3.55.47 \(\int \frac {20+40 x^2-30 x^3-30 x \log (x)}{16 x^3+24 x^4+9 x^5+(-160 x^3-120 x^4) \log (4)+400 x^3 \log ^2(4)+(16 x+24 x^2+9 x^3+(-160 x-120 x^2) \log (4)+400 x \log ^2(4)) \log (x)+(-16 x^3-12 x^4+80 x^3 \log (4)+(-16 x-12 x^2+80 x \log (4)) \log (x)) \log (x^2+\log (x))+(4 x^3+4 x \log (x)) \log ^2(x^2+\log (x))} \, dx\)

Optimal. Leaf size=29 \[ \frac {10}{-x+2 \left (2+2 (x-5 \log (4))-\log \left (x^2+\log (x)\right )\right )} \]

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Rubi [A]  time = 0.36, antiderivative size = 25, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, integrand size = 159, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6, 6688, 12, 6686} \begin {gather*} \frac {10}{-2 \log \left (x^2+\log (x)\right )+3 x+4 (1-5 \log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20 + 40*x^2 - 30*x^3 - 30*x*Log[x])/(16*x^3 + 24*x^4 + 9*x^5 + (-160*x^3 - 120*x^4)*Log[4] + 400*x^3*Log[
4]^2 + (16*x + 24*x^2 + 9*x^3 + (-160*x - 120*x^2)*Log[4] + 400*x*Log[4]^2)*Log[x] + (-16*x^3 - 12*x^4 + 80*x^
3*Log[4] + (-16*x - 12*x^2 + 80*x*Log[4])*Log[x])*Log[x^2 + Log[x]] + (4*x^3 + 4*x*Log[x])*Log[x^2 + Log[x]]^2
),x]

[Out]

10/(3*x + 4*(1 - 5*Log[4]) - 2*Log[x^2 + Log[x]])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -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 {20+40 x^2-30 x^3-30 x \log (x)}{24 x^4+9 x^5+\left (-160 x^3-120 x^4\right ) \log (4)+x^3 \left (16+400 \log ^2(4)\right )+\left (16 x+24 x^2+9 x^3+\left (-160 x-120 x^2\right ) \log (4)+400 x \log ^2(4)\right ) \log (x)+\left (-16 x^3-12 x^4+80 x^3 \log (4)+\left (-16 x-12 x^2+80 x \log (4)\right ) \log (x)\right ) \log \left (x^2+\log (x)\right )+\left (4 x^3+4 x \log (x)\right ) \log ^2\left (x^2+\log (x)\right )} \, dx\\ &=\int \frac {10 \left (2+4 x^2-3 x^3-3 x \log (x)\right )}{x \left (x^2+\log (x)\right ) \left (3 x+4 (1-5 \log (4))-2 \log \left (x^2+\log (x)\right )\right )^2} \, dx\\ &=10 \int \frac {2+4 x^2-3 x^3-3 x \log (x)}{x \left (x^2+\log (x)\right ) \left (3 x+4 (1-5 \log (4))-2 \log \left (x^2+\log (x)\right )\right )^2} \, dx\\ &=\frac {10}{3 x+4 (1-5 \log (4))-2 \log \left (x^2+\log (x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 22, normalized size = 0.76 \begin {gather*} \frac {10}{4+3 x-20 \log (4)-2 \log \left (x^2+\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 + 40*x^2 - 30*x^3 - 30*x*Log[x])/(16*x^3 + 24*x^4 + 9*x^5 + (-160*x^3 - 120*x^4)*Log[4] + 400*x^
3*Log[4]^2 + (16*x + 24*x^2 + 9*x^3 + (-160*x - 120*x^2)*Log[4] + 400*x*Log[4]^2)*Log[x] + (-16*x^3 - 12*x^4 +
80*x^3*Log[4] + (-16*x - 12*x^2 + 80*x*Log[4])*Log[x])*Log[x^2 + Log[x]] + (4*x^3 + 4*x*Log[x])*Log[x^2 + Log
[x]]^2),x]

[Out]

10/(4 + 3*x - 20*Log[4] - 2*Log[x^2 + Log[x]])

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fricas [A]  time = 0.74, size = 22, normalized size = 0.76 \begin {gather*} \frac {10}{3 \, x - 40 \, \log \relax (2) - 2 \, \log \left (x^{2} + \log \relax (x)\right ) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*x*log(x)-30*x^3+40*x^2+20)/((4*x*log(x)+4*x^3)*log(log(x)+x^2)^2+((160*x*log(2)-12*x^2-16*x)*lo
g(x)+160*x^3*log(2)-12*x^4-16*x^3)*log(log(x)+x^2)+(1600*x*log(2)^2+2*(-120*x^2-160*x)*log(2)+9*x^3+24*x^2+16*
x)*log(x)+1600*x^3*log(2)^2+2*(-120*x^4-160*x^3)*log(2)+9*x^5+24*x^4+16*x^3),x, algorithm="fricas")

[Out]

10/(3*x - 40*log(2) - 2*log(x^2 + log(x)) + 4)

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giac [A]  time = 0.24, size = 22, normalized size = 0.76 \begin {gather*} \frac {10}{3 \, x - 40 \, \log \relax (2) - 2 \, \log \left (x^{2} + \log \relax (x)\right ) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*x*log(x)-30*x^3+40*x^2+20)/((4*x*log(x)+4*x^3)*log(log(x)+x^2)^2+((160*x*log(2)-12*x^2-16*x)*lo
g(x)+160*x^3*log(2)-12*x^4-16*x^3)*log(log(x)+x^2)+(1600*x*log(2)^2+2*(-120*x^2-160*x)*log(2)+9*x^3+24*x^2+16*
x)*log(x)+1600*x^3*log(2)^2+2*(-120*x^4-160*x^3)*log(2)+9*x^5+24*x^4+16*x^3),x, algorithm="giac")

