3.94.20 \(\int \frac {x \log (\frac {4}{x^2})+(23+x) \log ^3(\frac {4}{x^2})+(4 x-25 \log ^3(\frac {4}{x^2})) \log (x)}{x^2 \log ^3(\frac {4}{x^2})} \, dx\)

Optimal. Leaf size=22 \[ \frac {2+\left (25+x+\frac {x}{\log ^2\left (\frac {4}{x^2}\right )}\right ) \log (x)}{x} \]

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Rubi [A]  time = 0.73, antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 18, number of rules used = 9, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.180, Rules used = {6688, 14, 43, 2302, 30, 6742, 2304, 2366, 12} \begin {gather*} \frac {\log (x)}{\log ^2\left (\frac {4}{x^2}\right )}+\frac {2}{x}+\frac {25 \log (x)}{x}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*Log[4/x^2] + (23 + x)*Log[4/x^2]^3 + (4*x - 25*Log[4/x^2]^3)*Log[x])/(x^2*Log[4/x^2]^3),x]

[Out]

2/x + Log[x] + (25*Log[x])/x + Log[x]/Log[4/x^2]^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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

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 6688

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

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 {23+x+\frac {x}{\log ^2\left (\frac {4}{x^2}\right )}-25 \log (x)+\frac {4 x \log (x)}{\log ^3\left (\frac {4}{x^2}\right )}}{x^2} \, dx\\ &=\int \left (\frac {x+23 \log ^2\left (\frac {4}{x^2}\right )+x \log ^2\left (\frac {4}{x^2}\right )}{x^2 \log ^2\left (\frac {4}{x^2}\right )}+\frac {\left (4 x-25 \log ^3\left (\frac {4}{x^2}\right )\right ) \log (x)}{x^2 \log ^3\left (\frac {4}{x^2}\right )}\right ) \, dx\\ &=\int \frac {x+23 \log ^2\left (\frac {4}{x^2}\right )+x \log ^2\left (\frac {4}{x^2}\right )}{x^2 \log ^2\left (\frac {4}{x^2}\right )} \, dx+\int \frac {\left (4 x-25 \log ^3\left (\frac {4}{x^2}\right )\right ) \log (x)}{x^2 \log ^3\left (\frac {4}{x^2}\right )} \, dx\\ &=\int \frac {23+x+\frac {x}{\log ^2\left (\frac {4}{x^2}\right )}}{x^2} \, dx+\int \frac {\left (-25+\frac {4 x}{\log ^3\left (\frac {4}{x^2}\right )}\right ) \log (x)}{x^2} \, dx\\ &=\int \left (\frac {23+x}{x^2}+\frac {1}{x \log ^2\left (\frac {4}{x^2}\right )}\right ) \, dx+\int \left (-\frac {25 \log (x)}{x^2}+\frac {4 \log (x)}{x \log ^3\left (\frac {4}{x^2}\right )}\right ) \, dx\\ &=4 \int \frac {\log (x)}{x \log ^3\left (\frac {4}{x^2}\right )} \, dx-25 \int \frac {\log (x)}{x^2} \, dx+\int \frac {23+x}{x^2} \, dx+\int \frac {1}{x \log ^2\left (\frac {4}{x^2}\right )} \, dx\\ &=\frac {25}{x}+\frac {25 \log (x)}{x}+\frac {\log (x)}{\log ^2\left (\frac {4}{x^2}\right )}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {4}{x^2}\right )\right )-4 \int \frac {1}{4 x \log ^2\left (\frac {4}{x^2}\right )} \, dx+\int \left (\frac {23}{x^2}+\frac {1}{x}\right ) \, dx\\ &=\frac {2}{x}+\frac {1}{2 \log \left (\frac {4}{x^2}\right )}+\log (x)+\frac {25 \log (x)}{x}+\frac {\log (x)}{\log ^2\left (\frac {4}{x^2}\right )}-\int \frac {1}{x \log ^2\left (\frac {4}{x^2}\right )} \, dx\\ &=\frac {2}{x}+\frac {1}{2 \log \left (\frac {4}{x^2}\right )}+\log (x)+\frac {25 \log (x)}{x}+\frac {\log (x)}{\log ^2\left (\frac {4}{x^2}\right )}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {4}{x^2}\right )\right )\\ &=\frac {2}{x}+\log (x)+\frac {25 \log (x)}{x}+\frac {\log (x)}{\log ^2\left (\frac {4}{x^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 24, normalized size = 1.09 \begin {gather*} \frac {2}{x}+\left (1+\frac {25}{x}+\frac {1}{\log ^2\left (\frac {4}{x^2}\right )}\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*Log[4/x^2] + (23 + x)*Log[4/x^2]^3 + (4*x - 25*Log[4/x^2]^3)*Log[x])/(x^2*Log[4/x^2]^3),x]

[Out]

2/x + (1 + 25/x + Log[4/x^2]^(-2))*Log[x]

