3.28.91 \(\int \frac {(68-74 x+22 x^2-x^3) \log (4-4 x+x^2)+(-320+132 x-30 x^2+2 x^3) \log (\frac {-10+x}{32-10 x+2 x^2})}{320-292 x+96 x^2-17 x^3+x^4} \, dx\)

Optimal. Leaf size=27 \[ \log \left ((-2+x)^2\right ) \log \left (\frac {5-\frac {x}{2}}{-16+(5-x) x}\right ) \]

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Rubi [C]  time = 0.70, antiderivative size = 204, normalized size of antiderivative = 7.56, number of steps used = 30, number of rules used = 7, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6742, 2418, 2394, 2393, 2391, 2524, 12} \begin {gather*} 2 \log \left (-\frac {10-x}{2 \left (x^2-5 x+16\right )}\right ) \log (x-2)+2 \log \left (\frac {-2 x-i \sqrt {39}+5}{1-i \sqrt {39}}\right ) \log (x-2)+2 \log \left (\frac {-2 x+i \sqrt {39}+5}{1+i \sqrt {39}}\right ) \log (x-2)-2 \log \left (\frac {10-x}{8}\right ) \log (x-2)-\log \left (\frac {-2 x-i \sqrt {39}+5}{1-i \sqrt {39}}\right ) \log \left ((x-2)^2\right )-\log \left (\frac {-2 x+i \sqrt {39}+5}{1+i \sqrt {39}}\right ) \log \left ((x-2)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((x-2)^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((68 - 74*x + 22*x^2 - x^3)*Log[4 - 4*x + x^2] + (-320 + 132*x - 30*x^2 + 2*x^3)*Log[(-10 + x)/(32 - 10*x
+ 2*x^2)])/(320 - 292*x + 96*x^2 - 17*x^3 + x^4),x]

[Out]

2*Log[(5 - I*Sqrt[39] - 2*x)/(1 - I*Sqrt[39])]*Log[-2 + x] + 2*Log[(5 + I*Sqrt[39] - 2*x)/(1 + I*Sqrt[39])]*Lo
g[-2 + x] - 2*Log[(10 - x)/8]*Log[-2 + x] - Log[(5 - I*Sqrt[39] - 2*x)/(1 - I*Sqrt[39])]*Log[(-2 + x)^2] - Log
[(5 + I*Sqrt[39] - 2*x)/(1 + I*Sqrt[39])]*Log[(-2 + x)^2] + Log[(10 - x)/8]*Log[(-2 + x)^2] + 2*Log[-2 + x]*Lo
g[-1/2*(10 - x)/(16 - 5*x + 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 2391

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

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 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 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 2524

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[d + e*x]*(a + b
*Log[c*RFx^p])^n)/e, x] - Dist[(b*n*p)/e, Int[(Log[d + e*x]*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x
] /; FreeQ[{a, b, c, d, e, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0]

