3.94.63 \(\int \frac {25 x^2+e^x (-100 e^{4/x}+96 x^2)+(10 x^2+e^x (-40 e^{4/x}+39 x^2)) \log (x)+(x^2+e^x (-4 e^{4/x}+4 x^2)) \log ^2(x)}{-25 x^2+e^x (25 e^{4/x} x^2+95 x^3)+(-10 x^2+e^x (10 e^{4/x} x^2+39 x^3)) \log (x)+(-x^2+e^x (e^{4/x} x^2+4 x^3)) \log ^2(x)} \, dx\)

Optimal. Leaf size=28 \[ \log \left (e^{4/x}-e^{-x}+4 x-\frac {x}{5+\log (x)}\right ) \]

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Rubi [F]  time = 26.04, 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 {25 x^2+e^x \left (-100 e^{4/x}+96 x^2\right )+\left (10 x^2+e^x \left (-40 e^{4/x}+39 x^2\right )\right ) \log (x)+\left (x^2+e^x \left (-4 e^{4/x}+4 x^2\right )\right ) \log ^2(x)}{-25 x^2+e^x \left (25 e^{4/x} x^2+95 x^3\right )+\left (-10 x^2+e^x \left (10 e^{4/x} x^2+39 x^3\right )\right ) \log (x)+\left (-x^2+e^x \left (e^{4/x} x^2+4 x^3\right )\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(25*x^2 + E^x*(-100*E^(4/x) + 96*x^2) + (10*x^2 + E^x*(-40*E^(4/x) + 39*x^2))*Log[x] + (x^2 + E^x*(-4*E^(4
/x) + 4*x^2))*Log[x]^2)/(-25*x^2 + E^x*(25*E^(4/x)*x^2 + 95*x^3) + (-10*x^2 + E^x*(10*E^(4/x)*x^2 + 39*x^3))*L
og[x] + (-x^2 + E^x*(E^(4/x)*x^2 + 4*x^3))*Log[x]^2),x]

[Out]

