3.10.17 \(\int \frac {-8-2 x+(7+9 x-15 x^2-2 x^3) \log (x)+(-x-2 x^2) \log ^2(x)+(1+2 x) \log (x) \log (\log (x))}{(64 x+32 x^2+4 x^3) \log (x)+(16 x+4 x^2) \log ^2(x)+x \log ^3(x)} \, dx\) [917]

Optimal. Leaf size=24 \[ \frac {x-x^2+\log (x)-\log (\log (x))}{8+2 x+\log (x)} \]

[Out]

(x-ln(ln(x))+ln(x)-x^2)/(8+2*x+ln(x))

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Rubi [F]
time = 1.04, 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 {-8-2 x+\left (7+9 x-15 x^2-2 x^3\right ) \log (x)+\left (-x-2 x^2\right ) \log ^2(x)+(1+2 x) \log (x) \log (\log (x))}{\left (64 x+32 x^2+4 x^3\right ) \log (x)+\left (16 x+4 x^2\right ) \log ^2(x)+x \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8 - 2*x + (7 + 9*x - 15*x^2 - 2*x^3)*Log[x] + (-x - 2*x^2)*Log[x]^2 + (1 + 2*x)*Log[x]*Log[Log[x]])/((64
*x + 32*x^2 + 4*x^3)*Log[x] + (16*x + 4*x^2)*Log[x]^2 + x*Log[x]^3),x]

[Out]

