3.88.60 \(\int -\frac {4-60 x-(4-60 x) \log (\frac {3}{2})+(-4+4 \log (\frac {3}{2})) \log (x)+(-2+2 \log (\frac {3}{2})) \log ^2(x)}{5 \log (\frac {3}{2})} \, dx\)

Optimal. Leaf size=29 \[ 2 \left (x-\frac {x}{\log \left (\frac {3}{2}\right )}\right ) \left (-3 x+\frac {1}{5} \left (2-\log ^2(x)\right )\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 53, normalized size of antiderivative = 1.83, number of steps used = 5, number of rules used = 3, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {12, 2295, 2296} \begin {gather*} \frac {6 x^2}{\log \left (\frac {3}{2}\right )}-\frac {2}{75} (1-15 x)^2-\frac {2}{5} x \left (1-\frac {1}{\log \left (\frac {3}{2}\right )}\right ) \log ^2(x)-\frac {4 x}{5 \log \left (\frac {3}{2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-1/5*(4 - 60*x - (4 - 60*x)*Log[3/2] + (-4 + 4*Log[3/2])*Log[x] + (-2 + 2*Log[3/2])*Log[x]^2)/Log[3/2],x]

[Out]

(-2*(1 - 15*x)^2)/75 - (4*x)/(5*Log[3/2]) + (6*x^2)/Log[3/2] - (2*x*(1 - Log[3/2]^(-1))*Log[x]^2)/5

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 {\int \left (4-60 x-(4-60 x) \log \left (\frac {3}{2}\right )+\left (-4+4 \log \left (\frac {3}{2}\right )\right ) \log (x)+\left (-2+2 \log \left (\frac {3}{2}\right )\right ) \log ^2(x)\right ) \, dx}{5 \log \left (\frac {3}{2}\right )}\\ &=-\frac {2}{75} (1-15 x)^2-\frac {4 x}{5 \log \left (\frac {3}{2}\right )}+\frac {6 x^2}{\log \left (\frac {3}{2}\right )}-\frac {1}{5} \left (2 \left (1-\frac {1}{\log \left (\frac {3}{2}\right )}\right )\right ) \int \log ^2(x) \, dx-\frac {1}{5} \left (4 \left (1-\frac {1}{\log \left (\frac {3}{2}\right )}\right )\right ) \int \log (x) \, dx\\ &=-\frac {2}{75} (1-15 x)^2+\frac {4}{5} x \left (1-\frac {1}{\log \left (\frac {3}{2}\right )}\right )-\frac {4 x}{5 \log \left (\frac {3}{2}\right )}+\frac {6 x^2}{\log \left (\frac {3}{2}\right )}-\frac {4}{5} x \left (1-\frac {1}{\log \left (\frac {3}{2}\right )}\right ) \log (x)-\frac {2}{5} x \left (1-\frac {1}{\log \left (\frac {3}{2}\right )}\right ) \log ^2(x)+\frac {1}{5} \left (4 \left (1-\frac {1}{\log \left (\frac {3}{2}\right )}\right )\right ) \int \log (x) \, dx\\ &=-\frac {2}{75} (1-15 x)^2-\frac {4 x}{5 \log \left (\frac {3}{2}\right )}+\frac {6 x^2}{\log \left (\frac {3}{2}\right )}-\frac {2}{5} x \left (1-\frac {1}{\log \left (\frac {3}{2}\right )}\right ) \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 31, normalized size = 1.07 \begin {gather*} -\frac {2 \left (-1+\log \left (\frac {3}{2}\right )\right ) \left (-2 x+15 x^2+x \log ^2(x)\right )}{5 \log \left (\frac {3}{2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-1/5*(4 - 60*x - (4 - 60*x)*Log[3/2] + (-4 + 4*Log[3/2])*Log[x] + (-2 + 2*Log[3/2])*Log[x]^2)/Log[3/
2],x]

[Out]

(-2*(-1 + Log[3/2])*(-2*x + 15*x^2 + x*Log[x]^2))/(5*Log[3/2])

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fricas [A]  time = 0.56, size = 38, normalized size = 1.31 \begin {gather*} -\frac {2 \, {\left ({\left (x \log \left (\frac {2}{3}\right ) + x\right )} \log \relax (x)^{2} + 15 \, x^{2} + {\left (15 \, x^{2} - 2 \, x\right )} \log \left (\frac {2}{3}\right ) - 2 \, x\right )}}{5 \, \log \left (\frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-2*log(2/3)-2)*log(x)^2+(-4*log(2/3)-4)*log(x)+(-60*x+4)*log(2/3)-60*x+4)/log(2/3),x, algorith
m="fricas")

[Out]

-2/5*((x*log(2/3) + x)*log(x)^2 + 15*x^2 + (15*x^2 - 2*x)*log(2/3) - 2*x)/log(2/3)

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giac [B]  time = 0.13, size = 61, normalized size = 2.10 \begin {gather*} -\frac {2 \, {\left (15 \, x^{2} + {\left (x \log \relax (x)^{2} - 2 \, x \log \relax (x) + 2 \, x\right )} {\left (\log \left (\frac {2}{3}\right ) + 1\right )} + 2 \, {\left (x \log \relax (x) - x\right )} {\left (\log \left (\frac {2}{3}\right ) + 1\right )} + {\left (15 \, x^{2} - 2 \, x\right )} \log \left (\frac {2}{3}\right ) - 2 \, x\right )}}{5 \, \log \left (\frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-2*log(2/3)-2)*log(x)^2+(-4*log(2/3)-4)*log(x)+(-60*x+4)*log(2/3)-60*x+4)/log(2/3),x, algorith
m="giac")

[Out]