4/x + 96*Defer[Int][1/((5 + Log[x])*(5*E^(4/x) + 19*x + E^(4/x)*Log[x] + 4*x*Log[x])), x] + 380*Defer[Int][1/(
x*(5 + Log[x])*(5*E^(4/x) + 19*x + E^(4/x)*Log[x] + 4*x*Log[x])), x] + 39*Defer[Int][Log[x]/((5 + Log[x])*(5*E
^(4/x) + 19*x + E^(4/x)*Log[x] + 4*x*Log[x])), x] + 156*Defer[Int][Log[x]/(x*(5 + Log[x])*(5*E^(4/x) + 19*x +
E^(4/x)*Log[x] + 4*x*Log[x])), x] + 4*Defer[Int][Log[x]^2/((5 + Log[x])*(5*E^(4/x) + 19*x + E^(4/x)*Log[x] + 4
*x*Log[x])), x] + 16*Defer[Int][Log[x]^2/(x*(5 + Log[x])*(5*E^(4/x) + 19*x + E^(4/x)*Log[x] + 4*x*Log[x])), x]
 + 96*Defer[Int][1/((5*E^(4/x) + 19*x + E^(4/x)*Log[x] + 4*x*Log[x])*(-5 + 5*E^(4/x + x) + 19*E^x*x - Log[x] +
 E^(4/x + x)*Log[x] + 4*E^x*x*Log[x])), x] + 25*Defer[Int][E^(4/x)/((5*E^(4/x) + 19*x + E^(4/x)*Log[x] + 4*x*L
og[x])*(-5 + 5*E^(4/x + x) + 19*E^x*x - Log[x] + E^(4/x + x)*Log[x] + 4*E^x*x*Log[x])), x] - 100*Defer[Int][E^
(4/x)/(x^2*(5*E^(4/x) + 19*x + E^(4/x)*Log[x] + 4*x*Log[x])*(-5 + 5*E^(4/x + x) + 19*E^x*x - Log[x] + E^(4/x +
 x)*Log[x] + 4*E^x*x*Log[x])), x] + 95*Defer[Int][x/((5*E^(4/x) + 19*x + E^(4/x)*Log[x] + 4*x*Log[x])*(-5 + 5*
E^(4/x + x) + 19*E^x*x - Log[x] + E^(4/x + x)*Log[x] + 4*E^x*x*Log[x])), x] + 39*Defer[Int][Log[x]/((5*E^(4/x)
 + 19*x + E^(4/x)*Log[x] + 4*x*Log[x])*(-5 + 5*E^(4/x + x) + 19*E^x*x - Log[x] + E^(4/x + x)*Log[x] + 4*E^x*x*
Log[x])), x] + 10*Defer[Int][(E^(4/x)*Log[x])/((5*E^(4/x) + 19*x + E^(4/x)*Log[x] + 4*x*Log[x])*(-5 + 5*E^(4/x
 + x) + 19*E^x*x - Log[x] + E^(4/x + x)*Log[x] + 4*E^x*x*Log[x])), x] - 40*Defer[Int][(E^(4/x)*Log[x])/(x^2*(5
*E^(4/x) + 19*x + E^(4/x)*Log[x] + 4*x*Log[x])*(-5 + 5*E^(4/x + x) + 19*E^x*x - Log[x] + E^(4/x + x)*Log[x] +
4*E^x*x*Log[x])), x] + 39*Defer[Int][(x*Log[x])/((5*E^(4/x) + 19*x + E^(4/x)*Log[x] + 4*x*Log[x])*(-5 + 5*E^(4
/x + x) + 19*E^x*x - Log[x] + E^(4/x + x)*Log[x] + 4*E^x*x*Log[x])), x] + 4*Defer[Int][Log[x]^2/((5*E^(4/x) +
19*x + E^(4/x)*Log[x] + 4*x*Log[x])*(-5 + 5*E^(4/x + x) + 19*E^x*x - Log[x] + E^(4/x + x)*Log[x] + 4*E^x*x*Log
[x])), x] + Defer[Int][(E^(4/x)*Log[x]^2)/((5*E^(4/x) + 19*x + E^(4/x)*Log[x] + 4*x*Log[x])*(-5 + 5*E^(4/x + x
) + 19*E^x*x - Log[x] + E^(4/x + x)*Log[x] + 4*E^x*x*Log[x])), x] - 4*Defer[Int][(E^(4/x)*Log[x]^2)/(x^2*(5*E^
(4/x) + 19*x + E^(4/x)*Log[x] + 4*x*Log[x])*(-5 + 5*E^(4/x + x) + 19*E^x*x - Log[x] + E^(4/x + x)*Log[x] + 4*E
^x*x*Log[x])), x] + 4*Defer[Int][(x*Log[x]^2)/((5*E^(4/x) + 19*x + E^(4/x)*Log[x] + 4*x*Log[x])*(-5 + 5*E^(4/x
 + x) + 19*E^x*x - Log[x] + E^(4/x + x)*Log[x] + 4*E^x*x*Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-25 x^2-e^x \left (-100 e^{4/x}+96 x^2\right )-\left (10 x^2+e^x \left (-40 e^{4/x}+39 x^2\right )\right ) \log (x)-\left (x^2+e^x \left (-4 e^{4/x}+4 x^2\right )\right ) \log ^2(x)}{x^2 (5+\log (x)) \left (5-5 e^{\frac {4}{x}+x}-19 e^x x+\log (x)-e^{\frac {4}{x}+x} \log (x)-4 e^x x \log (x)\right )} \, dx\\ &=\int \left (\frac {-100 e^{4/x}+96 x^2-40 e^{4/x} \log (x)+39 x^2 \log (x)-4 e^{4/x} \log ^2(x)+4 x^2 \log ^2(x)}{x^2 (5+\log (x)) \left (5 e^{4/x}+19 x+e^{4/x} \log (x)+4 x \log (x)\right )}+\frac {-100 e^{4/x}+96 x^2+25 e^{4/x} x^2+95 x^3-40 e^{4/x} \log (x)+39 x^2 \log (x)+10 e^{4/x} x^2 \log (x)+39 x^3 \log (x)-4 e^{4/x} \log ^2(x)+4 x^2 \log ^2(x)+e^{4/x} x^2 \log ^2(x)+4 x^3 \log ^2(x)}{x^2 \left (5 e^{4/x}+19 x+e^{4/x} \log (x)+4 x \log (x)\right ) \left (-5+5 e^{\frac {4}{x}+x}+19 e^x x-\log (x)+e^{\frac {4}{x}+x} \log (x)+4 e^x x \log (x)\right )}\right ) \, dx\\ &=\int \frac {-100 e^{4/x}+96 x^2-40 e^{4/x} \log (x)+39 x^2 \log (x)-4 e^{4/x} \log ^2(x)+4 x^2 \log ^2(x)}{x^2 (5+\log (x)) \left (5 e^{4/x}+19 x+e^{4/x} \log (x)+4 x \log (x)\right )} \, dx+\int \frac {-100 e^{4/x}+96 x^2+25 e^{4/x} x^2+95 x^3-40 e^{4/x} \log (x)+39 x^2 \log (x)+10 e^{4/x} x^2 \log (x)+39 x^3 \log (x)-4 e^{4/x} \log ^2(x)+4 x^2 \log ^2(x)+e^{4/x} x^2 \log ^2(x)+4 x^3 \log ^2(x)}{x^2 \left (5 e^{4/x}+19 x+e^{4/x} \log (x)+4 x \log (x)\right ) \left (-5+5 e^{\frac {4}{x}+x}+19 e^x x-\log (x)+e^{\frac {4}{x}+x} \log (x)+4 e^x x \log (x)\right )} \, 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 = 54, normalized size = 1.93 \begin {gather*} -x-\log (5+\log (x))+\log \left (5-5 e^{\frac {4}{x}+x}-19 e^x x+\log (x)-e^{\frac {4}{x}+x} \log (x)-4 e^x x \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*x^2 + E^x*(-100*E^(4/x) + 96*x^2) + (10*x^2 + E^x*(-40*E^(4/x) + 39*x^2))*Log[x] + (x^2 + E^x*(-
4*E^(4/x) + 4*x^2))*Log[x]^2)/(-25*x^2 + E^x*(25*E^(4/x)*x^2 + 95*x^3) + (-10*x^2 + E^x*(10*E^(4/x)*x^2 + 39*x
^3))*Log[x] + (-x^2 + E^x*(E^(4/x)*x^2 + 4*x^3))*Log[x]^2),x]