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 \left (-\frac {\left (34-20 x+x^2\right ) \log \left ((-2+x)^2\right )}{(-10+x) \left (16-5 x+x^2\right )}+\frac {2 \log \left (\frac {-10+x}{2 \left (16-5 x+x^2\right )}\right )}{-2+x}\right ) \, dx\\ &=2 \int \frac {\log \left (\frac {-10+x}{2 \left (16-5 x+x^2\right )}\right )}{-2+x} \, dx-\int \frac {\left (34-20 x+x^2\right ) \log \left ((-2+x)^2\right )}{(-10+x) \left (16-5 x+x^2\right )} \, dx\\ &=2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )-2 \int \frac {2 \left (16-5 x+x^2\right ) \left (-\frac {(-10+x) (-5+2 x)}{2 \left (16-5 x+x^2\right )^2}+\frac {1}{2 \left (16-5 x+x^2\right )}\right ) \log (-2+x)}{-10+x} \, dx-\int \left (\frac {\log \left ((-2+x)^2\right )}{10-x}+\frac {(-5+2 x) \log \left ((-2+x)^2\right )}{16-5 x+x^2}\right ) \, dx\\ &=2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )-4 \int \frac {\left (16-5 x+x^2\right ) \left (-\frac {(-10+x) (-5+2 x)}{2 \left (16-5 x+x^2\right )^2}+\frac {1}{2 \left (16-5 x+x^2\right )}\right ) \log (-2+x)}{-10+x} \, dx-\int \frac {\log \left ((-2+x)^2\right )}{10-x} \, dx-\int \frac {(-5+2 x) \log \left ((-2+x)^2\right )}{16-5 x+x^2} \, dx\\ &=\log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )-2 \int \frac {\log \left (\frac {10-x}{8}\right )}{-2+x} \, dx-4 \int \left (\frac {\log (-2+x)}{2 (-10+x)}+\frac {(5-2 x) \log (-2+x)}{2 \left (16-5 x+x^2\right )}\right ) \, dx-\int \left (\frac {2 \log \left ((-2+x)^2\right )}{-5-i \sqrt {39}+2 x}+\frac {2 \log \left ((-2+x)^2\right )}{-5+i \sqrt {39}+2 x}\right ) \, dx\\ &=\log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )-2 \int \frac {\log (-2+x)}{-10+x} \, dx-2 \int \frac {(5-2 x) \log (-2+x)}{16-5 x+x^2} \, dx-2 \int \frac {\log \left ((-2+x)^2\right )}{-5-i \sqrt {39}+2 x} \, dx-2 \int \frac {\log \left ((-2+x)^2\right )}{-5+i \sqrt {39}+2 x} \, dx-2 \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{8}\right )}{x} \, dx,x,-2+x\right )\\ &=-2 \log \left (\frac {10-x}{8}\right ) \log (-2+x)-\log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )-\log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )+2 \text {Li}_2\left (\frac {1}{8} (-2+x)\right )+2 \int \frac {\log \left (\frac {10-x}{8}\right )}{-2+x} \, dx-2 \int \left (-\frac {2 \log (-2+x)}{-5-i \sqrt {39}+2 x}-\frac {2 \log (-2+x)}{-5+i \sqrt {39}+2 x}\right ) \, dx+2 \int \frac {\log \left (\frac {-5-i \sqrt {39}+2 x}{-1-i \sqrt {39}}\right )}{-2+x} \, dx+2 \int \frac {\log \left (\frac {-5+i \sqrt {39}+2 x}{-1+i \sqrt {39}}\right )}{-2+x} \, dx\\ &=-2 \log \left (\frac {10-x}{8}\right ) \log (-2+x)-\log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )-\log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )+2 \text {Li}_2\left (\frac {1}{8} (-2+x)\right )+2 \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{8}\right )}{x} \, dx,x,-2+x\right )+2 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{-1-i \sqrt {39}}\right )}{x} \, dx,x,-2+x\right )+2 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{-1+i \sqrt {39}}\right )}{x} \, dx,x,-2+x\right )+4 \int \frac {\log (-2+x)}{-5-i \sqrt {39}+2 x} \, dx+4 \int \frac {\log (-2+x)}{-5+i \sqrt {39}+2 x} \, dx\\ &=2 \log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log (-2+x)+2 \log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log (-2+x)-2 \log \left (\frac {10-x}{8}\right ) \log (-2+x)-\log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )-\log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )-2 \text {Li}_2\left (-\frac {2 (2-x)}{1-i \sqrt {39}}\right )-2 \text {Li}_2\left (-\frac {2 (2-x)}{1+i \sqrt {39}}\right )-2 \int \frac {\log \left (\frac {-5-i \sqrt {39}+2 x}{-1-i \sqrt {39}}\right )}{-2+x} \, dx-2 \int \frac {\log \left (\frac {-5+i \sqrt {39}+2 x}{-1+i \sqrt {39}}\right )}{-2+x} \, dx\\ &=2 \log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log (-2+x)+2 \log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log (-2+x)-2 \log \left (\frac {10-x}{8}\right ) \log (-2+x)-\log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )-\log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )-2 \text {Li}_2\left (-\frac {2 (2-x)}{1-i \sqrt {39}}\right )-2 \text {Li}_2\left (-\frac {2 (2-x)}{1+i \sqrt {39}}\right )-2 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{-1-i \sqrt {39}}\right )}{x} \, dx,x,-2+x\right )-2 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{-1+i \sqrt {39}}\right )}{x} \, dx,x,-2+x\right )\\ &=2 \log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log (-2+x)+2 \log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log (-2+x)-2 \log \left (\frac {10-x}{8}\right ) \log (-2+x)-\log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )-\log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.18, size = 220, normalized size = 8.15 \begin {gather*} -2 \log (8) \log (-10+x)+2 \log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log (-2+x)+2 \log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log (-2+x)-\log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )-\log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )+2 \text {Li}_2\left (\frac {10-x}{8}\right )+2 \text {Li}_2\left (\frac {1}{8} (-2+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((68 - 74*x + 22*x^2 - x^3)*Log[4 - 4*x + x^2] + (-320 + 132*x - 30*x^2 + 2*x^3)*Log[(-10 + x)/(32 -
10*x + 2*x^2)])/(320 - 292*x + 96*x^2 - 17*x^3 + x^4),x]

