\(\int \frac {x}{\sqrt [4]{1+x^2}} \, dx\) [11]

Optimal result

Integrand size = 11, antiderivative size = 13 \[ \int \frac {x}{\sqrt [4]{1+x^2}} \, dx=\frac {2}{3} \left (1+x^2\right )^{3/4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {267} \[ \int \frac {x}{\sqrt [4]{1+x^2}} \, dx=\frac {2}{3} \left (x^2+1\right )^{3/4} \]

[In]

Int[x/(1 + x^2)^(1/4),x]

[Out]

(2*(1 + x^2)^(3/4))/3

Rule 267

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

Rubi steps \begin{align*} \text {integral}& = \frac {2}{3} \left (1+x^2\right )^{3/4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\sqrt [4]{1+x^2}} \, dx=\frac {2}{3} \left (1+x^2\right )^{3/4} \]

[In]

Integrate[x/(1 + x^2)^(1/4),x]

[Out]

(2*(1 + x^2)^(3/4))/3

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77

method	result	size
gosper	\(\frac {2 \left (x^{2}+1\right )^{\frac {3}{4}}}{3}\)	\(10\)
derivativedivides	\(\frac {2 \left (x^{2}+1\right )^{\frac {3}{4}}}{3}\)	\(10\)
default	\(\frac {2 \left (x^{2}+1\right )^{\frac {3}{4}}}{3}\)	\(10\)
trager	\(\frac {2 \left (x^{2}+1\right )^{\frac {3}{4}}}{3}\)	\(10\)
risch	\(\frac {2 \left (x^{2}+1\right )^{\frac {3}{4}}}{3}\)	\(10\)
pseudoelliptic	\(\frac {2 \left (x^{2}+1\right )^{\frac {3}{4}}}{3}\)	\(10\)
meijerg	\(\frac {x^{2} \operatorname {hypergeom}\left (\left [\frac {1}{4}, 1\right ], \left [2\right ], -x^{2}\right )}{2}\)	\(17\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {x}{\sqrt [4]{1+x^2}} \, dx=\frac {2}{3} \, {\left (x^{2} + 1\right )}^{\frac {3}{4}} \]

[In]

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

[Out]

2/3*(x^2 + 1)^(3/4)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {x}{\sqrt [4]{1+x^2}} \, dx=\frac {2 \left (x^{2} + 1\right )^{\frac {3}{4}}}{3} \]

[In]

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

[Out]

2*(x**2 + 1)**(3/4)/3

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {x}{\sqrt [4]{1+x^2}} \, dx=\frac {2}{3} \, {\left (x^{2} + 1\right )}^{\frac {3}{4}} \]

[In]

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

[Out]

2/3*(x^2 + 1)^(3/4)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {x}{\sqrt [4]{1+x^2}} \, dx=\frac {2}{3} \, {\left (x^{2} + 1\right )}^{\frac {3}{4}} \]

[In]

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

[Out]

2/3*(x^2 + 1)^(3/4)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {x}{\sqrt [4]{1+x^2}} \, dx=\frac {2\,{\left (x^2+1\right )}^{3/4}}{3} \]

[In]

int(x/(x^2 + 1)^(1/4),x)

[Out]

(2*(x^2 + 1)^(3/4))/3