3.7.96 \(\int \frac {-25-10 \log (3)-\log ^2(3)+(2426-3040 x+969 x^2-120 x^3+5 x^4+(10-40 x) \log (3)+(1-4 x) \log ^2(3)) \log (x)+(-25-10 \log (3)-\log ^2(3)) \log (x) \log (\log (x))}{(25+10 \log (3)+\log ^2(3)) \log (x)} \, dx\)

Optimal. Leaf size=32 \[ x \left (1-2 x+\frac {\left (-(7-x)^2+x\right )^2}{(5+\log (3))^2}-\log (\log (x))\right ) \]

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Rubi [B]  time = 0.36, antiderivative size = 78, normalized size of antiderivative = 2.44, number of steps used = 9, number of rules used = 6, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {12, 6741, 6742, 6688, 2298, 2520} \begin {gather*} \frac {x^5}{(5+\log (3))^2}-\frac {30 x^4}{(5+\log (3))^2}+\frac {323 x^3}{(5+\log (3))^2}-\frac {2 x^2 \left (760+\log ^2(3)+10 \log (3)\right )}{(5+\log (3))^2}+\frac {x \left (2426+\log ^2(3)+10 \log (3)\right )}{(5+\log (3))^2}-x \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25 - 10*Log[3] - Log[3]^2 + (2426 - 3040*x + 969*x^2 - 120*x^3 + 5*x^4 + (10 - 40*x)*Log[3] + (1 - 4*x)*
Log[3]^2)*Log[x] + (-25 - 10*Log[3] - Log[3]^2)*Log[x]*Log[Log[x]])/((25 + 10*Log[3] + Log[3]^2)*Log[x]),x]

[Out]

(323*x^3)/(5 + Log[3])^2 - (30*x^4)/(5 + Log[3])^2 + x^5/(5 + Log[3])^2 - (2*x^2*(760 + 10*Log[3] + Log[3]^2))
/(5 + Log[3])^2 + (x*(2426 + 10*Log[3] + Log[3]^2))/(5 + Log[3])^2 - x*Log[Log[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 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2520

Int[Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)], x_Symbol] :> Simp[x*Log[c*Log[d*x^n]^p], x] - Dist[n*p, Int[1/Log[
d*x^n], x], x] /; FreeQ[{c, d, n, p}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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} &=\frac {\int \frac {-25-10 \log (3)-\log ^2(3)+\left (2426-3040 x+969 x^2-120 x^3+5 x^4+(10-40 x) \log (3)+(1-4 x) \log ^2(3)\right ) \log (x)+\left (-25-10 \log (3)-\log ^2(3)\right ) \log (x) \log (\log (x))}{\log (x)} \, dx}{(5+\log (3))^2}\\ &=\frac {\int \frac {-25 \left (1+\frac {1}{25} \log (3) (10+\log (3))\right )+\left (2426-3040 x+969 x^2-120 x^3+5 x^4+(10-40 x) \log (3)+(1-4 x) \log ^2(3)\right ) \log (x)+\left (-25-10 \log (3)-\log ^2(3)\right ) \log (x) \log (\log (x))}{\log (x)} \, dx}{(5+\log (3))^2}\\ &=\frac {\int \left (\frac {-25 \left (1+\frac {1}{25} \log (3) (10+\log (3))\right )+969 x^2 \log (x)-120 x^3 \log (x)+5 x^4 \log (x)+2426 \left (1+\frac {\log (3) (10+\log (3))}{2426}\right ) \log (x)-3040 x \left (1+\frac {1}{760} \log (3) (10+\log (3))\right ) \log (x)}{\log (x)}-(5+\log (3))^2 \log (\log (x))\right ) \, dx}{(5+\log (3))^2}\\ &=\frac {\int \frac {-25 \left (1+\frac {1}{25} \log (3) (10+\log (3))\right )+969 x^2 \log (x)-120 x^3 \log (x)+5 x^4 \log (x)+2426 \left (1+\frac {\log (3) (10+\log (3))}{2426}\right ) \log (x)-3040 x \left (1+\frac {1}{760} \log (3) (10+\log (3))\right ) \log (x)}{\log (x)} \, dx}{(5+\log (3))^2}-\int \log (\log (x)) \, dx\\ &=-x \log (\log (x))+\frac {\int \left (2426+969 x^2-120 x^3+5 x^4+10 \log (3)+\log ^2(3)-4 x \left (760+10 \log (3)+\log ^2(3)\right )-\frac {(5+\log (3))^2}{\log (x)}\right ) \, dx}{(5+\log (3))^2}+\int \frac {1}{\log (x)} \, dx\\ &=\frac {323 x^3}{(5+\log (3))^2}-\frac {30 x^4}{(5+\log (3))^2}+\frac {x^5}{(5+\log (3))^2}-\frac {2 x^2 \left (760+10 \log (3)+\log ^2(3)\right )}{(5+\log (3))^2}+\frac {x \left (2426+10 \log (3)+\log ^2(3)\right )}{(5+\log (3))^2}-x \log (\log (x))+\text {li}(x)-\int \frac {1}{\log (x)} \, dx\\ &=\frac {323 x^3}{(5+\log (3))^2}-\frac {30 x^4}{(5+\log (3))^2}+\frac {x^5}{(5+\log (3))^2}-\frac {2 x^2 \left (760+10 \log (3)+\log ^2(3)\right )}{(5+\log (3))^2}+\frac {x \left (2426+10 \log (3)+\log ^2(3)\right )}{(5+\log (3))^2}-x \log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 55, normalized size = 1.72 \begin {gather*} \frac {x \left (2426+323 x^2-30 x^3+x^4+10 \log (3)+\log ^2(3)-2 x \left (760+10 \log (3)+\log ^2(3)\right )-(5+\log (3))^2 \log (\log (x))\right )}{(5+\log (3))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 - 10*Log[3] - Log[3]^2 + (2426 - 3040*x + 969*x^2 - 120*x^3 + 5*x^4 + (10 - 40*x)*Log[3] + (1 -
4*x)*Log[3]^2)*Log[x] + (-25 - 10*Log[3] - Log[3]^2)*Log[x]*Log[Log[x]])/((25 + 10*Log[3] + Log[3]^2)*Log[x])
,x]

