\(\int \frac {1}{(1+x^2) \sqrt [4]{x^2+x^4}} \, dx\) [224]

Optimal result

Integrand size = 19, antiderivative size = 23 \[ \int \frac {1}{\left (1+x^2\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {2 \left (x^2+x^4\right )^{3/4}}{x \left (1+x^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1160, 270} \[ \int \frac {1}{\left (1+x^2\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {2 x}{\sqrt [4]{x^4+x^2}} \]

[In]

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

[Out]

(2*x)/(x^2 + x^4)^(1/4)

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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt [4]{1+x^2}\right ) \int \frac {1}{\sqrt {x} \left (1+x^2\right )^{5/4}} \, dx}{\sqrt [4]{x^2+x^4}} \\ & = \frac {2 x}{\sqrt [4]{x^2+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61 \[ \int \frac {1}{\left (1+x^2\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {2 x}{\sqrt [4]{x^2+x^4}} \]

[In]

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

[Out]

(2*x)/(x^2 + x^4)^(1/4)

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57

method	result	size
gosper	\(\frac {2 x}{\left (x^{4}+x^{2}\right )^{\frac {1}{4}}}\)	\(13\)
meijerg	\(\frac {2 \sqrt {x}}{\left (x^{2}+1\right )^{\frac {1}{4}}}\)	\(13\)
risch	\(\frac {2 x}{\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\)	\(15\)
pseudoelliptic	\(\frac {2 x}{\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\)	\(15\)
trager	\(\frac {2 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}}{x \left (x^{2}+1\right )}\)	\(22\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (1+x^2\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {2 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} + x} \]

[In]

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

[Out]

2*(x^4 + x^2)^(3/4)/(x^3 + x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

-2/3*(x^3 + x)/((x^(5/2) + sqrt(x))*(x^2 + 1)^(1/4)) + integrate(4/3*(x^2 + 1)^(3/4)/(x^(9/2) + 2*x^(5/2) + sq
rt(x)), x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.39 \[ \int \frac {1}{\left (1+x^2\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {2}{{\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}} \]

[In]

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

[Out]

2/(1/x^2 + 1)^(1/4)

Mupad [B] (verification not implemented)

Time = 5.44 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (1+x^2\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {2\,{\left (x^4+x^2\right )}^{3/4}}{x\,\left (x^2+1\right )} \]

[In]

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

[Out]

(2*(x^2 + x^4)^(3/4))/(x*(x^2 + 1))