3.48.60 \(\int \frac {-50+6 e^4-6 \log (4)+6 \log (x^2)+(-8+e^4-\log (4)+\log (x^2)) \log (-\frac {5 x}{-8+e^4-\log (4)+\log (x^2)})}{-24+3 e^4-3 \log (4)+3 \log (x^2)} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{3} x \left (5+\log \left (\frac {5 x}{8-e^4+\log (4)-\log \left (x^2\right )}\right )\right ) \]

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Rubi [C]  time = 0.44, antiderivative size = 187, normalized size of antiderivative = 6.68, number of steps used = 15, number of rules used = 10, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6741, 12, 6742, 2300, 2178, 2361, 15, 6482, 2549, 2360} \begin {gather*} \frac {2 e^{4-\frac {e^4}{2}} x \text {Ei}\left (\frac {1}{2} \left (\log \left (\frac {x^2}{4}\right )+e^4-8\right )\right )}{3 \sqrt {x^2}}+\frac {2 e^{4-\frac {e^4}{2}} x \left (-\log \left (\frac {x^2}{4}\right )-e^4+8\right ) \text {Ei}\left (\frac {1}{2} \left (\log \left (\frac {x^2}{4}\right )+e^4-8\right )\right )}{\sqrt {x^2}}-\frac {2 e^{4-\frac {e^4}{2}} x \left (-3 \log \left (x^2\right )-3 e^4+25+\log (64)\right ) \text {Ei}\left (\frac {1}{2} \left (\log \left (\frac {x^2}{4}\right )+e^4-8\right )\right )}{3 \sqrt {x^2}}+\frac {1}{3} x \log \left (\frac {5 x}{-\log \left (\frac {x^2}{4}\right )-e^4+8}\right )+\frac {5 x}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-50 + 6*E^4 - 6*Log[4] + 6*Log[x^2] + (-8 + E^4 - Log[4] + Log[x^2])*Log[(-5*x)/(-8 + E^4 - Log[4] + Log[
x^2])])/(-24 + 3*E^4 - 3*Log[4] + 3*Log[x^2]),x]

[Out]

(5*x)/3 + (2*E^(4 - E^4/2)*x*ExpIntegralEi[(-8 + E^4 + Log[x^2/4])/2])/(3*Sqrt[x^2]) + (2*E^(4 - E^4/2)*x*ExpI
ntegralEi[(-8 + E^4 + Log[x^2/4])/2]*(8 - E^4 - Log[x^2/4]))/Sqrt[x^2] - (2*E^(4 - E^4/2)*x*ExpIntegralEi[(-8
+ E^4 + Log[x^2/4])/2]*(25 - 3*E^4 + Log[64] - 3*Log[x^2]))/(3*Sqrt[x^2]) + (x*Log[(5*x)/(8 - E^4 - Log[x^2/4]
)])/3

Rule 12

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

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

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 2300

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 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

Rule 2361

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.)), x_Symbol] :> With[{u =
IntHide[(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[SimplifyIntegrand[u/x, x], x],
x]] /; FreeQ[{a, b, c, d, e, f, n, p, r}, x]

Rule 2549

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*Simplify[D[u, x]/u], x], x] /; ProductQ[
u]

Rule 6482

Int[ExpIntegralEi[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[((a + b*x)*ExpIntegralEi[a + b*x])/b, x] - Simp[E^(a
+ b*x)/b, x] /; FreeQ[{a, b}, x]

