\(\int \frac {-1+x^4}{x^2 \sqrt {x+x^3}} \, dx\) [81]

Optimal result

Integrand size = 18, antiderivative size = 16 \[ \int \frac {-1+x^4}{x^2 \sqrt {x+x^3}} \, dx=\frac {2 \left (x+x^3\right )^{3/2}}{3 x^3} \]

[Out]

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

Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 4 vs. order 2 in optimal.

Time = 0.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 4.88, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2073, 2050, 2036, 335, 226} \[ \int \frac {-1+x^4}{x^2 \sqrt {x+x^3}} \, dx=\frac {\sqrt {x} (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{3 \sqrt {x^3+x}}+\frac {2 \sqrt {x^3+x}}{3}+\frac {2 \sqrt {x^3+x}}{3 x^2} \]

[In]

Int[(-1 + x^4)/(x^2*Sqrt[x + x^3]),x]

[Out]

(2*Sqrt[x + x^3])/3 + (2*Sqrt[x + x^3])/(3*x^2) + (Sqrt[x]*(1 + x)*Sqrt[(1 + x^2)/(1 + x)^2]*EllipticF[2*ArcTa
n[Sqrt[x]], 1/2])/(3*Sqrt[x + x^3])

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2050

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*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2073

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]},
With[{Pqq = Coeff[Pq, x, q]}, Int[(c*x)^m*ExpandToSum[Pq - Pqq*x^q - a*Pqq*(m + q - n + 1)*(x^(q - n)/(b*(m +
q + n*p + 1))), x]*(a*x^j + b*x^n)^p, x] + Simp[Pqq*(c*x)^(m + q - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*c^(q - n
+ 1)*(m + q + n*p + 1))), x]] /; GtQ[q, n - 1] && NeQ[m + q + n*p + 1, 0] && (IntegerQ[2*p] || IntegerQ[p + (
q + 1)/(2*n)])] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] &&  !IntegerQ[p] && IGtQ[j, 0] && IGtQ[n, 0] && L
tQ[j, n]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sqrt {x+x^3}}{3}-\int \frac {1}{x^2 \sqrt {x+x^3}} \, dx \\ & = \frac {2 \sqrt {x+x^3}}{3}+\frac {2 \sqrt {x+x^3}}{3 x^2}+\frac {1}{3} \int \frac {1}{\sqrt {x+x^3}} \, dx \\ & = \frac {2 \sqrt {x+x^3}}{3}+\frac {2 \sqrt {x+x^3}}{3 x^2}+\frac {\left (\sqrt {x} \sqrt {1+x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {1+x^2}} \, dx}{3 \sqrt {x+x^3}} \\ & = \frac {2 \sqrt {x+x^3}}{3}+\frac {2 \sqrt {x+x^3}}{3 x^2}+\frac {\left (2 \sqrt {x} \sqrt {1+x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt {x}\right )}{3 \sqrt {x+x^3}} \\ & = \frac {2 \sqrt {x+x^3}}{3}+\frac {2 \sqrt {x+x^3}}{3 x^2}+\frac {\sqrt {x} (1+x) \sqrt {\frac {1+x^2}{(1+x)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{3 \sqrt {x+x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44 \[ \int \frac {-1+x^4}{x^2 \sqrt {x+x^3}} \, dx=\frac {2 \left (1+x^2\right )^2}{3 x \sqrt {x+x^3}} \]

[In]

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

[Out]

(2*(1 + x^2)^2)/(3*x*Sqrt[x + x^3])

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12

method	result	size
trager	\(\frac {2 \left (x^{2}+1\right ) \sqrt {x^{3}+x}}{3 x^{2}}\)	\(18\)
gosper	\(\frac {2 \left (x^{2}+1\right )^{2}}{3 \sqrt {x^{3}+x}\, x}\)	\(20\)
pseudoelliptic	\(\frac {2 \left (x^{2}+1\right ) \sqrt {\left (x^{2}+1\right ) x}}{3 x^{2}}\)	\(20\)
default	\(\frac {2 \sqrt {x^{3}+x}}{3}+\frac {2 \sqrt {x^{3}+x}}{3 x^{2}}\)	\(23\)
elliptic	\(\frac {2 \sqrt {x^{3}+x}}{3}+\frac {2 \sqrt {x^{3}+x}}{3 x^{2}}\)	\(23\)
risch	\(\frac {\frac {2}{3} x^{4}+\frac {4}{3} x^{2}+\frac {2}{3}}{x \sqrt {\left (x^{2}+1\right ) x}}\)	\(25\)
meijerg	\(\frac {2 x^{\frac {5}{2}} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {5}{4}\right ], \left [\frac {9}{4}\right ], -x^{2}\right )}{5}+\frac {2 \operatorname {hypergeom}\left (\left [-\frac {3}{4}, \frac {1}{2}\right ], \left [\frac {1}{4}\right ], -x^{2}\right )}{3 x^{\frac {3}{2}}}\)	\(34\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {-1+x^4}{x^2 \sqrt {x+x^3}} \, dx=\frac {2 \, \sqrt {x^{3} + x} {\left (x^{2} + 1\right )}}{3 \, x^{2}} \]

[In]

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

[Out]

2/3*sqrt(x^3 + x)*(x^2 + 1)/x^2

Sympy [F]

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

[In]

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

[Out]

Integral((x - 1)*(x + 1)*(x**2 + 1)/(x**2*sqrt(x*(x**2 + 1))), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \frac {-1+x^4}{x^2 \sqrt {x+x^3}} \, dx=\frac {2}{3} \, \sqrt {x^{3} + x} + \frac {2}{3} \, \sqrt {\frac {1}{x} + \frac {1}{x^{3}}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.90 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.56 \[ \int \frac {-1+x^4}{x^2 \sqrt {x+x^3}} \, dx=\frac {4\,\sqrt {x^3+x}}{3} \]

[In]

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

[Out]

(4*(x + x^3)^(1/2))/3