3.103.65 \(\int \frac {e^{25/x} (-25-x)-x^2+e^{50/x} (x+8 x^2)+e^{25/x} (-2 x^2-16 x^3) \log (x \log (3))+(x^3+8 x^4) \log ^2(x \log (3))}{2 e^{50/x} x-4 e^{25/x} x^2 \log (x \log (3))+2 x^3 \log ^2(x \log (3))} \, dx\)

Optimal. Leaf size=31 \[ \frac {1}{2} \left (x+4 x^2+\frac {1}{-\frac {e^{25/x}}{x}+\log (x \log (3))}\right ) \]

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Rubi [F]  time = 1.99, 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 {e^{25/x} (-25-x)-x^2+e^{50/x} \left (x+8 x^2\right )+e^{25/x} \left (-2 x^2-16 x^3\right ) \log (x \log (3))+\left (x^3+8 x^4\right ) \log ^2(x \log (3))}{2 e^{50/x} x-4 e^{25/x} x^2 \log (x \log (3))+2 x^3 \log ^2(x \log (3))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(25/x)*(-25 - x) - x^2 + E^(50/x)*(x + 8*x^2) + E^(25/x)*(-2*x^2 - 16*x^3)*Log[x*Log[3]] + (x^3 + 8*x^4
)*Log[x*Log[3]]^2)/(2*E^(50/x)*x - 4*E^(25/x)*x^2*Log[x*Log[3]] + 2*x^3*Log[x*Log[3]]^2),x]

[Out]

x/2 + 2*x^2 - Defer[Int][(E^(25/x) - x*Log[x*Log[3]])^(-1), x]/2 - Defer[Int][x/(-E^(25/x) + x*Log[x*Log[3]])^
2, x]/2 - (25*Defer[Int][Log[x*Log[3]]/(-E^(25/x) + x*Log[x*Log[3]])^2, x])/2 - Defer[Int][(x*Log[x*Log[3]])/(
-E^(25/x) + x*Log[x*Log[3]])^2, x]/2 + (25*Defer[Int][1/(x*(-E^(25/x) + x*Log[x*Log[3]])), x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{25/x} (-25-x)-x^2+e^{50/x} \left (x+8 x^2\right )+e^{25/x} \left (-2 x^2-16 x^3\right ) \log (x \log (3))+\left (x^3+8 x^4\right ) \log ^2(x \log (3))}{2 x \left (e^{25/x}-x \log (x \log (3))\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{25/x} (-25-x)-x^2+e^{50/x} \left (x+8 x^2\right )+e^{25/x} \left (-2 x^2-16 x^3\right ) \log (x \log (3))+\left (x^3+8 x^4\right ) \log ^2(x \log (3))}{x \left (e^{25/x}-x \log (x \log (3))\right )^2} \, dx\\ &=\frac {1}{2} \int \left (1+8 x+\frac {25+x}{x \left (-e^{25/x}+x \log (x \log (3))\right )}-\frac {x+25 \log (x \log (3))+x \log (x \log (3))}{\left (e^{25/x}-x \log (x \log (3))\right )^2}\right ) \, dx\\ &=\frac {x}{2}+2 x^2+\frac {1}{2} \int \frac {25+x}{x \left (-e^{25/x}+x \log (x \log (3))\right )} \, dx-\frac {1}{2} \int \frac {x+25 \log (x \log (3))+x \log (x \log (3))}{\left (e^{25/x}-x \log (x \log (3))\right )^2} \, dx\\ &=\frac {x}{2}+2 x^2-\frac {1}{2} \int \frac {x+(25+x) \log (x \log (3))}{\left (e^{25/x}-x \log (x \log (3))\right )^2} \, dx+\frac {1}{2} \int \left (-\frac {1}{e^{25/x}-x \log (x \log (3))}+\frac {25}{x \left (-e^{25/x}+x \log (x \log (3))\right )}\right ) \, dx\\ &=\frac {x}{2}+2 x^2-\frac {1}{2} \int \frac {1}{e^{25/x}-x \log (x \log (3))} \, dx-\frac {1}{2} \int \left (\frac {x}{\left (-e^{25/x}+x \log (x \log (3))\right )^2}+\frac {25 \log (x \log (3))}{\left (-e^{25/x}+x \log (x \log (3))\right )^2}+\frac {x \log (x \log (3))}{\left (-e^{25/x}+x \log (x \log (3))\right )^2}\right ) \, dx+\frac {25}{2} \int \frac {1}{x \left (-e^{25/x}+x \log (x \log (3))\right )} \, dx\\ &=\frac {x}{2}+2 x^2-\frac {1}{2} \int \frac {1}{e^{25/x}-x \log (x \log (3))} \, dx-\frac {1}{2} \int \frac {x}{\left (-e^{25/x}+x \log (x \log (3))\right )^2} \, dx-\frac {1}{2} \int \frac {x \log (x \log (3))}{\left (-e^{25/x}+x \log (x \log (3))\right )^2} \, dx-\frac {25}{2} \int \frac {\log (x \log (3))}{\left (-e^{25/x}+x \log (x \log (3))\right )^2} \, dx+\frac {25}{2} \int \frac {1}{x \left (-e^{25/x}+x \log (x \log (3))\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 28, normalized size = 0.90 \begin {gather*} -\frac {1}{2} x \left (-1-4 x+\frac {1}{e^{25/x}-x \log (x \log (3))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(25/x)*(-25 - x) - x^2 + E^(50/x)*(x + 8*x^2) + E^(25/x)*(-2*x^2 - 16*x^3)*Log[x*Log[3]] + (x^3 +
8*x^4)*Log[x*Log[3]]^2)/(2*E^(50/x)*x - 4*E^(25/x)*x^2*Log[x*Log[3]] + 2*x^3*Log[x*Log[3]]^2),x]