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 {50 \left (1-\frac {3}{25} \left (e^4-\log (4)\right )\right )-6 \log \left (x^2\right )-\left (-8+e^4-\log (4)+\log \left (x^2\right )\right ) \log \left (-\frac {5 x}{-8+e^4-\log (4)+\log \left (x^2\right )}\right )}{3 \left (8 \left (1-\frac {e^4}{8}\right )-\log \left (\frac {x^2}{4}\right )\right )} \, dx\\ &=\frac {1}{3} \int \frac {50 \left (1-\frac {3}{25} \left (e^4-\log (4)\right )\right )-6 \log \left (x^2\right )-\left (-8+e^4-\log (4)+\log \left (x^2\right )\right ) \log \left (-\frac {5 x}{-8+e^4-\log (4)+\log \left (x^2\right )}\right )}{8 \left (1-\frac {e^4}{8}\right )-\log \left (\frac {x^2}{4}\right )} \, dx\\ &=\frac {1}{3} \int \left (\frac {2 \left (25 \left (1-\frac {3}{25} \left (e^4-\log (4)\right )\right )-3 \log \left (x^2\right )\right )}{8 \left (1-\frac {e^4}{8}\right )-\log \left (\frac {x^2}{4}\right )}+\log \left (-\frac {5 x}{-8 \left (1-\frac {e^4}{8}\right )+\log \left (\frac {x^2}{4}\right )}\right )\right ) \, dx\\ &=\frac {1}{3} \int \log \left (-\frac {5 x}{-8 \left (1-\frac {e^4}{8}\right )+\log \left (\frac {x^2}{4}\right )}\right ) \, dx+\frac {2}{3} \int \frac {25 \left (1-\frac {3}{25} \left (e^4-\log (4)\right )\right )-3 \log \left (x^2\right )}{8 \left (1-\frac {e^4}{8}\right )-\log \left (\frac {x^2}{4}\right )} \, dx\\ &=-\frac {2 e^{4-\frac {e^4}{2}} x \text {Ei}\left (\frac {1}{2} \left (-8+e^4+\log \left (\frac {x^2}{4}\right )\right )\right ) \left (25-3 e^4+\log (64)-3 \log \left (x^2\right )\right )}{3 \sqrt {x^2}}+\frac {1}{3} x \log \left (\frac {5 x}{8-e^4-\log \left (\frac {x^2}{4}\right )}\right )-\frac {1}{3} \int \frac {10 \left (1-\frac {e^4}{10}\right )-\log \left (\frac {x^2}{4}\right )}{8 \left (1-\frac {e^4}{8}\right )-\log \left (\frac {x^2}{4}\right )} \, dx-4 \int \frac {e^{4-\frac {e^4}{2}} \text {Ei}\left (\frac {1}{2} \left (-8 \left (1-\frac {e^4}{8}\right )+\log \left (\frac {x^2}{4}\right )\right )\right )}{\sqrt {x^2}} \, dx\\ &=-\frac {2 e^{4-\frac {e^4}{2}} x \text {Ei}\left (\frac {1}{2} \left (-8+e^4+\log \left (\frac {x^2}{4}\right )\right )\right ) \left (25-3 e^4+\log (64)-3 \log \left (x^2\right )\right )}{3 \sqrt {x^2}}+\frac {1}{3} x \log \left (\frac {5 x}{8-e^4-\log \left (\frac {x^2}{4}\right )}\right )-\frac {1}{3} \int \left (1+\frac {-8 \left (1-\frac {e^4}{8}\right )+10 \left (1-\frac {e^4}{10}\right )}{8 \left (1-\frac {e^4}{8}\right )-\log \left (\frac {x^2}{4}\right )}\right ) \, dx-\left (4 e^{4-\frac {e^4}{2}}\right ) \int \frac {\text {Ei}\left (\frac {1}{2} \left (-8 \left (1-\frac {e^4}{8}\right )+\log \left (\frac {x^2}{4}\right )\right )\right )}{\sqrt {x^2}} \, dx\\ &=-\frac {x}{3}-\frac {2 e^{4-\frac {e^4}{2}} x \text {Ei}\left (\frac {1}{2} \left (-8+e^4+\log \left (\frac {x^2}{4}\right )\right )\right ) \left (25-3 e^4+\log (64)-3 \log \left (x^2\right )\right )}{3 \sqrt {x^2}}+\frac {1}{3} x \log \left (\frac {5 x}{8-e^4-\log \left (\frac {x^2}{4}\right )}\right )-\frac {2}{3} \int \frac {1}{8 \left (1-\frac {e^4}{8}\right )-\log \left (\frac {x^2}{4}\right )} \, dx-\frac {\left (4 e^{4-\frac {e^4}{2}} x\right ) \int \frac {\text {Ei}\left (\frac {1}{2} \left (-8 \left (1-\frac {e^4}{8}\right )+\log \left (\frac {x^2}{4}\right )\right )\right )}{x} \, dx}{\sqrt {x^2}}\\ &=-\frac {x}{3}-\frac {2 e^{4-\frac {e^4}{2}} x \text {Ei}\left (\frac {1}{2} \left (-8+e^4+\log \left (\frac {x^2}{4}\right )\right )\right ) \left (25-3 e^4+\log (64)-3 \log \left (x^2\right )\right )}{3 \sqrt {x^2}}+\frac {1}{3} x \log \left (\frac {5 x}{8-e^4-\log \left (\frac {x^2}{4}\right )}\right )-\frac {(2 x) \operatorname {Subst}\left (\int \frac {e^{x/2}}{8 \left (1-\frac {e^4}{8}\right )-x} \, dx,x,\log \left (\frac {x^2}{4}\right )\right )}{3 \sqrt {x^2}}-\frac {\left (2 e^{4-\frac {e^4}{2}} x\right ) \operatorname {Subst}\left (\int \text {Ei}\left (\frac {1}{2} \left (-8 \left (1-\frac {e^4}{8}\right )+x\right )\right ) \, dx,x,\log \left (\frac {x^2}{4}\right )\right )}{\sqrt {x^2}}\\ &=-\frac {x}{3}+\frac {2 e^{4-\frac {e^4}{2}} x \text {Ei}\left (\frac {1}{2} \left (-8+e^4+\log \left (\frac {x^2}{4}\right )\right )\right )}{3 \sqrt {x^2}}-\frac {2 e^{4-\frac {e^4}{2}} x \text {Ei}\left (\frac {1}{2} \left (-8+e^4+\log \left (\frac {x^2}{4}\right )\right )\right ) \left (25-3 e^4+\log (64)-3 \log \left (x^2\right )\right )}{3 \sqrt {x^2}}+\frac {1}{3} x \log \left (\frac {5 x}{8-e^4-\log \left (\frac {x^2}{4}\right )}\right )-\frac {\left (4 e^{4-\frac {e^4}{2}} x\right ) \operatorname {Subst}\left (\int \text {Ei}(x) \, dx,x,-4 \left (1-\frac {e^4}{8}\right )+\frac {1}{2} \log \left (\frac {x^2}{4}\right )\right )}{\sqrt {x^2}}\\ &=\frac {5 x}{3}+\frac {2 e^{4-\frac {e^4}{2}} x \text {Ei}\left (\frac {1}{2} \left (-8+e^4+\log \left (\frac {x^2}{4}\right )\right )\right )}{3 \sqrt {x^2}}+\frac {2 e^{4-\frac {e^4}{2}} x \text {Ei}\left (\frac {1}{2} \left (-8+e^4\right )+\frac {1}{2} \log \left (\frac {x^2}{4}\right )\right ) \left (8-e^4-\log \left (\frac {x^2}{4}\right )\right )}{\sqrt {x^2}}-\frac {2 e^{4-\frac {e^4}{2}} x \text {Ei}\left (\frac {1}{2} \left (-8+e^4+\log \left (\frac {x^2}{4}\right )\right )\right ) \left (25-3 e^4+\log (64)-3 \log \left (x^2\right )\right )}{3 \sqrt {x^2}}+\frac {1}{3} x \log \left (\frac {5 x}{8-e^4-\log \left (\frac {x^2}{4}\right )}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 26, normalized size = 0.93 \begin {gather*} \frac {1}{3} x \left (5+\log \left (-\frac {5 x}{-8+e^4+\log \left (\frac {x^2}{4}\right )}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50 + 6*E^4 - 6*Log[4] + 6*Log[x^2] + (-8 + E^4 - Log[4] + Log[x^2])*Log[(-5*x)/(-8 + E^4 - Log[4]
+ Log[x^2])])/(-24 + 3*E^4 - 3*Log[4] + 3*Log[x^2]),x]

