3.57.76 \(\int \frac {-80+320 x-128 x^3-512 x^5+(96 x^2+384 x^4) \log (x)+(-24 x-96 x^3) \log ^2(x)+(2+8 x^2) \log ^3(x)}{-64 x^4+48 x^3 \log (x)-12 x^2 \log ^2(x)+x \log ^3(x)} \, dx\)

Optimal. Leaf size=25 \[ 2 \left (-1+2 x^2+\frac {5}{\left (2 x-\frac {\log (x)}{2}\right )^2}+\log (x)\right ) \]

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Rubi [A]  time = 0.44, antiderivative size = 22, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 4, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {6741, 6742, 14, 6686} \begin {gather*} 4 x^2+2 \log (x)+\frac {40}{(4 x-\log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-80 + 320*x - 128*x^3 - 512*x^5 + (96*x^2 + 384*x^4)*Log[x] + (-24*x - 96*x^3)*Log[x]^2 + (2 + 8*x^2)*Log
[x]^3)/(-64*x^4 + 48*x^3*Log[x] - 12*x^2*Log[x]^2 + x*Log[x]^3),x]

[Out]

4*x^2 + 40/(4*x - Log[x])^2 + 2*Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 {80-320 x+128 x^3+512 x^5-\left (96 x^2+384 x^4\right ) \log (x)-\left (-24 x-96 x^3\right ) \log ^2(x)-\left (2+8 x^2\right ) \log ^3(x)}{x (4 x-\log (x))^3} \, dx\\ &=\int \left (\frac {2 \left (1+4 x^2\right )}{x}-\frac {80 (-1+4 x)}{x (4 x-\log (x))^3}\right ) \, dx\\ &=2 \int \frac {1+4 x^2}{x} \, dx-80 \int \frac {-1+4 x}{x (4 x-\log (x))^3} \, dx\\ &=\frac {40}{(4 x-\log (x))^2}+2 \int \left (\frac {1}{x}+4 x\right ) \, dx\\ &=4 x^2+\frac {40}{(4 x-\log (x))^2}+2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 20, normalized size = 0.80 \begin {gather*} 4 x^2+2 \log (x)+\frac {40}{(-4 x+\log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-80 + 320*x - 128*x^3 - 512*x^5 + (96*x^2 + 384*x^4)*Log[x] + (-24*x - 96*x^3)*Log[x]^2 + (2 + 8*x^
2)*Log[x]^3)/(-64*x^4 + 48*x^3*Log[x] - 12*x^2*Log[x]^2 + x*Log[x]^3),x]

[Out]

4*x^2 + 2*Log[x] + 40/(-4*x + Log[x])^2

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fricas [B]  time = 1.35, size = 56, normalized size = 2.24 \begin {gather*} \frac {2 \, {\left (32 \, x^{4} + 2 \, {\left (x^{2} - 4 \, x\right )} \log \relax (x)^{2} + \log \relax (x)^{3} - 16 \, {\left (x^{3} - x^{2}\right )} \log \relax (x) + 20\right )}}{16 \, x^{2} - 8 \, x \log \relax (x) + \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+2)*log(x)^3+(-96*x^3-24*x)*log(x)^2+(384*x^4+96*x^2)*log(x)-512*x^5-128*x^3+320*x-80)/(x*log
(x)^3-12*x^2*log(x)^2+48*x^3*log(x)-64*x^4),x, algorithm="fricas")

[Out]

2*(32*x^4 + 2*(x^2 - 4*x)*log(x)^2 + log(x)^3 - 16*(x^3 - x^2)*log(x) + 20)/(16*x^2 - 8*x*log(x) + log(x)^2)

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giac [B]  time = 0.14, size = 55, normalized size = 2.20 \begin {gather*} 4 \, x^{2} + \frac {40 \, {\left (4 \, x - 1\right )}}{64 \, x^{3} - 32 \, x^{2} \log \relax (x) + 4 \, x \log \relax (x)^{2} - 16 \, x^{2} + 8 \, x \log \relax (x) - \log \relax (x)^{2}} + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+2)*log(x)^3+(-96*x^3-24*x)*log(x)^2+(384*x^4+96*x^2)*log(x)-512*x^5-128*x^3+320*x-80)/(x*log
(x)^3-12*x^2*log(x)^2+48*x^3*log(x)-64*x^4),x, algorithm="giac")

[Out]

4*x^2 + 40*(4*x - 1)/(64*x^3 - 32*x^2*log(x) + 4*x*log(x)^2 - 16*x^2 + 8*x*log(x) - log(x)^2) + 2*log(x)

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maple [A]  time = 0.06, size = 23, normalized size = 0.92

method	result	size
risch	\(4 x^{2}+2 \ln \relax (x )+\frac {40}{\left (4 x -\ln \relax (x )\right )^{2}}\)	\(23\)
norman	\(\frac {40+64 x^{4}+4 x^{2} \ln \relax (x )^{2}-32 x^{3} \ln \relax (x )}{\left (4 x -\ln \relax (x )\right )^{2}}+2 \ln \relax (x )\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^2+2)*ln(x)^3+(-96*x^3-24*x)*ln(x)^2+(384*x^4+96*x^2)*ln(x)-512*x^5-128*x^3+320*x-80)/(x*ln(x)^3-12*x
^2*ln(x)^2+48*x^3*ln(x)-64*x^4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 46, normalized size = 1.84 \begin {gather*} \frac {4 \, {\left (16 \, x^{4} - 8 \, x^{3} \log \relax (x) + x^{2} \log \relax (x)^{2} + 10\right )}}{16 \, x^{2} - 8 \, x \log \relax (x) + \log \relax (x)^{2}} + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+2)*log(x)^3+(-96*x^3-24*x)*log(x)^2+(384*x^4+96*x^2)*log(x)-512*x^5-128*x^3+320*x-80)/(x*log
(x)^3-12*x^2*log(x)^2+48*x^3*log(x)-64*x^4),x, algorithm="maxima")

[Out]

4*(16*x^4 - 8*x^3*log(x) + x^2*log(x)^2 + 10)/(16*x^2 - 8*x*log(x) + log(x)^2) + 2*log(x)

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mupad [B]  time = 3.71, size = 22, normalized size = 0.88 \begin {gather*} 2\,\ln \relax (x)+4\,x^2+\frac {40}{{\left (4\,x-\ln \relax (x)\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)^2*(24*x + 96*x^3) - 320*x - log(x)*(96*x^2 + 384*x^4) - log(x)^3*(8*x^2 + 2) + 128*x^3 + 512*x^5
+ 80)/(x*log(x)^3 + 48*x^3*log(x) - 12*x^2*log(x)^2 - 64*x^4),x)

[Out]

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

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sympy [A]  time = 0.14, size = 27, normalized size = 1.08 \begin {gather*} 4 x^{2} + 2 \log {\relax (x )} + \frac {40}{16 x^{2} - 8 x \log {\relax (x )} + \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**2+2)*ln(x)**3+(-96*x**3-24*x)*ln(x)**2+(384*x**4+96*x**2)*ln(x)-512*x**5-128*x**3+320*x-80)/(
x*ln(x)**3-12*x**2*ln(x)**2+48*x**3*ln(x)-64*x**4),x)

[Out]

4*x**2 + 2*log(x) + 40/(16*x**2 - 8*x*log(x) + log(x)**2)

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