\(\int \frac {72-18 x-10 x^3+3 x^4+(-72+4 x^2+12 x^3) \log (x)}{-4 x^3+x^4+4 x^3 \log (x)} \, dx\) [1100]

Optimal result

Integrand size = 49, antiderivative size = 22 \[ \int \frac {72-18 x-10 x^3+3 x^4+\left (-72+4 x^2+12 x^3\right ) \log (x)}{-4 x^3+x^4+4 x^3 \log (x)} \, dx=\frac {9}{x^2}+3 x+\log (x)+\log (x-4 (1-\log (x))) \]

[Out]

9/x^2+3*x+ln(x-4+4*ln(x))+ln(x)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {6874, 14, 6816} \[ \int \frac {72-18 x-10 x^3+3 x^4+\left (-72+4 x^2+12 x^3\right ) \log (x)}{-4 x^3+x^4+4 x^3 \log (x)} \, dx=\frac {9}{x^2}+3 x+\log (x)+\log (-x-4 \log (x)+4) \]

[In]

Int[(72 - 18*x - 10*x^3 + 3*x^4 + (-72 + 4*x^2 + 12*x^3)*Log[x])/(-4*x^3 + x^4 + 4*x^3*Log[x]),x]

[Out]

9/x^2 + 3*x + Log[x] + Log[4 - x - 4*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 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 \left (\frac {-18+x^2+3 x^3}{x^3}+\frac {4+x}{x (-4+x+4 \log (x))}\right ) \, dx \\ & = \int \frac {-18+x^2+3 x^3}{x^3} \, dx+\int \frac {4+x}{x (-4+x+4 \log (x))} \, dx \\ & = \log (4-x-4 \log (x))+\int \left (3-\frac {18}{x^3}+\frac {1}{x}\right ) \, dx \\ & = \frac {9}{x^2}+3 x+\log (x)+\log (4-x-4 \log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {72-18 x-10 x^3+3 x^4+\left (-72+4 x^2+12 x^3\right ) \log (x)}{-4 x^3+x^4+4 x^3 \log (x)} \, dx=\frac {9}{x^2}+3 x+\log (x)+\log (4-x-4 \log (x)) \]

[In]

Integrate[(72 - 18*x - 10*x^3 + 3*x^4 + (-72 + 4*x^2 + 12*x^3)*Log[x])/(-4*x^3 + x^4 + 4*x^3*Log[x]),x]

[Out]

9/x^2 + 3*x + Log[x] + Log[4 - x - 4*Log[x]]

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23

method	result	size
default	\(\frac {9+x^{2} \ln \left (x \right )+3 x^{3}}{x^{2}}+\ln \left (x -4+4 \ln \left (x \right )\right )\)	\(27\)
norman	\(\frac {9+x^{2} \ln \left (x \right )+3 x^{3}}{x^{2}}+\ln \left (x -4+4 \ln \left (x \right )\right )\)	\(27\)
risch	\(\frac {9+x^{2} \ln \left (x \right )+3 x^{3}}{x^{2}}+\ln \left (\frac {x}{4}+\ln \left (x \right )-1\right )\)	\(27\)
parallelrisch	\(\frac {\ln \left (x -4+4 \ln \left (x \right )\right ) x^{2}+3 x^{3}+x^{2} \ln \left (x \right )+9}{x^{2}}\)	\(30\)

[In]

int(((12*x^3+4*x^2-72)*ln(x)+3*x^4-10*x^3-18*x+72)/(4*x^3*ln(x)+x^4-4*x^3),x,method=_RETURNVERBOSE)

[Out]

(9+x^2*ln(x)+3*x^3)/x^2+ln(x-4+4*ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {72-18 x-10 x^3+3 x^4+\left (-72+4 x^2+12 x^3\right ) \log (x)}{-4 x^3+x^4+4 x^3 \log (x)} \, dx=\frac {3 \, x^{3} + x^{2} \log \left (x + 4 \, \log \left (x\right ) - 4\right ) + x^{2} \log \left (x\right ) + 9}{x^{2}} \]

[In]

integrate(((12*x^3+4*x^2-72)*log(x)+3*x^4-10*x^3-18*x+72)/(4*x^3*log(x)+x^4-4*x^3),x, algorithm="fricas")

[Out]

(3*x^3 + x^2*log(x + 4*log(x) - 4) + x^2*log(x) + 9)/x^2

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {72-18 x-10 x^3+3 x^4+\left (-72+4 x^2+12 x^3\right ) \log (x)}{-4 x^3+x^4+4 x^3 \log (x)} \, dx=3 x + \log {\left (x \right )} + \log {\left (\frac {x}{4} + \log {\left (x \right )} - 1 \right )} + \frac {9}{x^{2}} \]

[In]

integrate(((12*x**3+4*x**2-72)*ln(x)+3*x**4-10*x**3-18*x+72)/(4*x**3*ln(x)+x**4-4*x**3),x)

[Out]

3*x + log(x) + log(x/4 + log(x) - 1) + 9/x**2

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {72-18 x-10 x^3+3 x^4+\left (-72+4 x^2+12 x^3\right ) \log (x)}{-4 x^3+x^4+4 x^3 \log (x)} \, dx=\frac {3 \, {\left (x^{3} + 3\right )}}{x^{2}} + \log \left (x\right ) + \log \left (\frac {1}{4} \, x + \log \left (x\right ) - 1\right ) \]

[In]

integrate(((12*x^3+4*x^2-72)*log(x)+3*x^4-10*x^3-18*x+72)/(4*x^3*log(x)+x^4-4*x^3),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {72-18 x-10 x^3+3 x^4+\left (-72+4 x^2+12 x^3\right ) \log (x)}{-4 x^3+x^4+4 x^3 \log (x)} \, dx=3 \, x + \frac {9}{x^{2}} + \log \left (x + 4 \, \log \left (x\right ) - 4\right ) + \log \left (x\right ) \]

[In]

integrate(((12*x^3+4*x^2-72)*log(x)+3*x^4-10*x^3-18*x+72)/(4*x^3*log(x)+x^4-4*x^3),x, algorithm="giac")

[Out]

3*x + 9/x^2 + log(x + 4*log(x) - 4) + log(x)

Mupad [B] (verification not implemented)

Time = 8.64 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {72-18 x-10 x^3+3 x^4+\left (-72+4 x^2+12 x^3\right ) \log (x)}{-4 x^3+x^4+4 x^3 \log (x)} \, dx=3\,x+\ln \left (\frac {x}{4}+\ln \left (x\right )-1\right )+\ln \left (x\right )+\frac {9}{x^2} \]

[In]

int((log(x)*(4*x^2 + 12*x^3 - 72) - 18*x - 10*x^3 + 3*x^4 + 72)/(4*x^3*log(x) - 4*x^3 + x^4),x)

[Out]

3*x + log(x/4 + log(x) - 1) + log(x) + 9/x^2