3.56.41 \(\int \frac {-x+10 x^2-2 x^3+3 x^4+(10+3 x^2) \log (x)}{2 x^4+2 x^2 \log (x)} \, dx\)

Optimal. Leaf size=27 \[ 2 x+\frac {1}{2} \left (5-\frac {10}{x}-x-\log \left (x^2+\log (x)\right )\right ) \]

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Rubi [A]  time = 0.47, antiderivative size = 22, normalized size of antiderivative = 0.81, number of steps used = 8, number of rules used = 6, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2561, 6741, 12, 6742, 14, 6684} \begin {gather*} -\frac {1}{2} \log \left (x^2+\log (x)\right )+\frac {3 x}{2}-\frac {5}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-5/x + (3*x)/2 - Log[x^2 + Log[x]]/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2561

Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.))^(p_.), x_Symbol] :> Int[u*x^(p*r)*
(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]

Rule 6684

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

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 {-x+10 x^2-2 x^3+3 x^4+\left (10+3 x^2\right ) \log (x)}{x^2 \left (2 x^2+2 \log (x)\right )} \, dx\\ &=\int \frac {-x+10 x^2-2 x^3+3 x^4+\left (10+3 x^2\right ) \log (x)}{2 x^2 \left (x^2+\log (x)\right )} \, dx\\ &=\frac {1}{2} \int \frac {-x+10 x^2-2 x^3+3 x^4+\left (10+3 x^2\right ) \log (x)}{x^2 \left (x^2+\log (x)\right )} \, dx\\ &=\frac {1}{2} \int \left (\frac {10+3 x^2}{x^2}+\frac {-1-2 x^2}{x \left (x^2+\log (x)\right )}\right ) \, dx\\ &=\frac {1}{2} \int \frac {10+3 x^2}{x^2} \, dx+\frac {1}{2} \int \frac {-1-2 x^2}{x \left (x^2+\log (x)\right )} \, dx\\ &=-\frac {1}{2} \log \left (x^2+\log (x)\right )+\frac {1}{2} \int \left (3+\frac {10}{x^2}\right ) \, dx\\ &=-\frac {5}{x}+\frac {3 x}{2}-\frac {1}{2} \log \left (x^2+\log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 22, normalized size = 0.81 \begin {gather*} \frac {1}{2} \left (-\frac {10}{x}+3 x-\log \left (x^2+\log (x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-10/x + 3*x - Log[x^2 + Log[x]])/2

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fricas [A]  time = 0.57, size = 22, normalized size = 0.81 \begin {gather*} \frac {3 \, x^{2} - x \log \left (x^{2} + \log \relax (x)\right ) - 10}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+10)*log(x)+3*x^4-2*x^3+10*x^2-x)/(2*x^2*log(x)+2*x^4),x, algorithm="fricas")

[Out]

1/2*(3*x^2 - x*log(x^2 + log(x)) - 10)/x

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giac [A]  time = 0.16, size = 22, normalized size = 0.81 \begin {gather*} \frac {3}{2} \, x - \frac {5}{x} - \frac {1}{2} \, \log \left (-x^{2} - \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+10)*log(x)+3*x^4-2*x^3+10*x^2-x)/(2*x^2*log(x)+2*x^4),x, algorithm="giac")

[Out]

3/2*x - 5/x - 1/2*log(-x^2 - log(x))

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maple [A]  time = 0.02, size = 22, normalized size = 0.81

method	result	size
norman	\(\frac {-5+\frac {3 x^{2}}{2}}{x}-\frac {\ln \left (\ln \relax (x )+x^{2}\right )}{2}\)	\(22\)
risch	\(\frac {3 x^{2}-10}{2 x}-\frac {\ln \left (\ln \relax (x )+x^{2}\right )}{2}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^2+10)*ln(x)+3*x^4-2*x^3+10*x^2-x)/(2*x^2*ln(x)+2*x^4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 22, normalized size = 0.81 \begin {gather*} \frac {3 \, x^{2} - 10}{2 \, x} - \frac {1}{2} \, \log \left (x^{2} + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+10)*log(x)+3*x^4-2*x^3+10*x^2-x)/(2*x^2*log(x)+2*x^4),x, algorithm="maxima")

[Out]

1/2*(3*x^2 - 10)/x - 1/2*log(x^2 + log(x))

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mupad [B]  time = 3.53, size = 24, normalized size = 0.89 \begin {gather*} -\frac {\ln \left (\ln \relax (x)+x^2\right )}{2}-\frac {10\,x-3\,x^3}{2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.13, size = 17, normalized size = 0.63 \begin {gather*} \frac {3 x}{2} - \frac {\log {\left (x^{2} + \log {\relax (x )} \right )}}{2} - \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**2+10)*ln(x)+3*x**4-2*x**3+10*x**2-x)/(2*x**2*ln(x)+2*x**4),x)

[Out]

3*x/2 - log(x**2 + log(x))/2 - 5/x

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