3.54.4 \(\int \frac {(-8+2 x) \log (2)+4 \log (2) \log (\frac {8}{x^2})+(5 x^2-5 x^2 \log (2)) \log ^2(\frac {8}{x^2})}{5 x^2 \log (2) \log ^2(\frac {8}{x^2})} \, dx\)

Optimal. Leaf size=29 \[ -1-x+\frac {x}{\log (2)}+\frac {-4+x}{5 x \log \left (\frac {8}{x^2}\right )} \]

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Rubi [C]  time = 0.35, antiderivative size = 106, normalized size of antiderivative = 3.66, number of steps used = 12, number of rules used = 8, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 6742, 2353, 2306, 2310, 2178, 2302, 30} \begin {gather*} \frac {\text {Ei}\left (\frac {1}{2} \log \left (\frac {8}{x^2}\right )\right )}{5 \sqrt {2} \sqrt {\frac {1}{x^2}} x}-\frac {\log (16) \text {Ei}\left (\frac {1}{2} \log \left (\frac {8}{x^2}\right )\right )}{20 \sqrt {2} \sqrt {\frac {1}{x^2}} x \log (2)}+\frac {1}{5 \log \left (\frac {8}{x^2}\right )}-\frac {4}{5 x \log \left (\frac {8}{x^2}\right )}+\frac {x (1-\log (2))}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-8 + 2*x)*Log[2] + 4*Log[2]*Log[8/x^2] + (5*x^2 - 5*x^2*Log[2])*Log[8/x^2]^2)/(5*x^2*Log[2]*Log[8/x^2]^2
),x]

[Out]

ExpIntegralEi[Log[8/x^2]/2]/(5*Sqrt[2]*Sqrt[x^(-2)]*x) + (x*(1 - Log[2]))/Log[2] - (ExpIntegralEi[Log[8/x^2]/2
]*Log[16])/(20*Sqrt[2]*Sqrt[x^(-2)]*x*Log[2]) + 1/(5*Log[8/x^2]) - 4/(5*x*Log[8/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 30

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

Rule 2178

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

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 2306

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

Rule 2310

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

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

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 {\int \frac {(-8+2 x) \log (2)+4 \log (2) \log \left (\frac {8}{x^2}\right )+\left (5 x^2-5 x^2 \log (2)\right ) \log ^2\left (\frac {8}{x^2}\right )}{x^2 \log ^2\left (\frac {8}{x^2}\right )} \, dx}{5 \log (2)}\\ &=\frac {\int \left (-5 (-1+\log (2))+\frac {2 (-4+x) \log (2)}{x^2 \log ^2\left (\frac {8}{x^2}\right )}+\frac {\log (16)}{x^2 \log \left (\frac {8}{x^2}\right )}\right ) \, dx}{5 \log (2)}\\ &=\frac {x (1-\log (2))}{\log (2)}+\frac {2}{5} \int \frac {-4+x}{x^2 \log ^2\left (\frac {8}{x^2}\right )} \, dx+\frac {\log (16) \int \frac {1}{x^2 \log \left (\frac {8}{x^2}\right )} \, dx}{5 \log (2)}\\ &=\frac {x (1-\log (2))}{\log (2)}+\frac {2}{5} \int \left (-\frac {4}{x^2 \log ^2\left (\frac {8}{x^2}\right )}+\frac {1}{x \log ^2\left (\frac {8}{x^2}\right )}\right ) \, dx-\frac {\log (16) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (\frac {8}{x^2}\right )\right )}{20 \sqrt {2} \sqrt {\frac {1}{x^2}} x \log (2)}\\ &=\frac {x (1-\log (2))}{\log (2)}-\frac {\text {Ei}\left (\frac {1}{2} \log \left (\frac {8}{x^2}\right )\right ) \log (16)}{20 \sqrt {2} \sqrt {\frac {1}{x^2}} x \log (2)}+\frac {2}{5} \int \frac {1}{x \log ^2\left (\frac {8}{x^2}\right )} \, dx-\frac {8}{5} \int \frac {1}{x^2 \log ^2\left (\frac {8}{x^2}\right )} \, dx\\ &=\frac {x (1-\log (2))}{\log (2)}-\frac {\text {Ei}\left (\frac {1}{2} \log \left (\frac {8}{x^2}\right )\right ) \log (16)}{20 \sqrt {2} \sqrt {\frac {1}{x^2}} x \log (2)}-\frac {4}{5 x \log \left (\frac {8}{x^2}\right )}-\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {8}{x^2}\right )\right )-\frac {4}{5} \int \frac {1}{x^2 \log \left (\frac {8}{x^2}\right )} \, dx\\ &=\frac {x (1-\log (2))}{\log (2)}-\frac {\text {Ei}\left (\frac {1}{2} \log \left (\frac {8}{x^2}\right )\right ) \log (16)}{20 \sqrt {2} \sqrt {\frac {1}{x^2}} x \log (2)}+\frac {1}{5 \log \left (\frac {8}{x^2}\right )}-\frac {4}{5 x \log \left (\frac {8}{x^2}\right )}+\frac {\operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (\frac {8}{x^2}\right )\right )}{5 \sqrt {2} \sqrt {\frac {1}{x^2}} x}\\ &=\frac {\text {Ei}\left (\frac {1}{2} \log \left (\frac {8}{x^2}\right )\right )}{5 \sqrt {2} \sqrt {\frac {1}{x^2}} x}+\frac {x (1-\log (2))}{\log (2)}-\frac {\text {Ei}\left (\frac {1}{2} \log \left (\frac {8}{x^2}\right )\right ) \log (16)}{20 \sqrt {2} \sqrt {\frac {1}{x^2}} x \log (2)}+\frac {1}{5 \log \left (\frac {8}{x^2}\right )}-\frac {4}{5 x \log \left (\frac {8}{x^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 33, normalized size = 1.14 \begin {gather*} \frac {-5 x (-1+\log (2))+\frac {(-4+x) \log (2)}{x \log \left (\frac {8}{x^2}\right )}}{5 \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-8 + 2*x)*Log[2] + 4*Log[2]*Log[8/x^2] + (5*x^2 - 5*x^2*Log[2])*Log[8/x^2]^2)/(5*x^2*Log[2]*Log[8/
x^2]^2),x]

