3.85.98 \(\int \frac {75 x^3-30 x^4+3 x^5+e^{3/x} (15+2 x+24 x^2-10 x^3+x^4)+(-e^{3/x} x^2-3 x^3) \log (\frac {e^{3/x}+3 x}{x})}{75 x^5-30 x^6+3 x^7+e^{3/x} (25 x^4-10 x^5+x^6)+(-30 x^4+6 x^5+e^{3/x} (-10 x^3+2 x^4)) \log (\frac {e^{3/x}+3 x}{x})+(e^{3/x} x^2+3 x^3) \log ^2(\frac {e^{3/x}+3 x}{x})} \, dx\)

Optimal. Leaf size=30 \[ 4+\frac {1}{-x+\frac {\log \left (3+\frac {e^{3/x}}{x}\right )}{5-x}} \]

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Rubi [F]  time = 3.93, 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 {75 x^3-30 x^4+3 x^5+e^{3/x} \left (15+2 x+24 x^2-10 x^3+x^4\right )+\left (-e^{3/x} x^2-3 x^3\right ) \log \left (\frac {e^{3/x}+3 x}{x}\right )}{75 x^5-30 x^6+3 x^7+e^{3/x} \left (25 x^4-10 x^5+x^6\right )+\left (-30 x^4+6 x^5+e^{3/x} \left (-10 x^3+2 x^4\right )\right ) \log \left (\frac {e^{3/x}+3 x}{x}\right )+\left (e^{3/x} x^2+3 x^3\right ) \log ^2\left (\frac {e^{3/x}+3 x}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(75*x^3 - 30*x^4 + 3*x^5 + E^(3/x)*(15 + 2*x + 24*x^2 - 10*x^3 + x^4) + (-(E^(3/x)*x^2) - 3*x^3)*Log[(E^(3
/x) + 3*x)/x])/(75*x^5 - 30*x^6 + 3*x^7 + E^(3/x)*(25*x^4 - 10*x^5 + x^6) + (-30*x^4 + 6*x^5 + E^(3/x)*(-10*x^
3 + 2*x^4))*Log[(E^(3/x) + 3*x)/x] + (E^(3/x)*x^2 + 3*x^3)*Log[(E^(3/x) + 3*x)/x]^2),x]

[Out]

24*Defer[Int][(-5*x + x^2 + Log[3 + E^(3/x)/x])^(-2), x] + 15*Defer[Int][1/(x^2*(-5*x + x^2 + Log[3 + E^(3/x)/
x])^2), x] + 2*Defer[Int][1/(x*(-5*x + x^2 + Log[3 + E^(3/x)/x])^2), x] - 15*Defer[Int][x/(-5*x + x^2 + Log[3
+ E^(3/x)/x])^2, x] + 2*Defer[Int][x^2/(-5*x + x^2 + Log[3 + E^(3/x)/x])^2, x] - 6*Defer[Int][1/((E^(3/x) + 3*
x)*(-5*x + x^2 + Log[3 + E^(3/x)/x])^2), x] - 45*Defer[Int][1/(x*(E^(3/x) + 3*x)*(-5*x + x^2 + Log[3 + E^(3/x)
/x])^2), x] + 3*Defer[Int][x/((E^(3/x) + 3*x)*(-5*x + x^2 + Log[3 + E^(3/x)/x])^2), x] - Defer[Int][(-5*x + x^
2 + Log[3 + E^(3/x)/x])^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(-5+x) \left (3 (-5+x) x^3+e^{3/x} \left (-3-x-5 x^2+x^3\right )\right )-x^2 \left (e^{3/x}+3 x\right ) \log \left (3+\frac {e^{3/x}}{x}\right )}{x^2 \left (e^{3/x}+3 x\right ) \left ((-5+x) x+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx\\ &=\int \left (\frac {3 \left (-15-2 x+x^2\right )}{x \left (e^{3/x}+3 x\right ) \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2}+\frac {15+2 x+24 x^2-10 x^3+x^4-x^2 \log \left (3+\frac {e^{3/x}}{x}\right )}{x^2 \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2}\right ) \, dx\\ &=3 \int \frac {-15-2 x+x^2}{x \left (e^{3/x}+3 x\right ) \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx+\int \frac {15+2 x+24 x^2-10 x^3+x^4-x^2 \log \left (3+\frac {e^{3/x}}{x}\right )}{x^2 \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx\\ &=3 \int \left (-\frac {2}{\left (e^{3/x}+3 x\right ) \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2}-\frac {15}{x \left (e^{3/x}+3 x\right ) \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2}+\frac {x}{\left (e^{3/x}+3 x\right ) \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2}\right ) \, dx+\int \left (\frac {15+2 x+24 x^2-15 x^3+2 x^4}{x^2 \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2}-\frac {1}{-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )}\right ) \, dx\\ &=3 \int \frac {x}{\left (e^{3/x}+3 x\right ) \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx-6 \int \frac {1}{\left (e^{3/x}+3 x\right ) \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx-45 \int \frac {1}{x \left (e^{3/x}+3 x\right ) \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx+\int \frac {15+2 x+24 x^2-15 x^3+2 x^4}{x^2 \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx-\int \frac {1}{-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )} \, dx\\ &=3 \int \frac {x}{\left (e^{3/x}+3 x\right ) \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx-6 \int \frac {1}{\left (e^{3/x}+3 x\right ) \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx-45 \int \frac {1}{x \left (e^{3/x}+3 x\right ) \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx-\int \frac {1}{-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )} \, dx+\int \left (\frac {24}{\left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2}+\frac {15}{x^2 \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2}+\frac {2}{x \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2}-\frac {15 x}{\left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2}+\frac {2 x^2}{\left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2}\right ) \, dx\\ &=2 \int \frac {1}{x \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx+2 \int \frac {x^2}{\left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx+3 \int \frac {x}{\left (e^{3/x}+3 x\right ) \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx-6 \int \frac {1}{\left (e^{3/x}+3 x\right ) \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx+15 \int \frac {1}{x^2 \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx-15 \int \frac {x}{\left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx+24 \int \frac {1}{\left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx-45 \int \frac {1}{x \left (e^{3/x}+3 x\right ) \left (-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )\right )^2} \, dx-\int \frac {1}{-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 29, normalized size = 0.97 \begin {gather*} \frac {5-x}{-5 x+x^2+\log \left (3+\frac {e^{3/x}}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(75*x^3 - 30*x^4 + 3*x^5 + E^(3/x)*(15 + 2*x + 24*x^2 - 10*x^3 + x^4) + (-(E^(3/x)*x^2) - 3*x^3)*Log
[(E^(3/x) + 3*x)/x])/(75*x^5 - 30*x^6 + 3*x^7 + E^(3/x)*(25*x^4 - 10*x^5 + x^6) + (-30*x^4 + 6*x^5 + E^(3/x)*(
-10*x^3 + 2*x^4))*Log[(E^(3/x) + 3*x)/x] + (E^(3/x)*x^2 + 3*x^3)*Log[(E^(3/x) + 3*x)/x]^2),x]