[Out]

-x - Log[5 + Log[x]] + Log[5 - 5*E^(4/x + x) - 19*E^x*x + Log[x] - E^(4/x + x)*Log[x] - 4*E^x*x*Log[x]]

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fricas [B]  time = 0.94, size = 105, normalized size = 3.75 \begin {gather*} -x + \log \left (4 \, x + e^{\frac {4}{x}}\right ) + \log \left (\frac {{\left (19 \, x + 5 \, e^{\frac {4}{x}}\right )} e^{x} + {\left ({\left (4 \, x + e^{\frac {4}{x}}\right )} e^{x} - 1\right )} \log \relax (x) - 5}{{\left (4 \, x + e^{\frac {4}{x}}\right )} e^{x} - 1}\right ) + \log \left (\frac {{\left (4 \, x + e^{\frac {4}{x}}\right )} e^{x} - 1}{4 \, x + e^{\frac {4}{x}}}\right ) - \log \left (\log \relax (x) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*exp(4/x)+4*x^2)*exp(x)+x^2)*log(x)^2+((-40*exp(4/x)+39*x^2)*exp(x)+10*x^2)*log(x)+(-100*exp(4/
x)+96*x^2)*exp(x)+25*x^2)/(((x^2*exp(4/x)+4*x^3)*exp(x)-x^2)*log(x)^2+((10*x^2*exp(4/x)+39*x^3)*exp(x)-10*x^2)
*log(x)+(25*x^2*exp(4/x)+95*x^3)*exp(x)-25*x^2),x, algorithm="fricas")

