3.81.64 \(\int \frac {-20 \log (x)+x \log ^2(x)+10 \log (x^2)}{2 x \log ^2(x)} \, dx\)

Optimal. Leaf size=16 \[ \frac {1}{2} \left (x-\frac {10 \log \left (x^2\right )}{\log (x)}\right ) \]

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Rubi [A]  time = 0.21, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {12, 6742, 2302, 29, 30, 2366} \begin {gather*} \frac {x}{2}-\frac {5 \log \left (x^2\right )}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20*Log[x] + x*Log[x]^2 + 10*Log[x^2])/(2*x*Log[x]^2),x]

[Out]

x/2 - (5*Log[x^2])/Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {-20 \log (x)+x \log ^2(x)+10 \log \left (x^2\right )}{x \log ^2(x)} \, dx\\ &=\frac {1}{2} \int \left (\frac {-20+x \log (x)}{x \log (x)}+\frac {10 \log \left (x^2\right )}{x \log ^2(x)}\right ) \, dx\\ &=\frac {1}{2} \int \frac {-20+x \log (x)}{x \log (x)} \, dx+5 \int \frac {\log \left (x^2\right )}{x \log ^2(x)} \, dx\\ &=-\frac {5 \log \left (x^2\right )}{\log (x)}+\frac {1}{2} \int \left (1-\frac {20}{x \log (x)}\right ) \, dx+10 \int \frac {1}{x \log (x)} \, dx\\ &=\frac {x}{2}-\frac {5 \log \left (x^2\right )}{\log (x)}-10 \int \frac {1}{x \log (x)} \, dx+10 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=\frac {x}{2}-\frac {5 \log \left (x^2\right )}{\log (x)}+10 \log (\log (x))-10 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=\frac {x}{2}-\frac {5 \log \left (x^2\right )}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 1.31 \begin {gather*} \frac {x}{2}-\frac {5 \left (-2 \log (x)+\log \left (x^2\right )\right )}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20*Log[x] + x*Log[x]^2 + 10*Log[x^2])/(2*x*Log[x]^2),x]

[Out]

x/2 - (5*(-2*Log[x] + Log[x^2]))/Log[x]

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fricas [A]  time = 0.68, size = 3, normalized size = 0.19 \begin {gather*} \frac {1}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(10*log(x^2)+x*log(x)^2-20*log(x))/x/log(x)^2,x, algorithm="fricas")

[Out]

1/2*x

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giac [A]  time = 0.23, size = 3, normalized size = 0.19 \begin {gather*} \frac {1}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(10*log(x^2)+x*log(x)^2-20*log(x))/x/log(x)^2,x, algorithm="giac")

[Out]

1/2*x

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maple [A]  time = 0.06, size = 15, normalized size = 0.94

method	result	size
default	\(\frac {x}{2}-\frac {5 \ln \left (x^{2}\right )}{\ln \relax (x )}\)	\(15\)
norman	\(\frac {\frac {x \ln \relax (x )}{2}-5 \ln \left (x^{2}\right )}{\ln \relax (x )}\)	\(18\)
risch	\(\frac {x}{2}+\frac {5 i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (\mathrm {csgn}\left (i x \right )^{2}-2 \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )+\mathrm {csgn}\left (i x^{2}\right )^{2}\right )}{2 \ln \relax (x )}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(10*ln(x^2)+x*ln(x)^2-20*ln(x))/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*x-5/ln(x)*ln(x^2)

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maxima [A]  time = 0.36, size = 14, normalized size = 0.88 \begin {gather*} \frac {1}{2} \, x - \frac {5 \, \log \left (x^{2}\right )}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(10*log(x^2)+x*log(x)^2-20*log(x))/x/log(x)^2,x, algorithm="maxima")

[Out]

1/2*x - 5*log(x^2)/log(x)

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mupad [B]  time = 5.38, size = 14, normalized size = 0.88 \begin {gather*} \frac {x}{2}-\frac {5\,\ln \left (x^2\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*log(x^2) - 10*log(x) + (x*log(x)^2)/2)/(x*log(x)^2),x)

[Out]

x/2 - (5*log(x^2))/log(x)

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sympy [A]  time = 0.20, size = 2, normalized size = 0.12 \begin {gather*} \frac {x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(10*ln(x**2)+x*ln(x)**2-20*ln(x))/x/ln(x)**2,x)

[Out]

x/2

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