[Out]

-1/2*(x*(-1 - 4*x + (E^(25/x) - x*Log[x*Log[3]])^(-1)))

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fricas [A]  time = 0.54, size = 54, normalized size = 1.74 \begin {gather*} -\frac {{\left (4 \, x^{2} + x\right )} e^{\frac {25}{x}} - {\left (4 \, x^{3} + x^{2}\right )} \log \left (x \log \relax (3)\right ) - x}{2 \, {\left (x \log \left (x \log \relax (3)\right ) - e^{\frac {25}{x}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^4+x^3)*log(x*log(3))^2+(-16*x^3-2*x^2)*exp(25/x)*log(x*log(3))+(8*x^2+x)*exp(25/x)^2+(-x-25)*e
xp(25/x)-x^2)/(2*x^3*log(x*log(3))^2-4*x^2*exp(25/x)*log(x*log(3))+2*x*exp(25/x)^2),x, algorithm="fricas")

[Out]

-1/2*((4*x^2 + x)*e^(25/x) - (4*x^3 + x^2)*log(x*log(3)) - x)/(x*log(x*log(3)) - e^(25/x))

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giac [B]  time = 0.23, size = 72, normalized size = 2.32 \begin {gather*} \frac {4 \, x^{3} \log \relax (x) + 4 \, x^{3} \log \left (\log \relax (3)\right ) - 4 \, x^{2} e^{\frac {25}{x}} + x^{2} \log \relax (x) + x^{2} \log \left (\log \relax (3)\right ) - x e^{\frac {25}{x}} + x}{2 \, {\left (x \log \relax (x) + x \log \left (\log \relax (3)\right ) - e^{\frac {25}{x}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^4+x^3)*log(x*log(3))^2+(-16*x^3-2*x^2)*exp(25/x)*log(x*log(3))+(8*x^2+x)*exp(25/x)^2+(-x-25)*e
xp(25/x)-x^2)/(2*x^3*log(x*log(3))^2-4*x^2*exp(25/x)*log(x*log(3))+2*x*exp(25/x)^2),x, algorithm="giac")

[Out]

1/2*(4*x^3*log(x) + 4*x^3*log(log(3)) - 4*x^2*e^(25/x) + x^2*log(x) + x^2*log(log(3)) - x*e^(25/x) + x)/(x*log
(x) + x*log(log(3)) - e^(25/x))