[Out]

-x + log(4*x + e^(4/x)) + log(((19*x + 5*e^(4/x))*e^x + ((4*x + e^(4/x))*e^x - 1)*log(x) - 5)/((4*x + e^(4/x))
*e^x - 1)) + log(((4*x + e^(4/x))*e^x - 1)/(4*x + e^(4/x))) - log(log(x) + 5)

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giac [B]  time = 0.34, size = 55, normalized size = 1.96 \begin {gather*} -x + \log \left (4 \, x e^{x} \log \relax (x) + 19 \, x e^{x} + e^{\left (\frac {x^{2} + 4}{x}\right )} \log \relax (x) + 5 \, e^{\left (\frac {x^{2} + 4}{x}\right )} - \log \relax (x) - 5\right ) - \log \left (\log \relax (x) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*exp(4/x)+4*x^2)*exp(x)+x^2)*log(x)^2+((-40*exp(4/x)+39*x^2)*exp(x)+10*x^2)*log(x)+(-100*exp(4/
x)+96*x^2)*exp(x)+25*x^2)/(((x^2*exp(4/x)+4*x^3)*exp(x)-x^2)*log(x)^2+((10*x^2*exp(4/x)+39*x^3)*exp(x)-10*x^2)
*log(x)+(25*x^2*exp(4/x)+95*x^3)*exp(x)-25*x^2),x, algorithm="giac")

[Out]

-x + log(4*x*e^x*log(x) + 19*x*e^x + e^((x^2 + 4)/x)*log(x) + 5*e^((x^2 + 4)/x) - log(x) - 5) - log(log(x) + 5
)

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maple [B]  time = 0.06, size = 72, normalized size = 2.57




method result size



risch \(\ln \left ({\mathrm e}^{\frac {4}{x}}+\left (4 \,{\mathrm e}^{x} x -1\right ) {\mathrm e}^{-x}\right )+\ln \left (\ln \relax (x )+\frac {19 \,{\mathrm e}^{x} x +5 \,{\mathrm e}^{\frac {x^{2}+4}{x}}-5}{4 \,{\mathrm e}^{x} x +{\mathrm e}^{\frac {x^{2}+4}{x}}-1}\right )-\ln \left (5+\ln \relax (x )\right )\) \(72\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*exp(4/x)+4*x^2)*exp(x)+x^2)*ln(x)^2+((-40*exp(4/x)+39*x^2)*exp(x)+10*x^2)*ln(x)+(-100*exp(4/x)+96*x^
2)*exp(x)+25*x^2)/(((x^2*exp(4/x)+4*x^3)*exp(x)-x^2)*ln(x)^2+((10*x^2*exp(4/x)+39*x^3)*exp(x)-10*x^2)*ln(x)+(2
5*x^2*exp(4/x)+95*x^3)*exp(x)-25*x^2),x,method=_RETURNVERBOSE)

[Out]

ln(exp(4/x)+(4*exp(x)*x-1)*exp(-x))+ln(ln(x)+(19*exp(x)*x+5*exp(1/x*(x^2+4))-5)/(4*exp(x)*x+exp(1/x*(x^2+4))-1
))-ln(5+ln(x))

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maxima [A]  time = 0.48, size = 43, normalized size = 1.54 \begin {gather*} \log \left (\frac {{\left ({\left (\log \relax (x) + 5\right )} e^{\left (x + \frac {4}{x}\right )} + {\left (4 \, x \log \relax (x) + 19 \, x\right )} e^{x} - \log \relax (x) - 5\right )} e^{\left (-x\right )}}{\log \relax (x) + 5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*exp(4/x)+4*x^2)*exp(x)+x^2)*log(x)^2+((-40*exp(4/x)+39*x^2)*exp(x)+10*x^2)*log(x)+(-100*exp(4/
x)+96*x^2)*exp(x)+25*x^2)/(((x^2*exp(4/x)+4*x^3)*exp(x)-x^2)*log(x)^2+((10*x^2*exp(4/x)+39*x^3)*exp(x)-10*x^2)
*log(x)+(25*x^2*exp(4/x)+95*x^3)*exp(x)-25*x^2),x, algorithm="maxima")

