\(\int \frac {75 x^2+5 x^4}{25+10 x^2+x^4} \, dx\) [405]

Optimal result

Integrand size = 24, antiderivative size = 17 \[ \int \frac {75 x^2+5 x^4}{25+10 x^2+x^4} \, dx=2+\frac {x^3}{1+\frac {x^2}{5}} \]

[Out]

2+x^3/(1+1/5*x^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {28, 1602} \[ \int \frac {75 x^2+5 x^4}{25+10 x^2+x^4} \, dx=\frac {5 x^3}{x^2+5} \]

[In]

Int[(75*x^2 + 5*x^4)/(25 + 10*x^2 + x^4),x]

[Out]

(5*x^3)/(5 + x^2)

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \int \frac {75 x^2+5 x^4}{\left (5+x^2\right )^2} \, dx \\ & = \frac {5 x^3}{5+x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {75 x^2+5 x^4}{25+10 x^2+x^4} \, dx=5 \left (x-\frac {5 x}{5+x^2}\right ) \]

[In]

Integrate[(75*x^2 + 5*x^4)/(25 + 10*x^2 + x^4),x]

[Out]

5*(x - (5*x)/(5 + x^2))

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76

method	result	size
gosper	\(\frac {5 x^{3}}{x^{2}+5}\)	\(13\)
norman	\(\frac {5 x^{3}}{x^{2}+5}\)	\(13\)
parallelrisch	\(\frac {5 x^{3}}{x^{2}+5}\)	\(13\)
default	\(5 x -\frac {25 x}{x^{2}+5}\)	\(15\)
risch	\(5 x -\frac {25 x}{x^{2}+5}\)	\(15\)

[In]

int((5*x^4+75*x^2)/(x^4+10*x^2+25),x,method=_RETURNVERBOSE)

[Out]

5*x^3/(x^2+5)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {75 x^2+5 x^4}{25+10 x^2+x^4} \, dx=\frac {5 \, x^{3}}{x^{2} + 5} \]

[In]

integrate((5*x^4+75*x^2)/(x^4+10*x^2+25),x, algorithm="fricas")

[Out]

5*x^3/(x^2 + 5)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \frac {75 x^2+5 x^4}{25+10 x^2+x^4} \, dx=5 x - \frac {25 x}{x^{2} + 5} \]

[In]

integrate((5*x**4+75*x**2)/(x**4+10*x**2+25),x)

[Out]

5*x - 25*x/(x**2 + 5)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {75 x^2+5 x^4}{25+10 x^2+x^4} \, dx=5 \, x - \frac {25 \, x}{x^{2} + 5} \]

[In]

integrate((5*x^4+75*x^2)/(x^4+10*x^2+25),x, algorithm="maxima")

[Out]

5*x - 25*x/(x^2 + 5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {75 x^2+5 x^4}{25+10 x^2+x^4} \, dx=5 \, x - \frac {25 \, x}{x^{2} + 5} \]

[In]

integrate((5*x^4+75*x^2)/(x^4+10*x^2+25),x, algorithm="giac")

[Out]

5*x - 25*x/(x^2 + 5)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {75 x^2+5 x^4}{25+10 x^2+x^4} \, dx=\frac {5\,x^3}{x^2+5} \]

[In]

int((75*x^2 + 5*x^4)/(10*x^2 + x^4 + 25),x)

[Out]

(5*x^3)/(x^2 + 5)