3.5.35 \(\int \frac {1}{x^8 \sqrt [4]{x^2+x^4}} \, dx\)

Optimal. Leaf size=35 \[ \frac {2 \left (x^4+x^2\right )^{3/4} \left (128 x^6-96 x^4+84 x^2-77\right )}{1155 x^9} \]

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Rubi [B]  time = 0.10, antiderivative size = 73, normalized size of antiderivative = 2.09, number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2016, 2014} \begin {gather*} -\frac {2 \left (x^4+x^2\right )^{3/4}}{15 x^9}+\frac {8 \left (x^4+x^2\right )^{3/4}}{55 x^7}-\frac {64 \left (x^4+x^2\right )^{3/4}}{385 x^5}+\frac {256 \left (x^4+x^2\right )^{3/4}}{1155 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^8*(x^2 + x^4)^(1/4)),x]

[Out]

(-2*(x^2 + x^4)^(3/4))/(15*x^9) + (8*(x^2 + x^4)^(3/4))/(55*x^7) - (64*(x^2 + x^4)^(3/4))/(385*x^5) + (256*(x^
2 + x^4)^(3/4))/(1155*x^3)

Rule 2014

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*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2016

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, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
 0])

Rubi steps

\begin {align*} \int \frac {1}{x^8 \sqrt [4]{x^2+x^4}} \, dx &=-\frac {2 \left (x^2+x^4\right )^{3/4}}{15 x^9}-\frac {4}{5} \int \frac {1}{x^6 \sqrt [4]{x^2+x^4}} \, dx\\ &=-\frac {2 \left (x^2+x^4\right )^{3/4}}{15 x^9}+\frac {8 \left (x^2+x^4\right )^{3/4}}{55 x^7}+\frac {32}{55} \int \frac {1}{x^4 \sqrt [4]{x^2+x^4}} \, dx\\ &=-\frac {2 \left (x^2+x^4\right )^{3/4}}{15 x^9}+\frac {8 \left (x^2+x^4\right )^{3/4}}{55 x^7}-\frac {64 \left (x^2+x^4\right )^{3/4}}{385 x^5}-\frac {128}{385} \int \frac {1}{x^2 \sqrt [4]{x^2+x^4}} \, dx\\ &=-\frac {2 \left (x^2+x^4\right )^{3/4}}{15 x^9}+\frac {8 \left (x^2+x^4\right )^{3/4}}{55 x^7}-\frac {64 \left (x^2+x^4\right )^{3/4}}{385 x^5}+\frac {256 \left (x^2+x^4\right )^{3/4}}{1155 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 35, normalized size = 1.00 \begin {gather*} \frac {2 \left (x^4+x^2\right )^{3/4} \left (128 x^6-96 x^4+84 x^2-77\right )}{1155 x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^8*(x^2 + x^4)^(1/4)),x]

[Out]

(2*(x^2 + x^4)^(3/4)*(-77 + 84*x^2 - 96*x^4 + 128*x^6))/(1155*x^9)

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IntegrateAlgebraic [A]  time = 0.16, size = 35, normalized size = 1.00 \begin {gather*} \frac {2 \left (x^2+x^4\right )^{3/4} \left (-77+84 x^2-96 x^4+128 x^6\right )}{1155 x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^8*(x^2 + x^4)^(1/4)),x]

[Out]

(2*(x^2 + x^4)^(3/4)*(-77 + 84*x^2 - 96*x^4 + 128*x^6))/(1155*x^9)

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fricas [A]  time = 0.45, size = 31, normalized size = 0.89 \begin {gather*} \frac {2 \, {\left (128 \, x^{6} - 96 \, x^{4} + 84 \, x^{2} - 77\right )} {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{1155 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(x^4+x^2)^(1/4),x, algorithm="fricas")

[Out]

2/1155*(128*x^6 - 96*x^4 + 84*x^2 - 77)*(x^4 + x^2)^(3/4)/x^9

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giac [A]  time = 0.80, size = 37, normalized size = 1.06 \begin {gather*} -\frac {2}{15} \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {15}{4}} + \frac {6}{11} \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {11}{4}} - \frac {6}{7} \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {7}{4}} + \frac {2}{3} \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {3}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(x^4+x^2)^(1/4),x, algorithm="giac")

[Out]

-2/15*(1/x^2 + 1)^(15/4) + 6/11*(1/x^2 + 1)^(11/4) - 6/7*(1/x^2 + 1)^(7/4) + 2/3*(1/x^2 + 1)^(3/4)

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maple [A]  time = 0.07, size = 30, normalized size = 0.86

method result size
meijerg \(-\frac {2 \left (-\frac {128}{77} x^{6}+\frac {96}{77} x^{4}-\frac {12}{11} x^{2}+1\right ) \left (x^{2}+1\right )^{\frac {3}{4}}}{15 x^{\frac {15}{2}}}\) \(30\)
trager \(\frac {2 \left (x^{4}+x^{2}\right )^{\frac {3}{4}} \left (128 x^{6}-96 x^{4}+84 x^{2}-77\right )}{1155 x^{9}}\) \(32\)
gosper \(\frac {2 \left (x^{2}+1\right ) \left (128 x^{6}-96 x^{4}+84 x^{2}-77\right )}{1155 x^{7} \left (x^{4}+x^{2}\right )^{\frac {1}{4}}}\) \(37\)
risch \(\frac {\frac {2}{165} x^{2}-\frac {2}{15}-\frac {8}{385} x^{4}+\frac {64}{1155} x^{6}+\frac {256}{1155} x^{8}}{x^{7} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\) \(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^8/(x^4+x^2)^(1/4),x,method=_RETURNVERBOSE)

[Out]

-2/15*(-128/77*x^6+96/77*x^4-12/11*x^2+1)*(x^2+1)^(3/4)/x^(15/2)

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maxima [A]  time = 0.60, size = 36, normalized size = 1.03 \begin {gather*} \frac {2 \, {\left (128 \, x^{9} + 32 \, x^{7} - 12 \, x^{5} + 7 \, x^{3} - 77 \, x\right )}}{1155 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}} x^{\frac {17}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(x^4+x^2)^(1/4),x, algorithm="maxima")

[Out]

2/1155*(128*x^9 + 32*x^7 - 12*x^5 + 7*x^3 - 77*x)/((x^2 + 1)^(1/4)*x^(17/2))

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mupad [B]  time = 0.25, size = 57, normalized size = 1.63 \begin {gather*} \frac {256\,{\left (x^4+x^2\right )}^{3/4}}{1155\,x^3}-\frac {64\,{\left (x^4+x^2\right )}^{3/4}}{385\,x^5}+\frac {8\,{\left (x^4+x^2\right )}^{3/4}}{55\,x^7}-\frac {2\,{\left (x^4+x^2\right )}^{3/4}}{15\,x^9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^8*(x^2 + x^4)^(1/4)),x)

[Out]

(256*(x^2 + x^4)^(3/4))/(1155*x^3) - (64*(x^2 + x^4)^(3/4))/(385*x^5) + (8*(x^2 + x^4)^(3/4))/(55*x^7) - (2*(x
^2 + x^4)^(3/4))/(15*x^9)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{8} \sqrt [4]{x^{2} \left (x^{2} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**8/(x**4+x**2)**(1/4),x)

[Out]

Integral(1/(x**8*(x**2*(x**2 + 1))**(1/4)), x)

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