3.35.49 \(\int \frac {(-29 x-5 e^{16} x+5 x^2) \log (x)+(24+5 e^{16}-5 x) \log (\frac {1}{5} (-24-5 e^{16}+5 x))}{72 x+15 e^{16} x-15 x^2} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{4}+\frac {1}{3} \left (x+\log (x) \left (-x+\log \left (-\frac {24}{5}-e^{16}+x\right )\right )\right ) \]

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Rubi [B]  time = 0.27, antiderivative size = 67, normalized size of antiderivative = 2.39, number of steps used = 12, number of rules used = 9, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6, 1593, 6688, 12, 2357, 2295, 2316, 2315, 2394} \begin {gather*} \frac {x}{3}-\frac {1}{3} x \log (x)+\frac {1}{3} \log \left (\frac {24}{5}+e^{16}\right ) \log \left (-5 x+5 e^{16}+24\right )+\frac {1}{3} \log \left (\frac {5 x}{24+5 e^{16}}\right ) \log \left (x+\frac {1}{5} \left (-24-5 e^{16}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-29*x - 5*E^16*x + 5*x^2)*Log[x] + (24 + 5*E^16 - 5*x)*Log[(-24 - 5*E^16 + 5*x)/5])/(72*x + 15*E^16*x -
15*x^2),x]

[Out]

x/3 + (Log[24/5 + E^16]*Log[24 + 5*E^16 - 5*x])/3 - (x*Log[x])/3 + (Log[(5*x)/(24 + 5*E^16)]*Log[(-24 - 5*E^16
)/5 + x])/3

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2316

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[((a + b*Log[-((c*d)/e)])*Log[d + e*
x])/e, x] + Dist[b, Int[Log[-((e*x)/d)]/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[-((c*d)/e), 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (-29 x-5 e^{16} x+5 x^2\right ) \log (x)+\left (24+5 e^{16}-5 x\right ) \log \left (\frac {1}{5} \left (-24-5 e^{16}+5 x\right )\right )}{\left (72+15 e^{16}\right ) x-15 x^2} \, dx\\ &=\int \frac {\left (-29 x-5 e^{16} x+5 x^2\right ) \log (x)+\left (24+5 e^{16}-5 x\right ) \log \left (\frac {1}{5} \left (-24-5 e^{16}+5 x\right )\right )}{\left (72+15 e^{16}-15 x\right ) x} \, dx\\ &=\int \frac {1}{3} \left (\frac {\left (-29-5 e^{16}+5 x\right ) \log (x)}{24+5 e^{16}-5 x}+\frac {\log \left (-\frac {24}{5}-e^{16}+x\right )}{x}\right ) \, dx\\ &=\frac {1}{3} \int \left (\frac {\left (-29-5 e^{16}+5 x\right ) \log (x)}{24+5 e^{16}-5 x}+\frac {\log \left (-\frac {24}{5}-e^{16}+x\right )}{x}\right ) \, dx\\ &=\frac {1}{3} \int \frac {\left (-29-5 e^{16}+5 x\right ) \log (x)}{24+5 e^{16}-5 x} \, dx+\frac {1}{3} \int \frac {\log \left (-\frac {24}{5}-e^{16}+x\right )}{x} \, dx\\ &=\frac {1}{3} \log \left (\frac {5 x}{24+5 e^{16}}\right ) \log \left (\frac {1}{5} \left (-24-5 e^{16}\right )+x\right )+\frac {1}{3} \int \left (-\log (x)-\frac {5 \log (x)}{24+5 e^{16}-5 x}\right ) \, dx-\frac {1}{3} \int \frac {\log \left (\frac {x}{\frac {24}{5}+e^{16}}\right )}{-\frac {24}{5}-e^{16}+x} \, dx\\ &=\frac {1}{3} \log \left (\frac {5 x}{24+5 e^{16}}\right ) \log \left (\frac {1}{5} \left (-24-5 e^{16}\right )+x\right )+\frac {1}{3} \text {Li}_2\left (1-\frac {5 x}{24+5 e^{16}}\right )-\frac {1}{3} \int \log (x) \, dx-\frac {5}{3} \int \frac {\log (x)}{24+5 e^{16}-5 x} \, dx\\ &=\frac {x}{3}+\frac {1}{3} \log \left (\frac {24}{5}+e^{16}\right ) \log \left (24+5 e^{16}-5 x\right )-\frac {1}{3} x \log (x)+\frac {1}{3} \log \left (\frac {5 x}{24+5 e^{16}}\right ) \log \left (\frac {1}{5} \left (-24-5 e^{16}\right )+x\right )+\frac {1}{3} \text {Li}_2\left (1-\frac {5 x}{24+5 e^{16}}\right )-\frac {5}{3} \int \frac {\log \left (\frac {5 x}{24+5 e^{16}}\right )}{24+5 e^{16}-5 x} \, dx\\ &=\frac {x}{3}+\frac {1}{3} \log \left (\frac {24}{5}+e^{16}\right ) \log \left (24+5 e^{16}-5 x\right )-\frac {1}{3} x \log (x)+\frac {1}{3} \log \left (\frac {5 x}{24+5 e^{16}}\right ) \log \left (\frac {1}{5} \left (-24-5 e^{16}\right )+x\right )\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.04, size = 102, normalized size = 3.64 \begin {gather*} \frac {1}{3} \left (x+\log \left (\frac {24}{5}+e^{16}\right ) \log \left (24+5 e^{16}-5 x\right )-x \log (x)+\log \left (\frac {x}{\frac {24}{5}+e^{16}}\right ) \log \left (-\frac {24}{5}-e^{16}+x\right )-\text {Li}_2\left (\frac {24+5 e^{16}-5 x}{24+5 e^{16}}\right )+\text {Li}_2\left (-\frac {5 \left (-\frac {24}{5}-e^{16}+x\right )}{24+5 e^{16}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-29*x - 5*E^16*x + 5*x^2)*Log[x] + (24 + 5*E^16 - 5*x)*Log[(-24 - 5*E^16 + 5*x)/5])/(72*x + 15*E^1
6*x - 15*x^2),x]

