I*Pi*csgn(I/(x+1/4*ln(x))*(-1+ln(x)+x))*cs
gn(I/(x+1/4*ln(x)))*csgn(I*(-1+ln(x)+x))+4*ln(2)+2*ln(x+1/4*ln(x))-2*ln(-1+ln(x)+x)+2*x+10)/x)

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maxima [A]  time = 0.48, size = 31, normalized size = 1.07 \begin {gather*} e^{\left (\frac {3 \, \log \left (4 \, x + \log \relax (x)\right )}{x} - \frac {3 \, \log \left (x + \log \relax (x) - 1\right )}{x} + \frac {15}{x} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*log(x)^2+(15*x-3)*log(x)+12*x^2-12*x)*log((-1+log(x)+x)/(4*x+log(x)))-15*log(x)^2+(-66*x+15)*log
(x)-60*x^2+39*x-3)*exp((-3*log((-1+log(x)+x)/(4*x+log(x)))+15+3*x)/x)/(x^2*log(x)^2+(5*x^3-x^2)*log(x)+4*x^4-4
*x^3),x, algorithm="maxima")

[Out]

e^(3*log(4*x + log(x))/x - 3*log(x + log(x) - 1)/x + 15/x + 3)

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mupad [B]  time = 8.76, size = 31, normalized size = 1.07 \begin {gather*} \frac {{\mathrm {e}}^3\,{\mathrm {e}}^{15/x}}{{\left (\frac {x+\ln \relax (x)-1}{4\,x+\ln \relax (x)}\right )}^{3/x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((3*x - 3*log((x + log(x) - 1)/(4*x + log(x))) + 15)/x)*(15*log(x)^2 - 39*x + log(x)*(66*x - 15) + 60
*x^2 - log((x + log(x) - 1)/(4*x + log(x)))*(3*log(x)^2 - 12*x + log(x)*(15*x - 3) + 12*x^2) + 3))/(x^2*log(x)
^2 - log(x)*(x^2 - 5*x^3) - 4*x^3 + 4*x^4),x)

[Out]

(exp(3)*exp(15/x))/((x + log(x) - 1)/(4*x + log(x)))^(3/x)

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sympy [A]  time = 1.70, size = 24, normalized size = 0.83 \begin {gather*} e^{\frac {3 x - 3 \log {\left (\frac {x + \log {\relax (x )} - 1}{4 x + \log {\relax (x )}} \right )} + 15}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*ln(x)**2+(15*x-3)*ln(x)+12*x**2-12*x)*ln((-1+ln(x)+x)/(4*x+ln(x)))-15*ln(x)**2+(-66*x+15)*ln(x)-
60*x**2+39*x-3)*exp((-3*ln((-1+ln(x)+x)/(4*x+ln(x)))+15+3*x)/x)/(x**2*ln(x)**2+(5*x**3-x**2)*ln(x)+4*x**4-4*x*
*3),x)

[Out]

exp((3*x - 3*log((x + log(x) - 1)/(4*x + log(x))) + 15)/x)

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