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

Optimal. Leaf size=25 \[ -16-x \left (x+\frac {\log ^2\left (\frac {x}{2}\right )}{x}\right )^2-\log (x) \]

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Rubi [B]  time = 0.14, antiderivative size = 52, normalized size of antiderivative = 2.08, number of steps used = 14, number of rules used = 5, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {14, 2296, 2295, 2305, 2304} \begin {gather*} -x^3-\frac {\log ^4\left (\frac {x}{2}\right )}{x}-2 x \log ^2\left (\frac {x}{2}\right )+4 x \log \left (\frac {x}{2}\right )-4 x \log (x)-\log (x)+x \log (16) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x^3 + x*Log[16] + 4*x*Log[x/2] - 2*x*Log[x/2]^2 - Log[x/2]^4/x - Log[x] - 4*x*Log[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 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 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
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-1-3 x^3}{x}+\log (16)-2 \log ^2\left (\frac {x}{2}\right )-\frac {4 \log ^3\left (\frac {x}{2}\right )}{x^2}+\frac {\log ^4\left (\frac {x}{2}\right )}{x^2}-4 \log (x)\right ) \, dx\\ &=x \log (16)-2 \int \log ^2\left (\frac {x}{2}\right ) \, dx-4 \int \frac {\log ^3\left (\frac {x}{2}\right )}{x^2} \, dx-4 \int \log (x) \, dx+\int \frac {-1-3 x^3}{x} \, dx+\int \frac {\log ^4\left (\frac {x}{2}\right )}{x^2} \, dx\\ &=4 x+x \log (16)-2 x \log ^2\left (\frac {x}{2}\right )+\frac {4 \log ^3\left (\frac {x}{2}\right )}{x}-\frac {\log ^4\left (\frac {x}{2}\right )}{x}-4 x \log (x)+4 \int \log \left (\frac {x}{2}\right ) \, dx+4 \int \frac {\log ^3\left (\frac {x}{2}\right )}{x^2} \, dx-12 \int \frac {\log ^2\left (\frac {x}{2}\right )}{x^2} \, dx+\int \left (-\frac {1}{x}-3 x^2\right ) \, dx\\ &=-x^3+x \log (16)+4 x \log \left (\frac {x}{2}\right )+\frac {12 \log ^2\left (\frac {x}{2}\right )}{x}-2 x \log ^2\left (\frac {x}{2}\right )-\frac {\log ^4\left (\frac {x}{2}\right )}{x}-\log (x)-4 x \log (x)+12 \int \frac {\log ^2\left (\frac {x}{2}\right )}{x^2} \, dx-24 \int \frac {\log \left (\frac {x}{2}\right )}{x^2} \, dx\\ &=\frac {24}{x}-x^3+x \log (16)+\frac {24 \log \left (\frac {x}{2}\right )}{x}+4 x \log \left (\frac {x}{2}\right )-2 x \log ^2\left (\frac {x}{2}\right )-\frac {\log ^4\left (\frac {x}{2}\right )}{x}-\log (x)-4 x \log (x)+24 \int \frac {\log \left (\frac {x}{2}\right )}{x^2} \, dx\\ &=-x^3+x \log (16)+4 x \log \left (\frac {x}{2}\right )-2 x \log ^2\left (\frac {x}{2}\right )-\frac {\log ^4\left (\frac {x}{2}\right )}{x}-\log (x)-4 x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 34, normalized size = 1.36 \begin {gather*} -x^3-2 x \log ^2\left (\frac {x}{2}\right )-\frac {\log ^4\left (\frac {x}{2}\right )}{x}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x^3 - 2*x*Log[x/2]^2 - Log[x/2]^4/x - Log[x]

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fricas [A]  time = 0.54, size = 32, normalized size = 1.28 \begin {gather*} -\frac {x^{4} + 2 \, x^{2} \log \left (\frac {1}{2} \, x\right )^{2} + \log \left (\frac {1}{2} \, x\right )^{4} + x \log \left (\frac {1}{2} \, x\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.53, size = 30, normalized size = 1.20 \begin {gather*} -x^{3} - 2 \, x \log \left (\frac {1}{2} \, x\right )^{2} - \frac {\log \left (\frac {1}{2} \, x\right )^{4}}{x} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 31, normalized size = 1.24

method	result	size
risch	\(-\frac {\ln \left (\frac {x}{2}\right )^{4}}{x}-2 x \ln \left (\frac {x}{2}\right )^{2}-x^{3}-\ln \relax (x )\)	\(31\)
derivativedivides	\(-\frac {\ln \left (\frac {x}{2}\right )^{4}}{x}-2 x \ln \left (\frac {x}{2}\right )^{2}-x^{3}-\ln \left (\frac {x}{2}\right )\)	\(33\)
default	\(-\frac {\ln \left (\frac {x}{2}\right )^{4}}{x}-2 x \ln \left (\frac {x}{2}\right )^{2}-x^{3}-\ln \left (\frac {x}{2}\right )\)	\(33\)
norman	\(\frac {-x^{4}-\ln \left (\frac {x}{2}\right )^{4}-2 x^{2} \ln \left (\frac {x}{2}\right )^{2}}{x}-\ln \relax (x )\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.54, size = 99, normalized size = 3.96 \begin {gather*} -x^{3} - 2 \, {\left (\log \left (\frac {1}{2} \, x\right )^{2} - 2 \, \log \left (\frac {1}{2} \, x\right ) + 2\right )} x - 4 \, x \log \left (\frac {1}{2} \, x\right ) + 4 \, x - \frac {\log \left (\frac {1}{2} \, x\right )^{4} + 4 \, \log \left (\frac {1}{2} \, x\right )^{3} + 12 \, \log \left (\frac {1}{2} \, x\right )^{2} + 24 \, \log \left (\frac {1}{2} \, x\right ) + 24}{x} + \frac {4 \, {\left (\log \left (\frac {1}{2} \, x\right )^{3} + 3 \, \log \left (\frac {1}{2} \, x\right )^{2} + 6 \, \log \left (\frac {1}{2} \, x\right ) + 6\right )}}{x} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^3 - 2*(log(1/2*x)^2 - 2*log(1/2*x) + 2)*x - 4*x*log(1/2*x) + 4*x - (log(1/2*x)^4 + 4*log(1/2*x)^3 + 12*log(
1/2*x)^2 + 24*log(1/2*x) + 24)/x + 4*(log(1/2*x)^3 + 3*log(1/2*x)^2 + 6*log(1/2*x) + 6)/x - log(x)

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mupad [B]  time = 1.26, size = 30, normalized size = 1.20 \begin {gather*} -\ln \relax (x)-2\,x\,{\ln \left (\frac {x}{2}\right )}^2-x^3-\frac {{\ln \left (\frac {x}{2}\right )}^4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- log(x) - 2*x*log(x/2)^2 - x^3 - log(x/2)^4/x

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sympy [A]  time = 0.15, size = 26, normalized size = 1.04 \begin {gather*} - x^{3} - 2 x \log {\left (\frac {x}{2} \right )}^{2} - \log {\relax (x )} - \frac {\log {\left (\frac {x}{2} \right )}^{4}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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