3.12.12 \(\int \frac {450+4608 x+(-100-2688 x) \log (x^3)+(5+584 x) \log ^2(x^3)-56 x \log ^3(x^3)+2 x \log ^4(x^3)}{2304-1344 \log (x^3)+292 \log ^2(x^3)-28 \log ^3(x^3)+\log ^4(x^3)} \, dx\)

Optimal. Leaf size=20 \[ x \left (x+\frac {5}{-1+\left (7-\log \left (x^3\right )\right )^2}\right ) \]

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Rubi [A]  time = 0.37, antiderivative size = 34, normalized size of antiderivative = 1.70, number of steps used = 13, number of rules used = 5, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {6741, 6742, 2297, 2300, 2178} \begin {gather*} \frac {5 x}{2 \left (6-\log \left (x^3\right )\right )}-\frac {5 x}{2 \left (8-\log \left (x^3\right )\right )}+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(450 + 4608*x + (-100 - 2688*x)*Log[x^3] + (5 + 584*x)*Log[x^3]^2 - 56*x*Log[x^3]^3 + 2*x*Log[x^3]^4)/(230
4 - 1344*Log[x^3] + 292*Log[x^3]^2 - 28*Log[x^3]^3 + Log[x^3]^4),x]

[Out]

x^2 + (5*x)/(2*(6 - Log[x^3])) - (5*x)/(2*(8 - Log[x^3]))

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2300

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {450+4608 x+(-100-2688 x) \log \left (x^3\right )+(5+584 x) \log ^2\left (x^3\right )-56 x \log ^3\left (x^3\right )+2 x \log ^4\left (x^3\right )}{\left (48-14 \log \left (x^3\right )+\log ^2\left (x^3\right )\right )^2} \, dx\\ &=\int \left (2 x-\frac {15}{2 \left (-8+\log \left (x^3\right )\right )^2}+\frac {5}{2 \left (-8+\log \left (x^3\right )\right )}+\frac {15}{2 \left (-6+\log \left (x^3\right )\right )^2}-\frac {5}{2 \left (-6+\log \left (x^3\right )\right )}\right ) \, dx\\ &=x^2+\frac {5}{2} \int \frac {1}{-8+\log \left (x^3\right )} \, dx-\frac {5}{2} \int \frac {1}{-6+\log \left (x^3\right )} \, dx-\frac {15}{2} \int \frac {1}{\left (-8+\log \left (x^3\right )\right )^2} \, dx+\frac {15}{2} \int \frac {1}{\left (-6+\log \left (x^3\right )\right )^2} \, dx\\ &=x^2+\frac {5 x}{2 \left (6-\log \left (x^3\right )\right )}-\frac {5 x}{2 \left (8-\log \left (x^3\right )\right )}-\frac {5}{2} \int \frac {1}{-8+\log \left (x^3\right )} \, dx+\frac {5}{2} \int \frac {1}{-6+\log \left (x^3\right )} \, dx+\frac {(5 x) \operatorname {Subst}\left (\int \frac {e^{x/3}}{-8+x} \, dx,x,\log \left (x^3\right )\right )}{6 \sqrt [3]{x^3}}-\frac {(5 x) \operatorname {Subst}\left (\int \frac {e^{x/3}}{-6+x} \, dx,x,\log \left (x^3\right )\right )}{6 \sqrt [3]{x^3}}\\ &=x^2+\frac {5 e^{8/3} x \text {Ei}\left (\frac {1}{3} \left (-8+\log \left (x^3\right )\right )\right )}{6 \sqrt [3]{x^3}}-\frac {5 e^2 x \text {Ei}\left (\frac {1}{3} \left (-6+\log \left (x^3\right )\right )\right )}{6 \sqrt [3]{x^3}}+\frac {5 x}{2 \left (6-\log \left (x^3\right )\right )}-\frac {5 x}{2 \left (8-\log \left (x^3\right )\right )}-\frac {(5 x) \operatorname {Subst}\left (\int \frac {e^{x/3}}{-8+x} \, dx,x,\log \left (x^3\right )\right )}{6 \sqrt [3]{x^3}}+\frac {(5 x) \operatorname {Subst}\left (\int \frac {e^{x/3}}{-6+x} \, dx,x,\log \left (x^3\right )\right )}{6 \sqrt [3]{x^3}}\\ &=x^2+\frac {5 x}{2 \left (6-\log \left (x^3\right )\right )}-\frac {5 x}{2 \left (8-\log \left (x^3\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 23, normalized size = 1.15 \begin {gather*} x^2+\frac {5 x}{48-14 \log \left (x^3\right )+\log ^2\left (x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(450 + 4608*x + (-100 - 2688*x)*Log[x^3] + (5 + 584*x)*Log[x^3]^2 - 56*x*Log[x^3]^3 + 2*x*Log[x^3]^4
)/(2304 - 1344*Log[x^3] + 292*Log[x^3]^2 - 28*Log[x^3]^3 + Log[x^3]^4),x]

