\(\int \frac {-24+2 x^2+10 x^4}{-4 x+5 x^3} \, dx\) [10217]

Optimal result

Integrand size = 24, antiderivative size = 22 \[ \int \frac {-24+2 x^2+10 x^4}{-4 x+5 x^3} \, dx=-4+x^2+\log \left (\frac {25 x^4}{\left (-\frac {4}{5 x}+x\right )^2}\right ) \]

[Out]

ln(25*x^4/(x-4/5/x)^2)+x^2-4

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1607, 1265, 907} \[ \int \frac {-24+2 x^2+10 x^4}{-4 x+5 x^3} \, dx=x^2-2 \log \left (4-5 x^2\right )+6 \log (x) \]

[In]

Int[(-24 + 2*x^2 + 10*x^4)/(-4*x + 5*x^3),x]

[Out]

x^2 + 6*Log[x] - 2*Log[4 - 5*x^2]

Rule 907

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1265

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

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 {-24+2 x^2+10 x^4}{x \left (-4+5 x^2\right )} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {-24+2 x+10 x^2}{x (-4+5 x)} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (2+\frac {6}{x}-\frac {20}{-4+5 x}\right ) \, dx,x,x^2\right ) \\ & = x^2+6 \log (x)-2 \log \left (4-5 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {-24+2 x^2+10 x^4}{-4 x+5 x^3} \, dx=2 \left (\frac {x^2}{2}+3 \log (x)-\log \left (4-5 x^2\right )\right ) \]

[In]

Integrate[(-24 + 2*x^2 + 10*x^4)/(-4*x + 5*x^3),x]

[Out]

2*(x^2/2 + 3*Log[x] - Log[4 - 5*x^2])

Maple [A] (verified)

Time = 1.83 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77

method	result	size
parallelrisch	\(x^{2}+6 \ln \left (x \right )-2 \ln \left (x^{2}-\frac {4}{5}\right )\)	\(17\)
default	\(x^{2}-2 \ln \left (5 x^{2}-4\right )+6 \ln \left (x \right )\)	\(19\)
norman	\(x^{2}-2 \ln \left (5 x^{2}-4\right )+6 \ln \left (x \right )\)	\(19\)
risch	\(x^{2}-2 \ln \left (5 x^{2}-4\right )+6 \ln \left (x \right )\)	\(19\)
meijerg	\(6 \ln \left (x \right )+3 \ln \left (5\right )-6 \ln \left (2\right )+3 i \pi -2 \ln \left (1-\frac {5 x^{2}}{4}\right )+x^{2}\)	\(31\)

[In]

int((10*x^4+2*x^2-24)/(5*x^3-4*x),x,method=_RETURNVERBOSE)

[Out]

x^2+6*ln(x)-2*ln(x^2-4/5)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {-24+2 x^2+10 x^4}{-4 x+5 x^3} \, dx=x^{2} - 2 \, \log \left (5 \, x^{2} - 4\right ) + 6 \, \log \left (x\right ) \]

[In]

integrate((10*x^4+2*x^2-24)/(5*x^3-4*x),x, algorithm="fricas")

[Out]

x^2 - 2*log(5*x^2 - 4) + 6*log(x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {-24+2 x^2+10 x^4}{-4 x+5 x^3} \, dx=x^{2} + 6 \log {\left (x \right )} - 2 \log {\left (5 x^{2} - 4 \right )} \]

[In]

integrate((10*x**4+2*x**2-24)/(5*x**3-4*x),x)

[Out]

x**2 + 6*log(x) - 2*log(5*x**2 - 4)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {-24+2 x^2+10 x^4}{-4 x+5 x^3} \, dx=x^{2} - 2 \, \log \left (5 \, x^{2} - 4\right ) + 6 \, \log \left (x\right ) \]

[In]

integrate((10*x^4+2*x^2-24)/(5*x^3-4*x),x, algorithm="maxima")

[Out]

x^2 - 2*log(5*x^2 - 4) + 6*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-24+2 x^2+10 x^4}{-4 x+5 x^3} \, dx=x^{2} + 3 \, \log \left (x^{2}\right ) - 2 \, \log \left ({\left | 5 \, x^{2} - 4 \right |}\right ) \]

[In]

integrate((10*x^4+2*x^2-24)/(5*x^3-4*x),x, algorithm="giac")

[Out]

x^2 + 3*log(x^2) - 2*log(abs(5*x^2 - 4))

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {-24+2 x^2+10 x^4}{-4 x+5 x^3} \, dx=6\,\ln \left (x\right )-2\,\ln \left (x^2-\frac {4}{5}\right )+x^2 \]

[In]

int(-(2*x^2 + 10*x^4 - 24)/(4*x - 5*x^3),x)

[Out]

6*log(x) - 2*log(x^2 - 4/5) + x^2