3.94.77 \(\int \frac {1}{4} ((3-10 x+(3-2 x) \log (3)) \log (4)+(-4-4 \log (3)) \log (4) \log (x)+(1+\log (3)) \log (4) \log ^2(x)) \, dx\)

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

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Rubi [B]  time = 0.05, antiderivative size = 67, normalized size of antiderivative = 2.39, number of steps used = 5, number of rules used = 3, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 2295, 2296} \begin {gather*} \frac {1}{4} x (1+\log (3)) \log (4) \log ^2(x)-\frac {\log (4) (-10 x+(3-2 x) \log (3)+3)^2}{8 (10+\log (9))}-\frac {3}{2} x (1+\log (3)) \log (4) \log (x)+\frac {3}{2} x (1+\log (3)) \log (4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((3 - 10*x + (3 - 2*x)*Log[3])*Log[4] + (-4 - 4*Log[3])*Log[4]*Log[x] + (1 + Log[3])*Log[4]*Log[x]^2)/4,x]

[Out]

(3*x*(1 + Log[3])*Log[4])/2 - ((3 - 10*x + (3 - 2*x)*Log[3])^2*Log[4])/(8*(10 + Log[9])) - (3*x*(1 + Log[3])*L
og[4]*Log[x])/2 + (x*(1 + Log[3])*Log[4]*Log[x]^2)/4

Rule 12

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

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \left ((3-10 x+(3-2 x) \log (3)) \log (4)+(-4-4 \log (3)) \log (4) \log (x)+(1+\log (3)) \log (4) \log ^2(x)\right ) \, dx\\ &=-\frac {(3-10 x+(3-2 x) \log (3))^2 \log (4)}{8 (10+\log (9))}+\frac {1}{4} ((1+\log (3)) \log (4)) \int \log ^2(x) \, dx-((1+\log (3)) \log (4)) \int \log (x) \, dx\\ &=x (1+\log (3)) \log (4)-\frac {(3-10 x+(3-2 x) \log (3))^2 \log (4)}{8 (10+\log (9))}-x (1+\log (3)) \log (4) \log (x)+\frac {1}{4} x (1+\log (3)) \log (4) \log ^2(x)-\frac {1}{2} ((1+\log (3)) \log (4)) \int \log (x) \, dx\\ &=\frac {3}{2} x (1+\log (3)) \log (4)-\frac {(3-10 x+(3-2 x) \log (3))^2 \log (4)}{8 (10+\log (9))}-\frac {3}{2} x (1+\log (3)) \log (4) \log (x)+\frac {1}{4} x (1+\log (3)) \log (4) \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 43, normalized size = 1.54 \begin {gather*} \frac {1}{4} \log (4) \left (-\frac {1}{2} x^2 (10+\log (9))+x (9+\log (19683))-6 x (1+\log (3)) \log (x)+x (1+\log (3)) \log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((3 - 10*x + (3 - 2*x)*Log[3])*Log[4] + (-4 - 4*Log[3])*Log[4]*Log[x] + (1 + Log[3])*Log[4]*Log[x]^2
)/4,x]

[Out]

(Log[4]*(-1/2*(x^2*(10 + Log[9])) + x*(9 + Log[19683]) - 6*x*(1 + Log[3])*Log[x] + x*(1 + Log[3])*Log[x]^2))/4

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fricas [B]  time = 0.70, size = 59, normalized size = 2.11 \begin {gather*} -\frac {1}{2} \, {\left (x^{2} - 9 \, x\right )} \log \relax (3) \log \relax (2) + \frac {1}{2} \, {\left (x \log \relax (3) \log \relax (2) + x \log \relax (2)\right )} \log \relax (x)^{2} - \frac {1}{2} \, {\left (5 \, x^{2} - 9 \, x\right )} \log \relax (2) - 3 \, {\left (x \log \relax (3) \log \relax (2) + x \log \relax (2)\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(log(3)+1)*log(2)*log(x)^2+1/2*(-4*log(3)-4)*log(2)*log(x)+1/2*((3-2*x)*log(3)-10*x+3)*log(2),x,
algorithm="fricas")

[Out]

-1/2*(x^2 - 9*x)*log(3)*log(2) + 1/2*(x*log(3)*log(2) + x*log(2))*log(x)^2 - 1/2*(5*x^2 - 9*x)*log(2) - 3*(x*l
og(3)*log(2) + x*log(2))*log(x)

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giac [B]  time = 0.19, size = 63, normalized size = 2.25 \begin {gather*} \frac {1}{2} \, {\left (x \log \relax (x)^{2} - 2 \, x \log \relax (x) + 2 \, x\right )} {\left (\log \relax (3) + 1\right )} \log \relax (2) - 2 \, {\left (x \log \relax (x) - x\right )} {\left (\log \relax (3) + 1\right )} \log \relax (2) - \frac {1}{2} \, {\left (5 \, x^{2} + {\left (x^{2} - 3 \, x\right )} \log \relax (3) - 3 \, x\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(log(3)+1)*log(2)*log(x)^2+1/2*(-4*log(3)-4)*log(2)*log(x)+1/2*((3-2*x)*log(3)-10*x+3)*log(2),x,
algorithm="giac")

[Out]