[Out]

(5 - x)/(-5*x + x^2 + Log[3 + E^(3/x)/x])

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fricas [A]  time = 0.49, size = 29, normalized size = 0.97 \begin {gather*} -\frac {x - 5}{x^{2} - 5 \, x + \log \left (\frac {3 \, x + e^{\frac {3}{x}}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2*exp(3/x)-3*x^3)*log((exp(3/x)+3*x)/x)+(x^4-10*x^3+24*x^2+2*x+15)*exp(3/x)+3*x^5-30*x^4+75*x^3
)/((x^2*exp(3/x)+3*x^3)*log((exp(3/x)+3*x)/x)^2+((2*x^4-10*x^3)*exp(3/x)+6*x^5-30*x^4)*log((exp(3/x)+3*x)/x)+(
x^6-10*x^5+25*x^4)*exp(3/x)+3*x^7-30*x^6+75*x^5),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.39, size = 29, normalized size = 0.97 \begin {gather*} -\frac {x - 5}{x^{2} - 5 \, x + \log \left (\frac {3 \, x + e^{\frac {3}{x}}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2*exp(3/x)-3*x^3)*log((exp(3/x)+3*x)/x)+(x^4-10*x^3+24*x^2+2*x+15)*exp(3/x)+3*x^5-30*x^4+75*x^3
)/((x^2*exp(3/x)+3*x^3)*log((exp(3/x)+3*x)/x)^2+((2*x^4-10*x^3)*exp(3/x)+6*x^5-30*x^4)*log((exp(3/x)+3*x)/x)+(
x^6-10*x^5+25*x^4)*exp(3/x)+3*x^7-30*x^6+75*x^5),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.15, size = 170, normalized size = 5.67