-2/5*(15*x^2 + (x*log(x)^2 - 2*x*log(x) + 2*x)*(log(2/3) + 1) + 2*(x*log(x) - x)*(log(2/3) + 1) + (15*x^2 - 2*
x)*log(2/3) - 2*x)/log(2/3)

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maple [B]  time = 0.06, size = 44, normalized size = 1.52

method	result	size
default	\(\frac {4 x -30 \ln \left (\frac {2}{3}\right ) x^{2}+4 \ln \left (\frac {2}{3}\right ) x -2 \ln \left (\frac {2}{3}\right ) \ln \relax (x )^{2} x -2 x \ln \relax (x )^{2}-30 x^{2}}{5 \ln \left (\frac {2}{3}\right )}\)	\(44\)
norman	\(\frac {4 \left (\ln \relax (3)-\ln \relax (2)-1\right ) x}{5 \left (\ln \relax (3)-\ln \relax (2)\right )}-\frac {6 \left (\ln \relax (3)-\ln \relax (2)-1\right ) x^{2}}{\ln \relax (3)-\ln \relax (2)}-\frac {2 \left (\ln \relax (3)-\ln \relax (2)-1\right ) x \ln \relax (x )^{2}}{5 \left (\ln \relax (3)-\ln \relax (2)\right )}\)	\(68\)
risch	\(\frac {2 \left (\ln \relax (3)-\ln \relax (2)-1\right ) x \ln \relax (x )^{2}}{5 \left (\ln \relax (2)-\ln \relax (3)\right )}+\frac {6 x^{2} \ln \relax (3)}{\ln \relax (2)-\ln \relax (3)}-\frac {6 x^{2} \ln \relax (2)}{\ln \relax (2)-\ln \relax (3)}-\frac {4 x \ln \relax (3)}{5 \left (\ln \relax (2)-\ln \relax (3)\right )}+\frac {4 x \ln \relax (2)}{5 \left (\ln \relax (2)-\ln \relax (3)\right )}-\frac {6 x^{2}}{\ln \relax (2)-\ln \relax (3)}+\frac {4 x}{5 \left (\ln \relax (2)-\ln \relax (3)\right )}\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-2*ln(2/3)-2)*ln(x)^2+(-4*ln(2/3)-4)*ln(x)+(-60*x+4)*ln(2/3)-60*x+4)/ln(2/3),x,method=_RETURNVERBOSE
)

[Out]

1/5/ln(2/3)*(4*x-30*ln(2/3)*x^2+4*ln(2/3)*x-2*ln(2/3)*ln(x)^2*x-2*x*ln(x)^2-30*x^2)

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maxima [B]  time = 0.35, size = 57, normalized size = 1.97 \begin {gather*} -\frac {2 \, {\left ({\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x {\left (\log \left (\frac {2}{3}\right ) + 1\right )} + 15 \, x^{2} + 2 \, {\left (x \log \relax (x) - x\right )} {\left (\log \left (\frac {2}{3}\right ) + 1\right )} + {\left (15 \, x^{2} - 2 \, x\right )} \log \left (\frac {2}{3}\right ) - 2 \, x\right )}}{5 \, \log \left (\frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-2*log(2/3)-2)*log(x)^2+(-4*log(2/3)-4)*log(x)+(-60*x+4)*log(2/3)-60*x+4)/log(2/3),x, algorith
m="maxima")

[Out]

-2/5*((log(x)^2 - 2*log(x) + 2)*x*(log(2/3) + 1) + 15*x^2 + 2*(x*log(x) - x)*(log(2/3) + 1) + (15*x^2 - 2*x)*l
og(2/3) - 2*x)/log(2/3)

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mupad [B]  time = 5.50, size = 38, normalized size = 1.31 \begin {gather*} \frac {x\,\left (\frac {4\,\ln \left (\frac {2}{3}\right )}{5}+\frac {{\ln \relax (x)}^2\,\left (\ln \left (\frac {9}{4}\right )-2\right )}{5}+\frac {4}{5}\right )}{\ln \left (\frac {2}{3}\right )}-\frac {x^2\,\left (6\,\ln \left (\frac {2}{3}\right )+6\right )}{\ln \left (\frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(12*x + (log(2/3)*(60*x - 4))/5 + (log(x)*(4*log(2/3) + 4))/5 + (log(x)^2*(2*log(2/3) + 2))/5 - 4/5)/log(
2/3),x)

[Out]

(x*((4*log(2/3))/5 + (log(x)^2*(log(9/4) - 2))/5 + 4/5))/log(2/3) - (x^2*(6*log(2/3) + 6))/log(2/3)

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sympy [B]  time = 0.21, size = 76, normalized size = 2.62 \begin {gather*} \frac {x^{2} \left (-6 - 6 \log {\relax (2 )} + 6 \log {\relax (3 )}\right )}{- \log {\relax (3 )} + \log {\relax (2 )}} + \frac {x \left (- 4 \log {\relax (3 )} + 4 \log {\relax (2 )} + 4\right )}{- 5 \log {\relax (3 )} + 5 \log {\relax (2 )}} + \frac {\left (- 2 x - 2 x \log {\relax (2 )} + 2 x \log {\relax (3 )}\right ) \log {\relax (x )}^{2}}{- 5 \log {\relax (3 )} + 5 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-2*ln(2/3)-2)*ln(x)**2+(-4*ln(2/3)-4)*ln(x)+(-60*x+4)*ln(2/3)-60*x+4)/ln(2/3),x)

[Out]

x**2*(-6 - 6*log(2) + 6*log(3))/(-log(3) + log(2)) + x*(-4*log(3) + 4*log(2) + 4)/(-5*log(3) + 5*log(2)) + (-2
*x - 2*x*log(2) + 2*x*log(3))*log(x)**2/(-5*log(3) + 5*log(2))

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