3.62.18 \(\int \frac {(-1250-500 x+250 x^2) \log ^2(2)+(10+127 x) \log ^2(2) \log (x)+(3+x) \log ^2(2) \log ^2(x)}{125 x} \, dx\)

Optimal. Leaf size=21 \[ \log ^2(2) (5+x+\log (x)) \left (-10+x+\frac {\log ^2(x)}{125}\right ) \]

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Rubi [B]  time = 0.11, antiderivative size = 69, normalized size of antiderivative = 3.29, number of steps used = 13, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {12, 14, 2346, 2301, 2295, 2302, 30, 2296} \begin {gather*} x^2 \log ^2(2)+\frac {1}{125} x \log ^2(2) \log ^2(x)+\frac {1}{25} \log ^2(2) \log ^2(x)+x \log ^2(2) \log (x)-10 \log ^2(2) \log (x)-5 x \log ^2(2)+\frac {1}{125} \log ^2(2) \log ^3(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-1250 - 500*x + 250*x^2)*Log[2]^2 + (10 + 127*x)*Log[2]^2*Log[x] + (3 + x)*Log[2]^2*Log[x]^2)/(125*x),x]

[Out]

-5*x*Log[2]^2 + x^2*Log[2]^2 - 10*Log[2]^2*Log[x] + x*Log[2]^2*Log[x] + (Log[2]^2*Log[x]^2)/25 + (x*Log[2]^2*L
og[x]^2)/125 + (Log[2]^2*Log[x]^3)/125

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

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

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 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

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 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{125} \int \frac {\left (-1250-500 x+250 x^2\right ) \log ^2(2)+(10+127 x) \log ^2(2) \log (x)+(3+x) \log ^2(2) \log ^2(x)}{x} \, dx\\ &=\frac {1}{125} \int \left (\frac {250 \left (-5-2 x+x^2\right ) \log ^2(2)}{x}+\frac {(10+127 x) \log ^2(2) \log (x)}{x}+\frac {(3+x) \log ^2(2) \log ^2(x)}{x}\right ) \, dx\\ &=\frac {1}{125} \log ^2(2) \int \frac {(10+127 x) \log (x)}{x} \, dx+\frac {1}{125} \log ^2(2) \int \frac {(3+x) \log ^2(x)}{x} \, dx+\left (2 \log ^2(2)\right ) \int \frac {-5-2 x+x^2}{x} \, dx\\ &=\frac {1}{125} \log ^2(2) \int \log ^2(x) \, dx+\frac {1}{125} \left (3 \log ^2(2)\right ) \int \frac {\log ^2(x)}{x} \, dx+\frac {1}{25} \left (2 \log ^2(2)\right ) \int \frac {\log (x)}{x} \, dx+\frac {1}{125} \left (127 \log ^2(2)\right ) \int \log (x) \, dx+\left (2 \log ^2(2)\right ) \int \left (-2-\frac {5}{x}+x\right ) \, dx\\ &=-\frac {627}{125} x \log ^2(2)+x^2 \log ^2(2)-10 \log ^2(2) \log (x)+\frac {127}{125} x \log ^2(2) \log (x)+\frac {1}{25} \log ^2(2) \log ^2(x)+\frac {1}{125} x \log ^2(2) \log ^2(x)-\frac {1}{125} \left (2 \log ^2(2)\right ) \int \log (x) \, dx+\frac {1}{125} \left (3 \log ^2(2)\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,\log (x)\right )\\ &=-5 x \log ^2(2)+x^2 \log ^2(2)-10 \log ^2(2) \log (x)+x \log ^2(2) \log (x)+\frac {1}{25} \log ^2(2) \log ^2(x)+\frac {1}{125} x \log ^2(2) \log ^2(x)+\frac {1}{125} \log ^2(2) \log ^3(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 42, normalized size = 2.00 \begin {gather*} \frac {1}{125} \log ^2(2) \left (-625 x+125 x^2-1250 \log (x)+125 x \log (x)+5 \log ^2(x)+x \log ^2(x)+\log ^3(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-1250 - 500*x + 250*x^2)*Log[2]^2 + (10 + 127*x)*Log[2]^2*Log[x] + (3 + x)*Log[2]^2*Log[x]^2)/(125
*x),x]

[Out]

(Log[2]^2*(-625*x + 125*x^2 - 1250*Log[x] + 125*x*Log[x] + 5*Log[x]^2 + x*Log[x]^2 + Log[x]^3))/125

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fricas [B]  time = 0.69, size = 46, normalized size = 2.19 \begin {gather*} \frac {1}{125} \, {\left (x + 5\right )} \log \relax (2)^{2} \log \relax (x)^{2} + \frac {1}{125} \, \log \relax (2)^{2} \log \relax (x)^{3} + {\left (x - 10\right )} \log \relax (2)^{2} \log \relax (x) + {\left (x^{2} - 5 \, x\right )} \log \relax (2)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*((3+x)*log(2)^2*log(x)^2+(127*x+10)*log(2)^2*log(x)+(250*x^2-500*x-1250)*log(2)^2)/x,x, algori
thm="fricas")

[Out]

1/125*(x + 5)*log(2)^2*log(x)^2 + 1/125*log(2)^2*log(x)^3 + (x - 10)*log(2)^2*log(x) + (x^2 - 5*x)*log(2)^2

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giac [B]  time = 0.12, size = 61, normalized size = 2.90 \begin {gather*} \frac {1}{125} \, \log \relax (2)^{2} \log \relax (x)^{3} + x^{2} \log \relax (2)^{2} + x \log \relax (2)^{2} \log \relax (x) - 5 \, x \log \relax (2)^{2} - 10 \, \log \relax (2)^{2} \log \relax (x) + \frac {1}{125} \, {\left (x \log \relax (2)^{2} + 5 \, \log \relax (2)^{2}\right )} \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*((3+x)*log(2)^2*log(x)^2+(127*x+10)*log(2)^2*log(x)+(250*x^2-500*x-1250)*log(2)^2)/x,x, algori
thm="giac")

