3.74.6 \(\int \frac {4-12 x-15 x^2-4 x^3+e^3 (-4-4 x-x^2)+(-20 x-18 x^2-4 x^3) \log (\frac {2}{x})+(-4 x-2 x^2-4 x \log (\frac {2}{x})) \log (x)}{4 x-28 x^2+33 x^3+56 x^4+16 x^5+e^6 (4 x+4 x^2+x^3)+e^3 (-8 x+24 x^2+30 x^3+8 x^4)+(-8 x^2+28 x^3+16 x^4+e^3 (8 x^2+4 x^3)) \log (x)+4 x^3 \log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {\log \left (\frac {2}{x}\right )}{-1+e^3+2 x \left (2+\frac {\log (x)}{2+x}\right )} \]

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Rubi [F]  time = 10.01, 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 {4-12 x-15 x^2-4 x^3+e^3 \left (-4-4 x-x^2\right )+\left (-20 x-18 x^2-4 x^3\right ) \log \left (\frac {2}{x}\right )+\left (-4 x-2 x^2-4 x \log \left (\frac {2}{x}\right )\right ) \log (x)}{4 x-28 x^2+33 x^3+56 x^4+16 x^5+e^6 \left (4 x+4 x^2+x^3\right )+e^3 \left (-8 x+24 x^2+30 x^3+8 x^4\right )+\left (-8 x^2+28 x^3+16 x^4+e^3 \left (8 x^2+4 x^3\right )\right ) \log (x)+4 x^3 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(4 - 12*x - 15*x^2 - 4*x^3 + E^3*(-4 - 4*x - x^2) + (-20*x - 18*x^2 - 4*x^3)*Log[2/x] + (-4*x - 2*x^2 - 4*
x*Log[2/x])*Log[x])/(4*x - 28*x^2 + 33*x^3 + 56*x^4 + 16*x^5 + E^6*(4*x + 4*x^2 + x^3) + E^3*(-8*x + 24*x^2 +
30*x^3 + 8*x^4) + (-8*x^2 + 28*x^3 + 16*x^4 + E^3*(8*x^2 + 4*x^3))*Log[x] + 4*x^3*Log[x]^2),x]

[Out]

