\(\int \frac {\sqrt [4]{x^2+x^4}}{x^4} \, dx\) [141]

Optimal result

Integrand size = 15, antiderivative size = 18 \[ \int \frac {\sqrt [4]{x^2+x^4}}{x^4} \, dx=-\frac {2 \left (x^2+x^4\right )^{5/4}}{5 x^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2039} \[ \int \frac {\sqrt [4]{x^2+x^4}}{x^4} \, dx=-\frac {2 \left (x^4+x^2\right )^{5/4}}{5 x^5} \]

[In]

Int[(x^2 + x^4)^(1/4)/x^4,x]

[Out]

(-2*(x^2 + x^4)^(5/4))/(5*x^5)

Rule 2039

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (x^2+x^4\right )^{5/4}}{5 x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {\sqrt [4]{x^2+x^4}}{x^4} \, dx=-\frac {2 \left (1+x^2\right ) \sqrt [4]{x^2+x^4}}{5 x^3} \]

[In]

Integrate[(x^2 + x^4)^(1/4)/x^4,x]

[Out]

(-2*(1 + x^2)*(x^2 + x^4)^(1/4))/(5*x^3)

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72

method	result	size
meijerg	\(-\frac {2 \left (x^{2}+1\right )^{\frac {5}{4}}}{5 x^{\frac {5}{2}}}\)	\(13\)
gosper	\(-\frac {2 \left (x^{2}+1\right ) \left (x^{4}+x^{2}\right )^{\frac {1}{4}}}{5 x^{3}}\)	\(20\)
trager	\(-\frac {2 \left (x^{2}+1\right ) \left (x^{4}+x^{2}\right )^{\frac {1}{4}}}{5 x^{3}}\)	\(20\)
pseudoelliptic	\(-\frac {2 \left (x^{2}+1\right ) \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{5 x^{3}}\)	\(22\)
risch	\(-\frac {2 \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}} \left (x^{4}+2 x^{2}+1\right )}{5 x^{3} \left (x^{2}+1\right )}\)	\(34\)

[In]

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

[Out]

-2/5/x^(5/2)*(x^2+1)^(5/4)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt [4]{x^2+x^4}}{x^4} \, dx=-\frac {2 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )}}{5 \, x^{3}} \]

[In]

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

[Out]

-2/5*(x^4 + x^2)^(1/4)*(x^2 + 1)/x^3

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt [4]{x^2+x^4}}{x^4} \, dx=-\frac {2 \, {\left (x^{3} + x\right )} {\left (x^{2} + 1\right )}^{\frac {1}{4}}}{5 \, x^{\frac {7}{2}}} \]

[In]

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

[Out]

-2/5*(x^3 + x)*(x^2 + 1)^(1/4)/x^(7/2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.50 \[ \int \frac {\sqrt [4]{x^2+x^4}}{x^4} \, dx=-\frac {2}{5} \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {5}{4}} \]

[In]

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

[Out]

-2/5*(1/x^2 + 1)^(5/4)

Mupad [B] (verification not implemented)

Time = 5.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.61 \[ \int \frac {\sqrt [4]{x^2+x^4}}{x^4} \, dx=-\frac {2\,{\left (x^4+x^2\right )}^{1/4}}{5\,x}-\frac {2\,{\left (x^4+x^2\right )}^{1/4}}{5\,x^3} \]

[In]

int((x^2 + x^4)^(1/4)/x^4,x)

[Out]

- (2*(x^2 + x^4)^(1/4))/(5*x) - (2*(x^2 + x^4)^(1/4))/(5*x^3)