3.3.74 \(\int \frac {-60 x+30 x^3+e^x (-20 x^3+10 x^5) \log (2)+(-120 x+40 e^x x^2 \log (2)) \log (\frac {3-e^x x \log (2)}{x \log (2)})}{(-12+12 x^2-3 x^4+e^x (4 x-4 x^3+x^5) \log (2)) \log ^2(\frac {3-e^x x \log (2)}{x \log (2)})} \, dx\)

Optimal. Leaf size=32 \[ \frac {5 x}{\left (\frac {1}{x}-\frac {x}{2}\right ) \log \left (-e^x+\frac {3}{x \log (2)}\right )} \]

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Rubi [F]  time = 8.51, 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 {-60 x+30 x^3+e^x \left (-20 x^3+10 x^5\right ) \log (2)+\left (-120 x+40 e^x x^2 \log (2)\right ) \log \left (\frac {3-e^x x \log (2)}{x \log (2)}\right )}{\left (-12+12 x^2-3 x^4+e^x \left (4 x-4 x^3+x^5\right ) \log (2)\right ) \log ^2\left (\frac {3-e^x x \log (2)}{x \log (2)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-60*x + 30*x^3 + E^x*(-20*x^3 + 10*x^5)*Log[2] + (-120*x + 40*E^x*x^2*Log[2])*Log[(3 - E^x*x*Log[2])/(x*L
og[2])])/((-12 + 12*x^2 - 3*x^4 + E^x*(4*x - 4*x^3 + x^5)*Log[2])*Log[(3 - E^x*x*Log[2])/(x*Log[2])]^2),x]

[Out]

10*Defer[Int][Log[-E^x + 3/(x*Log[2])]^(-2), x] - 5*Sqrt[2]*Defer[Int][1/((Sqrt[2] - x)*Log[-E^x + 3/(x*Log[2]
)]^2), x] - 5*Sqrt[2]*Defer[Int][1/((Sqrt[2] + x)*Log[-E^x + 3/(x*Log[2])]^2), x] + 30*Defer[Int][1/((-3 + E^x
*x*Log[2])*Log[-E^x + 3/(x*Log[2])]^2), x] - 15*Defer[Int][1/((Sqrt[2] - x)*(-3 + E^x*x*Log[2])*Log[-E^x + 3/(
x*Log[2])]^2), x] - 15*Sqrt[2]*Defer[Int][1/((Sqrt[2] - x)*(-3 + E^x*x*Log[2])*Log[-E^x + 3/(x*Log[2])]^2), x]
+ 15*Defer[Int][1/((Sqrt[2] + x)*(-3 + E^x*x*Log[2])*Log[-E^x + 3/(x*Log[2])]^2), x] - 15*Sqrt[2]*Defer[Int][
1/((Sqrt[2] + x)*(-3 + E^x*x*Log[2])*Log[-E^x + 3/(x*Log[2])]^2), x] + 40*Defer[Int][x/((-2 + x^2)^2*Log[-E^x
+ 3/(x*Log[2])]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {60 x-30 x^3-e^x \left (-20 x^3+10 x^5\right ) \log (2)-\left (-120 x+40 e^x x^2 \log (2)\right ) \log \left (\frac {3-e^x x \log (2)}{x \log (2)}\right )}{\left (2-x^2\right )^2 \left (3-e^x x \log (2)\right ) \log ^2\left (\frac {3-e^x x \log (2)}{x \log (2)}\right )} \, dx\\ &=\int \left (\frac {30 x (1+x)}{\left (-2+x^2\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )}+\frac {10 x \left (-2 x+x^3+4 \log \left (-e^x+\frac {3}{x \log (2)}\right )\right )}{\left (-2+x^2\right )^2 \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )}\right ) \, dx\\ &=10 \int \frac {x \left (-2 x+x^3+4 \log \left (-e^x+\frac {3}{x \log (2)}\right )\right )}{\left (-2+x^2\right )^2 \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+30 \int \frac {x (1+x)}{\left (-2+x^2\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx\\ &=10 \int \left (\frac {x^2}{\left (-2+x^2\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )}+\frac {4 x}{\left (-2+x^2\right )^2 \log \left (-e^x+\frac {3}{x \log (2)}\right )}\right ) \, dx+30 \int \left (\frac {1}{\left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )}+\frac {2+x}{\left (-2+x^2\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )}\right ) \, dx\\ &=10 \int \frac {x^2}{\left (-2+x^2\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+30 \int \frac {1}{\left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+30 \int \frac {2+x}{\left (-2+x^2\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+40 \int \frac {x}{\left (-2+x^2\right )^2 \log \left (-e^x+\frac {3}{x \log (2)}\right )} \, dx\\ &=10 \int \left (\frac {1}{\log ^2\left (-e^x+\frac {3}{x \log (2)}\right )}+\frac {2}{\left (-2+x^2\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )}\right ) \, dx+30 \int \left (\frac {2}{\left (-2+x^2\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )}+\frac {x}{\left (-2+x^2\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )}\right ) \, dx+30 \int \frac {1}{\left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+40 \int \frac {x}{\left (-2+x^2\right )^2 \log \left (-e^x+\frac {3}{x \log (2)}\right )} \, dx\\ &=10 \int \frac {1}{\log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+20 \int \frac {1}{\left (-2+x^2\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+30 \int \frac {1}{\left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+30 \int \frac {x}{\left (-2+x^2\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+40 \int \frac {x}{\left (-2+x^2\right )^2 \log \left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+60 \int \frac {1}{\left (-2+x^2\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx\\ &=10 \int \frac {1}{\log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+20 \int \left (-\frac {1}{2 \sqrt {2} \left (\sqrt {2}-x\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )}-\frac {1}{2 \sqrt {2} \left (\sqrt {2}+x\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )}\right ) \, dx+30 \int \left (-\frac {1}{2 \left (\sqrt {2}-x\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )}+\frac {1}{2 \left (\sqrt {2}+x\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )}\right ) \, dx+30 \int \frac {1}{\left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+40 \int \frac {x}{\left (-2+x^2\right )^2 \log \left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+60 \int \left (-\frac {1}{2 \sqrt {2} \left (\sqrt {2}-x\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )}-\frac {1}{2 \sqrt {2} \left (\sqrt {2}+x\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )}\right ) \, dx\\ &=10 \int \frac {1}{\log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx-15 \int \frac {1}{\left (\sqrt {2}-x\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+15 \int \frac {1}{\left (\sqrt {2}+x\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+30 \int \frac {1}{\left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx+40 \int \frac {x}{\left (-2+x^2\right )^2 \log \left (-e^x+\frac {3}{x \log (2)}\right )} \, dx-\left (5 \sqrt {2}\right ) \int \frac {1}{\left (\sqrt {2}-x\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx-\left (5 \sqrt {2}\right ) \int \frac {1}{\left (\sqrt {2}+x\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx-\left (15 \sqrt {2}\right ) \int \frac {1}{\left (\sqrt {2}-x\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx-\left (15 \sqrt {2}\right ) \int \frac {1}{\left (\sqrt {2}+x\right ) \left (-3+e^x x \log (2)\right ) \log ^2\left (-e^x+\frac {3}{x \log (2)}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.15, size = 69, normalized size = 2.16 \begin {gather*} -\frac {10 x^2 \left (-6+3 x^2+e^x x^4 \log (2)-e^x x^2 \log (4)\right )}{\left (-2+x^2\right )^2 \left (3+e^x x^2 \log (2)\right ) \log \left (-e^x+\frac {3}{x \log (2)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-60*x + 30*x^3 + E^x*(-20*x^3 + 10*x^5)*Log[2] + (-120*x + 40*E^x*x^2*Log[2])*Log[(3 - E^x*x*Log[2]
)/(x*Log[2])])/((-12 + 12*x^2 - 3*x^4 + E^x*(4*x - 4*x^3 + x^5)*Log[2])*Log[(3 - E^x*x*Log[2])/(x*Log[2])]^2),
x]