[Out]

10/(3*x - 40*log(2) - 2*log(x^2 + log(x)) + 4)

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maple [A]  time = 0.05, size = 23, normalized size = 0.79

method	result	size
risch	\(-\frac {10}{2 \ln \left (\ln \relax (x )+x^{2}\right )+40 \ln \relax (2)-3 x -4}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-30*x*ln(x)-30*x^3+40*x^2+20)/((4*x*ln(x)+4*x^3)*ln(ln(x)+x^2)^2+((160*x*ln(2)-12*x^2-16*x)*ln(x)+160*x^3
*ln(2)-12*x^4-16*x^3)*ln(ln(x)+x^2)+(1600*x*ln(2)^2+2*(-120*x^2-160*x)*ln(2)+9*x^3+24*x^2+16*x)*ln(x)+1600*x^3
*ln(2)^2+2*(-120*x^4-160*x^3)*ln(2)+9*x^5+24*x^4+16*x^3),x,method=_RETURNVERBOSE)

[Out]

-10/(2*ln(ln(x)+x^2)+40*ln(2)-3*x-4)

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maxima [A]  time = 0.50, size = 22, normalized size = 0.76 \begin {gather*} \frac {10}{3 \, x - 40 \, \log \relax (2) - 2 \, \log \left (x^{2} + \log \relax (x)\right ) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*x*log(x)-30*x^3+40*x^2+20)/((4*x*log(x)+4*x^3)*log(log(x)+x^2)^2+((160*x*log(2)-12*x^2-16*x)*lo
g(x)+160*x^3*log(2)-12*x^4-16*x^3)*log(log(x)+x^2)+(1600*x*log(2)^2+2*(-120*x^2-160*x)*log(2)+9*x^3+24*x^2+16*
x)*log(x)+1600*x^3*log(2)^2+2*(-120*x^4-160*x^3)*log(2)+9*x^5+24*x^4+16*x^3),x, algorithm="maxima")

[Out]

10/(3*x - 40*log(2) - 2*log(x^2 + log(x)) + 4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {30\,x\,\ln \relax (x)-40\,x^2+30\,x^3-20}{1600\,x^3\,{\ln \relax (2)}^2-\ln \left (\ln \relax (x)+x^2\right )\,\left (16\,x^3-160\,x^3\,\ln \relax (2)+12\,x^4+\ln \relax (x)\,\left (16\,x-160\,x\,\ln \relax (2)+12\,x^2\right )\right )+{\ln \left (\ln \relax (x)+x^2\right )}^2\,\left (4\,x\,\ln \relax (x)+4\,x^3\right )-2\,\ln \relax (2)\,\left (120\,x^4+160\,x^3\right )+\ln \relax (x)\,\left (16\,x-2\,\ln \relax (2)\,\left (120\,x^2+160\,x\right )+1600\,x\,{\ln \relax (2)}^2+24\,x^2+9\,x^3\right )+16\,x^3+24\,x^4+9\,x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(30*x*log(x) - 40*x^2 + 30*x^3 - 20)/(1600*x^3*log(2)^2 - log(log(x) + x^2)*(16*x^3 - 160*x^3*log(2) + 12
*x^4 + log(x)*(16*x - 160*x*log(2) + 12*x^2)) + log(log(x) + x^2)^2*(4*x*log(x) + 4*x^3) - 2*log(2)*(160*x^3 +
120*x^4) + log(x)*(16*x - 2*log(2)*(160*x + 120*x^2) + 1600*x*log(2)^2 + 24*x^2 + 9*x^3) + 16*x^3 + 24*x^4 +
9*x^5),x)

[Out]

-int((30*x*log(x) - 40*x^2 + 30*x^3 - 20)/(1600*x^3*log(2)^2 - log(log(x) + x^2)*(16*x^3 - 160*x^3*log(2) + 12
*x^4 + log(x)*(16*x - 160*x*log(2) + 12*x^2)) + log(log(x) + x^2)^2*(4*x*log(x) + 4*x^3) - 2*log(2)*(160*x^3 +
120*x^4) + log(x)*(16*x - 2*log(2)*(160*x + 120*x^2) + 1600*x*log(2)^2 + 24*x^2 + 9*x^3) + 16*x^3 + 24*x^4 +
9*x^5), x)

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sympy [A]  time = 0.36, size = 22, normalized size = 0.76 \begin {gather*} - \frac {5}{- \frac {3 x}{2} + \log {\left (x^{2} + \log {\relax (x )} \right )} - 2 + 20 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*x*ln(x)-30*x**3+40*x**2+20)/((4*x*ln(x)+4*x**3)*ln(ln(x)+x**2)**2+((160*x*ln(2)-12*x**2-16*x)*l
n(x)+160*x**3*ln(2)-12*x**4-16*x**3)*ln(ln(x)+x**2)+(1600*x*ln(2)**2+2*(-120*x**2-160*x)*ln(2)+9*x**3+24*x**2+
16*x)*ln(x)+1600*x**3*ln(2)**2+2*(-120*x**4-160*x**3)*ln(2)+9*x**5+24*x**4+16*x**3),x)

[Out]

-5/(-3*x/2 + log(x**2 + log(x)) - 2 + 20*log(2))

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