[Out]

(x + Log[24/5 + E^16]*Log[24 + 5*E^16 - 5*x] - x*Log[x] + Log[x/(24/5 + E^16)]*Log[-24/5 - E^16 + x] - PolyLog
[2, (24 + 5*E^16 - 5*x)/(24 + 5*E^16)] + PolyLog[2, (-5*(-24/5 - E^16 + x))/(24 + 5*E^16)])/3

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fricas [A]  time = 0.64, size = 20, normalized size = 0.71 \begin {gather*} -\frac {1}{3} \, {\left (x - \log \left (x - e^{16} - \frac {24}{5}\right )\right )} \log \relax (x) + \frac {1}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x*exp(16)+5*x^2-29*x)*log(x)+(5*exp(16)-5*x+24)*log(-exp(16)+x-24/5))/(15*x*exp(16)-15*x^2+72*x
),x, algorithm="fricas")

[Out]

-1/3*(x - log(x - e^16 - 24/5))*log(x) + 1/3*x

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giac [A]  time = 0.16, size = 29, normalized size = 1.04 \begin {gather*} -\frac {1}{3} \, x \log \relax (x) - \frac {1}{3} \, \log \relax (5) \log \relax (x) + \frac {1}{3} \, \log \left (5 \, x - 5 \, e^{16} - 24\right ) \log \relax (x) + \frac {1}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x*exp(16)+5*x^2-29*x)*log(x)+(5*exp(16)-5*x+24)*log(-exp(16)+x-24/5))/(15*x*exp(16)-15*x^2+72*x
),x, algorithm="giac")

[Out]

-1/3*x*log(x) - 1/3*log(5)*log(x) + 1/3*log(5*x - 5*e^16 - 24)*log(x) + 1/3*x

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maple [A]  time = 0.30, size = 22, normalized size = 0.79

method	result	size
norman	\(\frac {x}{3}-\frac {x \ln \relax (x )}{3}+\frac {\ln \relax (x ) \ln \left (-{\mathrm e}^{16}+x -\frac {24}{5}\right )}{3}\)	\(22\)
risch	\(\frac {x}{3}-\frac {x \ln \relax (x )}{3}+\frac {\ln \relax (x ) \ln \left (-{\mathrm e}^{16}+x -\frac {24}{5}\right )}{3}\)	\(22\)
default	\(\frac {\ln \left (-5 \,{\mathrm e}^{16}+5 x -24\right ) \ln \left (\frac {5 x}{5 \,{\mathrm e}^{16}+24}\right )}{3}-\frac {\ln \relax (5) \ln \relax (x )}{3}-\frac {x \ln \relax (x )}{3}+\frac {x}{3}+\frac {\left (\ln \relax (x )-\ln \left (\frac {5 x}{5 \,{\mathrm e}^{16}+24}\right )\right ) \ln \left (\frac {5 \,{\mathrm e}^{16}-5 x +24}{5 \,{\mathrm e}^{16}+24}\right )}{3}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x*exp(16)+5*x^2-29*x)*ln(x)+(5*exp(16)-5*x+24)*ln(-exp(16)+x-24/5))/(15*x*exp(16)-15*x^2+72*x),x,meth
od=_RETURNVERBOSE)

[Out]

1/3*x-1/3*x*ln(x)+1/3*ln(x)*ln(-exp(16)+x-24/5)

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maxima [A]  time = 0.68, size = 26, normalized size = 0.93 \begin {gather*} -\frac {1}{3} \, {\left (x + \log \relax (5)\right )} \log \relax (x) + \frac {1}{3} \, \log \left (5 \, x - 5 \, e^{16} - 24\right ) \log \relax (x) + \frac {1}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x*exp(16)+5*x^2-29*x)*log(x)+(5*exp(16)-5*x+24)*log(-exp(16)+x-24/5))/(15*x*exp(16)-15*x^2+72*x
),x, algorithm="maxima")

[Out]

-1/3*(x + log(5))*log(x) + 1/3*log(5*x - 5*e^16 - 24)*log(x) + 1/3*x

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mupad [B]  time = 2.38, size = 21, normalized size = 0.75 \begin {gather*} \frac {x}{3}+\frac {\ln \left (x-{\mathrm {e}}^{16}-\frac {24}{5}\right )\,\ln \relax (x)}{3}-\frac {x\,\ln \relax (x)}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x - exp(16) - 24/5)*(5*exp(16) - 5*x + 24) - log(x)*(29*x + 5*x*exp(16) - 5*x^2))/(72*x + 15*x*exp(16
) - 15*x^2),x)

[Out]

x/3 + (log(x - exp(16) - 24/5)*log(x))/3 - (x*log(x))/3

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sympy [A]  time = 0.39, size = 24, normalized size = 0.86 \begin {gather*} - \frac {x \log {\relax (x )}}{3} + \frac {x}{3} + \frac {\log {\relax (x )} \log {\left (x - e^{16} - \frac {24}{5} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x*exp(16)+5*x**2-29*x)*ln(x)+(5*exp(16)-5*x+24)*ln(-exp(16)+x-24/5))/(15*x*exp(16)-15*x**2+72*x
),x)

[Out]

-x*log(x)/3 + x/3 + log(x)*log(x - exp(16) - 24/5)/3

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