[Out]

x^2 + (5*x)/(48 - 14*Log[x^3] + Log[x^3]^2)

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fricas [B]  time = 0.59, size = 45, normalized size = 2.25 \begin {gather*} \frac {x^{2} \log \left (x^{3}\right )^{2} - 14 \, x^{2} \log \left (x^{3}\right ) + 48 \, x^{2} + 5 \, x}{\log \left (x^{3}\right )^{2} - 14 \, \log \left (x^{3}\right ) + 48} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(x^3)^4-56*x*log(x^3)^3+(584*x+5)*log(x^3)^2+(-2688*x-100)*log(x^3)+4608*x+450)/(log(x^3)^4-
28*log(x^3)^3+292*log(x^3)^2-1344*log(x^3)+2304),x, algorithm="fricas")

[Out]

(x^2*log(x^3)^2 - 14*x^2*log(x^3) + 48*x^2 + 5*x)/(log(x^3)^2 - 14*log(x^3) + 48)

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giac [A]  time = 0.41, size = 23, normalized size = 1.15 \begin {gather*} x^{2} + \frac {5 \, x}{\log \left (x^{3}\right )^{2} - 14 \, \log \left (x^{3}\right ) + 48} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(x^3)^4-56*x*log(x^3)^3+(584*x+5)*log(x^3)^2+(-2688*x-100)*log(x^3)+4608*x+450)/(log(x^3)^4-
28*log(x^3)^3+292*log(x^3)^2-1344*log(x^3)+2304),x, algorithm="giac")

[Out]

x^2 + 5*x/(log(x^3)^2 - 14*log(x^3) + 48)

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maple [A]  time = 0.03, size = 24, normalized size = 1.20

method	result	size
risch	\(x^{2}+\frac {5 x}{\ln \left (x^{3}\right )^{2}-14 \ln \left (x^{3}\right )+48}\)	\(24\)
norman	\(\frac {x^{2} \ln \left (x^{3}\right )^{2}+5 x +48 x^{2}-14 x^{2} \ln \left (x^{3}\right )}{\ln \left (x^{3}\right )^{2}-14 \ln \left (x^{3}\right )+48}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*ln(x^3)^4-56*x*ln(x^3)^3+(584*x+5)*ln(x^3)^2+(-2688*x-100)*ln(x^3)+4608*x+450)/(ln(x^3)^4-28*ln(x^3)^
3+292*ln(x^3)^2-1344*ln(x^3)+2304),x,method=_RETURNVERBOSE)

[Out]

x^2+5*x/(ln(x^3)^2-14*ln(x^3)+48)

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maxima [B]  time = 0.50, size = 41, normalized size = 2.05 \begin {gather*} \frac {9 \, x^{2} \log \relax (x)^{2} - 42 \, x^{2} \log \relax (x) + 48 \, x^{2} + 5 \, x}{3 \, {\left (3 \, \log \relax (x)^{2} - 14 \, \log \relax (x) + 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(x^3)^4-56*x*log(x^3)^3+(584*x+5)*log(x^3)^2+(-2688*x-100)*log(x^3)+4608*x+450)/(log(x^3)^4-
28*log(x^3)^3+292*log(x^3)^2-1344*log(x^3)+2304),x, algorithm="maxima")

[Out]

1/3*(9*x^2*log(x)^2 - 42*x^2*log(x) + 48*x^2 + 5*x)/(3*log(x)^2 - 14*log(x) + 16)

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mupad [B]  time = 1.06, size = 34, normalized size = 1.70 \begin {gather*} x^2+\frac {x\,\left (48\,x+5\right )-48\,x^2}{\left (\ln \left (x^3\right )-6\right )\,\left (\ln \left (x^3\right )-8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4608*x + log(x^3)^2*(584*x + 5) - 56*x*log(x^3)^3 + 2*x*log(x^3)^4 - log(x^3)*(2688*x + 100) + 450)/(292*
log(x^3)^2 - 1344*log(x^3) - 28*log(x^3)^3 + log(x^3)^4 + 2304),x)

[Out]

x^2 + (x*(48*x + 5) - 48*x^2)/((log(x^3) - 6)*(log(x^3) - 8))

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sympy [A]  time = 0.12, size = 20, normalized size = 1.00 \begin {gather*} x^{2} + \frac {5 x}{\log {\left (x^{3} \right )}^{2} - 14 \log {\left (x^{3} \right )} + 48} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*ln(x**3)**4-56*x*ln(x**3)**3+(584*x+5)*ln(x**3)**2+(-2688*x-100)*ln(x**3)+4608*x+450)/(ln(x**3)
**4-28*ln(x**3)**3+292*ln(x**3)**2-1344*ln(x**3)+2304),x)

[Out]

x**2 + 5*x/(log(x**3)**2 - 14*log(x**3) + 48)

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