-2*(3 - E^3)*Defer[Int][Log[2/x]/(2 - 7*x - 4*x^2 - E^3*(2 + x) - 2*x*Log[x])^2, x] - 4*(1 - E^3)*Defer[Int][L
og[2/x]/(x*(2 - 7*x - 4*x^2 - E^3*(2 + x) - 2*x*Log[x])^2), x] - 10*Defer[Int][(x*Log[2/x])/(2 - 7*x - 4*x^2 -
E^3*(2 + x) - 2*x*Log[x])^2, x] - 4*Defer[Int][(x^2*Log[2/x])/(2 - 7*x - 4*x^2 - E^3*(2 + x) - 2*x*Log[x])^2,
x] + Defer[Int][(2 - 7*x - 4*x^2 - E^3*(2 + x) - 2*x*Log[x])^(-1), x] + 2*Defer[Int][1/(x*(2 - 7*x - 4*x^2 -
E^3*(2 + x) - 2*x*Log[x])), x] + 2*Defer[Int][Log[2/x]/(x*(2 - 7*x - 4*x^2 - E^3*(2 + x) - 2*x*Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 x \log \left (\frac {2}{x}\right ) \left (10+9 x+2 x^2+2 \log (x)\right )-(2+x) \left (-2+7 x+4 x^2+e^3 (2+x)+2 x \log (x)\right )}{x \left (2-7 x-4 x^2-e^3 (2+x)-2 x \log (x)\right )^2} \, dx\\ &=\int \left (\frac {2 (2+x) \left (-1+e^3-x-2 x^2\right ) \log \left (\frac {2}{x}\right )}{x \left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )^2}+\frac {2+x+2 \log \left (\frac {2}{x}\right )}{x \left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )}\right ) \, dx\\ &=2 \int \frac {(2+x) \left (-1+e^3-x-2 x^2\right ) \log \left (\frac {2}{x}\right )}{x \left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )^2} \, dx+\int \frac {2+x+2 \log \left (\frac {2}{x}\right )}{x \left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )} \, dx\\ &=2 \int \frac {(2+x) \left (-1+e^3-x-2 x^2\right ) \log \left (\frac {2}{x}\right )}{x \left (2-7 x-4 x^2-e^3 (2+x)-2 x \log (x)\right )^2} \, dx+\int \frac {2+x+2 \log \left (\frac {2}{x}\right )}{x \left (2-7 x-4 x^2-e^3 (2+x)-2 x \log (x)\right )} \, dx\\ &=2 \int \left (-\frac {3 \left (1-\frac {e^3}{3}\right ) \log \left (\frac {2}{x}\right )}{\left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )^2}+\frac {2 \left (-1+e^3\right ) \log \left (\frac {2}{x}\right )}{x \left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )^2}-\frac {5 x \log \left (\frac {2}{x}\right )}{\left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )^2}-\frac {2 x^2 \log \left (\frac {2}{x}\right )}{\left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )^2}\right ) \, dx+\int \left (\frac {1}{2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)}+\frac {2}{x \left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )}+\frac {2 \log \left (\frac {2}{x}\right )}{x \left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )}\right ) \, dx\\ &=2 \int \frac {1}{x \left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )} \, dx+2 \int \frac {\log \left (\frac {2}{x}\right )}{x \left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )} \, dx-4 \int \frac {x^2 \log \left (\frac {2}{x}\right )}{\left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )^2} \, dx-10 \int \frac {x \log \left (\frac {2}{x}\right )}{\left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )^2} \, dx-\left (4 \left (1-e^3\right )\right ) \int \frac {\log \left (\frac {2}{x}\right )}{x \left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )^2} \, dx-\left (2 \left (3-e^3\right )\right ) \int \frac {\log \left (\frac {2}{x}\right )}{\left (2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)\right )^2} \, dx+\int \frac {1}{2 \left (1-e^3\right )-7 \left (1+\frac {e^3}{7}\right ) x-4 x^2-2 x \log (x)} \, dx\\ &=2 \int \frac {1}{x \left (2-7 x-4 x^2-e^3 (2+x)-2 x \log (x)\right )} \, dx+2 \int \frac {\log \left (\frac {2}{x}\right )}{x \left (2-7 x-4 x^2-e^3 (2+x)-2 x \log (x)\right )} \, dx-4 \int \frac {x^2 \log \left (\frac {2}{x}\right )}{\left (2-7 x-4 x^2-e^3 (2+x)-2 x \log (x)\right )^2} \, dx-10 \int \frac {x \log \left (\frac {2}{x}\right )}{\left (2-7 x-4 x^2-e^3 (2+x)-2 x \log (x)\right )^2} \, dx-\left (4 \left (1-e^3\right )\right ) \int \frac {\log \left (\frac {2}{x}\right )}{x \left (2-7 x-4 x^2-e^3 (2+x)-2 x \log (x)\right )^2} \, dx-\left (2 \left (3-e^3\right )\right ) \int \frac {\log \left (\frac {2}{x}\right )}{\left (2-7 x-4 x^2-e^3 (2+x)-2 x \log (x)\right )^2} \, dx+\int \frac {1}{2-7 x-4 x^2-e^3 (2+x)-2 x \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.61, size = 64, normalized size = 2.37 \begin {gather*} \frac {-2+2 e^3+7 x+e^3 x+4 x^2+2 (2+x) \log \left (\frac {2}{x}\right )+2 x \log (x)}{2 \left (-2+7 x+4 x^2+e^3 (2+x)+2 x \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 12*x - 15*x^2 - 4*x^3 + E^3*(-4 - 4*x - x^2) + (-20*x - 18*x^2 - 4*x^3)*Log[2/x] + (-4*x - 2*x^
2 - 4*x*Log[2/x])*Log[x])/(4*x - 28*x^2 + 33*x^3 + 56*x^4 + 16*x^5 + E^6*(4*x + 4*x^2 + x^3) + E^3*(-8*x + 24*
x^2 + 30*x^3 + 8*x^4) + (-8*x^2 + 28*x^3 + 16*x^4 + E^3*(8*x^2 + 4*x^3))*Log[x] + 4*x^3*Log[x]^2),x]

[Out]

(-2 + 2*E^3 + 7*x + E^3*x + 4*x^2 + 2*(2 + x)*Log[2/x] + 2*x*Log[x])/(2*(-2 + 7*x + 4*x^2 + E^3*(2 + x) + 2*x*
Log[x]))