method	result	size
risch	\(-\frac {2 \left (x -5\right )}{-i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (\frac {{\mathrm e}^{\frac {3}{x}}}{3}+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{\frac {3}{x}}}{3}+x \right )}{x}\right )+i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{\frac {3}{x}}}{3}+x \right )}{x}\right )^{2}+i \pi \,\mathrm {csgn}\left (i \left (\frac {{\mathrm e}^{\frac {3}{x}}}{3}+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{\frac {3}{x}}}{3}+x \right )}{x}\right )^{2}-i \pi \mathrm {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{\frac {3}{x}}}{3}+x \right )}{x}\right )^{3}+2 x^{2}+2 \ln \relax (3)-10 x -2 \ln \relax (x )+2 \ln \left (\frac {{\mathrm e}^{\frac {3}{x}}}{3}+x \right )}\)	\(170\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2*exp(3/x)-3*x^3)*ln((exp(3/x)+3*x)/x)+(x^4-10*x^3+24*x^2+2*x+15)*exp(3/x)+3*x^5-30*x^4+75*x^3)/((x^2
*exp(3/x)+3*x^3)*ln((exp(3/x)+3*x)/x)^2+((2*x^4-10*x^3)*exp(3/x)+6*x^5-30*x^4)*ln((exp(3/x)+3*x)/x)+(x^6-10*x^
5+25*x^4)*exp(3/x)+3*x^7-30*x^6+75*x^5),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.42, size = 29, normalized size = 0.97 \begin {gather*} -\frac {x - 5}{x^{2} - 5 \, x + \log \left (3 \, x + e^{\frac {3}{x}}\right ) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2*exp(3/x)-3*x^3)*log((exp(3/x)+3*x)/x)+(x^4-10*x^3+24*x^2+2*x+15)*exp(3/x)+3*x^5-30*x^4+75*x^3
)/((x^2*exp(3/x)+3*x^3)*log((exp(3/x)+3*x)/x)^2+((2*x^4-10*x^3)*exp(3/x)+6*x^5-30*x^4)*log((exp(3/x)+3*x)/x)+(
x^6-10*x^5+25*x^4)*exp(3/x)+3*x^7-30*x^6+75*x^5),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{3/x}\,\left (x^4-10\,x^3+24\,x^2+2\,x+15\right )-\ln \left (\frac {3\,x+{\mathrm {e}}^{3/x}}{x}\right )\,\left (x^2\,{\mathrm {e}}^{3/x}+3\,x^3\right )+75\,x^3-30\,x^4+3\,x^5}{{\ln \left (\frac {3\,x+{\mathrm {e}}^{3/x}}{x}\right )}^2\,\left (x^2\,{\mathrm {e}}^{3/x}+3\,x^3\right )-\ln \left (\frac {3\,x+{\mathrm {e}}^{3/x}}{x}\right )\,\left ({\mathrm {e}}^{3/x}\,\left (10\,x^3-2\,x^4\right )+30\,x^4-6\,x^5\right )+{\mathrm {e}}^{3/x}\,\left (x^6-10\,x^5+25\,x^4\right )+75\,x^5-30\,x^6+3\,x^7} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3/x)*(2*x + 24*x^2 - 10*x^3 + x^4 + 15) - log((3*x + exp(3/x))/x)*(x^2*exp(3/x) + 3*x^3) + 75*x^3 - 3
0*x^4 + 3*x^5)/(log((3*x + exp(3/x))/x)^2*(x^2*exp(3/x) + 3*x^3) - log((3*x + exp(3/x))/x)*(exp(3/x)*(10*x^3 -
2*x^4) + 30*x^4 - 6*x^5) + exp(3/x)*(25*x^4 - 10*x^5 + x^6) + 75*x^5 - 30*x^6 + 3*x^7),x)

[Out]

int((exp(3/x)*(2*x + 24*x^2 - 10*x^3 + x^4 + 15) - log((3*x + exp(3/x))/x)*(x^2*exp(3/x) + 3*x^3) + 75*x^3 - 3
0*x^4 + 3*x^5)/(log((3*x + exp(3/x))/x)^2*(x^2*exp(3/x) + 3*x^3) - log((3*x + exp(3/x))/x)*(exp(3/x)*(10*x^3 -
2*x^4) + 30*x^4 - 6*x^5) + exp(3/x)*(25*x^4 - 10*x^5 + x^6) + 75*x^5 - 30*x^6 + 3*x^7), x)

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sympy [A]  time = 0.55, size = 20, normalized size = 0.67 \begin {gather*} \frac {5 - x}{x^{2} - 5 x + \log {\left (\frac {3 x + e^{\frac {3}{x}}}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2*exp(3/x)-3*x**3)*ln((exp(3/x)+3*x)/x)+(x**4-10*x**3+24*x**2+2*x+15)*exp(3/x)+3*x**5-30*x**4+
75*x**3)/((x**2*exp(3/x)+3*x**3)*ln((exp(3/x)+3*x)/x)**2+((2*x**4-10*x**3)*exp(3/x)+6*x**5-30*x**4)*ln((exp(3/
x)+3*x)/x)+(x**6-10*x**5+25*x**4)*exp(3/x)+3*x**7-30*x**6+75*x**5),x)

[Out]

(5 - x)/(x**2 - 5*x + log((3*x + exp(3/x))/x))

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