[Out]

(-10*x^2*(-6 + 3*x^2 + E^x*x^4*Log[2] - E^x*x^2*Log[4]))/((-2 + x^2)^2*(3 + E^x*x^2*Log[2])*Log[-E^x + 3/(x*Lo
g[2])])

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fricas [A]  time = 0.83, size = 32, normalized size = 1.00 \begin {gather*} -\frac {10 \, x^{2}}{{\left (x^{2} - 2\right )} \log \left (-\frac {x e^{x} \log \relax (2) - 3}{x \log \relax (2)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x^2*log(2)*exp(x)-120*x)*log((-x*log(2)*exp(x)+3)/x/log(2))+(10*x^5-20*x^3)*log(2)*exp(x)+30*x^
3-60*x)/((x^5-4*x^3+4*x)*log(2)*exp(x)-3*x^4+12*x^2-12)/log((-x*log(2)*exp(x)+3)/x/log(2))^2,x, algorithm="fri
cas")

[Out]

-10*x^2/((x^2 - 2)*log(-(x*e^x*log(2) - 3)/(x*log(2))))

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giac [A]  time = 1.17, size = 51, normalized size = 1.59 \begin {gather*} -\frac {10 \, x^{2}}{x^{2} \log \left (-x e^{x} \log \relax (2) + 3\right ) - x^{2} \log \left (x \log \relax (2)\right ) - 2 \, \log \left (-x e^{x} \log \relax (2) + 3\right ) + 2 \, \log \left (x \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x^2*log(2)*exp(x)-120*x)*log((-x*log(2)*exp(x)+3)/x/log(2))+(10*x^5-20*x^3)*log(2)*exp(x)+30*x^
3-60*x)/((x^5-4*x^3+4*x)*log(2)*exp(x)-3*x^4+12*x^2-12)/log((-x*log(2)*exp(x)+3)/x/log(2))^2,x, algorithm="gia
c")

[Out]

-10*x^2/(x^2*log(-x*e^x*log(2) + 3) - x^2*log(x*log(2)) - 2*log(-x*e^x*log(2) + 3) + 2*log(x*log(2)))

