3.58.63 \(\int \frac {-11+409 x+420 x^2+100 x^3+(200+520 x+280 x^2+20 x^3) \log (x)+(160+180 x+40 x^2) \log ^2(x)+(20+20 x) \log ^3(x)}{-11 x+100 x^2+100 x^3+(220 x^2+20 x^3) \log (x)+(100 x+40 x^2) \log ^2(x)+20 x \log ^3(x)} \, dx\)

Optimal. Leaf size=28 \[ x+\log \left (\frac {9 x}{20}-x \left (1-(5+\log (x)) \left (x+(x+\log (x))^2\right )\right )\right ) \]

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Rubi [A]  time = 0.49, antiderivative size = 46, normalized size of antiderivative = 1.64, number of steps used = 5, number of rules used = 3, integrand size = 110, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6742, 43, 6684} \begin {gather*} \log \left (-100 x^2-20 x^2 \log (x)-100 x-20 \log ^3(x)-40 x \log ^2(x)-100 \log ^2(x)-220 x \log (x)+11\right )+x+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-11 + 409*x + 420*x^2 + 100*x^3 + (200 + 520*x + 280*x^2 + 20*x^3)*Log[x] + (160 + 180*x + 40*x^2)*Log[x]
^2 + (20 + 20*x)*Log[x]^3)/(-11*x + 100*x^2 + 100*x^3 + (220*x^2 + 20*x^3)*Log[x] + (100*x + 40*x^2)*Log[x]^2
+ 20*x*Log[x]^3),x]

[Out]

x + Log[x] + Log[11 - 100*x - 100*x^2 - 220*x*Log[x] - 20*x^2*Log[x] - 100*Log[x]^2 - 40*x*Log[x]^2 - 20*Log[x
]^3]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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 \left (\frac {1+x}{x}+\frac {20 \left (16 x+11 x^2+10 \log (x)+15 x \log (x)+2 x^2 \log (x)+3 \log ^2(x)+2 x \log ^2(x)\right )}{x \left (-11+100 x+100 x^2+220 x \log (x)+20 x^2 \log (x)+100 \log ^2(x)+40 x \log ^2(x)+20 \log ^3(x)\right )}\right ) \, dx\\ &=20 \int \frac {16 x+11 x^2+10 \log (x)+15 x \log (x)+2 x^2 \log (x)+3 \log ^2(x)+2 x \log ^2(x)}{x \left (-11+100 x+100 x^2+220 x \log (x)+20 x^2 \log (x)+100 \log ^2(x)+40 x \log ^2(x)+20 \log ^3(x)\right )} \, dx+\int \frac {1+x}{x} \, dx\\ &=\log \left (11-100 x-100 x^2-220 x \log (x)-20 x^2 \log (x)-100 \log ^2(x)-40 x \log ^2(x)-20 \log ^3(x)\right )+\int \left (1+\frac {1}{x}\right ) \, dx\\ &=x+\log (x)+\log \left (11-100 x-100 x^2-220 x \log (x)-20 x^2 \log (x)-100 \log ^2(x)-40 x \log ^2(x)-20 \log ^3(x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 46, normalized size = 1.64 \begin {gather*} x+\log (x)+\log \left (11-100 x-100 x^2-220 x \log (x)-20 x^2 \log (x)-100 \log ^2(x)-40 x \log ^2(x)-20 \log ^3(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-11 + 409*x + 420*x^2 + 100*x^3 + (200 + 520*x + 280*x^2 + 20*x^3)*Log[x] + (160 + 180*x + 40*x^2)*
Log[x]^2 + (20 + 20*x)*Log[x]^3)/(-11*x + 100*x^2 + 100*x^3 + (220*x^2 + 20*x^3)*Log[x] + (100*x + 40*x^2)*Log
[x]^2 + 20*x*Log[x]^3),x]

[Out]

x + Log[x] + Log[11 - 100*x - 100*x^2 - 220*x*Log[x] - 20*x^2*Log[x] - 100*Log[x]^2 - 40*x*Log[x]^2 - 20*Log[x
]^3]

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fricas [A]  time = 0.67, size = 43, normalized size = 1.54 \begin {gather*} x + \log \left (20 \, {\left (2 \, x + 5\right )} \log \relax (x)^{2} + 20 \, \log \relax (x)^{3} + 100 \, x^{2} + 20 \, {\left (x^{2} + 11 \, x\right )} \log \relax (x) + 100 \, x - 11\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x+20)*log(x)^3+(40*x^2+180*x+160)*log(x)^2+(20*x^3+280*x^2+520*x+200)*log(x)+100*x^3+420*x^2+40
9*x-11)/(20*x*log(x)^3+(40*x^2+100*x)*log(x)^2+(20*x^3+220*x^2)*log(x)+100*x^3+100*x^2-11*x),x, algorithm="fri
cas")

[Out]

x + log(20*(2*x + 5)*log(x)^2 + 20*log(x)^3 + 100*x^2 + 20*(x^2 + 11*x)*log(x) + 100*x - 11) + log(x)