-1/2*Defer[Int][1/(x*(4 + x)*Log[x]), x] + 17*Defer[Int][(8 + 2*x + Log[x])^(-2), x] + 8*Defer[Int][1/(x*(8 +
2*x + Log[x])^2), x] + 3*Defer[Int][x/(8 + 2*x + Log[x])^2, x] + 2*Defer[Int][x^2/(8 + 2*x + Log[x])^2, x] - D
efer[Int][(8 + 2*x + Log[x])^(-1), x] + Defer[Int][1/(x*(8 + 2*x + Log[x])), x]/8 - 2*Defer[Int][x/(8 + 2*x +
Log[x]), x] - Defer[Int][1/((4 + x)*(8 + 2*x + Log[x])), x]/8 + 2*Defer[Int][Log[Log[x]]/(8 + 2*x + Log[x])^2,
x] + Defer[Int][Log[Log[x]]/(x*(8 + 2*x + Log[x])^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 (4+x)-x (1+2 x) \log ^2(x)+\log (x) \left (7+9 x-15 x^2-2 x^3+(1+2 x) \log (\log (x))\right )}{x \log (x) (8+2 x+\log (x))^2} \, dx\\ &=\int \left (\frac {9}{(8+2 x+\log (x))^2}+\frac {7}{x (8+2 x+\log (x))^2}-\frac {15 x}{(8+2 x+\log (x))^2}-\frac {2 x^2}{(8+2 x+\log (x))^2}-\frac {2 (4+x)}{x \log (x) (8+2 x+\log (x))^2}-\frac {(1+2 x) \log (x)}{(8+2 x+\log (x))^2}+\frac {(1+2 x) \log (\log (x))}{x (8+2 x+\log (x))^2}\right ) \, dx\\ &=-\left (2 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx\right )-2 \int \frac {4+x}{x \log (x) (8+2 x+\log (x))^2} \, dx+7 \int \frac {1}{x (8+2 x+\log (x))^2} \, dx+9 \int \frac {1}{(8+2 x+\log (x))^2} \, dx-15 \int \frac {x}{(8+2 x+\log (x))^2} \, dx-\int \frac {(1+2 x) \log (x)}{(8+2 x+\log (x))^2} \, dx+\int \frac {(1+2 x) \log (\log (x))}{x (8+2 x+\log (x))^2} \, dx\\ &=-\left (2 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx\right )-2 \int \left (\frac {1}{4 x (4+x) \log (x)}-\frac {1}{2 x (8+2 x+\log (x))^2}-\frac {1}{4 x (4+x) (8+2 x+\log (x))}\right ) \, dx+7 \int \frac {1}{x (8+2 x+\log (x))^2} \, dx+9 \int \frac {1}{(8+2 x+\log (x))^2} \, dx-15 \int \frac {x}{(8+2 x+\log (x))^2} \, dx-\int \left (-\frac {2 \left (4+9 x+2 x^2\right )}{(8+2 x+\log (x))^2}+\frac {1+2 x}{8+2 x+\log (x)}\right ) \, dx+\int \left (\frac {2 \log (\log (x))}{(8+2 x+\log (x))^2}+\frac {\log (\log (x))}{x (8+2 x+\log (x))^2}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {1}{x (4+x) \log (x)} \, dx\right )+\frac {1}{2} \int \frac {1}{x (4+x) (8+2 x+\log (x))} \, dx-2 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx+2 \int \frac {4+9 x+2 x^2}{(8+2 x+\log (x))^2} \, dx+2 \int \frac {\log (\log (x))}{(8+2 x+\log (x))^2} \, dx+7 \int \frac {1}{x (8+2 x+\log (x))^2} \, dx+9 \int \frac {1}{(8+2 x+\log (x))^2} \, dx-15 \int \frac {x}{(8+2 x+\log (x))^2} \, dx+\int \frac {1}{x (8+2 x+\log (x))^2} \, dx-\int \frac {1+2 x}{8+2 x+\log (x)} \, dx+\int \frac {\log (\log (x))}{x (8+2 x+\log (x))^2} \, dx\\ &=-\left (\frac {1}{2} \int \frac {1}{x (4+x) \log (x)} \, dx\right )+\frac {1}{2} \int \left (\frac {1}{4 x (8+2 x+\log (x))}-\frac {1}{4 (4+x) (8+2 x+\log (x))}\right ) \, dx-2 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx+2 \int \left (\frac {4}{(8+2 x+\log (x))^2}+\frac {9 x}{(8+2 x+\log (x))^2}+\frac {2 x^2}{(8+2 x+\log (x))^2}\right ) \, dx+2 \int \frac {\log (\log (x))}{(8+2 x+\log (x))^2} \, dx+7 \int \frac {1}{x (8+2 x+\log (x))^2} \, dx+9 \int \frac {1}{(8+2 x+\log (x))^2} \, dx-15 \int \frac {x}{(8+2 x+\log (x))^2} \, dx+\int \frac {1}{x (8+2 x+\log (x))^2} \, dx-\int \left (\frac {1}{8+2 x+\log (x)}+\frac {2 x}{8+2 x+\log (x)}\right ) \, dx+\int \frac {\log (\log (x))}{x (8+2 x+\log (x))^2} \, dx\\ &=\frac {1}{8} \int \frac {1}{x (8+2 x+\log (x))} \, dx-\frac {1}{8} \int \frac {1}{(4+x) (8+2 x+\log (x))} \, dx-\frac {1}{2} \int \frac {1}{x (4+x) \log (x)} \, dx-2 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx-2 \int \frac {x}{8+2 x+\log (x)} \, dx+2 \int \frac {\log (\log (x))}{(8+2 x+\log (x))^2} \, dx+4 \int \frac {x^2}{(8+2 x+\log (x))^2} \, dx+7 \int \frac {1}{x (8+2 x+\log (x))^2} \, dx+8 \int \frac {1}{(8+2 x+\log (x))^2} \, dx+9 \int \frac {1}{(8+2 x+\log (x))^2} \, dx-15 \int \frac {x}{(8+2 x+\log (x))^2} \, dx+18 \int \frac {x}{(8+2 x+\log (x))^2} \, dx+\int \frac {1}{x (8+2 x+\log (x))^2} \, dx-\int \frac {1}{8+2 x+\log (x)} \, dx+\int \frac {\log (\log (x))}{x (8+2 x+\log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.11, size = 20, normalized size = 0.83 \begin {gather*} -\frac {8+x+x^2+\log (\log (x))}{8+2 x+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 - 2*x + (7 + 9*x - 15*x^2 - 2*x^3)*Log[x] + (-x - 2*x^2)*Log[x]^2 + (1 + 2*x)*Log[x]*Log[Log[x]]
)/((64*x + 32*x^2 + 4*x^3)*Log[x] + (16*x + 4*x^2)*Log[x]^2 + x*Log[x]^3),x]

[Out]

-((8 + x + x^2 + Log[Log[x]])/(8 + 2*x + Log[x]))