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fricas [B]  time = 0.85, size = 55, normalized size = 2.50 \begin {gather*} -\frac {{\left (x + 25\right )} \log \left (\frac {4}{x^{2}}\right )^{3} - 2 \, {\left (25 \, \log \relax (2) + 2\right )} \log \left (\frac {4}{x^{2}}\right )^{2} - 2 \, x \log \relax (2) + x \log \left (\frac {4}{x^{2}}\right )}{2 \, x \log \left (\frac {4}{x^{2}}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*log(4/x^2)^3+4*x)*log(x)+(x+23)*log(4/x^2)^3+x*log(4/x^2))/x^2/log(4/x^2)^3,x, algorithm="fric
as")

[Out]

-1/2*((x + 25)*log(4/x^2)^3 - 2*(25*log(2) + 2)*log(4/x^2)^2 - 2*x*log(2) + x*log(4/x^2))/(x*log(4/x^2)^2)

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giac [A]  time = 0.41, size = 36, normalized size = 1.64 \begin {gather*} \frac {\log \relax (x)}{4 \, {\left (\log \relax (2)^{2} - 2 \, \log \relax (2) \log \relax (x) + \log \relax (x)^{2}\right )}} + \frac {25 \, \log \relax (x)}{x} + \frac {2}{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*log(4/x^2)^3+4*x)*log(x)+(x+23)*log(4/x^2)^3+x*log(4/x^2))/x^2/log(4/x^2)^3,x, algorithm="giac
")

[Out]

1/4*log(x)/(log(2)^2 - 2*log(2)*log(x) + log(x)^2) + 25*log(x)/x + 2/x + log(x)

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maple [C]  time = 0.27, size = 83, normalized size = 3.77

method	result	size
risch	\(\frac {25 \ln \relax (x )}{x}+\frac {x \ln \relax (x )+2}{x}+\frac {4 \ln \relax (x )}{\left (-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-4 \ln \relax (2)+4 \ln \relax (x )\right )^{2}}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-25*ln(4/x^2)^3+4*x)*ln(x)+(x+23)*ln(4/x^2)^3+x*ln(4/x^2))/x^2/ln(4/x^2)^3,x,method=_RETURNVERBOSE)

[Out]

25*ln(x)/x+(x*ln(x)+2)/x+4*ln(x)/(-I*Pi*csgn(I*x)^2*csgn(I*x^2)+2*I*Pi*csgn(I*x)*csgn(I*x^2)^2-I*Pi*csgn(I*x^2
)^3-4*ln(2)+4*ln(x))^2

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maxima [A]  time = 0.42, size = 26, normalized size = 1.18 \begin {gather*} \frac {25 \, \log \relax (x)}{x} + \frac {2}{x} + \frac {\log \relax (x)}{\log \left (\frac {4}{x^{2}}\right )^{2}} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*log(4/x^2)^3+4*x)*log(x)+(x+23)*log(4/x^2)^3+x*log(4/x^2))/x^2/log(4/x^2)^3,x, algorithm="maxi
ma")

[Out]

25*log(x)/x + 2/x + log(x)/log(4/x^2)^2 + log(x)

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mupad [B]  time = 7.26, size = 74, normalized size = 3.36 \begin {gather*} \ln \relax (x)-\frac {1}{4\,\ln \left (\frac {4}{x^2}\right )}+\frac {25\,\ln \relax (x)}{x}+\frac {\frac {\ln \left (\frac {4}{x^2}\right )}{4}+\ln \relax (x)}{4\,{\ln \relax (x)}^2-4\,\ln \relax (x)\,\left (\ln \left (\frac {4}{x^2}\right )+2\,\ln \relax (x)\right )+{\left (\ln \left (\frac {4}{x^2}\right )+2\,\ln \relax (x)\right )}^2}+\frac {2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(4/x^2)^3*(x + 23) + x*log(4/x^2) + log(x)*(4*x - 25*log(4/x^2)^3))/(x^2*log(4/x^2)^3),x)

[Out]

log(x) - 1/(4*log(4/x^2)) + (25*log(x))/x + (log(4/x^2)/4 + log(x))/(4*log(x)^2 - 4*log(x)*(log(4/x^2) + 2*log
(x)) + (log(4/x^2) + 2*log(x))^2) + 2/x

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sympy [A]  time = 0.33, size = 37, normalized size = 1.68 \begin {gather*} \log {\relax (x )} + \frac {\log {\relax (x )}}{4 \log {\relax (x )}^{2} - 8 \log {\relax (2 )} \log {\relax (x )} + 4 \log {\relax (2 )}^{2}} + \frac {25 \log {\relax (x )}}{x} + \frac {2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*ln(4/x**2)**3+4*x)*ln(x)+(x+23)*ln(4/x**2)**3+x*ln(4/x**2))/x**2/ln(4/x**2)**3,x)

[Out]

log(x) + log(x)/(4*log(x)**2 - 8*log(2)*log(x) + 4*log(2)**2) + 25*log(x)/x + 2/x

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