[Out]

1/125*log(2)^2*log(x)^3 + x^2*log(2)^2 + x*log(2)^2*log(x) - 5*x*log(2)^2 - 10*log(2)^2*log(x) + 1/125*(x*log(
2)^2 + 5*log(2)^2)*log(x)^2

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maple [B]  time = 0.05, size = 62, normalized size = 2.95

method	result	size
risch	\(\frac {\ln \relax (2)^{2} \ln \relax (x )^{3}}{125}+\frac {\left (x \ln \relax (2)^{2}+5 \ln \relax (2)^{2}\right ) \ln \relax (x )^{2}}{125}+\ln \relax (x ) \ln \relax (2)^{2} x +x^{2} \ln \relax (2)^{2}-5 x \ln \relax (2)^{2}-10 \ln \relax (2)^{2} \ln \relax (x )\)	\(62\)
norman	\(x^{2} \ln \relax (2)^{2}-10 \ln \relax (2)^{2} \ln \relax (x )+\ln \relax (x ) \ln \relax (2)^{2} x -5 x \ln \relax (2)^{2}+\frac {\ln \relax (2)^{2} \ln \relax (x )^{2}}{25}+\frac {\ln \relax (2)^{2} \ln \relax (x )^{3}}{125}+\frac {x \ln \relax (2)^{2} \ln \relax (x )^{2}}{125}\)	\(64\)
default	\(\frac {\ln \relax (2)^{2} \left (x \ln \relax (x )^{2}-2 x \ln \relax (x )+2 x \right )}{125}+\frac {\ln \relax (2)^{2} \ln \relax (x )^{3}}{125}+\frac {127 \ln \relax (2)^{2} \left (x \ln \relax (x )-x \right )}{125}+x^{2} \ln \relax (2)^{2}+\frac {\ln \relax (2)^{2} \ln \relax (x )^{2}}{25}-4 x \ln \relax (2)^{2}-10 \ln \relax (2)^{2} \ln \relax (x )\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/125*((3+x)*ln(2)^2*ln(x)^2+(127*x+10)*ln(2)^2*ln(x)+(250*x^2-500*x-1250)*ln(2)^2)/x,x,method=_RETURNVERB
OSE)

[Out]

1/125*ln(2)^2*ln(x)^3+1/125*(x*ln(2)^2+5*ln(2)^2)*ln(x)^2+ln(x)*ln(2)^2*x+x^2*ln(2)^2-5*x*ln(2)^2-10*ln(2)^2*l
n(x)

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maxima [B]  time = 0.34, size = 75, normalized size = 3.57 \begin {gather*} \frac {1}{125} \, \log \relax (2)^{2} \log \relax (x)^{3} + \frac {1}{125} \, {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x \log \relax (2)^{2} + x^{2} \log \relax (2)^{2} + \frac {1}{25} \, \log \relax (2)^{2} \log \relax (x)^{2} + \frac {127}{125} \, {\left (x \log \relax (x) - x\right )} \log \relax (2)^{2} - 4 \, x \log \relax (2)^{2} - 10 \, \log \relax (2)^{2} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*((3+x)*log(2)^2*log(x)^2+(127*x+10)*log(2)^2*log(x)+(250*x^2-500*x-1250)*log(2)^2)/x,x, algori
thm="maxima")

[Out]

1/125*log(2)^2*log(x)^3 + 1/125*(log(x)^2 - 2*log(x) + 2)*x*log(2)^2 + x^2*log(2)^2 + 1/25*log(2)^2*log(x)^2 +
127/125*(x*log(x) - x)*log(2)^2 - 4*x*log(2)^2 - 10*log(2)^2*log(x)

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mupad [B]  time = 4.67, size = 40, normalized size = 1.90 \begin {gather*} \frac {{\ln \relax (2)}^2\,\left (125\,x^2+x\,{\ln \relax (x)}^2+125\,x\,\ln \relax (x)-625\,x+{\ln \relax (x)}^3+5\,{\ln \relax (x)}^2-1250\,\ln \relax (x)\right )}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(2)^2*log(x)*(127*x + 10))/125 - (log(2)^2*(500*x - 250*x^2 + 1250))/125 + (log(2)^2*log(x)^2*(x + 3)
)/125)/x,x)

[Out]

(log(2)^2*(x*log(x)^2 - 1250*log(x) - 625*x + 5*log(x)^2 + log(x)^3 + 125*x*log(x) + 125*x^2))/125

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sympy [B]  time = 0.21, size = 68, normalized size = 3.24 \begin {gather*} x^{2} \log {\relax (2 )}^{2} + x \log {\relax (2 )}^{2} \log {\relax (x )} - 5 x \log {\relax (2 )}^{2} + \left (\frac {x \log {\relax (2 )}^{2}}{125} + \frac {\log {\relax (2 )}^{2}}{25}\right ) \log {\relax (x )}^{2} + \frac {\log {\relax (2 )}^{2} \log {\relax (x )}^{3}}{125} - 10 \log {\relax (2 )}^{2} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*((3+x)*ln(2)**2*ln(x)**2+(127*x+10)*ln(2)**2*ln(x)+(250*x**2-500*x-1250)*ln(2)**2)/x,x)

[Out]

x**2*log(2)**2 + x*log(2)**2*log(x) - 5*x*log(2)**2 + (x*log(2)**2/125 + log(2)**2/25)*log(x)**2 + log(2)**2*l
og(x)**3/125 - 10*log(2)**2*log(x)

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