[Out]

(x*(5 + Log[(-5*x)/(-8 + E^4 + Log[x^2/4])]))/3

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fricas [A]  time = 0.70, size = 25, normalized size = 0.89 \begin {gather*} \frac {1}{3} \, x \log \left (-\frac {5 \, x}{e^{4} - 2 \, \log \relax (2) + \log \left (x^{2}\right ) - 8}\right ) + \frac {5}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x^2)-2*log(2)+exp(4)-8)*log(-5*x/(log(x^2)-2*log(2)+exp(4)-8))+6*log(x^2)-12*log(2)+6*exp(4)-5
0)/(3*log(x^2)-6*log(2)+3*exp(4)-24),x, algorithm="fricas")

[Out]

1/3*x*log(-5*x/(e^4 - 2*log(2) + log(x^2) - 8)) + 5/3*x

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giac [B]  time = 0.37, size = 123, normalized size = 4.39 \begin {gather*} \frac {1}{3} \, x \log \relax (5) - \frac {1}{6} \, x \log \left (4 \, \pi ^{2} \left \lfloor -\frac {1}{2} \, \mathrm {sgn}\relax (x) + 1 \right \rfloor ^{2} + 4 \, \pi ^{2} \left \lfloor -\frac {1}{2} \, \mathrm {sgn}\relax (x) + 1 \right \rfloor \mathrm {sgn}\relax (x) - 4 \, \pi ^{2} \left \lfloor -\frac {1}{2} \, \mathrm {sgn}\relax (x) + 1 \right \rfloor - 2 \, \pi ^{2} \mathrm {sgn}\relax (x) + 2 \, \pi ^{2} - 4 \, e^{4} \log \relax (2) + 4 \, \log \relax (2)^{2} + 2 \, e^{4} \log \left (x^{2}\right ) - 4 \, \log \relax (2) \log \left (x^{2}\right ) + \log \left (x^{2}\right )^{2} + e^{8} - 16 \, e^{4} + 32 \, \log \relax (2) - 16 \, \log \left (x^{2}\right ) + 64\right ) + \frac {1}{3} \, x \log \left ({\left | x \right |}\right ) + \frac {5}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x^2)-2*log(2)+exp(4)-8)*log(-5*x/(log(x^2)-2*log(2)+exp(4)-8))+6*log(x^2)-12*log(2)+6*exp(4)-5
0)/(3*log(x^2)-6*log(2)+3*exp(4)-24),x, algorithm="giac")

