\(\int \frac {1+8 x^5+4 x^6+(-1+2 x) \log (x)}{2 x^6+x \log (x)} \, dx\) [7753]

Optimal result

Integrand size = 33, antiderivative size = 20 \[ \int \frac {1+8 x^5+4 x^6+(-1+2 x) \log (x)}{2 x^6+x \log (x)} \, dx=2 (-25+x)+\log \left (x^4+\frac {\log (x)}{2 x}\right ) \]

[Out]

2*x-50+ln(x^4+1/2*ln(x)/x)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {2641, 6874, 45, 6816} \[ \int \frac {1+8 x^5+4 x^6+(-1+2 x) \log (x)}{2 x^6+x \log (x)} \, dx=\log \left (2 x^5+\log (x)\right )+2 x-\log (x) \]

[In]

Int[(1 + 8*x^5 + 4*x^6 + (-1 + 2*x)*Log[x])/(2*x^6 + x*Log[x]),x]

[Out]

2*x - Log[x] + Log[2*x^5 + Log[x]]

Rule 45

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 2641

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 6816

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

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1+8 x^5+4 x^6+(-1+2 x) \log (x)}{x \left (2 x^5+\log (x)\right )} \, dx \\ & = \int \left (\frac {-1+2 x}{x}+\frac {1+10 x^5}{x \left (2 x^5+\log (x)\right )}\right ) \, dx \\ & = \int \frac {-1+2 x}{x} \, dx+\int \frac {1+10 x^5}{x \left (2 x^5+\log (x)\right )} \, dx \\ & = \log \left (2 x^5+\log (x)\right )+\int \left (2-\frac {1}{x}\right ) \, dx \\ & = 2 x-\log (x)+\log \left (2 x^5+\log (x)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {1+8 x^5+4 x^6+(-1+2 x) \log (x)}{2 x^6+x \log (x)} \, dx=2 x-\log (x)+\log \left (2 x^5+\log (x)\right ) \]

[In]

Integrate[(1 + 8*x^5 + 4*x^6 + (-1 + 2*x)*Log[x])/(2*x^6 + x*Log[x]),x]

[Out]

2*x - Log[x] + Log[2*x^5 + Log[x]]

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90

method	result	size
default	\(-\ln \left (x \right )+2 x +\ln \left (2 x^{5}+\ln \left (x \right )\right )\)	\(18\)
norman	\(-\ln \left (x \right )+2 x +\ln \left (2 x^{5}+\ln \left (x \right )\right )\)	\(18\)
risch	\(-\ln \left (x \right )+2 x +\ln \left (2 x^{5}+\ln \left (x \right )\right )\)	\(18\)
parallelrisch	\(\ln \left (x^{5}+\frac {\ln \left (x \right )}{2}\right )+2 x -\ln \left (x \right )\)	\(18\)

[In]

int(((-1+2*x)*ln(x)+4*x^6+8*x^5+1)/(x*ln(x)+2*x^6),x,method=_RETURNVERBOSE)

[Out]

-ln(x)+2*x+ln(2*x^5+ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {1+8 x^5+4 x^6+(-1+2 x) \log (x)}{2 x^6+x \log (x)} \, dx=2 \, x + \log \left (2 \, x^{5} + \log \left (x\right )\right ) - \log \left (x\right ) \]

[In]

integrate(((-1+2*x)*log(x)+4*x^6+8*x^5+1)/(x*log(x)+2*x^6),x, algorithm="fricas")

[Out]

2*x + log(2*x^5 + log(x)) - log(x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {1+8 x^5+4 x^6+(-1+2 x) \log (x)}{2 x^6+x \log (x)} \, dx=2 x - \log {\left (x \right )} + \log {\left (2 x^{5} + \log {\left (x \right )} \right )} \]

[In]

integrate(((-1+2*x)*ln(x)+4*x**6+8*x**5+1)/(x*ln(x)+2*x**6),x)

[Out]

2*x - log(x) + log(2*x**5 + log(x))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {1+8 x^5+4 x^6+(-1+2 x) \log (x)}{2 x^6+x \log (x)} \, dx=2 \, x + \log \left (2 \, x^{5} + \log \left (x\right )\right ) - \log \left (x\right ) \]

[In]

integrate(((-1+2*x)*log(x)+4*x^6+8*x^5+1)/(x*log(x)+2*x^6),x, algorithm="maxima")

[Out]

2*x + log(2*x^5 + log(x)) - log(x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {1+8 x^5+4 x^6+(-1+2 x) \log (x)}{2 x^6+x \log (x)} \, dx=2 \, x + \log \left (2 \, x^{5} + \log \left (x\right )\right ) - \log \left (x\right ) \]

[In]

integrate(((-1+2*x)*log(x)+4*x^6+8*x^5+1)/(x*log(x)+2*x^6),x, algorithm="giac")

[Out]

2*x + log(2*x^5 + log(x)) - log(x)

Mupad [B] (verification not implemented)

Time = 12.70 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {1+8 x^5+4 x^6+(-1+2 x) \log (x)}{2 x^6+x \log (x)} \, dx=2\,x-\ln \left (x\right )+\ln \left (\ln \left (x\right )+2\,x^5\right ) \]

[In]

int((log(x)*(2*x - 1) + 8*x^5 + 4*x^6 + 1)/(x*log(x) + 2*x^6),x)

[Out]

2*x - log(x) + log(log(x) + 2*x^5)