[Out]

(-5*x*(-1 + Log[2]) + ((-4 + x)*Log[2])/(x*Log[8/x^2]))/(5*Log[2])

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fricas [A]  time = 0.91, size = 44, normalized size = 1.52 \begin {gather*} \frac {{\left (x - 4\right )} \log \relax (2) - 5 \, {\left (x^{2} \log \relax (2) - x^{2}\right )} \log \left (\frac {8}{x^{2}}\right )}{5 \, x \log \relax (2) \log \left (\frac {8}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x^2*log(2)+5*x^2)*log(8/x^2)^2+4*log(2)*log(8/x^2)+(2*x-8)*log(2))/x^2/log(2)/log(8/x^2)^2,
x, algorithm="fricas")

[Out]

1/5*((x - 4)*log(2) - 5*(x^2*log(2) - x^2)*log(8/x^2))/(x*log(2)*log(8/x^2))

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giac [A]  time = 1.17, size = 40, normalized size = 1.38 \begin {gather*} -\frac {5 \, x {\left (\log \relax (2) - 1\right )} - \frac {x \log \relax (2) - 4 \, \log \relax (2)}{3 \, x \log \relax (2) - x \log \left (x^{2}\right )}}{5 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x^2*log(2)+5*x^2)*log(8/x^2)^2+4*log(2)*log(8/x^2)+(2*x-8)*log(2))/x^2/log(2)/log(8/x^2)^2,
x, algorithm="giac")

[Out]

-1/5*(5*x*(log(2) - 1) - (x*log(2) - 4*log(2))/(3*x*log(2) - x*log(x^2)))/log(2)

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maple [A]  time = 0.19, size = 27, normalized size = 0.93

method	result	size
risch	\(-x +\frac {x}{\ln \relax (2)}+\frac {x -4}{5 x \ln \left (\frac {8}{x^{2}}\right )}\)	\(27\)
norman	\(\frac {-\frac {4}{5}+\frac {x}{5}-\frac {\left (\ln \relax (2)-1\right ) x^{2} \ln \left (\frac {8}{x^{2}}\right )}{\ln \relax (2)}}{\ln \left (\frac {8}{x^{2}}\right ) x}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-5*x^2*ln(2)+5*x^2)*ln(8/x^2)^2+4*ln(2)*ln(8/x^2)+(2*x-8)*ln(2))/x^2/ln(2)/ln(8/x^2)^2,x,method=_RET
URNVERBOSE)

[Out]

-x+x/ln(2)+1/5*(x-4)/x/ln(8/x^2)

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maxima [A]  time = 0.48, size = 39, normalized size = 1.34 \begin {gather*} -\frac {5 \, x \log \relax (2) - 5 \, x - \frac {x \log \relax (2) - 4 \, \log \relax (2)}{3 \, x \log \relax (2) - 2 \, x \log \relax (x)}}{5 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x^2*log(2)+5*x^2)*log(8/x^2)^2+4*log(2)*log(8/x^2)+(2*x-8)*log(2))/x^2/log(2)/log(8/x^2)^2,
x, algorithm="maxima")

[Out]

-1/5*(5*x*log(2) - 5*x - (x*log(2) - 4*log(2))/(3*x*log(2) - 2*x*log(x)))/log(2)

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mupad [B]  time = 3.53, size = 42, normalized size = 1.45 \begin {gather*} -\frac {x\,\left (\ln \left (32\right )-5\right )}{5\,\ln \relax (2)}-\frac {\frac {x\,\ln \left (16\right )}{5}-\frac {x^2\,\ln \relax (2)}{5}}{x^2\,\ln \relax (2)\,\ln \left (\frac {8}{x^2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(2)*(2*x - 8))/5 + (4*log(2)*log(8/x^2))/5 - (log(8/x^2)^2*(5*x^2*log(2) - 5*x^2))/5)/(x^2*log(2)*log
(8/x^2)^2),x)

[Out]

- (x*(log(32) - 5))/(5*log(2)) - ((x*log(16))/5 - (x^2*log(2))/5)/(x^2*log(2)*log(8/x^2))

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sympy [A]  time = 0.13, size = 22, normalized size = 0.76 \begin {gather*} \frac {x \left (1 - \log {\relax (2 )}\right )}{\log {\relax (2 )}} + \frac {x - 4}{5 x \log {\left (\frac {8}{x^{2}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-5*x**2*ln(2)+5*x**2)*ln(8/x**2)**2+4*ln(2)*ln(8/x**2)+(2*x-8)*ln(2))/x**2/ln(2)/ln(8/x**2)**2
,x)

[Out]

x*(1 - log(2))/log(2) + (x - 4)/(5*x*log(8/x**2))

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