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fricas [B]  time = 0.63, size = 63, normalized size = 2.33 \begin {gather*} \frac {4 \, x^{2} + {\left (x + 2\right )} e^{3} + 2 \, x \log \relax (2) + 7 \, x + 4 \, \log \left (\frac {2}{x}\right ) - 2}{2 \, {\left (4 \, x^{2} + {\left (x + 2\right )} e^{3} + 2 \, x \log \relax (2) - 2 \, x \log \left (\frac {2}{x}\right ) + 7 \, x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(2/x)-2*x^2-4*x)*log(x)+(-4*x^3-18*x^2-20*x)*log(2/x)+(-x^2-4*x-4)*exp(3)-4*x^3-15*x^2-12*
x+4)/(4*x^3*log(x)^2+((4*x^3+8*x^2)*exp(3)+16*x^4+28*x^3-8*x^2)*log(x)+(x^3+4*x^2+4*x)*exp(3)^2+(8*x^4+30*x^3+
24*x^2-8*x)*exp(3)+16*x^5+56*x^4+33*x^3-28*x^2+4*x),x, algorithm="fricas")

[Out]

1/2*(4*x^2 + (x + 2)*e^3 + 2*x*log(2) + 7*x + 4*log(2/x) - 2)/(4*x^2 + (x + 2)*e^3 + 2*x*log(2) - 2*x*log(2/x)
+ 7*x - 2)

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giac [B]  time = 0.33, size = 58, normalized size = 2.15 \begin {gather*} \frac {4 \, x^{2} + x e^{3} + 2 \, x \log \relax (2) + 7 \, x + 2 \, e^{3} + 4 \, \log \relax (2) - 4 \, \log \relax (x) - 2}{2 \, {\left (4 \, x^{2} + x e^{3} + 2 \, x \log \relax (x) + 7 \, x + 2 \, e^{3} - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(2/x)-2*x^2-4*x)*log(x)+(-4*x^3-18*x^2-20*x)*log(2/x)+(-x^2-4*x-4)*exp(3)-4*x^3-15*x^2-12*
x+4)/(4*x^3*log(x)^2+((4*x^3+8*x^2)*exp(3)+16*x^4+28*x^3-8*x^2)*log(x)+(x^3+4*x^2+4*x)*exp(3)^2+(8*x^4+30*x^3+
24*x^2-8*x)*exp(3)+16*x^5+56*x^4+33*x^3-28*x^2+4*x),x, algorithm="giac")

[Out]

1/2*(4*x^2 + x*e^3 + 2*x*log(2) + 7*x + 2*e^3 + 4*log(2) - 4*log(x) - 2)/(4*x^2 + x*e^3 + 2*x*log(x) + 7*x + 2
*e^3 - 2)

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maple [B]  time = 0.40, size = 79, normalized size = 2.93

method	result	size
risch	\(-\frac {1}{x}+\frac {-4+2 x^{2} \ln \relax (2)+x^{2} {\mathrm e}^{3}+4 x^{3}+4 x \ln \relax (2)+4 x \,{\mathrm e}^{3}+15 x^{2}+4 \,{\mathrm e}^{3}+12 x}{2 x \left (2 x \ln \relax (x )+x \,{\mathrm e}^{3}+4 x^{2}+2 \,{\mathrm e}^{3}+7 x -2\right )}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x*ln(2/x)-2*x^2-4*x)*ln(x)+(-4*x^3-18*x^2-20*x)*ln(2/x)+(-x^2-4*x-4)*exp(3)-4*x^3-15*x^2-12*x+4)/(4*x
^3*ln(x)^2+((4*x^3+8*x^2)*exp(3)+16*x^4+28*x^3-8*x^2)*ln(x)+(x^3+4*x^2+4*x)*exp(3)^2+(8*x^4+30*x^3+24*x^2-8*x)
*exp(3)+16*x^5+56*x^4+33*x^3-28*x^2+4*x),x,method=_RETURNVERBOSE)

[Out]

-1/x+1/2*(-4+2*x^2*ln(2)+x^2*exp(3)+4*x^3+4*x*ln(2)+4*x*exp(3)+15*x^2+4*exp(3)+12*x)/x/(2*x*ln(x)+x*exp(3)+4*x
^2+2*exp(3)+7*x-2)

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maxima [B]  time = 0.51, size = 55, normalized size = 2.04 \begin {gather*} \frac {4 \, x^{2} + x {\left (e^{3} + 2 \, \log \relax (2) + 7\right )} + 2 \, e^{3} + 4 \, \log \relax (2) - 4 \, \log \relax (x) - 2}{2 \, {\left (4 \, x^{2} + x {\left (e^{3} + 7\right )} + 2 \, x \log \relax (x) + 2 \, e^{3} - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(2/x)-2*x^2-4*x)*log(x)+(-4*x^3-18*x^2-20*x)*log(2/x)+(-x^2-4*x-4)*exp(3)-4*x^3-15*x^2-12*
x+4)/(4*x^3*log(x)^2+((4*x^3+8*x^2)*exp(3)+16*x^4+28*x^3-8*x^2)*log(x)+(x^3+4*x^2+4*x)*exp(3)^2+(8*x^4+30*x^3+
24*x^2-8*x)*exp(3)+16*x^5+56*x^4+33*x^3-28*x^2+4*x),x, algorithm="maxima")

