3.94.99 \(\int \frac {2+2 x+5 x^3-10 x^4+15 x^5-20 x^6+(10 x^2-20 x^3+30 x^4-40 x^5) \log (x)+(5 x-10 x^2+15 x^3-20 x^4) \log ^2(x)}{5 x^3+10 x^2 \log (x)+5 x \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ -3+x-x^2+x^3-x^4-\frac {2}{5 (x+\log (x))} \]

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Rubi [A]  time = 0.56, antiderivative size = 25, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 4, integrand size = 96, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {6741, 12, 6742, 6686} \begin {gather*} -x^4+x^3-x^2+x-\frac {2}{5 (x+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 2*x + 5*x^3 - 10*x^4 + 15*x^5 - 20*x^6 + (10*x^2 - 20*x^3 + 30*x^4 - 40*x^5)*Log[x] + (5*x - 10*x^2 +
15*x^3 - 20*x^4)*Log[x]^2)/(5*x^3 + 10*x^2*Log[x] + 5*x*Log[x]^2),x]

[Out]

x - x^2 + x^3 - x^4 - 2/(5*(x + Log[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 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 {2+2 x+5 x^3-10 x^4+15 x^5-20 x^6+\left (10 x^2-20 x^3+30 x^4-40 x^5\right ) \log (x)+\left (5 x-10 x^2+15 x^3-20 x^4\right ) \log ^2(x)}{5 x (x+\log (x))^2} \, dx\\ &=\frac {1}{5} \int \frac {2+2 x+5 x^3-10 x^4+15 x^5-20 x^6+\left (10 x^2-20 x^3+30 x^4-40 x^5\right ) \log (x)+\left (5 x-10 x^2+15 x^3-20 x^4\right ) \log ^2(x)}{x (x+\log (x))^2} \, dx\\ &=\frac {1}{5} \int \left (-5 \left (-1+2 x-3 x^2+4 x^3\right )+\frac {2 (1+x)}{x (x+\log (x))^2}\right ) \, dx\\ &=\frac {2}{5} \int \frac {1+x}{x (x+\log (x))^2} \, dx-\int \left (-1+2 x-3 x^2+4 x^3\right ) \, dx\\ &=x-x^2+x^3-x^4-\frac {2}{5 (x+\log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 31, normalized size = 1.19 \begin {gather*} \frac {1}{5} \left (5 x-5 x^2+5 x^3-5 x^4-\frac {2}{x+\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 2*x + 5*x^3 - 10*x^4 + 15*x^5 - 20*x^6 + (10*x^2 - 20*x^3 + 30*x^4 - 40*x^5)*Log[x] + (5*x - 10
*x^2 + 15*x^3 - 20*x^4)*Log[x]^2)/(5*x^3 + 10*x^2*Log[x] + 5*x*Log[x]^2),x]

[Out]

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

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fricas [B]  time = 0.76, size = 49, normalized size = 1.88 \begin {gather*} -\frac {5 \, x^{5} - 5 \, x^{4} + 5 \, x^{3} - 5 \, x^{2} + 5 \, {\left (x^{4} - x^{3} + x^{2} - x\right )} \log \relax (x) + 2}{5 \, {\left (x + \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^4+15*x^3-10*x^2+5*x)*log(x)^2+(-40*x^5+30*x^4-20*x^3+10*x^2)*log(x)-20*x^6+15*x^5-10*x^4+5*x
^3+2*x+2)/(5*x*log(x)^2+10*x^2*log(x)+5*x^3),x, algorithm="fricas")

[Out]

-1/5*(5*x^5 - 5*x^4 + 5*x^3 - 5*x^2 + 5*(x^4 - x^3 + x^2 - x)*log(x) + 2)/(x + log(x))

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giac [A]  time = 0.15, size = 23, normalized size = 0.88 \begin {gather*} -x^{4} + x^{3} - x^{2} + x - \frac {2}{5 \, {\left (x + \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^4+15*x^3-10*x^2+5*x)*log(x)^2+(-40*x^5+30*x^4-20*x^3+10*x^2)*log(x)-20*x^6+15*x^5-10*x^4+5*x
^3+2*x+2)/(5*x*log(x)^2+10*x^2*log(x)+5*x^3),x, algorithm="giac")

[Out]

-x^4 + x^3 - x^2 + x - 2/5/(x + log(x))

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maple [A]  time = 0.04, size = 24, normalized size = 0.92

method	result	size
risch	\(-x^{4}+x^{3}-x^{2}+x -\frac {2}{5 \left (x +\ln \relax (x )\right )}\)	\(24\)
norman	\(\frac {-\frac {2}{5}+x^{2}+x^{4}+x^{3} \ln \relax (x )+x \ln \relax (x )-x^{3}-x^{5}-x^{2} \ln \relax (x )-x^{4} \ln \relax (x )}{x +\ln \relax (x )}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-20*x^4+15*x^3-10*x^2+5*x)*ln(x)^2+(-40*x^5+30*x^4-20*x^3+10*x^2)*ln(x)-20*x^6+15*x^5-10*x^4+5*x^3+2*x+2
)/(5*x*ln(x)^2+10*x^2*ln(x)+5*x^3),x,method=_RETURNVERBOSE)

[Out]

-x^4+x^3-x^2+x-2/5/(x+ln(x))

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maxima [B]  time = 0.38, size = 49, normalized size = 1.88 \begin {gather*} -\frac {5 \, x^{5} - 5 \, x^{4} + 5 \, x^{3} - 5 \, x^{2} + 5 \, {\left (x^{4} - x^{3} + x^{2} - x\right )} \log \relax (x) + 2}{5 \, {\left (x + \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^4+15*x^3-10*x^2+5*x)*log(x)^2+(-40*x^5+30*x^4-20*x^3+10*x^2)*log(x)-20*x^6+15*x^5-10*x^4+5*x
^3+2*x+2)/(5*x*log(x)^2+10*x^2*log(x)+5*x^3),x, algorithm="maxima")

[Out]

-1/5*(5*x^5 - 5*x^4 + 5*x^3 - 5*x^2 + 5*(x^4 - x^3 + x^2 - x)*log(x) + 2)/(x + log(x))

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mupad [B]  time = 7.50, size = 27, normalized size = 1.04 \begin {gather*} x-\frac {2}{5\,\left (x+\ln \relax (x)\right )}-x^2+x^3-x^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + log(x)^2*(5*x - 10*x^2 + 15*x^3 - 20*x^4) + log(x)*(10*x^2 - 20*x^3 + 30*x^4 - 40*x^5) + 5*x^3 - 10
*x^4 + 15*x^5 - 20*x^6 + 2)/(5*x*log(x)^2 + 10*x^2*log(x) + 5*x^3),x)

[Out]

x - 2/(5*(x + log(x))) - x^2 + x^3 - x^4

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sympy [A]  time = 0.12, size = 20, normalized size = 0.77 \begin {gather*} - x^{4} + x^{3} - x^{2} + x - \frac {2}{5 x + 5 \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x**4+15*x**3-10*x**2+5*x)*ln(x)**2+(-40*x**5+30*x**4-20*x**3+10*x**2)*ln(x)-20*x**6+15*x**5-10
*x**4+5*x**3+2*x+2)/(5*x*ln(x)**2+10*x**2*ln(x)+5*x**3),x)

[Out]

-x**4 + x**3 - x**2 + x - 2/(5*x + 5*log(x))

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