3.38.3 \(\int \frac {-2+\log (x^2)+16 x \log ^2(x^2)}{2 \log ^2(x^2)} \, dx\) [3703]

Optimal. Leaf size=19 \[ x^2 \left (4+\frac {1}{2 x \log \left (x^2\right )}\right ) \]

[Out]

x^2*(1/2/x/ln(x^2)+4)

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Rubi [A]
time = 0.08, antiderivative size = 17, normalized size of antiderivative = 0.89, number of steps used = 8, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 6874, 2334, 2337, 2209} \begin {gather*} 4 x^2+\frac {x}{2 \log \left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + Log[x^2] + 16*x*Log[x^2]^2)/(2*Log[x^2]^2),x]

[Out]

4*x^2 + x/(2*Log[x^2])

Rule 12

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

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b*Log[c*x^n])^(p + 1)/(b*n*(p + 1)))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2337

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

Rule 6874

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 {-2+\log \left (x^2\right )+16 x \log ^2\left (x^2\right )}{\log ^2\left (x^2\right )} \, dx\\ &=\frac {1}{2} \int \left (16 x-\frac {2}{\log ^2\left (x^2\right )}+\frac {1}{\log \left (x^2\right )}\right ) \, dx\\ &=4 x^2+\frac {1}{2} \int \frac {1}{\log \left (x^2\right )} \, dx-\int \frac {1}{\log ^2\left (x^2\right )} \, dx\\ &=4 x^2+\frac {x}{2 \log \left (x^2\right )}-\frac {1}{2} \int \frac {1}{\log \left (x^2\right )} \, dx+\frac {x \text {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (x^2\right )\right )}{4 \sqrt {x^2}}\\ &=4 x^2+\frac {x \text {Ei}\left (\frac {\log \left (x^2\right )}{2}\right )}{4 \sqrt {x^2}}+\frac {x}{2 \log \left (x^2\right )}-\frac {x \text {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (x^2\right )\right )}{4 \sqrt {x^2}}\\ &=4 x^2+\frac {x}{2 \log \left (x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.04, size = 17, normalized size = 0.89 \begin {gather*} 4 x^2+\frac {x}{2 \log \left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + Log[x^2] + 16*x*Log[x^2]^2)/(2*Log[x^2]^2),x]

[Out]

4*x^2 + x/(2*Log[x^2])

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Maple [A]
time = 2.80, size = 16, normalized size = 0.84

method	result	size
risch	\(4 x^{2}+\frac {x}{2 \ln \left (x^{2}\right )}\)	\(16\)
norman	\(\frac {\frac {x}{2}+4 x^{2} \ln \left (x^{2}\right )}{\ln \left (x^{2}\right )}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(16*x*ln(x^2)^2+ln(x^2)-2)/ln(x^2)^2,x,method=_RETURNVERBOSE)

[Out]

4*x^2+1/2*x/ln(x^2)

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Maxima [A]
time = 0.30, size = 13, normalized size = 0.68 \begin {gather*} 4 \, x^{2} + \frac {x}{4 \, \log \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(16*x*log(x^2)^2+log(x^2)-2)/log(x^2)^2,x, algorithm="maxima")

[Out]

4*x^2 + 1/4*x/log(x)

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Fricas [A]
time = 0.33, size = 19, normalized size = 1.00 \begin {gather*} \frac {8 \, x^{2} \log \left (x^{2}\right ) + x}{2 \, \log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(16*x*log(x^2)^2+log(x^2)-2)/log(x^2)^2,x, algorithm="fricas")

[Out]

1/2*(8*x^2*log(x^2) + x)/log(x^2)

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Sympy [A]
time = 0.03, size = 12, normalized size = 0.63 \begin {gather*} 4 x^{2} + \frac {x}{2 \log {\left (x^{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(16*x*ln(x**2)**2+ln(x**2)-2)/ln(x**2)**2,x)

[Out]

4*x**2 + x/(2*log(x**2))

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Giac [A]
time = 0.40, size = 15, normalized size = 0.79 \begin {gather*} 4 \, x^{2} + \frac {x}{2 \, \log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(16*x*log(x^2)^2+log(x^2)-2)/log(x^2)^2,x, algorithm="giac")

[Out]

4*x^2 + 1/2*x/log(x^2)

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Mupad [B]
time = 2.22, size = 15, normalized size = 0.79 \begin {gather*} \frac {x}{2\,\ln \left (x^2\right )}+4\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^2)/2 + 8*x*log(x^2)^2 - 1)/log(x^2)^2,x)

[Out]

x/(2*log(x^2)) + 4*x^2

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