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maple [A]  time = 0.04, size = 31, normalized size = 1.00

method	result	size
risch	\(2 x^{2}+\frac {x}{2}+\frac {x}{2 \ln \left (x \ln \relax (3)\right ) x -2 \,{\mathrm e}^{\frac {25}{x}}}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^4+x^3)*ln(x*ln(3))^2+(-16*x^3-2*x^2)*exp(25/x)*ln(x*ln(3))+(8*x^2+x)*exp(25/x)^2+(-x-25)*exp(25/x)-x
^2)/(2*x^3*ln(x*ln(3))^2-4*x^2*exp(25/x)*ln(x*ln(3))+2*x*exp(25/x)^2),x,method=_RETURNVERBOSE)

[Out]

2*x^2+1/2*x+1/2*x/(ln(x*ln(3))*x-exp(25/x))

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maxima [B]  time = 0.49, size = 66, normalized size = 2.13 \begin {gather*} \frac {4 \, x^{3} \log \left (\log \relax (3)\right ) + x^{2} \log \left (\log \relax (3)\right ) - {\left (4 \, x^{2} + x\right )} e^{\frac {25}{x}} + {\left (4 \, x^{3} + x^{2}\right )} \log \relax (x) + x}{2 \, {\left (x \log \relax (x) + x \log \left (\log \relax (3)\right ) - e^{\frac {25}{x}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^4+x^3)*log(x*log(3))^2+(-16*x^3-2*x^2)*exp(25/x)*log(x*log(3))+(8*x^2+x)*exp(25/x)^2+(-x-25)*e
xp(25/x)-x^2)/(2*x^3*log(x*log(3))^2-4*x^2*exp(25/x)*log(x*log(3))+2*x*exp(25/x)^2),x, algorithm="maxima")

[Out]

1/2*(4*x^3*log(log(3)) + x^2*log(log(3)) - (4*x^2 + x)*e^(25/x) + (4*x^3 + x^2)*log(x) + x)/(x*log(x) + x*log(
log(3)) - e^(25/x))

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mupad [B]  time = 6.88, size = 58, normalized size = 1.87 \begin {gather*} -\frac {x\,\left (x\,\ln \left (x\,\ln \relax (3)\right )-4\,x\,{\mathrm {e}}^{25/x}-{\mathrm {e}}^{25/x}+4\,x^2\,\ln \left (x\,\ln \relax (3)\right )+1\right )}{2\,\left ({\mathrm {e}}^{25/x}-x\,\ln \left (x\,\ln \relax (3)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(25/x)*(x + 25) - exp(50/x)*(x + 8*x^2) + x^2 - log(x*log(3))^2*(x^3 + 8*x^4) + exp(25/x)*log(x*log(3
))*(2*x^2 + 16*x^3))/(2*x^3*log(x*log(3))^2 + 2*x*exp(50/x) - 4*x^2*exp(25/x)*log(x*log(3))),x)

[Out]

-(x*(x*log(x*log(3)) - 4*x*exp(25/x) - exp(25/x) + 4*x^2*log(x*log(3)) + 1))/(2*(exp(25/x) - x*log(x*log(3))))

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sympy [A]  time = 0.42, size = 26, normalized size = 0.84 \begin {gather*} 2 x^{2} + \frac {x}{2} - \frac {x}{- 2 x \log {\left (x \log {\relax (3 )} \right )} + 2 e^{\frac {25}{x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**4+x**3)*ln(x*ln(3))**2+(-16*x**3-2*x**2)*exp(25/x)*ln(x*ln(3))+(8*x**2+x)*exp(25/x)**2+(-x-25
)*exp(25/x)-x**2)/(2*x**3*ln(x*ln(3))**2-4*x**2*exp(25/x)*ln(x*ln(3))+2*x*exp(25/x)**2),x)

[Out]

2*x**2 + x/2 - x/(-2*x*log(x*log(3)) + 2*exp(25/x))

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