3.96.12 \(\int \frac {-5 \log (x)+(1+x) \log ^2(x)+(-20-4 x \log ^2(x)) \log ^3(\frac {-5+(1+x) \log (x)}{\log (x)})}{-5 x \log (x)+(x+x^2) \log ^2(x)} \, dx\)

Optimal. Leaf size=28 \[ 2 \left (-3+\frac {1}{2 e}\right )+\log (x)-\log ^4\left (1+x-\frac {5}{\log (x)}\right ) \]

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Rubi [A]  time = 0.49, antiderivative size = 17, normalized size of antiderivative = 0.61, number of steps used = 4, number of rules used = 3, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {6741, 6742, 6686} \begin {gather*} \log (x)-\log ^4\left (x-\frac {5}{\log (x)}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5*Log[x] + (1 + x)*Log[x]^2 + (-20 - 4*x*Log[x]^2)*Log[(-5 + (1 + x)*Log[x])/Log[x]]^3)/(-5*x*Log[x] + (
x + x^2)*Log[x]^2),x]

[Out]

Log[x] - Log[1 + x - 5/Log[x]]^4

Rule 6686

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

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 \log (x)-(1+x) \log ^2(x)-\left (-20-4 x \log ^2(x)\right ) \log ^3\left (\frac {-5+(1+x) \log (x)}{\log (x)}\right )}{x \log (x) (5-\log (x)-x \log (x))} \, dx\\ &=\int \left (\frac {1}{x}-\frac {4 \left (5+x \log ^2(x)\right ) \log ^3\left (1+x-\frac {5}{\log (x)}\right )}{x \log (x) (-5+\log (x)+x \log (x))}\right ) \, dx\\ &=\log (x)-4 \int \frac {\left (5+x \log ^2(x)\right ) \log ^3\left (1+x-\frac {5}{\log (x)}\right )}{x \log (x) (-5+\log (x)+x \log (x))} \, dx\\ &=\log (x)-\log ^4\left (1+x-\frac {5}{\log (x)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 17, normalized size = 0.61 \begin {gather*} \log (x)-\log ^4\left (1+x-\frac {5}{\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*Log[x] + (1 + x)*Log[x]^2 + (-20 - 4*x*Log[x]^2)*Log[(-5 + (1 + x)*Log[x])/Log[x]]^3)/(-5*x*Log[
x] + (x + x^2)*Log[x]^2),x]

[Out]

Log[x] - Log[1 + x - 5/Log[x]]^4

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fricas [A]  time = 0.62, size = 21, normalized size = 0.75 \begin {gather*} -\log \left (\frac {{\left (x + 1\right )} \log \relax (x) - 5}{\log \relax (x)}\right )^{4} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(x)^2-20)*log((log(x)*(x+1)-5)/log(x))^3+(x+1)*log(x)^2-5*log(x))/((x^2+x)*log(x)^2-5*x*lo
g(x)),x, algorithm="fricas")

[Out]

-log(((x + 1)*log(x) - 5)/log(x))^4 + log(x)

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giac [B]  time = 0.42, size = 72, normalized size = 2.57 \begin {gather*} -4 \, {\left (\log \left (x \log \relax (x) + \log \relax (x) - 5\right ) - \log \left (\log \relax (x)\right )\right )} \log \left (x \log \relax (x) + \log \relax (x) - 5\right )^{3} - 6 \, \log \left (x \log \relax (x) + \log \relax (x) - 5\right )^{2} \log \left (\log \relax (x)\right )^{2} + 4 \, \log \left (x \log \relax (x) + \log \relax (x) - 5\right ) \log \left (\log \relax (x)\right )^{3} - \log \left (\log \relax (x)\right )^{4} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(x)^2-20)*log((log(x)*(x+1)-5)/log(x))^3+(x+1)*log(x)^2-5*log(x))/((x^2+x)*log(x)^2-5*x*lo
g(x)),x, algorithm="giac")

[Out]

-4*(log(x*log(x) + log(x) - 5) - log(log(x)))*log(x*log(x) + log(x) - 5)^3 - 6*log(x*log(x) + log(x) - 5)^2*lo
g(log(x))^2 + 4*log(x*log(x) + log(x) - 5)*log(log(x))^3 - log(log(x))^4 + log(x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (-4 x \ln \relax (x )^{2}-20\right ) \ln \left (\frac {\ln \relax (x ) \left (x +1\right )-5}{\ln \relax (x )}\right )^{3}+\left (x +1\right ) \ln \relax (x )^{2}-5 \ln \relax (x )}{\left (x^{2}+x \right ) \ln \relax (x )^{2}-5 x \ln \relax (x )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x*ln(x)^2-20)*ln((ln(x)*(x+1)-5)/ln(x))^3+(x+1)*ln(x)^2-5*ln(x))/((x^2+x)*ln(x)^2-5*x*ln(x)),x)

[Out]

int(((-4*x*ln(x)^2-20)*ln((ln(x)*(x+1)-5)/ln(x))^3+(x+1)*ln(x)^2-5*ln(x))/((x^2+x)*ln(x)^2-5*x*ln(x)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {4 \, {\left (x \log \relax (x)^{2} + 5\right )} \log \left (\frac {{\left (x + 1\right )} \log \relax (x) - 5}{\log \relax (x)}\right )^{3} - {\left (x + 1\right )} \log \relax (x)^{2} + 5 \, \log \relax (x)}{{\left (x^{2} + x\right )} \log \relax (x)^{2} - 5 \, x \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(x)^2-20)*log((log(x)*(x+1)-5)/log(x))^3+(x+1)*log(x)^2-5*log(x))/((x^2+x)*log(x)^2-5*x*lo
g(x)),x, algorithm="maxima")

[Out]

-integrate((4*(x*log(x)^2 + 5)*log(((x + 1)*log(x) - 5)/log(x))^3 - (x + 1)*log(x)^2 + 5*log(x))/((x^2 + x)*lo
g(x)^2 - 5*x*log(x)), x)

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mupad [B]  time = 7.52, size = 21, normalized size = 0.75 \begin {gather*} \ln \relax (x)-{\ln \left (\frac {\ln \relax (x)\,\left (x+1\right )-5}{\ln \relax (x)}\right )}^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*log(x) - log(x)^2*(x + 1) + log((log(x)*(x + 1) - 5)/log(x))^3*(4*x*log(x)^2 + 20))/(log(x)^2*(x + x^2
) - 5*x*log(x)),x)

[Out]

log(x) - log((log(x)*(x + 1) - 5)/log(x))^4

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sympy [A]  time = 0.41, size = 17, normalized size = 0.61 \begin {gather*} \log {\relax (x )} - \log {\left (\frac {\left (x + 1\right ) \log {\relax (x )} - 5}{\log {\relax (x )}} \right )}^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*ln(x)**2-20)*ln((ln(x)*(x+1)-5)/ln(x))**3+(x+1)*ln(x)**2-5*ln(x))/((x**2+x)*ln(x)**2-5*x*ln(x
)),x)

[Out]

log(x) - log(((x + 1)*log(x) - 5)/log(x))**4

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