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

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maple [B]  time = 0.04, size = 62, normalized size = 2.21

method	result	size
norman	\(\left (-\frac {\ln \relax (2) \ln \relax (3)}{2}-\frac {5 \ln \relax (2)}{2}\right ) x^{2}+\left (\frac {9 \ln \relax (2) \ln \relax (3)}{2}+\frac {9 \ln \relax (2)}{2}\right ) x +\left (-3 \ln \relax (2) \ln \relax (3)-3 \ln \relax (2)\right ) x \ln \relax (x )+\left (\frac {\ln \relax (2) \ln \relax (3)}{2}+\frac {\ln \relax (2)}{2}\right ) x \ln \relax (x )^{2}\)	\(62\)
default	\(-\frac {\ln \relax (2) \ln \relax (3) x^{2}}{2}+\frac {9 x \ln \relax (2) \ln \relax (3)}{2}-\frac {5 x^{2} \ln \relax (2)}{2}+\frac {9 x \ln \relax (2)}{2}-3 \ln \relax (2) \ln \relax (3) \ln \relax (x ) x -3 x \ln \relax (2) \ln \relax (x )+\frac {\ln \relax (2) \ln \relax (3) \ln \relax (x )^{2} x}{2}+\frac {\ln \relax (2) \ln \relax (x )^{2} x}{2}\)	\(66\)
risch	\(-\frac {\ln \relax (2) \ln \relax (3) x^{2}}{2}+\frac {9 x \ln \relax (2) \ln \relax (3)}{2}-\frac {5 x^{2} \ln \relax (2)}{2}+\frac {9 x \ln \relax (2)}{2}-3 \ln \relax (2) \ln \relax (3) \ln \relax (x ) x -3 x \ln \relax (2) \ln \relax (x )+\frac {\ln \relax (2) \ln \relax (3) \ln \relax (x )^{2} x}{2}+\frac {\ln \relax (2) \ln \relax (x )^{2} x}{2}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(ln(3)+1)*ln(2)*ln(x)^2+1/2*(-4*ln(3)-4)*ln(2)*ln(x)+1/2*((3-2*x)*ln(3)-10*x+3)*ln(2),x,method=_RETURN
VERBOSE)

[Out]

(-1/2*ln(2)*ln(3)-5/2*ln(2))*x^2+(9/2*ln(2)*ln(3)+9/2*ln(2))*x+(-3*ln(2)*ln(3)-3*ln(2))*x*ln(x)+(1/2*ln(2)*ln(
3)+1/2*ln(2))*x*ln(x)^2

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maxima [B]  time = 0.36, size = 59, normalized size = 2.11 \begin {gather*} \frac {1}{2} \, {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x {\left (\log \relax (3) + 1\right )} \log \relax (2) - 2 \, {\left (x \log \relax (x) - x\right )} {\left (\log \relax (3) + 1\right )} \log \relax (2) - \frac {1}{2} \, {\left (5 \, x^{2} + {\left (x^{2} - 3 \, x\right )} \log \relax (3) - 3 \, x\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(log(3)+1)*log(2)*log(x)^2+1/2*(-4*log(3)-4)*log(2)*log(x)+1/2*((3-2*x)*log(3)-10*x+3)*log(2),x,
algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.48, size = 55, normalized size = 1.96 \begin {gather*} x\,\left (\frac {9\,\ln \relax (2)\,\left (\ln \relax (3)+1\right )}{2}-\frac {\ln \relax (x)\,\left (6\,\ln \relax (2)+6\,\ln \relax (2)\,\ln \relax (3)\right )}{2}+\frac {\ln \relax (2)\,{\ln \relax (x)}^2\,\left (\ln \relax (3)+1\right )}{2}\right )-x^2\,\left (\frac {\ln \left (32\right )}{2}+\frac {\ln \relax (2)\,\ln \relax (3)}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(2)*log(x)^2*(log(3) + 1))/2 - (log(2)*log(x)*(4*log(3) + 4))/2 - (log(2)*(10*x + log(3)*(2*x - 3) - 3
))/2,x)

[Out]

x*((9*log(2)*(log(3) + 1))/2 - (log(x)*(6*log(2) + 6*log(2)*log(3)))/2 + (log(2)*log(x)^2*(log(3) + 1))/2) - x
^2*(log(32)/2 + (log(2)*log(3))/2)

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sympy [B]  time = 0.18, size = 82, normalized size = 2.93 \begin {gather*} x^{2} \left (- \frac {5 \log {\relax (2 )}}{2} - \frac {\log {\relax (2 )} \log {\relax (3 )}}{2}\right ) + x \left (\frac {9 \log {\relax (2 )}}{2} + \frac {9 \log {\relax (2 )} \log {\relax (3 )}}{2}\right ) + \left (\frac {x \log {\relax (2 )}}{2} + \frac {x \log {\relax (2 )} \log {\relax (3 )}}{2}\right ) \log {\relax (x )}^{2} + \left (- 3 x \log {\relax (2 )} \log {\relax (3 )} - 3 x \log {\relax (2 )}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(ln(3)+1)*ln(2)*ln(x)**2+1/2*(-4*ln(3)-4)*ln(2)*ln(x)+1/2*((3-2*x)*ln(3)-10*x+3)*ln(2),x)

[Out]

x**2*(-5*log(2)/2 - log(2)*log(3)/2) + x*(9*log(2)/2 + 9*log(2)*log(3)/2) + (x*log(2)/2 + x*log(2)*log(3)/2)*l
og(x)**2 + (-3*x*log(2)*log(3) - 3*x*log(2))*log(x)

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