[Out]

(x*(2426 + 323*x^2 - 30*x^3 + x^4 + 10*Log[3] + Log[3]^2 - 2*x*(760 + 10*Log[3] + Log[3]^2) - (5 + Log[3])^2*L
og[Log[x]]))/(5 + Log[3])^2

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fricas [B]  time = 0.51, size = 83, normalized size = 2.59 \begin {gather*} \frac {x^{5} - 30 \, x^{4} + 323 \, x^{3} - {\left (2 \, x^{2} - x\right )} \log \relax (3)^{2} - 1520 \, x^{2} - 10 \, {\left (2 \, x^{2} - x\right )} \log \relax (3) - {\left (x \log \relax (3)^{2} + 10 \, x \log \relax (3) + 25 \, x\right )} \log \left (\log \relax (x)\right ) + 2426 \, x}{\log \relax (3)^{2} + 10 \, \log \relax (3) + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(3)^2-10*log(3)-25)*log(x)*log(log(x))+((-4*x+1)*log(3)^2+(-40*x+10)*log(3)+5*x^4-120*x^3+969*
x^2-3040*x+2426)*log(x)-log(3)^2-10*log(3)-25)/(log(3)^2+10*log(3)+25)/log(x),x, algorithm="fricas")

[Out]

(x^5 - 30*x^4 + 323*x^3 - (2*x^2 - x)*log(3)^2 - 1520*x^2 - 10*(2*x^2 - x)*log(3) - (x*log(3)^2 + 10*x*log(3)
+ 25*x)*log(log(x)) + 2426*x)/(log(3)^2 + 10*log(3) + 25)

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giac [B]  time = 0.33, size = 86, normalized size = 2.69 \begin {gather*} \frac {x^{5} - 30 \, x^{4} - 2 \, x^{2} \log \relax (3)^{2} - x \log \relax (3)^{2} \log \left (\log \relax (x)\right ) + 323 \, x^{3} - 20 \, x^{2} \log \relax (3) + x \log \relax (3)^{2} - 10 \, x \log \relax (3) \log \left (\log \relax (x)\right ) - 1520 \, x^{2} + 10 \, x \log \relax (3) - 25 \, x \log \left (\log \relax (x)\right ) + 2426 \, x}{\log \relax (3)^{2} + 10 \, \log \relax (3) + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(3)^2-10*log(3)-25)*log(x)*log(log(x))+((-4*x+1)*log(3)^2+(-40*x+10)*log(3)+5*x^4-120*x^3+969*
x^2-3040*x+2426)*log(x)-log(3)^2-10*log(3)-25)/(log(3)^2+10*log(3)+25)/log(x),x, algorithm="giac")

