3.79.6 \(\int \frac {1}{10} (10-39 x+9 x^2-4 x^3+2 x \log (\frac {x}{\log ^2(4)})) \, dx\)

Optimal. Leaf size=31 \[ x-\frac {1}{2} x^2 \left (4+\frac {1}{5} \left (-3 x+x^2-\log \left (\frac {x}{\log ^2(4)}\right )\right )\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 35, normalized size of antiderivative = 1.13, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {12, 2304} \begin {gather*} -\frac {x^4}{10}+\frac {3 x^3}{10}-2 x^2+\frac {1}{10} x^2 \log \left (\frac {x}{\log ^2(4)}\right )+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 - 39*x + 9*x^2 - 4*x^3 + 2*x*Log[x/Log[4]^2])/10,x]

[Out]

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

Rule 12

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

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{10} \int \left (10-39 x+9 x^2-4 x^3+2 x \log \left (\frac {x}{\log ^2(4)}\right )\right ) \, dx\\ &=x-\frac {39 x^2}{20}+\frac {3 x^3}{10}-\frac {x^4}{10}+\frac {1}{5} \int x \log \left (\frac {x}{\log ^2(4)}\right ) \, dx\\ &=x-2 x^2+\frac {3 x^3}{10}-\frac {x^4}{10}+\frac {1}{10} x^2 \log \left (\frac {x}{\log ^2(4)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 35, normalized size = 1.13 \begin {gather*} x-2 x^2+\frac {3 x^3}{10}-\frac {x^4}{10}+\frac {1}{10} x^2 \log \left (\frac {x}{\log ^2(4)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 - 39*x + 9*x^2 - 4*x^3 + 2*x*Log[x/Log[4]^2])/10,x]

[Out]

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

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fricas [A]  time = 1.11, size = 30, normalized size = 0.97 \begin {gather*} -\frac {1}{10} \, x^{4} + \frac {3}{10} \, x^{3} + \frac {1}{10} \, x^{2} \log \left (\frac {x}{4 \, \log \relax (2)^{2}}\right ) - 2 \, x^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*x*log(1/4*x/log(2)^2)-2/5*x^3+9/10*x^2-39/10*x+1,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 30, normalized size = 0.97 \begin {gather*} -\frac {1}{10} \, x^{4} + \frac {3}{10} \, x^{3} + \frac {1}{10} \, x^{2} \log \left (\frac {x}{4 \, \log \relax (2)^{2}}\right ) - 2 \, x^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*x*log(1/4*x/log(2)^2)-2/5*x^3+9/10*x^2-39/10*x+1,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 31, normalized size = 1.00

method	result	size
default	\(x -2 x^{2}+\frac {3 x^{3}}{10}-\frac {x^{4}}{10}+\frac {x^{2} \ln \left (\frac {x}{4 \ln \relax (2)^{2}}\right )}{10}\)	\(31\)
norman	\(x -2 x^{2}+\frac {3 x^{3}}{10}-\frac {x^{4}}{10}+\frac {x^{2} \ln \left (\frac {x}{4 \ln \relax (2)^{2}}\right )}{10}\)	\(31\)
risch	\(x -2 x^{2}+\frac {3 x^{3}}{10}-\frac {x^{4}}{10}+\frac {x^{2} \ln \left (\frac {x}{4 \ln \relax (2)^{2}}\right )}{10}\)	\(31\)
derivativedivides	\(\frac {4 \ln \relax (2)^{2} \left (\frac {5 x}{4 \ln \relax (2)^{2}}-\frac {5 x^{2}}{2 \ln \relax (2)^{2}}+\frac {3 x^{3}}{8 \ln \relax (2)^{2}}-\frac {x^{4}}{8 \ln \relax (2)^{2}}+\frac {x^{2} \ln \left (\frac {x}{4 \ln \relax (2)^{2}}\right )}{8 \ln \relax (2)^{2}}\right )}{5}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*x*ln(1/4*x/ln(2)^2)-2/5*x^3+9/10*x^2-39/10*x+1,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.36, size = 30, normalized size = 0.97 \begin {gather*} -\frac {1}{10} \, x^{4} + \frac {3}{10} \, x^{3} + \frac {1}{10} \, x^{2} \log \left (\frac {x}{4 \, \log \relax (2)^{2}}\right ) - 2 \, x^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*x*log(1/4*x/log(2)^2)-2/5*x^3+9/10*x^2-39/10*x+1,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.15, size = 32, normalized size = 1.03 \begin {gather*} -\frac {x\,\left (20\,x+2\,x\,\ln \relax (2)+2\,x\,\ln \left (\ln \relax (2)\right )-x\,\ln \relax (x)-3\,x^2+x^3-10\right )}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*x^2)/10 - (39*x)/10 - (2*x^3)/5 + (x*log(x/(4*log(2)^2)))/5 + 1,x)

[Out]

-(x*(20*x + 2*x*log(2) + 2*x*log(log(2)) - x*log(x) - 3*x^2 + x^3 - 10))/10

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sympy [A]  time = 0.11, size = 32, normalized size = 1.03 \begin {gather*} - \frac {x^{4}}{10} + \frac {3 x^{3}}{10} + \frac {x^{2} \log {\left (\frac {x}{4 \log {\relax (2 )}^{2}} \right )}}{10} - 2 x^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*x*ln(1/4*x/ln(2)**2)-2/5*x**3+9/10*x**2-39/10*x+1,x)

[Out]

-x**4/10 + 3*x**3/10 + x**2*log(x/(4*log(2)**2))/10 - 2*x**2 + x

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