[Out]

log(((log(x) + 5)*e^(x + 4/x) + (4*x*log(x) + 19*x)*e^x - log(x) - 5)*e^(-x)/(log(x) + 5))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int \frac {{\ln \relax (x)}^2\,\left ({\mathrm {e}}^x\,\left (4\,{\mathrm {e}}^{4/x}-4\,x^2\right )-x^2\right )+\ln \relax (x)\,\left ({\mathrm {e}}^x\,\left (40\,{\mathrm {e}}^{4/x}-39\,x^2\right )-10\,x^2\right )+{\mathrm {e}}^x\,\left (100\,{\mathrm {e}}^{4/x}-96\,x^2\right )-25\,x^2}{\ln \relax (x)\,\left ({\mathrm {e}}^x\,\left (10\,x^2\,{\mathrm {e}}^{4/x}+39\,x^3\right )-10\,x^2\right )+{\mathrm {e}}^x\,\left (25\,x^2\,{\mathrm {e}}^{4/x}+95\,x^3\right )-25\,x^2+{\ln \relax (x)}^2\,\left ({\mathrm {e}}^x\,\left (x^2\,{\mathrm {e}}^{4/x}+4\,x^3\right )-x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)^2*(exp(x)*(4*exp(4/x) - 4*x^2) - x^2) + log(x)*(exp(x)*(40*exp(4/x) - 39*x^2) - 10*x^2) + exp(x)*
(100*exp(4/x) - 96*x^2) - 25*x^2)/(log(x)*(exp(x)*(10*x^2*exp(4/x) + 39*x^3) - 10*x^2) + exp(x)*(25*x^2*exp(4/
x) + 95*x^3) - 25*x^2 + log(x)^2*(exp(x)*(x^2*exp(4/x) + 4*x^3) - x^2)),x)

[Out]

-int((log(x)^2*(exp(x)*(4*exp(4/x) - 4*x^2) - x^2) + log(x)*(exp(x)*(40*exp(4/x) - 39*x^2) - 10*x^2) + exp(x)*
(100*exp(4/x) - 96*x^2) - 25*x^2)/(log(x)*(exp(x)*(10*x^2*exp(4/x) + 39*x^3) - 10*x^2) + exp(x)*(25*x^2*exp(4/
x) + 95*x^3) - 25*x^2 + log(x)^2*(exp(x)*(x^2*exp(4/x) + 4*x^3) - x^2)), x)

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sympy [B]  time = 6.37, size = 60, normalized size = 2.14 \begin {gather*} - x + \log {\left (\frac {4 x \log {\relax (x )} + 19 x}{\log {\relax (x )} + 5} + e^{\frac {4}{x}} \right )} + \log {\left (\frac {- \log {\relax (x )} - 5}{4 x \log {\relax (x )} + 19 x + e^{\frac {4}{x}} \log {\relax (x )} + 5 e^{\frac {4}{x}}} + e^{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*exp(4/x)+4*x**2)*exp(x)+x**2)*ln(x)**2+((-40*exp(4/x)+39*x**2)*exp(x)+10*x**2)*ln(x)+(-100*exp
(4/x)+96*x**2)*exp(x)+25*x**2)/(((x**2*exp(4/x)+4*x**3)*exp(x)-x**2)*ln(x)**2+((10*x**2*exp(4/x)+39*x**3)*exp(
x)-10*x**2)*ln(x)+(25*x**2*exp(4/x)+95*x**3)*exp(x)-25*x**2),x)

[Out]

-x + log((4*x*log(x) + 19*x)/(log(x) + 5) + exp(4/x)) + log((-log(x) - 5)/(4*x*log(x) + 19*x + exp(4/x)*log(x)
 + 5*exp(4/x)) + exp(x))

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