\(\int \frac {(-1+x^2) \sqrt [4]{x^3+x^5}}{x^2 (1+x^2)} \, dx\) [98]

Optimal result

Integrand size = 27, antiderivative size = 16 \[ \int \frac {\left (-1+x^2\right ) \sqrt [4]{x^3+x^5}}{x^2 \left (1+x^2\right )} \, dx=\frac {4 \sqrt [4]{x^3+x^5}}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2081, 460} \[ \int \frac {\left (-1+x^2\right ) \sqrt [4]{x^3+x^5}}{x^2 \left (1+x^2\right )} \, dx=\frac {4 \sqrt [4]{x^5+x^3}}{x} \]

[In]

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

[Out]

(4*(x^3 + x^5)^(1/4))/x

Rule 460

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

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [4]{x^3+x^5} \int \frac {-1+x^2}{x^{5/4} \left (1+x^2\right )^{3/4}} \, dx}{x^{3/4} \sqrt [4]{1+x^2}} \\ & = \frac {4 \sqrt [4]{x^3+x^5}}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-1+x^2\right ) \sqrt [4]{x^3+x^5}}{x^2 \left (1+x^2\right )} \, dx=\frac {4 \sqrt [4]{x^3+x^5}}{x} \]

[In]

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

[Out]

(4*(x^3 + x^5)^(1/4))/x

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
gosper	\(\frac {4 \left (x^{5}+x^{3}\right )^{\frac {1}{4}}}{x}\)	\(15\)
trager	\(\frac {4 \left (x^{5}+x^{3}\right )^{\frac {1}{4}}}{x}\)	\(15\)
risch	\(\frac {4 \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{x}\)	\(17\)
pseudoelliptic	\(\frac {4 \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{x}\)	\(17\)
meijerg	\(\frac {4 \operatorname {hypergeom}\left (\left [-\frac {1}{8}, \frac {3}{4}\right ], \left [\frac {7}{8}\right ], -x^{2}\right )}{x^{\frac {1}{4}}}+\frac {4 x^{\frac {7}{4}} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {7}{8}\right ], \left [\frac {15}{8}\right ], -x^{2}\right )}{7}\)	\(34\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {\left (-1+x^2\right ) \sqrt [4]{x^3+x^5}}{x^2 \left (1+x^2\right )} \, dx=\frac {4 \, {\left (x^{5} + x^{3}\right )}^{\frac {1}{4}}}{x} \]

[In]

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

[Out]

4*(x^5 + x^3)^(1/4)/x

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.98 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {\left (-1+x^2\right ) \sqrt [4]{x^3+x^5}}{x^2 \left (1+x^2\right )} \, dx=\frac {4\,{\left (x^5+x^3\right )}^{1/4}}{x} \]

[In]

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

[Out]

(4*(x^3 + x^5)^(1/4))/x