\(\int \frac {1}{-2 x^3+x^4} \, dx\) [134]

Optimal result

Integrand size = 11, antiderivative size = 31 \[ \int \frac {1}{-2 x^3+x^4} \, dx=\frac {1}{4 x^2}+\frac {1}{4 x}+\frac {1}{8} \log (2-x)-\frac {\log (x)}{8} \]

[Out]

1/4/x^2+1/4/x+1/8*ln(2-x)-1/8*ln(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1607, 46} \[ \int \frac {1}{-2 x^3+x^4} \, dx=\frac {1}{4 x^2}+\frac {1}{4 x}+\frac {1}{8} \log (2-x)-\frac {\log (x)}{8} \]

[In]

Int[(-2*x^3 + x^4)^(-1),x]

[Out]

1/(4*x^2) + 1/(4*x) + Log[2 - x]/8 - Log[x]/8

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{(-2+x) x^3} \, dx \\ & = \int \left (\frac {1}{8 (-2+x)}-\frac {1}{2 x^3}-\frac {1}{4 x^2}-\frac {1}{8 x}\right ) \, dx \\ & = \frac {1}{4 x^2}+\frac {1}{4 x}+\frac {1}{8} \log (2-x)-\frac {\log (x)}{8} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {1}{-2 x^3+x^4} \, dx=\frac {1}{4 x^2}+\frac {1}{4 x}+\frac {1}{8} \log (2-x)-\frac {\log (x)}{8} \]

[In]

Integrate[(-2*x^3 + x^4)^(-1),x]

[Out]

1/(4*x^2) + 1/(4*x) + Log[2 - x]/8 - Log[x]/8

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68

method	result	size
norman	\(\frac {\frac {1}{4}+\frac {x}{4}}{x^{2}}-\frac {\ln \left (x \right )}{8}+\frac {\ln \left (-2+x \right )}{8}\)	\(21\)
risch	\(\frac {\frac {1}{4}+\frac {x}{4}}{x^{2}}-\frac {\ln \left (x \right )}{8}+\frac {\ln \left (-2+x \right )}{8}\)	\(21\)
default	\(\frac {1}{4 x^{2}}+\frac {1}{4 x}-\frac {\ln \left (x \right )}{8}+\frac {\ln \left (-2+x \right )}{8}\)	\(22\)
parallelrisch	\(-\frac {x^{2} \ln \left (x \right )-\ln \left (-2+x \right ) x^{2}-2-2 x}{8 x^{2}}\)	\(26\)
meijerg	\(\frac {1}{4 x^{2}}+\frac {1}{4 x}-\frac {\ln \left (x \right )}{8}+\frac {\ln \left (2\right )}{8}-\frac {i \pi }{8}+\frac {\ln \left (1-\frac {x}{2}\right )}{8}\)	\(32\)

[In]

int(1/(x^4-2*x^3),x,method=_RETURNVERBOSE)

[Out]

(1/4+1/4*x)/x^2-1/8*ln(x)+1/8*ln(-2+x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {1}{-2 x^3+x^4} \, dx=\frac {x^{2} \log \left (x - 2\right ) - x^{2} \log \left (x\right ) + 2 \, x + 2}{8 \, x^{2}} \]

[In]

integrate(1/(x^4-2*x^3),x, algorithm="fricas")

[Out]

1/8*(x^2*log(x - 2) - x^2*log(x) + 2*x + 2)/x^2

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.61 \[ \int \frac {1}{-2 x^3+x^4} \, dx=- \frac {\log {\left (x \right )}}{8} + \frac {\log {\left (x - 2 \right )}}{8} + \frac {x + 1}{4 x^{2}} \]

[In]

integrate(1/(x**4-2*x**3),x)

[Out]

-log(x)/8 + log(x - 2)/8 + (x + 1)/(4*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.61 \[ \int \frac {1}{-2 x^3+x^4} \, dx=\frac {x + 1}{4 \, x^{2}} + \frac {1}{8} \, \log \left (x - 2\right ) - \frac {1}{8} \, \log \left (x\right ) \]

[In]

integrate(1/(x^4-2*x^3),x, algorithm="maxima")

[Out]

1/4*(x + 1)/x^2 + 1/8*log(x - 2) - 1/8*log(x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {1}{-2 x^3+x^4} \, dx=\frac {x + 1}{4 \, x^{2}} + \frac {1}{8} \, \log \left ({\left | x - 2 \right |}\right ) - \frac {1}{8} \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/(x^4-2*x^3),x, algorithm="giac")

[Out]

1/4*(x + 1)/x^2 + 1/8*log(abs(x - 2)) - 1/8*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.52 \[ \int \frac {1}{-2 x^3+x^4} \, dx=\frac {\frac {x}{4}+\frac {1}{4}}{x^2}-\frac {\mathrm {atanh}\left (x-1\right )}{4} \]

[In]

int(-1/(2*x^3 - x^4),x)

[Out]

(x/4 + 1/4)/x^2 - atanh(x - 1)/4