3.25.62 \(\int \frac {120 x^2-50 x^3 \log (x)+50 x^3 \log (2 x)}{64+80 x^3+25 x^6+(-80 x-50 x^4) \log (x)+25 x^2 \log ^2(x)+(80 x+50 x^4-50 x^2 \log (x)) \log (2 x)+25 x^2 \log ^2(2 x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {x^2}{\frac {8}{5 x}+x^2-\log (x)+\log (2 x)} \]

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Rubi [A]  time = 0.23, antiderivative size = 34, normalized size of antiderivative = 1.36, number of steps used = 4, number of rules used = 4, integrand size = 90, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {6688, 12, 2102, 1588} \begin {gather*} -\frac {x \log (32)}{5 x^3+x \log (32)+8}-\frac {8}{5 x^3+x \log (32)+8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(120*x^2 - 50*x^3*Log[x] + 50*x^3*Log[2*x])/(64 + 80*x^3 + 25*x^6 + (-80*x - 50*x^4)*Log[x] + 25*x^2*Log[x
]^2 + (80*x + 50*x^4 - 50*x^2*Log[x])*Log[2*x] + 25*x^2*Log[2*x]^2),x]

[Out]

-8/(8 + 5*x^3 + x*Log[32]) - (x*Log[32])/(8 + 5*x^3 + x*Log[32])

Rule 12

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

Rule 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rule 2102

Int[(Pm_)*(Qn_)^(p_.), x_Symbol] :> With[{m = Expon[Pm, x], n = Expon[Qn, x]}, Simp[(Coeff[Pm, x, m]*x^(m - n
+ 1)*Qn^(p + 1))/((m + n*p + 1)*Coeff[Qn, x, n]), x] + Dist[1/((m + n*p + 1)*Coeff[Qn, x, n]), Int[ExpandToSum
[(m + n*p + 1)*Coeff[Qn, x, n]*Pm - Coeff[Pm, x, m]*x^(m - n)*((m - n + 1)*Qn + (p + 1)*x*D[Qn, x]), x]*Qn^p,
x], x] /; LtQ[1, n, m + 1] && m + n*p + 1 < 0] /; FreeQ[p, x] && PolyQ[Pm, x] && PolyQ[Qn, x] && LtQ[p, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 x^2 (12+x \log (32))}{\left (8+5 x^3+x \log (32)\right )^2} \, dx\\ &=10 \int \frac {x^2 (12+x \log (32))}{\left (8+5 x^3+x \log (32)\right )^2} \, dx\\ &=-\frac {x \log (32)}{8+5 x^3+x \log (32)}-\int \frac {-120 x^2-8 \log (32)}{\left (8+5 x^3+x \log (32)\right )^2} \, dx\\ &=-\frac {8}{8+5 x^3+x \log (32)}-\frac {x \log (32)}{8+5 x^3+x \log (32)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 0.84 \begin {gather*} \frac {-8-x \log (32)}{8+5 x^3+x \log (32)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(120*x^2 - 50*x^3*Log[x] + 50*x^3*Log[2*x])/(64 + 80*x^3 + 25*x^6 + (-80*x - 50*x^4)*Log[x] + 25*x^2
*Log[x]^2 + (80*x + 50*x^4 - 50*x^2*Log[x])*Log[2*x] + 25*x^2*Log[2*x]^2),x]

[Out]

(-8 - x*Log[32])/(8 + 5*x^3 + x*Log[32])

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fricas [A]  time = 0.69, size = 23, normalized size = 0.92 \begin {gather*} -\frac {5 \, x \log \relax (2) + 8}{5 \, x^{3} + 5 \, x \log \relax (2) + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*x^3*log(2*x)-50*x^3*log(x)+120*x^2)/(25*x^2*log(2*x)^2+(-50*x^2*log(x)+50*x^4+80*x)*log(2*x)+25*
x^2*log(x)^2+(-50*x^4-80*x)*log(x)+25*x^6+80*x^3+64),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 23, normalized size = 0.92 \begin {gather*} -\frac {5 \, x \log \relax (2) + 8}{5 \, x^{3} + 5 \, x \log \relax (2) + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*x^3*log(2*x)-50*x^3*log(x)+120*x^2)/(25*x^2*log(2*x)^2+(-50*x^2*log(x)+50*x^4+80*x)*log(2*x)+25*
x^2*log(x)^2+(-50*x^4-80*x)*log(x)+25*x^6+80*x^3+64),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.16, size = 21, normalized size = 0.84

method	result	size
default	\(\frac {-\frac {8}{5}-x \ln \relax (2)}{x^{3}+x \ln \relax (2)+\frac {8}{5}}\)	\(21\)
risch	\(\frac {-2 i x \ln \relax (2)-\frac {16 i}{5}}{2 i x^{3}+2 i x \ln \relax (2)+\frac {16 i}{5}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*x^3*ln(2*x)-50*x^3*ln(x)+120*x^2)/(25*x^2*ln(2*x)^2+(-50*x^2*ln(x)+50*x^4+80*x)*ln(2*x)+25*x^2*ln(x)^2
+(-50*x^4-80*x)*ln(x)+25*x^6+80*x^3+64),x,method=_RETURNVERBOSE)

[Out]

10*(-4/25-1/10*x*ln(2))/(x^3+x*ln(2)+8/5)

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maxima [A]  time = 0.94, size = 23, normalized size = 0.92 \begin {gather*} -\frac {5 \, x \log \relax (2) + 8}{5 \, x^{3} + 5 \, x \log \relax (2) + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*x^3*log(2*x)-50*x^3*log(x)+120*x^2)/(25*x^2*log(2*x)^2+(-50*x^2*log(x)+50*x^4+80*x)*log(2*x)+25*
x^2*log(x)^2+(-50*x^4-80*x)*log(x)+25*x^6+80*x^3+64),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.47, size = 23, normalized size = 0.92 \begin {gather*} -\frac {5\,x\,\ln \relax (2)+8}{5\,x^3+5\,\ln \relax (2)\,x+8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*x^3*log(2*x) - 50*x^3*log(x) + 120*x^2)/(log(2*x)*(80*x - 50*x^2*log(x) + 50*x^4) + 25*x^2*log(x)^2 -
log(x)*(80*x + 50*x^4) + 80*x^3 + 25*x^6 + 25*x^2*log(2*x)^2 + 64),x)

[Out]

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

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sympy [A]  time = 0.63, size = 22, normalized size = 0.88 \begin {gather*} \frac {- 5 x \log {\relax (2 )} - 8}{5 x^{3} + 5 x \log {\relax (2 )} + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*x**3*ln(2*x)-50*x**3*ln(x)+120*x**2)/(25*x**2*ln(2*x)**2+(-50*x**2*ln(x)+50*x**4+80*x)*ln(2*x)+2
5*x**2*ln(x)**2+(-50*x**4-80*x)*ln(x)+25*x**6+80*x**3+64),x)

[Out]

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

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