3.2.65 \(\int \frac {1+x^8}{x^4 \sqrt [4]{-1+x^4+x^8}} \, dx\)

Optimal. Leaf size=19 \[ \frac {\left (x^8+x^4-1\right )^{3/4}}{3 x^3} \]

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Rubi [A]  time = 0.02, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {1590} \begin {gather*} \frac {\left (x^8+x^4-1\right )^{3/4}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^8)/(x^4*(-1 + x^4 + x^8)^(1/4)),x]

[Out]

(-1 + x^4 + x^8)^(3/4)/(3*x^3)

Rule 1590

Int[(Pp_)*(Qq_)^(m_.)*(Rr_)^(n_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x], r = Expon[Rr, x]}, S
imp[(Coeff[Pp, x, p]*x^(p - q - r + 1)*Qq^(m + 1)*Rr^(n + 1))/((p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x
, r]), x] /; NeQ[p + m*q + n*r + 1, 0] && EqQ[(p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x, r]*Pp, Coeff[Pp
, x, p]*x^(p - q - r)*((p - q - r + 1)*Qq*Rr + (m + 1)*x*Rr*D[Qq, x] + (n + 1)*x*Qq*D[Rr, x])]] /; FreeQ[{m, n
}, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && PolyQ[Rr, x] && NeQ[m, -1] && NeQ[n, -1]

Rubi steps

\begin {align*} \int \frac {1+x^8}{x^4 \sqrt [4]{-1+x^4+x^8}} \, dx &=\frac {\left (-1+x^4+x^8\right )^{3/4}}{3 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 19, normalized size = 1.00 \begin {gather*} \frac {\left (x^8+x^4-1\right )^{3/4}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^8)/(x^4*(-1 + x^4 + x^8)^(1/4)),x]

[Out]

(-1 + x^4 + x^8)^(3/4)/(3*x^3)

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IntegrateAlgebraic [A]  time = 0.40, size = 19, normalized size = 1.00 \begin {gather*} \frac {\left (-1+x^4+x^8\right )^{3/4}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + x^8)/(x^4*(-1 + x^4 + x^8)^(1/4)),x]

[Out]

(-1 + x^4 + x^8)^(3/4)/(3*x^3)

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fricas [A]  time = 0.44, size = 15, normalized size = 0.79 \begin {gather*} \frac {{\left (x^{8} + x^{4} - 1\right )}^{\frac {3}{4}}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^8+1)/x^4/(x^8+x^4-1)^(1/4),x, algorithm="fricas")

[Out]

1/3*(x^8 + x^4 - 1)^(3/4)/x^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8} + 1}{{\left (x^{8} + x^{4} - 1\right )}^{\frac {1}{4}} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^8+1)/x^4/(x^8+x^4-1)^(1/4),x, algorithm="giac")

[Out]

integrate((x^8 + 1)/((x^8 + x^4 - 1)^(1/4)*x^4), x)

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maple [A]  time = 0.07, size = 16, normalized size = 0.84

method	result	size
gosper	\(\frac {\left (x^{8}+x^{4}-1\right )^{\frac {3}{4}}}{3 x^{3}}\)	\(16\)
trager	\(\frac {\left (x^{8}+x^{4}-1\right )^{\frac {3}{4}}}{3 x^{3}}\)	\(16\)
risch	\(\frac {\left (x^{8}+x^{4}-1\right )^{\frac {3}{4}}}{3 x^{3}}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8+1)/x^4/(x^8+x^4-1)^(1/4),x,method=_RETURNVERBOSE)

[Out]

1/3*(x^8+x^4-1)^(3/4)/x^3

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maxima [A]  time = 0.53, size = 15, normalized size = 0.79 \begin {gather*} \frac {{\left (x^{8} + x^{4} - 1\right )}^{\frac {3}{4}}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^8+1)/x^4/(x^8+x^4-1)^(1/4),x, algorithm="maxima")

[Out]

1/3*(x^8 + x^4 - 1)^(3/4)/x^3

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mupad [B]  time = 0.22, size = 15, normalized size = 0.79 \begin {gather*} \frac {{\left (x^8+x^4-1\right )}^{3/4}}{3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8 + 1)/(x^4*(x^4 + x^8 - 1)^(1/4)),x)

[Out]

(x^4 + x^8 - 1)^(3/4)/(3*x^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8} + 1}{x^{4} \sqrt [4]{x^{8} + x^{4} - 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**8+1)/x**4/(x**8+x**4-1)**(1/4),x)

[Out]

Integral((x**8 + 1)/(x**4*(x**8 + x**4 - 1)**(1/4)), x)

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