3.52.31 \(\int \frac {-4-9 x+8 x^2+(-7+8 x) \log (2 x^2)+(-2 x-2 \log (2 x^2)) \log (x+\log (2 x^2))}{49 x^3-56 x^4+16 x^5+(49 x^2-56 x^3+16 x^4) \log (2 x^2)+(28 x^3-16 x^4+(28 x^2-16 x^3) \log (2 x^2)) \log (x+\log (2 x^2))+(4 x^3+4 x^2 \log (2 x^2)) \log ^2(x+\log (2 x^2))} \, dx\) [5131]

Optimal. Leaf size=25 \[ \frac {1}{2 x \left (\frac {7}{2}-2 x+\log \left (x+\log \left (2 x^2\right )\right )\right )} \]

[Out]

1/2/x/(ln(ln(2*x^2)+x)-2*x+7/2)

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Rubi [A]
time = 0.37, antiderivative size = 22, normalized size of antiderivative = 0.88, number of steps used = 2, number of rules used = 2, integrand size = 154, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {6820, 6819} \begin {gather*} \frac {1}{x \left (2 \log \left (\log \left (2 x^2\right )+x\right )-4 x+7\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 - 9*x + 8*x^2 + (-7 + 8*x)*Log[2*x^2] + (-2*x - 2*Log[2*x^2])*Log[x + Log[2*x^2]])/(49*x^3 - 56*x^4 +
16*x^5 + (49*x^2 - 56*x^3 + 16*x^4)*Log[2*x^2] + (28*x^3 - 16*x^4 + (28*x^2 - 16*x^3)*Log[2*x^2])*Log[x + Log[
2*x^2]] + (4*x^3 + 4*x^2*Log[2*x^2])*Log[x + Log[2*x^2]]^2),x]

[Out]

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

Rule 6819

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[q*y^(m +
1)*(z^(m + 1)/(m + 1)), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, x] && NeQ[m, -1]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

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

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Mathematica [A]
time = 0.21, size = 22, normalized size = 0.88 \begin {gather*} \frac {1}{x \left (7-4 x+2 \log \left (x+\log \left (2 x^2\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - 9*x + 8*x^2 + (-7 + 8*x)*Log[2*x^2] + (-2*x - 2*Log[2*x^2])*Log[x + Log[2*x^2]])/(49*x^3 - 56*
x^4 + 16*x^5 + (49*x^2 - 56*x^3 + 16*x^4)*Log[2*x^2] + (28*x^3 - 16*x^4 + (28*x^2 - 16*x^3)*Log[2*x^2])*Log[x
+ Log[2*x^2]] + (4*x^3 + 4*x^2*Log[2*x^2])*Log[x + Log[2*x^2]]^2),x]

[Out]

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

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Maple [F]
time = 7.38, size = 0, normalized size = 0.00 \[\int \frac {\left (-2 \ln \left (2 x^{2}\right )-2 x \right ) \ln \left (\ln \left (2 x^{2}\right )+x \right )+\left (8 x -7\right ) \ln \left (2 x^{2}\right )+8 x^{2}-9 x -4}{\left (4 x^{2} \ln \left (2 x^{2}\right )+4 x^{3}\right ) \ln \left (\ln \left (2 x^{2}\right )+x \right )^{2}+\left (\left (-16 x^{3}+28 x^{2}\right ) \ln \left (2 x^{2}\right )-16 x^{4}+28 x^{3}\right ) \ln \left (\ln \left (2 x^{2}\right )+x \right )+\left (16 x^{4}-56 x^{3}+49 x^{2}\right ) \ln \left (2 x^{2}\right )+16 x^{5}-56 x^{4}+49 x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*ln(2*x^2)-2*x)*ln(ln(2*x^2)+x)+(8*x-7)*ln(2*x^2)+8*x^2-9*x-4)/((4*x^2*ln(2*x^2)+4*x^3)*ln(ln(2*x^2)+x
)^2+((-16*x^3+28*x^2)*ln(2*x^2)-16*x^4+28*x^3)*ln(ln(2*x^2)+x)+(16*x^4-56*x^3+49*x^2)*ln(2*x^2)+16*x^5-56*x^4+
49*x^3),x)