[Out]

1/2*(4*x^2 + x*(e^3 + 2*log(2) + 7) + 2*e^3 + 4*log(2) - 4*log(x) - 2)/(4*x^2 + x*(e^3 + 7) + 2*x*log(x) + 2*e
^3 - 2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {12\,x+\ln \relax (x)\,\left (4\,x+4\,x\,\ln \left (\frac {2}{x}\right )+2\,x^2\right )+\ln \left (\frac {2}{x}\right )\,\left (4\,x^3+18\,x^2+20\,x\right )+{\mathrm {e}}^3\,\left (x^2+4\,x+4\right )+15\,x^2+4\,x^3-4}{4\,x+{\mathrm {e}}^6\,\left (x^3+4\,x^2+4\,x\right )+4\,x^3\,{\ln \relax (x)}^2+{\mathrm {e}}^3\,\left (8\,x^4+30\,x^3+24\,x^2-8\,x\right )-28\,x^2+33\,x^3+56\,x^4+16\,x^5+\ln \relax (x)\,\left ({\mathrm {e}}^3\,\left (4\,x^3+8\,x^2\right )-8\,x^2+28\,x^3+16\,x^4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(12*x + log(x)*(4*x + 4*x*log(2/x) + 2*x^2) + log(2/x)*(20*x + 18*x^2 + 4*x^3) + exp(3)*(4*x + x^2 + 4) +
15*x^2 + 4*x^3 - 4)/(4*x + exp(6)*(4*x + 4*x^2 + x^3) + 4*x^3*log(x)^2 + exp(3)*(24*x^2 - 8*x + 30*x^3 + 8*x^
4) - 28*x^2 + 33*x^3 + 56*x^4 + 16*x^5 + log(x)*(exp(3)*(8*x^2 + 4*x^3) - 8*x^2 + 28*x^3 + 16*x^4)),x)

[Out]

int(-(12*x + log(x)*(4*x + 4*x*log(2/x) + 2*x^2) + log(2/x)*(20*x + 18*x^2 + 4*x^3) + exp(3)*(4*x + x^2 + 4) +
15*x^2 + 4*x^3 - 4)/(4*x + exp(6)*(4*x + 4*x^2 + x^3) + 4*x^3*log(x)^2 + exp(3)*(24*x^2 - 8*x + 30*x^3 + 8*x^
4) - 28*x^2 + 33*x^3 + 56*x^4 + 16*x^5 + log(x)*(exp(3)*(8*x^2 + 4*x^3) - 8*x^2 + 28*x^3 + 16*x^4)), x)

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sympy [B]  time = 0.43, size = 88, normalized size = 3.26 \begin {gather*} \frac {4 x^{3} + 2 x^{2} \log {\relax (2 )} + 15 x^{2} + x^{2} e^{3} + 4 x \log {\relax (2 )} + 12 x + 4 x e^{3} - 4 + 4 e^{3}}{8 x^{3} + 4 x^{2} \log {\relax (x )} + 14 x^{2} + 2 x^{2} e^{3} - 4 x + 4 x e^{3}} - \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*ln(2/x)-2*x**2-4*x)*ln(x)+(-4*x**3-18*x**2-20*x)*ln(2/x)+(-x**2-4*x-4)*exp(3)-4*x**3-15*x**2-
12*x+4)/(4*x**3*ln(x)**2+((4*x**3+8*x**2)*exp(3)+16*x**4+28*x**3-8*x**2)*ln(x)+(x**3+4*x**2+4*x)*exp(3)**2+(8*
x**4+30*x**3+24*x**2-8*x)*exp(3)+16*x**5+56*x**4+33*x**3-28*x**2+4*x),x)

[Out]

(4*x**3 + 2*x**2*log(2) + 15*x**2 + x**2*exp(3) + 4*x*log(2) + 12*x + 4*x*exp(3) - 4 + 4*exp(3))/(8*x**3 + 4*x
**2*log(x) + 14*x**2 + 2*x**2*exp(3) - 4*x + 4*x*exp(3)) - 1/x

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