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giac [A]  time = 0.19, size = 46, normalized size = 1.64 \begin {gather*} x + \log \left (20 \, x^{2} \log \relax (x) + 40 \, x \log \relax (x)^{2} + 20 \, \log \relax (x)^{3} + 100 \, x^{2} + 220 \, x \log \relax (x) + 100 \, \log \relax (x)^{2} + 100 \, x - 11\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x+20)*log(x)^3+(40*x^2+180*x+160)*log(x)^2+(20*x^3+280*x^2+520*x+200)*log(x)+100*x^3+420*x^2+40
9*x-11)/(20*x*log(x)^3+(40*x^2+100*x)*log(x)^2+(20*x^3+220*x^2)*log(x)+100*x^3+100*x^2-11*x),x, algorithm="gia
c")

[Out]

x + log(20*x^2*log(x) + 40*x*log(x)^2 + 20*log(x)^3 + 100*x^2 + 220*x*log(x) + 100*log(x)^2 + 100*x - 11) + lo
g(x)

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maple [A]  time = 0.06, size = 40, normalized size = 1.43

method	result	size
risch	\(x +\ln \relax (x )+\ln \left (\ln \relax (x )^{3}+\left (5+2 x \right ) \ln \relax (x )^{2}+\left (x^{2}+11 x \right ) \ln \relax (x )+5 x^{2}+5 x -\frac {11}{20}\right )\)	\(40\)
norman	\(x +\ln \relax (x )+\ln \left (20 \ln \relax (x )^{3}+40 x \ln \relax (x )^{2}+20 x^{2} \ln \relax (x )+100 \ln \relax (x )^{2}+220 x \ln \relax (x )+100 x^{2}+100 x -11\right )\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((20*x+20)*ln(x)^3+(40*x^2+180*x+160)*ln(x)^2+(20*x^3+280*x^2+520*x+200)*ln(x)+100*x^3+420*x^2+409*x-11)/(
20*x*ln(x)^3+(40*x^2+100*x)*ln(x)^2+(20*x^3+220*x^2)*ln(x)+100*x^3+100*x^2-11*x),x,method=_RETURNVERBOSE)

[Out]

x+ln(x)+ln(ln(x)^3+(5+2*x)*ln(x)^2+(x^2+11*x)*ln(x)+5*x^2+5*x-11/20)

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maxima [A]  time = 0.41, size = 39, normalized size = 1.39 \begin {gather*} x + \log \left ({\left (2 \, x + 5\right )} \log \relax (x)^{2} + \log \relax (x)^{3} + 5 \, x^{2} + {\left (x^{2} + 11 \, x\right )} \log \relax (x) + 5 \, x - \frac {11}{20}\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x+20)*log(x)^3+(40*x^2+180*x+160)*log(x)^2+(20*x^3+280*x^2+520*x+200)*log(x)+100*x^3+420*x^2+40
9*x-11)/(20*x*log(x)^3+(40*x^2+100*x)*log(x)^2+(20*x^3+220*x^2)*log(x)+100*x^3+100*x^2-11*x),x, algorithm="max
ima")

[Out]

x + log((2*x + 5)*log(x)^2 + log(x)^3 + 5*x^2 + (x^2 + 11*x)*log(x) + 5*x - 11/20) + log(x)

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mupad [B]  time = 3.76, size = 57, normalized size = 2.04 \begin {gather*} \ln \left (20\,x^2\,\ln \relax (x)+100\,x^2+40\,x\,{\ln \relax (x)}^2+220\,x\,\ln \relax (x)+100\,x+20\,{\ln \relax (x)}^3+100\,{\ln \relax (x)}^2-11\right )+\frac {x^2\,\ln \relax (x)+x^3}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((409*x + log(x)^2*(180*x + 40*x^2 + 160) + 420*x^2 + 100*x^3 + log(x)^3*(20*x + 20) + log(x)*(520*x + 280*
x^2 + 20*x^3 + 200) - 11)/(log(x)^2*(100*x + 40*x^2) - 11*x + log(x)*(220*x^2 + 20*x^3) + 20*x*log(x)^3 + 100*
x^2 + 100*x^3),x)

[Out]

log(100*x + 40*x*log(x)^2 + 20*x^2*log(x) + 100*log(x)^2 + 20*log(x)^3 + 220*x*log(x) + 100*x^2 - 11) + (x^2*l
og(x) + x^3)/x^2

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sympy [A]  time = 0.30, size = 42, normalized size = 1.50 \begin {gather*} x + \log {\relax (x )} + \log {\left (5 x^{2} + 5 x + \left (2 x + 5\right ) \log {\relax (x )}^{2} + \left (x^{2} + 11 x\right ) \log {\relax (x )} + \log {\relax (x )}^{3} - \frac {11}{20} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x+20)*ln(x)**3+(40*x**2+180*x+160)*ln(x)**2+(20*x**3+280*x**2+520*x+200)*ln(x)+100*x**3+420*x**
2+409*x-11)/(20*x*ln(x)**3+(40*x**2+100*x)*ln(x)**2+(20*x**3+220*x**2)*ln(x)+100*x**3+100*x**2-11*x),x)

[Out]

x + log(x) + log(5*x**2 + 5*x + (2*x + 5)*log(x)**2 + (x**2 + 11*x)*log(x) + log(x)**3 - 11/20)

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