\(\int \frac {-1+x^2}{x \sqrt [4]{x^3+x^5}} \, dx\) [150]

Optimal result

Integrand size = 20, antiderivative size = 18 \[ \int \frac {-1+x^2}{x \sqrt [4]{x^3+x^5}} \, dx=\frac {4 \left (x^3+x^5\right )^{3/4}}{3 x^3} \]

[Out]

4/3*(x^5+x^3)^(3/4)/x^3

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {2061} \[ \int \frac {-1+x^2}{x \sqrt [4]{x^3+x^5}} \, dx=\frac {4 \left (x^5+x^3\right )^{3/4}}{3 x^3} \]

[In]

Int[(-1 + x^2)/(x*(x^3 + x^5)^(1/4)),x]

[Out]

(4*(x^3 + x^5)^(3/4))/(3*x^3)

Rule 2061

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

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {-1+x^2}{x \sqrt [4]{x^3+x^5}} \, dx=\frac {4 \left (1+x^2\right )}{3 \sqrt [4]{x^3+x^5}} \]

[In]

Integrate[(-1 + x^2)/(x*(x^3 + x^5)^(1/4)),x]

[Out]

(4*(1 + x^2))/(3*(x^3 + x^5)^(1/4))

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
trager	\(\frac {4 \left (x^{5}+x^{3}\right )^{\frac {3}{4}}}{3 x^{3}}\)	\(15\)
gosper	\(\frac {\frac {4 x^{2}}{3}+\frac {4}{3}}{\left (x^{5}+x^{3}\right )^{\frac {1}{4}}}\)	\(17\)
pseudoelliptic	\(\frac {4 \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {3}{4}}}{3 x^{3}}\)	\(17\)
risch	\(\frac {\frac {4 x^{2}}{3}+\frac {4}{3}}{\left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\)	\(19\)
meijerg	\(\frac {4 \operatorname {hypergeom}\left (\left [-\frac {3}{8}, \frac {1}{4}\right ], \left [\frac {5}{8}\right ], -x^{2}\right )}{3 x^{\frac {3}{4}}}+\frac {4 x^{\frac {5}{4}} \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {5}{8}\right ], \left [\frac {13}{8}\right ], -x^{2}\right )}{5}\)	\(34\)

[In]

int((x^2-1)/x/(x^5+x^3)^(1/4),x,method=_RETURNVERBOSE)

[Out]

4/3*(x^5+x^3)^(3/4)/x^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {-1+x^2}{x \sqrt [4]{x^3+x^5}} \, dx=\frac {4 \, {\left (x^{5} + x^{3}\right )}^{\frac {3}{4}}}{3 \, x^{3}} \]

[In]

integrate((x^2-1)/x/(x^5+x^3)^(1/4),x, algorithm="fricas")

[Out]

4/3*(x^5 + x^3)^(3/4)/x^3

Sympy [F]

\[ \int \frac {-1+x^2}{x \sqrt [4]{x^3+x^5}} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right )}{x \sqrt [4]{x^{3} \left (x^{2} + 1\right )}}\, dx \]

[In]

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

[Out]

Integral((x - 1)*(x + 1)/(x*(x**3*(x**2 + 1))**(1/4)), x)

Maxima [F]

\[ \int \frac {-1+x^2}{x \sqrt [4]{x^3+x^5}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} x} \,d x } \]

[In]

integrate((x^2-1)/x/(x^5+x^3)^(1/4),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {-1+x^2}{x \sqrt [4]{x^3+x^5}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} x} \,d x } \]

[In]

integrate((x^2-1)/x/(x^5+x^3)^(1/4),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {-1+x^2}{x \sqrt [4]{x^3+x^5}} \, dx=\frac {4\,{\left (x^5+x^3\right )}^{3/4}}{3\,x^3} \]

[In]

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

[Out]

(4*(x^3 + x^5)^(3/4))/(3*x^3)