[Out]

-2*Log[8]*Log[-10 + x] + 2*Log[(5 - I*Sqrt[39] - 2*x)/(1 - I*Sqrt[39])]*Log[-2 + x] + 2*Log[(5 + I*Sqrt[39] -
2*x)/(1 + I*Sqrt[39])]*Log[-2 + x] - Log[(5 - I*Sqrt[39] - 2*x)/(1 - I*Sqrt[39])]*Log[(-2 + x)^2] - Log[(5 + I
*Sqrt[39] - 2*x)/(1 + I*Sqrt[39])]*Log[(-2 + x)^2] + Log[(10 - x)/8]*Log[(-2 + x)^2] + 2*Log[-2 + x]*Log[-1/2*
(10 - x)/(16 - 5*x + x^2)] + 2*PolyLog[2, (10 - x)/8] + 2*PolyLog[2, (-2 + x)/8]

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fricas [A]  time = 0.93, size = 26, normalized size = 0.96 \begin {gather*} \log \left (x^{2} - 4 \, x + 4\right ) \log \left (\frac {x - 10}{2 \, {\left (x^{2} - 5 \, x + 16\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3+22*x^2-74*x+68)*log(x^2-4*x+4)+(2*x^3-30*x^2+132*x-320)*log((x-10)/(2*x^2-10*x+32)))/(x^4-17*
x^3+96*x^2-292*x+320),x, algorithm="fricas")

[Out]

log(x^2 - 4*x + 4)*log(1/2*(x - 10)/(x^2 - 5*x + 16))

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giac [B]  time = 0.58, size = 60, normalized size = 2.22 \begin {gather*} -{\left (\log \left (x^{2} - 5 \, x + 16\right ) - \log \left (x - 10\right )\right )} \log \left (x^{2} - 4 \, x + 4\right ) - 2 \, \log \left (2 \, x^{2} - 10 \, x + 32\right ) \log \left (x - 2\right ) + 2 \, \log \left (x^{2} - 5 \, x + 16\right ) \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3+22*x^2-74*x+68)*log(x^2-4*x+4)+(2*x^3-30*x^2+132*x-320)*log((x-10)/(2*x^2-10*x+32)))/(x^4-17*
x^3+96*x^2-292*x+320),x, algorithm="giac")

[Out]

-(log(x^2 - 5*x + 16) - log(x - 10))*log(x^2 - 4*x + 4) - 2*log(2*x^2 - 10*x + 32)*log(x - 2) + 2*log(x^2 - 5*
x + 16)*log(x - 2)

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maple [C]  time = 2.54, size = 247, normalized size = 9.15