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Maple [A]
time = 0.20, size = 33, normalized size = 1.38

method	result	size
risch	\(-\frac {\ln \left (\ln \left (x \right )\right )}{8+2 x +\ln \left (x \right )}-\frac {x^{2}+x +8}{8+2 x +\ln \left (x \right )}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x+1)*ln(x)*ln(ln(x))+(-2*x^2-x)*ln(x)^2+(-2*x^3-15*x^2+9*x+7)*ln(x)-2*x-8)/(x*ln(x)^3+(4*x^2+16*x)*ln(
x)^2+(4*x^3+32*x^2+64*x)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

-1/(8+2*x+ln(x))*ln(ln(x))-(x^2+x+8)/(8+2*x+ln(x))

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Maxima [A]
time = 0.30, size = 20, normalized size = 0.83 \begin {gather*} -\frac {x^{2} + x + \log \left (\log \left (x\right )\right ) + 8}{2 \, x + \log \left (x\right ) + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+2*x)*log(x)*log(log(x))+(-2*x^2-x)*log(x)^2+(-2*x^3-15*x^2+9*x+7)*log(x)-2*x-8)/(x*log(x)^3+(4*x
^2+16*x)*log(x)^2+(4*x^3+32*x^2+64*x)*log(x)),x, algorithm="maxima")

[Out]

-(x^2 + x + log(log(x)) + 8)/(2*x + log(x) + 8)

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Fricas [A]
time = 0.36, size = 20, normalized size = 0.83 \begin {gather*} -\frac {x^{2} + x + \log \left (\log \left (x\right )\right ) + 8}{2 \, x + \log \left (x\right ) + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+2*x)*log(x)*log(log(x))+(-2*x^2-x)*log(x)^2+(-2*x^3-15*x^2+9*x+7)*log(x)-2*x-8)/(x*log(x)^3+(4*x
^2+16*x)*log(x)^2+(4*x^3+32*x^2+64*x)*log(x)),x, algorithm="fricas")

[Out]

-(x^2 + x + log(log(x)) + 8)/(2*x + log(x) + 8)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+2*x)*ln(x)*ln(ln(x))+(-2*x**2-x)*ln(x)**2+(-2*x**3-15*x**2+9*x+7)*ln(x)-2*x-8)/(x*ln(x)**3+(4*x*
*2+16*x)*ln(x)**2+(4*x**3+32*x**2+64*x)*ln(x)),x)

[Out]

Exception raised: TypeError >> '>' not supported between instances of 'Poly' and 'int'

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Giac [A]
time = 0.41, size = 32, normalized size = 1.33 \begin {gather*} -\frac {x^{2} + x + 8}{2 \, x + \log \left (x\right ) + 8} - \frac {\log \left (\log \left (x\right )\right )}{2 \, x + \log \left (x\right ) + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+2*x)*log(x)*log(log(x))+(-2*x^2-x)*log(x)^2+(-2*x^3-15*x^2+9*x+7)*log(x)-2*x-8)/(x*log(x)^3+(4*x
^2+16*x)*log(x)^2+(4*x^3+32*x^2+64*x)*log(x)),x, algorithm="giac")

[Out]

-(x^2 + x + 8)/(2*x + log(x) + 8) - log(log(x))/(2*x + log(x) + 8)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {2\,x+{\ln \left (x\right )}^2\,\left (2\,x^2+x\right )-\ln \left (x\right )\,\left (-2\,x^3-15\,x^2+9\,x+7\right )-\ln \left (\ln \left (x\right )\right )\,\ln \left (x\right )\,\left (2\,x+1\right )+8}{x\,{\ln \left (x\right )}^3+\left (4\,x^2+16\,x\right )\,{\ln \left (x\right )}^2+\left (4\,x^3+32\,x^2+64\,x\right )\,\ln \left (x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + log(x)^2*(x + 2*x^2) - log(x)*(9*x - 15*x^2 - 2*x^3 + 7) - log(log(x))*log(x)*(2*x + 1) + 8)/(log(
x)^2*(16*x + 4*x^2) + x*log(x)^3 + log(x)*(64*x + 32*x^2 + 4*x^3)),x)

[Out]

int(-(2*x + log(x)^2*(x + 2*x^2) - log(x)*(9*x - 15*x^2 - 2*x^3 + 7) - log(log(x))*log(x)*(2*x + 1) + 8)/(log(
x)^2*(16*x + 4*x^2) + x*log(x)^3 + log(x)*(64*x + 32*x^2 + 4*x^3)), x)

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