[Out]

(x^5 - 30*x^4 - 2*x^2*log(3)^2 - x*log(3)^2*log(log(x)) + 323*x^3 - 20*x^2*log(3) + x*log(3)^2 - 10*x*log(3)*l
og(log(x)) - 1520*x^2 + 10*x*log(3) - 25*x*log(log(x)) + 2426*x)/(log(3)^2 + 10*log(3) + 25)

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maple [B]  time = 0.04, size = 87, normalized size = 2.72

method	result	size
default	\(\frac {2426 x -10 \ln \left (\ln \relax (x )\right ) \ln \relax (3) x -20 x^{2} \ln \relax (3)+10 x \ln \relax (3)-x \ln \relax (3)^{2} \ln \left (\ln \relax (x )\right )-2 x^{2} \ln \relax (3)^{2}+x \ln \relax (3)^{2}-1520 x^{2}+323 x^{3}-30 x^{4}+x^{5}-25 x \ln \left (\ln \relax (x )\right )}{\ln \relax (3)^{2}+10 \ln \relax (3)+25}\)	\(87\)
risch	\(-x \ln \left (\ln \relax (x )\right )+\frac {x^{5}}{\ln \relax (3)^{2}+10 \ln \relax (3)+25}-\frac {2 x^{2} \ln \relax (3)^{2}}{\ln \relax (3)^{2}+10 \ln \relax (3)+25}-\frac {30 x^{4}}{\ln \relax (3)^{2}+10 \ln \relax (3)+25}+\frac {x \ln \relax (3)^{2}}{\ln \relax (3)^{2}+10 \ln \relax (3)+25}-\frac {20 x^{2} \ln \relax (3)}{\ln \relax (3)^{2}+10 \ln \relax (3)+25}+\frac {323 x^{3}}{\ln \relax (3)^{2}+10 \ln \relax (3)+25}+\frac {10 x \ln \relax (3)}{\ln \relax (3)^{2}+10 \ln \relax (3)+25}-\frac {1520 x^{2}}{\ln \relax (3)^{2}+10 \ln \relax (3)+25}+\frac {2426 x}{\ln \relax (3)^{2}+10 \ln \relax (3)+25}\)	\(165\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-ln(3)^2-10*ln(3)-25)*ln(x)*ln(ln(x))+((-4*x+1)*ln(3)^2+(-40*x+10)*ln(3)+5*x^4-120*x^3+969*x^2-3040*x+24
26)*ln(x)-ln(3)^2-10*ln(3)-25)/(ln(3)^2+10*ln(3)+25)/ln(x),x,method=_RETURNVERBOSE)

[Out]

1/(ln(3)^2+10*ln(3)+25)*(2426*x-10*ln(ln(x))*ln(3)*x-20*x^2*ln(3)+10*x*ln(3)-x*ln(3)^2*ln(ln(x))-2*x^2*ln(3)^2
+x*ln(3)^2-1520*x^2+323*x^3-30*x^4+x^5-25*x*ln(ln(x)))

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maxima [C]  time = 0.57, size = 116, normalized size = 3.62 \begin {gather*} \frac {x^{5} - 30 \, x^{4} - 2 \, x^{2} \log \relax (3)^{2} + 323 \, x^{3} - 20 \, x^{2} \log \relax (3) - {\left (x \log \left (\log \relax (x)\right ) - {\rm Ei}\left (\log \relax (x)\right )\right )} \log \relax (3)^{2} + x \log \relax (3)^{2} - {\rm Ei}\left (\log \relax (x)\right ) \log \relax (3)^{2} - 1520 \, x^{2} - 10 \, {\left (x \log \left (\log \relax (x)\right ) - {\rm Ei}\left (\log \relax (x)\right )\right )} \log \relax (3) + 10 \, x \log \relax (3) - 10 \, {\rm Ei}\left (\log \relax (x)\right ) \log \relax (3) - 25 \, x \log \left (\log \relax (x)\right ) + 2426 \, x}{\log \relax (3)^{2} + 10 \, \log \relax (3) + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(3)^2-10*log(3)-25)*log(x)*log(log(x))+((-4*x+1)*log(3)^2+(-40*x+10)*log(3)+5*x^4-120*x^3+969*
x^2-3040*x+2426)*log(x)-log(3)^2-10*log(3)-25)/(log(3)^2+10*log(3)+25)/log(x),x, algorithm="maxima")

