3.87.29 \(\int \frac {(-48 x+2 x^2+2 x^3) \log (x)+(-48 x+3 x^2+4 x^3) \log ^2(x)}{-64+(-24 x^2+x^3+x^4) \log ^2(x)} \, dx\)

Optimal. Leaf size=24 \[ \log \left (16+\frac {1}{2} x^2 \left (12-\frac {1}{2} x (1+x)\right ) \log ^2(x)\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps used = 2, number of rules used = 2, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6708, 31} \begin {gather*} \log \left (64-\left (x^4+x^3-24 x^2\right ) \log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-48*x + 2*x^2 + 2*x^3)*Log[x] + (-48*x + 3*x^2 + 4*x^3)*Log[x]^2)/(-64 + (-24*x^2 + x^3 + x^4)*Log[x]^2)
,x]

[Out]

Log[64 - (-24*x^2 + x^3 + x^4)*Log[x]^2]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 6708

Int[(u_)*((a_) + (b_.)*(v_)^(p_.)*(w_)^(p_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(w*D[v, x] + v*D[w, x])
]}, Dist[c, Subst[Int[(a + b*x^p)^m, x], x, v*w], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p}, x] && IntegerQ[p]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {1}{-64+x} \, dx,x,\left (-24 x^2+x^3+x^4\right ) \log ^2(x)\right )\\ &=\log \left (64-\left (-24 x^2+x^3+x^4\right ) \log ^2(x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 2.87, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-48 x+2 x^2+2 x^3\right ) \log (x)+\left (-48 x+3 x^2+4 x^3\right ) \log ^2(x)}{-64+\left (-24 x^2+x^3+x^4\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((-48*x + 2*x^2 + 2*x^3)*Log[x] + (-48*x + 3*x^2 + 4*x^3)*Log[x]^2)/(-64 + (-24*x^2 + x^3 + x^4)*Log
[x]^2),x]

[Out]

Integrate[((-48*x + 2*x^2 + 2*x^3)*Log[x] + (-48*x + 3*x^2 + 4*x^3)*Log[x]^2)/(-64 + (-24*x^2 + x^3 + x^4)*Log
[x]^2), x]

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fricas [B]  time = 0.46, size = 47, normalized size = 1.96 \begin {gather*} \log \left (x^{2} + x - 24\right ) + 2 \, \log \relax (x) + \log \left (\frac {{\left (x^{4} + x^{3} - 24 \, x^{2}\right )} \log \relax (x)^{2} - 64}{x^{4} + x^{3} - 24 \, x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+3*x^2-48*x)*log(x)^2+(2*x^3+2*x^2-48*x)*log(x))/((x^4+x^3-24*x^2)*log(x)^2-64),x, algorithm=
"fricas")

[Out]

log(x^2 + x - 24) + 2*log(x) + log(((x^4 + x^3 - 24*x^2)*log(x)^2 - 64)/(x^4 + x^3 - 24*x^2))

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giac [A]  time = 0.22, size = 28, normalized size = 1.17 \begin {gather*} \log \left (x^{4} \log \relax (x)^{2} + x^{3} \log \relax (x)^{2} - 24 \, x^{2} \log \relax (x)^{2} - 64\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+3*x^2-48*x)*log(x)^2+(2*x^3+2*x^2-48*x)*log(x))/((x^4+x^3-24*x^2)*log(x)^2-64),x, algorithm=
"giac")

[Out]

log(x^4*log(x)^2 + x^3*log(x)^2 - 24*x^2*log(x)^2 - 64)

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maple [A]  time = 0.05, size = 29, normalized size = 1.21

method	result	size
norman	\(\ln \left (x^{4} \ln \relax (x )^{2}+x^{3} \ln \relax (x )^{2}-24 x^{2} \ln \relax (x )^{2}-64\right )\)	\(29\)
risch	\(2 \ln \relax (x )+\ln \left (x^{2}+x -24\right )+\ln \left (\ln \relax (x )^{2}-\frac {64}{x^{2} \left (x^{2}+x -24\right )}\right )\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^3+3*x^2-48*x)*ln(x)^2+(2*x^3+2*x^2-48*x)*ln(x))/((x^4+x^3-24*x^2)*ln(x)^2-64),x,method=_RETURNVERBOS
E)

[Out]

ln(x^4*ln(x)^2+x^3*ln(x)^2-24*x^2*ln(x)^2-64)

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maxima [B]  time = 0.40, size = 47, normalized size = 1.96 \begin {gather*} \log \left (x^{2} + x - 24\right ) + 2 \, \log \relax (x) + \log \left (\frac {{\left (x^{4} + x^{3} - 24 \, x^{2}\right )} \log \relax (x)^{2} - 64}{x^{4} + x^{3} - 24 \, x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+3*x^2-48*x)*log(x)^2+(2*x^3+2*x^2-48*x)*log(x))/((x^4+x^3-24*x^2)*log(x)^2-64),x, algorithm=
"maxima")

[Out]

log(x^2 + x - 24) + 2*log(x) + log(((x^4 + x^3 - 24*x^2)*log(x)^2 - 64)/(x^4 + x^3 - 24*x^2))

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mupad [B]  time = 5.29, size = 28, normalized size = 1.17 \begin {gather*} \ln \left (x^4\,{\ln \relax (x)}^2+x^3\,{\ln \relax (x)}^2-24\,x^2\,{\ln \relax (x)}^2-64\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)^2*(3*x^2 - 48*x + 4*x^3) + log(x)*(2*x^2 - 48*x + 2*x^3))/(log(x)^2*(x^3 - 24*x^2 + x^4) - 64),x)

[Out]

log(x^3*log(x)^2 - 24*x^2*log(x)^2 + x^4*log(x)^2 - 64)

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sympy [A]  time = 0.66, size = 32, normalized size = 1.33 \begin {gather*} 2 \log {\relax (x )} + \log {\left (\log {\relax (x )}^{2} - \frac {64}{x^{4} + x^{3} - 24 x^{2}} \right )} + \log {\left (x^{2} + x - 24 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**3+3*x**2-48*x)*ln(x)**2+(2*x**3+2*x**2-48*x)*ln(x))/((x**4+x**3-24*x**2)*ln(x)**2-64),x)

[Out]

2*log(x) + log(log(x)**2 - 64/(x**4 + x**3 - 24*x**2)) + log(x**2 + x - 24)

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