\(\int \frac {-5764801+24559829 x-12395625 x^2+1562500 x^3}{5764801 x-3001250 x^2+390625 x^3} \, dx\) [2456]

Optimal result

Integrand size = 32, antiderivative size = 24 \[ \int \frac {-5764801+24559829 x-12395625 x^2+1562500 x^3}{5764801 x-3001250 x^2+390625 x^3} \, dx=-\left (\left (-4+\frac {10}{x}\right ) x\right )+\frac {x}{-\frac {2401}{625}+x}-\log (x) \]

[Out]

x/(x-2401/625)-x*(10/x-4)-ln(x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {1608, 27, 1634} \[ \int \frac {-5764801+24559829 x-12395625 x^2+1562500 x^3}{5764801 x-3001250 x^2+390625 x^3} \, dx=4 x-\frac {2401}{2401-625 x}-\log (x) \]

[In]

Int[(-5764801 + 24559829*x - 12395625*x^2 + 1562500*x^3)/(5764801*x - 3001250*x^2 + 390625*x^3),x]

[Out]

-2401/(2401 - 625*x) + 4*x - Log[x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1608

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

Rule 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-5764801+24559829 x-12395625 x^2+1562500 x^3}{x \left (5764801-3001250 x+390625 x^2\right )} \, dx \\ & = \int \frac {-5764801+24559829 x-12395625 x^2+1562500 x^3}{x (-2401+625 x)^2} \, dx \\ & = \int \left (4-\frac {1}{x}-\frac {1500625}{(-2401+625 x)^2}\right ) \, dx \\ & = -\frac {2401}{2401-625 x}+4 x-\log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-5764801+24559829 x-12395625 x^2+1562500 x^3}{5764801 x-3001250 x^2+390625 x^3} \, dx=4 x+\frac {2401}{-2401+625 x}-\log (x) \]

[In]

Integrate[(-5764801 + 24559829*x - 12395625*x^2 + 1562500*x^3)/(5764801*x - 3001250*x^2 + 390625*x^3),x]

[Out]

4*x + 2401/(-2401 + 625*x) - Log[x]

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67

method	result	size
risch	\(4 x +\frac {2401}{625 \left (x -\frac {2401}{625}\right )}-\ln \left (x \right )\)	\(16\)
default	\(4 x -\ln \left (x \right )+\frac {2401}{625 x -2401}\)	\(18\)
norman	\(\frac {2500 x^{2}-\frac {21558579}{625}}{625 x -2401}-\ln \left (x \right )\)	\(21\)
parallelrisch	\(-\frac {390625 x \ln \left (x \right )-1562500 x^{2}[phone] \ln \left (x \right )}{625 \left (625 x -2401\right )}\)	\(26\)

[In]

int((1562500*x^3-12395625*x^2+24559829*x-5764801)/(390625*x^3-3001250*x^2+5764801*x),x,method=_RETURNVERBOSE)

[Out]

4*x+2401/625/(x-2401/625)-ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {-5764801+24559829 x-12395625 x^2+1562500 x^3}{5764801 x-3001250 x^2+390625 x^3} \, dx=\frac {2500 \, x^{2} - {\left (625 \, x - 2401\right )} \log \left (x\right ) - 9604 \, x + 2401}{625 \, x - 2401} \]

[In]

integrate((1562500*x^3-12395625*x^2+24559829*x-5764801)/(390625*x^3-3001250*x^2+5764801*x),x, algorithm="frica
s")

[Out]

(2500*x^2 - (625*x - 2401)*log(x) - 9604*x + 2401)/(625*x - 2401)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.50 \[ \int \frac {-5764801+24559829 x-12395625 x^2+1562500 x^3}{5764801 x-3001250 x^2+390625 x^3} \, dx=4 x - \log {\left (x \right )} + \frac {2401}{625 x - 2401} \]

[In]

integrate((1562500*x**3-12395625*x**2+24559829*x-5764801)/(390625*x**3-3001250*x**2+5764801*x),x)

[Out]

4*x - log(x) + 2401/(625*x - 2401)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-5764801+24559829 x-12395625 x^2+1562500 x^3}{5764801 x-3001250 x^2+390625 x^3} \, dx=4 \, x + \frac {2401}{625 \, x - 2401} - \log \left (x\right ) \]

[In]

integrate((1562500*x^3-12395625*x^2+24559829*x-5764801)/(390625*x^3-3001250*x^2+5764801*x),x, algorithm="maxim
a")

[Out]

4*x + 2401/(625*x - 2401) - log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {-5764801+24559829 x-12395625 x^2+1562500 x^3}{5764801 x-3001250 x^2+390625 x^3} \, dx=4 \, x + \frac {2401}{625 \, x - 2401} - \log \left ({\left | x \right |}\right ) \]

[In]

integrate((1562500*x^3-12395625*x^2+24559829*x-5764801)/(390625*x^3-3001250*x^2+5764801*x),x, algorithm="giac"
)

[Out]

4*x + 2401/(625*x - 2401) - log(abs(x))

Mupad [B] (verification not implemented)

Time = 8.99 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-5764801+24559829 x-12395625 x^2+1562500 x^3}{5764801 x-3001250 x^2+390625 x^3} \, dx=4\,x-\ln \left (x\right )+\frac {2401}{625\,\left (x-\frac {2401}{625}\right )} \]

[In]

int((24559829*x - 12395625*x^2 + 1562500*x^3 - 5764801)/(5764801*x - 3001250*x^2 + 390625*x^3),x)

[Out]

4*x - log(x) + 2401/(625*(x - 2401/625))