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maple [C]  time = 0.16, size = 178, normalized size = 5.56

method	result	size
risch	\(-\frac {20 i x^{2}}{\left (x^{2}-2\right ) \left (2 \pi \mathrm {csgn}\left (\frac {i \left (x \ln \relax (2) {\mathrm e}^{x}-3\right )}{x}\right )^{2}+\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x \ln \relax (2) {\mathrm e}^{x}-3\right )\right ) \mathrm {csgn}\left (\frac {i \left (x \ln \relax (2) {\mathrm e}^{x}-3\right )}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (x \ln \relax (2) {\mathrm e}^{x}-3\right )}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x \ln \relax (2) {\mathrm e}^{x}-3\right )\right ) \mathrm {csgn}\left (\frac {i \left (x \ln \relax (2) {\mathrm e}^{x}-3\right )}{x}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i \left (x \ln \relax (2) {\mathrm e}^{x}-3\right )}{x}\right )^{3}-2 \pi -2 i \ln \left (\ln \relax (2)\right )-2 i \ln \relax (x )+2 i \ln \left (x \ln \relax (2) {\mathrm e}^{x}-3\right )\right )}\)	\(178\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((40*x^2*ln(2)*exp(x)-120*x)*ln((-x*ln(2)*exp(x)+3)/x/ln(2))+(10*x^5-20*x^3)*ln(2)*exp(x)+30*x^3-60*x)/((x
^5-4*x^3+4*x)*ln(2)*exp(x)-3*x^4+12*x^2-12)/ln((-x*ln(2)*exp(x)+3)/x/ln(2))^2,x,method=_RETURNVERBOSE)

[Out]

-20*I*x^2/(x^2-2)/(2*Pi*csgn(I/x*(x*ln(2)*exp(x)-3))^2+Pi*csgn(I/x)*csgn(I*(x*ln(2)*exp(x)-3))*csgn(I/x*(x*ln(
2)*exp(x)-3))-Pi*csgn(I/x)*csgn(I/x*(x*ln(2)*exp(x)-3))^2-Pi*csgn(I*(x*ln(2)*exp(x)-3))*csgn(I/x*(x*ln(2)*exp(
x)-3))^2-Pi*csgn(I/x*(x*ln(2)*exp(x)-3))^3-2*Pi-2*I*ln(ln(2))-2*I*ln(x)+2*I*ln(x*ln(2)*exp(x)-3))

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maxima [A]  time = 0.82, size = 45, normalized size = 1.41 \begin {gather*} \frac {10 \, x^{2}}{x^{2} \log \left (\log \relax (2)\right ) - {\left (x^{2} - 2\right )} \log \left (-x e^{x} \log \relax (2) + 3\right ) + {\left (x^{2} - 2\right )} \log \relax (x) - 2 \, \log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x^2*log(2)*exp(x)-120*x)*log((-x*log(2)*exp(x)+3)/x/log(2))+(10*x^5-20*x^3)*log(2)*exp(x)+30*x^
3-60*x)/((x^5-4*x^3+4*x)*log(2)*exp(x)-3*x^4+12*x^2-12)/log((-x*log(2)*exp(x)+3)/x/log(2))^2,x, algorithm="max
ima")

[Out]

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

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mupad [B]  time = 1.00, size = 34, normalized size = 1.06 \begin {gather*} -\frac {10\,x^2}{\left (x^2-2\right )\,\left (\ln \left (-\frac {x\,{\mathrm {e}}^x\,\ln \relax (2)-3}{x}\right )-\ln \left (\ln \relax (2)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(60*x + log(-(x*exp(x)*log(2) - 3)/(x*log(2)))*(120*x - 40*x^2*exp(x)*log(2)) - 30*x^3 + exp(x)*log(2)*(2
0*x^3 - 10*x^5))/(log(-(x*exp(x)*log(2) - 3)/(x*log(2)))^2*(12*x^2 - 3*x^4 + exp(x)*log(2)*(4*x - 4*x^3 + x^5)
- 12)),x)

[Out]

-(10*x^2)/((x^2 - 2)*(log(-(x*exp(x)*log(2) - 3)/x) - log(log(2))))

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sympy [A]  time = 0.28, size = 27, normalized size = 0.84 \begin {gather*} - \frac {10 x^{2}}{\left (x^{2} - 2\right ) \log {\left (\frac {- x e^{x} \log {\relax (2 )} + 3}{x \log {\relax (2 )}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x**2*ln(2)*exp(x)-120*x)*ln((-x*ln(2)*exp(x)+3)/x/ln(2))+(10*x**5-20*x**3)*ln(2)*exp(x)+30*x**3
-60*x)/((x**5-4*x**3+4*x)*ln(2)*exp(x)-3*x**4+12*x**2-12)/ln((-x*ln(2)*exp(x)+3)/x/ln(2))**2,x)

[Out]

-10*x**2/((x**2 - 2)*log((-x*exp(x)*log(2) + 3)/(x*log(2))))

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