\(\int \frac {\sqrt [4]{x^2+x^4}}{x^2 (1+x^2)} \, dx\) [92]

Optimal result

Integrand size = 22, antiderivative size = 16 \[ \int \frac {\sqrt [4]{x^2+x^4}}{x^2 \left (1+x^2\right )} \, dx=-\frac {2 \sqrt [4]{x^2+x^4}}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1325, 2025, 1160, 277, 270} \[ \int \frac {\sqrt [4]{x^2+x^4}}{x^2 \left (1+x^2\right )} \, dx=-\frac {2 \sqrt [4]{x^4+x^2}}{x} \]

[In]

Int[(x^2 + x^4)^(1/4)/(x^2*(1 + x^2)),x]

[Out]

(-2*(x^2 + x^4)^(1/4))/x

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 1160

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

Rule 1325

Int[(((f_.)*(x_))^(m_)*((a_.) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.))/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Dist[
1/(d*e), Int[(f*x)^m*(a*e + c*d*x^2)*(a + b*x^2 + c*x^4)^(p - 1), x], x] - Dist[(c*d^2 - b*d*e + a*e^2)/(d*e*f
^2), Int[(f*x)^(m + 2)*((a + b*x^2 + c*x^4)^(p - 1)/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && Ne
Q[b^2 - 4*a*c, 0] && GtQ[p, 0] && LtQ[m, 0]

Rule 2025

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1)*x
^(n - 1)), x] /; FreeQ[{a, b, j, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && EqQ[j*p - n + j + 1, 0]

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{x^2+x^4}}{x^2 \left (1+x^2\right )} \, dx=-\frac {2 \sqrt [4]{x^2+x^4}}{x} \]

[In]

Integrate[(x^2 + x^4)^(1/4)/(x^2*(1 + x^2)),x]

[Out]

(-2*(x^2 + x^4)^(1/4))/x

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
meijerg	\(-\frac {2 \left (x^{2}+1\right )^{\frac {1}{4}}}{\sqrt {x}}\)	\(13\)
gosper	\(-\frac {2 \left (x^{4}+x^{2}\right )^{\frac {1}{4}}}{x}\)	\(15\)
trager	\(-\frac {2 \left (x^{4}+x^{2}\right )^{\frac {1}{4}}}{x}\)	\(15\)
risch	\(-\frac {2 \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{x}\)	\(17\)
pseudoelliptic	\(-\frac {2 \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{x}\)	\(17\)

[In]

int((x^4+x^2)^(1/4)/x^2/(x^2+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt [4]{x^2+x^4}}{x^2 \left (1+x^2\right )} \, dx=-\frac {2 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}}}{x} \]

[In]

integrate((x^4+x^2)^(1/4)/x^2/(x^2+1),x, algorithm="fricas")

[Out]

-2*(x^4 + x^2)^(1/4)/x

Sympy [F]

\[ \int \frac {\sqrt [4]{x^2+x^4}}{x^2 \left (1+x^2\right )} \, dx=\int \frac {\sqrt [4]{x^{2} \left (x^{2} + 1\right )}}{x^{2} \left (x^{2} + 1\right )}\, dx \]

[In]

integrate((x**4+x**2)**(1/4)/x**2/(x**2+1),x)

[Out]

Integral((x**2*(x**2 + 1))**(1/4)/(x**2*(x**2 + 1)), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt [4]{x^2+x^4}}{x^2 \left (1+x^2\right )} \, dx=-\frac {2 \, {\left (x^{3} + x\right )}}{{\left (x^{2} + 1\right )}^{\frac {3}{4}} x^{\frac {3}{2}}} \]

[In]

integrate((x^4+x^2)^(1/4)/x^2/(x^2+1),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.56 \[ \int \frac {\sqrt [4]{x^2+x^4}}{x^2 \left (1+x^2\right )} \, dx=-2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} \]

[In]

integrate((x^4+x^2)^(1/4)/x^2/(x^2+1),x, algorithm="giac")

[Out]

-2*(1/x^2 + 1)^(1/4)

Mupad [B] (verification not implemented)

Time = 4.97 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt [4]{x^2+x^4}}{x^2 \left (1+x^2\right )} \, dx=-\frac {2\,{\left (x^4+x^2\right )}^{1/4}}{x} \]

[In]

int((x^2 + x^4)^(1/4)/(x^2*(x^2 + 1)),x)

[Out]

-(2*(x^2 + x^4)^(1/4))/x