3.14.74 \(\int \frac {x^2+e^{5+x} (4+x^2)+(-4-4 e^{5+x}-x^3) \log (x)+(-e^{5+x} x^2+(x^2+e^{5+x} x^2) \log (x)) \log (e^{-5-x} (-e^{5+x}+(1+e^{5+x}) \log (x)))}{-e^{5+x} x^2+(x^2+e^{5+x} x^2) \log (x)} \, dx\)

Optimal. Leaf size=24 \[ -1+\frac {4}{x}+x \log \left (-1+\log (x)+e^{-5-x} \log (x)\right ) \]

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Rubi [F]  time = 2.16, 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 {x^2+e^{5+x} \left (4+x^2\right )+\left (-4-4 e^{5+x}-x^3\right ) \log (x)+\left (-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)\right ) \log \left (e^{-5-x} \left (-e^{5+x}+\left (1+e^{5+x}\right ) \log (x)\right )\right )}{-e^{5+x} x^2+\left (x^2+e^{5+x} x^2\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 23, normalized size = 0.96 \begin {gather*} \frac {4}{x}+x \log \left (-1+\left (1+e^{-5-x}\right ) \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + E^(5 + x)*(4 + x^2) + (-4 - 4*E^(5 + x) - x^3)*Log[x] + (-(E^(5 + x)*x^2) + (x^2 + E^(5 + x)*
x^2)*Log[x])*Log[E^(-5 - x)*(-E^(5 + x) + (1 + E^(5 + x))*Log[x])])/(-(E^(5 + x)*x^2) + (x^2 + E^(5 + x)*x^2)*
Log[x]),x]

[Out]

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

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fricas [A]  time = 0.75, size = 34, normalized size = 1.42 \begin {gather*} \frac {x^{2} \log \left ({\left ({\left (e^{\left (x + 5\right )} + 1\right )} \log \relax (x) - e^{\left (x + 5\right )}\right )} e^{\left (-x - 5\right )}\right ) + 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.36, size = 37, normalized size = 1.54 \begin {gather*} -\frac {x^{3} - x^{2} \log \left (e^{\left (x + 5\right )} \log \relax (x) - e^{\left (x + 5\right )} + \log \relax (x)\right ) + 5 \, x^{2} - 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.19, size = 200, normalized size = 8.33

method	result	size
risch	\(-x \ln \left ({\mathrm e}^{x}\right )+\frac {-i \pi \,x^{2} \mathrm {csgn}\left (i \left (\left (\ln \relax (x )-1\right ) {\mathrm e}^{5+x}+\ln \relax (x )\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} \left (\left (\ln \relax (x )-1\right ) {\mathrm e}^{5+x}+\ln \relax (x )\right )\right )+i \pi \,x^{2} \mathrm {csgn}\left (i \left (\left (\ln \relax (x )-1\right ) {\mathrm e}^{5+x}+\ln \relax (x )\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} \left (\left (\ln \relax (x )-1\right ) {\mathrm e}^{5+x}+\ln \relax (x )\right )\right )^{2}+8+i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} \left (\left (\ln \relax (x )-1\right ) {\mathrm e}^{5+x}+\ln \relax (x )\right )\right )^{2}-i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x} \left (\left (\ln \relax (x )-1\right ) {\mathrm e}^{5+x}+\ln \relax (x )\right )\right )^{3}+2 x^{2} \ln \left (\left (\ln \relax (x )-1\right ) {\mathrm e}^{5+x}+\ln \relax (x )\right )-10 x^{2}}{2 x}\)	\(200\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*ln(exp(x))+1/2*(-I*Pi*x^2*csgn(I*((ln(x)-1)*exp(5+x)+ln(x)))*csgn(I*exp(-x))*csgn(I*exp(-x)*((ln(x)-1)*exp(
5+x)+ln(x)))+I*Pi*x^2*csgn(I*((ln(x)-1)*exp(5+x)+ln(x)))*csgn(I*exp(-x)*((ln(x)-1)*exp(5+x)+ln(x)))^2+8+I*Pi*x
^2*csgn(I*exp(-x))*csgn(I*exp(-x)*((ln(x)-1)*exp(5+x)+ln(x)))^2-I*Pi*x^2*csgn(I*exp(-x)*((ln(x)-1)*exp(5+x)+ln
(x)))^3+2*x^2*ln((ln(x)-1)*exp(5+x)+ln(x))-10*x^2)/x

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maxima [A]  time = 0.56, size = 37, normalized size = 1.54 \begin {gather*} -\frac {x^{3} - x^{2} \log \left ({\left (e^{\left (x + 5\right )} + 1\right )} \log \relax (x) - e^{\left (x + 5\right )}\right ) + 5 \, x^{2} - 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.25, size = 22, normalized size = 0.92 \begin {gather*} x\,\ln \left (\ln \relax (x)+{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-5}\,\ln \relax (x)-1\right )+\frac {4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - log(x)*(4*exp(5)*exp(x) + x^3 + 4) + log(exp(-x)*exp(-5)*(log(x)*(exp(5)*exp(x) + 1) - exp(5)*exp(x
)))*(log(x)*(x^2 + x^2*exp(5)*exp(x)) - x^2*exp(5)*exp(x)) + exp(5)*exp(x)*(x^2 + 4))/(log(x)*(x^2 + x^2*exp(5
)*exp(x)) - x^2*exp(5)*exp(x)),x)

[Out]

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

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sympy [A]  time = 0.97, size = 31, normalized size = 1.29 \begin {gather*} x \log {\left (\frac {\left (\left (e^{5} e^{x} + 1\right ) \log {\relax (x )} - e^{5} e^{x}\right ) e^{- x}}{e^{5}} \right )} + \frac {4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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