[Out]

(x^5 - 30*x^4 - 2*x^2*log(3)^2 + 323*x^3 - 20*x^2*log(3) - (x*log(log(x)) - Ei(log(x)))*log(3)^2 + x*log(3)^2
- Ei(log(x))*log(3)^2 - 1520*x^2 - 10*(x*log(log(x)) - Ei(log(x)))*log(3) + 10*x*log(3) - 10*Ei(log(x))*log(3)
- 25*x*log(log(x)) + 2426*x)/(log(3)^2 + 10*log(3) + 25)

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mupad [B]  time = 0.80, size = 112, normalized size = 3.50 \begin {gather*} \frac {323\,x^3}{10\,\ln \relax (3)+{\ln \relax (3)}^2+25}-x\,\ln \left (\ln \relax (x)\right )-\frac {30\,x^4}{10\,\ln \relax (3)+{\ln \relax (3)}^2+25}+\frac {x^5}{10\,\ln \relax (3)+{\ln \relax (3)}^2+25}+\frac {x\,\left (10\,\ln \relax (3)+{\ln \relax (3)}^2+2426\right )}{10\,\ln \relax (3)+{\ln \relax (3)}^2+25}-\frac {x^2\,\left (40\,\ln \relax (3)+4\,{\ln \relax (3)}^2+3040\right )}{2\,\left (10\,\ln \relax (3)+{\ln \relax (3)}^2+25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*log(3) + log(x)*(3040*x + log(3)*(40*x - 10) + log(3)^2*(4*x - 1) - 969*x^2 + 120*x^3 - 5*x^4 - 2426)
+ log(3)^2 + log(log(x))*log(x)*(10*log(3) + log(3)^2 + 25) + 25)/(log(x)*(10*log(3) + log(3)^2 + 25)),x)

[Out]

(323*x^3)/(10*log(3) + log(3)^2 + 25) - x*log(log(x)) - (30*x^4)/(10*log(3) + log(3)^2 + 25) + x^5/(10*log(3)
+ log(3)^2 + 25) + (x*(10*log(3) + log(3)^2 + 2426))/(10*log(3) + log(3)^2 + 25) - (x^2*(40*log(3) + 4*log(3)^
2 + 3040))/(2*(10*log(3) + log(3)^2 + 25))

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sympy [B]  time = 0.39, size = 110, normalized size = 3.44 \begin {gather*} \frac {x^{5}}{\log {\relax (3 )}^{2} + 10 \log {\relax (3 )} + 25} - \frac {30 x^{4}}{\log {\relax (3 )}^{2} + 10 \log {\relax (3 )} + 25} + \frac {323 x^{3}}{\log {\relax (3 )}^{2} + 10 \log {\relax (3 )} + 25} + \frac {x^{2} \left (-1520 - 20 \log {\relax (3 )} - 2 \log {\relax (3 )}^{2}\right )}{\log {\relax (3 )}^{2} + 10 \log {\relax (3 )} + 25} - x \log {\left (\log {\relax (x )} \right )} + \frac {x \left (\log {\relax (3 )}^{2} + 10 \log {\relax (3 )} + 2426\right )}{\log {\relax (3 )}^{2} + 10 \log {\relax (3 )} + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-ln(3)**2-10*ln(3)-25)*ln(x)*ln(ln(x))+((-4*x+1)*ln(3)**2+(-40*x+10)*ln(3)+5*x**4-120*x**3+969*x**
2-3040*x+2426)*ln(x)-ln(3)**2-10*ln(3)-25)/(ln(3)**2+10*ln(3)+25)/ln(x),x)

[Out]

x**5/(log(3)**2 + 10*log(3) + 25) - 30*x**4/(log(3)**2 + 10*log(3) + 25) + 323*x**3/(log(3)**2 + 10*log(3) + 2
5) + x**2*(-1520 - 20*log(3) - 2*log(3)**2)/(log(3)**2 + 10*log(3) + 25) - x*log(log(x)) + x*(log(3)**2 + 10*l
og(3) + 2426)/(log(3)**2 + 10*log(3) + 25)

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