[Out]

1/3*x*log(5) - 1/6*x*log(4*pi^2*floor(-1/2*sgn(x) + 1)^2 + 4*pi^2*floor(-1/2*sgn(x) + 1)*sgn(x) - 4*pi^2*floor
(-1/2*sgn(x) + 1) - 2*pi^2*sgn(x) + 2*pi^2 - 4*e^4*log(2) + 4*log(2)^2 + 2*e^4*log(x^2) - 4*log(2)*log(x^2) +
log(x^2)^2 + e^8 - 16*e^4 + 32*log(2) - 16*log(x^2) + 64) + 1/3*x*log(abs(x)) + 5/3*x

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maple [A]  time = 0.13, size = 26, normalized size = 0.93

method	result	size
norman	\(\frac {5 x}{3}+\frac {x \ln \left (-\frac {5 x}{\ln \left (x^{2}\right )-2 \ln \relax (2)+{\mathrm e}^{4}-8}\right )}{3}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((ln(x^2)-2*ln(2)+exp(4)-8)*ln(-5*x/(ln(x^2)-2*ln(2)+exp(4)-8))+6*ln(x^2)-12*ln(2)+6*exp(4)-50)/(3*ln(x^2)
-6*ln(2)+3*exp(4)-24),x,method=_RETURNVERBOSE)

[Out]

5/3*x+1/3*x*ln(-5*x/(ln(x^2)-2*ln(2)+exp(4)-8))

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maxima [A]  time = 0.48, size = 31, normalized size = 1.11 \begin {gather*} \frac {1}{3} \, x {\left (\log \relax (5) + 5\right )} + \frac {1}{3} \, x \log \relax (x) - \frac {1}{3} \, x \log \left (-e^{4} + 2 \, \log \relax (2) - 2 \, \log \relax (x) + 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((log(x^2)-2*log(2)+exp(4)-8)*log(-5*x/(log(x^2)-2*log(2)+exp(4)-8))+6*log(x^2)-12*log(2)+6*exp(4)-5
0)/(3*log(x^2)-6*log(2)+3*exp(4)-24),x, algorithm="maxima")

[Out]

1/3*x*(log(5) + 5) + 1/3*x*log(x) - 1/3*x*log(-e^4 + 2*log(2) - 2*log(x) + 8)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {6\,\ln \left (x^2\right )+6\,{\mathrm {e}}^4-12\,\ln \relax (2)+\ln \left (-\frac {5\,x}{\ln \left (x^2\right )+{\mathrm {e}}^4-2\,\ln \relax (2)-8}\right )\,\left (\ln \left (x^2\right )+{\mathrm {e}}^4-2\,\ln \relax (2)-8\right )-50}{3\,\ln \left (x^2\right )+3\,{\mathrm {e}}^4-6\,\ln \relax (2)-24} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*log(x^2) + 6*exp(4) - 12*log(2) + log(-(5*x)/(log(x^2) + exp(4) - 2*log(2) - 8))*(log(x^2) + exp(4) - 2
*log(2) - 8) - 50)/(3*log(x^2) + 3*exp(4) - 6*log(2) - 24),x)

[Out]

int((6*log(x^2) + 6*exp(4) - 12*log(2) + log(-(5*x)/(log(x^2) + exp(4) - 2*log(2) - 8))*(log(x^2) + exp(4) - 2
*log(2) - 8) - 50)/(3*log(x^2) + 3*exp(4) - 6*log(2) - 24), x)

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sympy [A]  time = 0.45, size = 29, normalized size = 1.04 \begin {gather*} \frac {x \log {\left (- \frac {5 x}{\log {\left (x^{2} \right )} - 8 - 2 \log {\relax (2 )} + e^{4}} \right )}}{3} + \frac {5 x}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((ln(x**2)-2*ln(2)+exp(4)-8)*ln(-5*x/(ln(x**2)-2*ln(2)+exp(4)-8))+6*ln(x**2)-12*ln(2)+6*exp(4)-50)/(
3*ln(x**2)-6*ln(2)+3*exp(4)-24),x)

[Out]

x*log(-5*x/(log(x**2) - 8 - 2*log(2) + exp(4)))/3 + 5*x/3

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