3.1.84 \(\int \frac {(-1+x^2) (1+x^2) \sqrt {1+x^4}}{x^4} \, dx\)

Optimal. Leaf size=16 \[ \frac {\left (x^4+1\right )^{3/2}}{3 x^3} \]

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Rubi [A]  time = 0.02, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {517, 449} \begin {gather*} \frac {\left (x^4+1\right )^{3/2}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + x^4)^(3/2)/(3*x^3)

Rule 449

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 517

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^
(p_.), x_Symbol] :> Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n)^q, x] /; FreeQ[{a1, b1, a2, b2, c, d, n, p, q}, x]
&& EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[p] || (GtQ[a1, 0] && GtQ[a2, 0]))

Rubi steps

\begin {align*} \int \frac {\left (-1+x^2\right ) \left (1+x^2\right ) \sqrt {1+x^4}}{x^4} \, dx &=\int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{x^4} \, dx\\ &=\frac {\left (1+x^4\right )^{3/2}}{3 x^3}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 40, normalized size = 2.50 \begin {gather*} x \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-x^4\right )+\frac {\, _2F_1\left (-\frac {3}{4},-\frac {1}{2};\frac {1}{4};-x^4\right )}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Hypergeometric2F1[-3/4, -1/2, 1/4, -x^4]/(3*x^3) + x*Hypergeometric2F1[-1/2, 1/4, 5/4, -x^4]

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IntegrateAlgebraic [A]  time = 0.13, size = 16, normalized size = 1.00 \begin {gather*} \frac {\left (1+x^4\right )^{3/2}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + x^4)^(3/2)/(3*x^3)

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fricas [A]  time = 0.44, size = 12, normalized size = 0.75 \begin {gather*} \frac {{\left (x^{4} + 1\right )}^{\frac {3}{2}}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x^4 + 1)^(3/2)/x^3

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giac [A]  time = 0.50, size = 23, normalized size = 1.44 \begin {gather*} \frac {1}{3} \, \sqrt {x^{4} + 1} x + \frac {\sqrt {\frac {1}{x^{4}} + 1}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 13, normalized size = 0.81

method	result	size
gosper	\(\frac {\left (x^{4}+1\right )^{\frac {3}{2}}}{3 x^{3}}\)	\(13\)
trager	\(\frac {\left (x^{4}+1\right )^{\frac {3}{2}}}{3 x^{3}}\)	\(13\)
elliptic	\(\frac {\left (x^{4}+1\right )^{\frac {3}{2}}}{3 x^{3}}\)	\(13\)
risch	\(\frac {x^{8}+2 x^{4}+1}{3 x^{3} \sqrt {x^{4}+1}}\)	\(23\)
default	\(\frac {\sqrt {x^{4}+1}\, x}{3}+\frac {\sqrt {x^{4}+1}}{3 x^{3}}\)	\(24\)
meijerg	\(\frac {\hypergeom \left (\left [-\frac {3}{4}, -\frac {1}{2}\right ], \left [\frac {1}{4}\right ], -x^{4}\right )}{3 x^{3}}+\hypergeom \left (\left [-\frac {1}{2}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], -x^{4}\right ) x\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.54, size = 12, normalized size = 0.75 \begin {gather*} \frac {{\left (x^{4} + 1\right )}^{\frac {3}{2}}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x^4 + 1)^(3/2)/x^3

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mupad [B]  time = 0.18, size = 12, normalized size = 0.75 \begin {gather*} \frac {{\left (x^4+1\right )}^{3/2}}{3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^4 + 1)^(3/2)/(3*x^3)

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sympy [C]  time = 2.05, size = 65, normalized size = 4.06 \begin {gather*} \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} - \frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), x**4*exp_polar(I*pi))/(4*gamma(5/4)) - gamma(-3/4)*hyper((-3/4, -1/2),
(1/4,), x**4*exp_polar(I*pi))/(4*x**3*gamma(1/4))

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