[Out]

int(((-2*ln(2*x^2)-2*x)*ln(ln(2*x^2)+x)+(8*x-7)*ln(2*x^2)+8*x^2-9*x-4)/((4*x^2*ln(2*x^2)+4*x^3)*ln(ln(2*x^2)+x
)^2+((-16*x^3+28*x^2)*ln(2*x^2)-16*x^4+28*x^3)*ln(ln(2*x^2)+x)+(16*x^4-56*x^3+49*x^2)*ln(2*x^2)+16*x^5-56*x^4+
49*x^3),x)

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Maxima [A]
time = 0.53, size = 25, normalized size = 1.00 \begin {gather*} -\frac {1}{4 \, x^{2} - 2 \, x \log \left (x + \log \left (2\right ) + 2 \, \log \left (x\right )\right ) - 7 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(2*x^2)-2*x)*log(log(2*x^2)+x)+(8*x-7)*log(2*x^2)+8*x^2-9*x-4)/((4*x^2*log(2*x^2)+4*x^3)*log
(log(2*x^2)+x)^2+((-16*x^3+28*x^2)*log(2*x^2)-16*x^4+28*x^3)*log(log(2*x^2)+x)+(16*x^4-56*x^3+49*x^2)*log(2*x^
2)+16*x^5-56*x^4+49*x^3),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 25, normalized size = 1.00 \begin {gather*} -\frac {1}{4 \, x^{2} - 2 \, x \log \left (x + \log \left (2 \, x^{2}\right )\right ) - 7 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(2*x^2)-2*x)*log(log(2*x^2)+x)+(8*x-7)*log(2*x^2)+8*x^2-9*x-4)/((4*x^2*log(2*x^2)+4*x^3)*log
(log(2*x^2)+x)^2+((-16*x^3+28*x^2)*log(2*x^2)-16*x^4+28*x^3)*log(log(2*x^2)+x)+(16*x^4-56*x^3+49*x^2)*log(2*x^
2)+16*x^5-56*x^4+49*x^3),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.12, size = 22, normalized size = 0.88 \begin {gather*} \frac {1}{- 4 x^{2} + 2 x \log {\left (x + \log {\left (2 x^{2} \right )} \right )} + 7 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*ln(2*x**2)-2*x)*ln(ln(2*x**2)+x)+(8*x-7)*ln(2*x**2)+8*x**2-9*x-4)/((4*x**2*ln(2*x**2)+4*x**3)*l
n(ln(2*x**2)+x)**2+((-16*x**3+28*x**2)*ln(2*x**2)-16*x**4+28*x**3)*ln(ln(2*x**2)+x)+(16*x**4-56*x**3+49*x**2)*
ln(2*x**2)+16*x**5-56*x**4+49*x**3),x)

[Out]

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

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Giac [A]
time = 0.54, size = 25, normalized size = 1.00 \begin {gather*} -\frac {1}{4 \, x^{2} - 2 \, x \log \left (x + \log \left (2 \, x^{2}\right )\right ) - 7 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(2*x^2)-2*x)*log(log(2*x^2)+x)+(8*x-7)*log(2*x^2)+8*x^2-9*x-4)/((4*x^2*log(2*x^2)+4*x^3)*log
(log(2*x^2)+x)^2+((-16*x^3+28*x^2)*log(2*x^2)-16*x^4+28*x^3)*log(log(2*x^2)+x)+(16*x^4-56*x^3+49*x^2)*log(2*x^
2)+16*x^5-56*x^4+49*x^3),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 3.48, size = 22, normalized size = 0.88 \begin {gather*} \frac {1}{x\,\left (2\,\ln \left (x+\ln \left (2\,x^2\right )\right )-4\,x+7\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(9*x - log(2*x^2)*(8*x - 7) + log(x + log(2*x^2))*(2*x + 2*log(2*x^2)) - 8*x^2 + 4)/(log(2*x^2)*(49*x^2 -
56*x^3 + 16*x^4) + log(x + log(2*x^2))^2*(4*x^3 + 4*x^2*log(2*x^2)) + 49*x^3 - 56*x^4 + 16*x^5 + log(x + log(
2*x^2))*(log(2*x^2)*(28*x^2 - 16*x^3) + 28*x^3 - 16*x^4)),x)

[Out]

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

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