method	result	size
default	\(-2 \ln \relax (2) \ln \left (x -2\right )+2 \ln \left (x -2\right ) \ln \left (\frac {x -10}{x^{2}-5 x +16}\right )-2 \left (\ln \left (x -2\right )-\ln \left (-\frac {1}{4}+\frac {x}{8}\right )\right ) \ln \left (\frac {5}{4}-\frac {x}{8}\right )+2 \ln \left (x -2\right ) \ln \left (\frac {i \sqrt {39}+5-2 x}{1+i \sqrt {39}}\right )+2 \ln \left (x -2\right ) \ln \left (\frac {i \sqrt {39}-5+2 x}{i \sqrt {39}-1}\right )+2 \dilog \left (\frac {i \sqrt {39}+5-2 x}{1+i \sqrt {39}}\right )+2 \dilog \left (\frac {i \sqrt {39}-5+2 x}{i \sqrt {39}-1}\right )-\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{2}-5 \textit {\_Z} +16\right )}{\sum }\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (x^{2}-4 x +4\right )-2 \dilog \left (\frac {x -2}{-2+\underline {\hspace {1.25 ex}}\alpha }\right )-2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x -2}{-2+\underline {\hspace {1.25 ex}}\alpha }\right )\right )\right )+\ln \left (x -10\right ) \ln \left (x^{2}-4 x +4\right )-2 \ln \left (x -10\right ) \ln \left (-\frac {1}{4}+\frac {x}{8}\right )\)	\(247\)
risch	\(\text {Expression too large to display}\)	\(50326\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^3+22*x^2-74*x+68)*ln(x^2-4*x+4)+(2*x^3-30*x^2+132*x-320)*ln((x-10)/(2*x^2-10*x+32)))/(x^4-17*x^3+96*x
^2-292*x+320),x,method=_RETURNVERBOSE)

[Out]

-2*ln(2)*ln(x-2)+2*ln(x-2)*ln((x-10)/(x^2-5*x+16))-2*(ln(x-2)-ln(-1/4+1/8*x))*ln(5/4-1/8*x)+2*ln(x-2)*ln((I*39
^(1/2)+5-2*x)/(1+I*39^(1/2)))+2*ln(x-2)*ln((I*39^(1/2)-5+2*x)/(I*39^(1/2)-1))+2*dilog((I*39^(1/2)+5-2*x)/(1+I*
39^(1/2)))+2*dilog((I*39^(1/2)-5+2*x)/(I*39^(1/2)-1))-Sum(ln(x-_alpha)*ln(x^2-4*x+4)-2*dilog((x-2)/(-2+_alpha)
)-2*ln(x-_alpha)*ln((x-2)/(-2+_alpha)),_alpha=RootOf(_Z^2-5*_Z+16))+ln(x-10)*ln(x^2-4*x+4)-2*ln(x-10)*ln(-1/4+
1/8*x)

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maxima [A]  time = 0.87, size = 31, normalized size = 1.15 \begin {gather*} -2 \, {\left (\log \relax (2) - \log \left (x - 10\right )\right )} \log \left (x - 2\right ) - 2 \, \log \left (x^{2} - 5 \, x + 16\right ) \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3+22*x^2-74*x+68)*log(x^2-4*x+4)+(2*x^3-30*x^2+132*x-320)*log((x-10)/(2*x^2-10*x+32)))/(x^4-17*
x^3+96*x^2-292*x+320),x, algorithm="maxima")

[Out]

-2*(log(2) - log(x - 10))*log(x - 2) - 2*log(x^2 - 5*x + 16)*log(x - 2)

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mupad [B]  time = 2.09, size = 28, normalized size = 1.04 \begin {gather*} \ln \left (x^2-4\,x+4\right )\,\left (\ln \left (x-10\right )-\ln \left (2\,x^2-10\,x+32\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x^2 - 4*x + 4)*(74*x - 22*x^2 + x^3 - 68) - log((x - 10)/(2*x^2 - 10*x + 32))*(132*x - 30*x^2 + 2*x^
3 - 320))/(96*x^2 - 292*x - 17*x^3 + x^4 + 320),x)

[Out]

log(x^2 - 4*x + 4)*(log(x - 10) - log(2*x^2 - 10*x + 32))

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sympy [A]  time = 0.41, size = 24, normalized size = 0.89 \begin {gather*} \log {\left (\frac {x - 10}{2 x^{2} - 10 x + 32} \right )} \log {\left (x^{2} - 4 x + 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**3+22*x**2-74*x+68)*ln(x**2-4*x+4)+(2*x**3-30*x**2+132*x-320)*ln((x-10)/(2*x**2-10*x+32)))/(x**
4-17*x**3+96*x**2-292*x+320),x)

[Out]

log((x - 10)/(2*x**2 - 10*x + 32))*log(x**2 - 4*x + 4)

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