3.1524 \(\int \frac {x^{13}}{\sqrt {1+x^8}} \, dx\)

Optimal. Leaf size=130 \[ \frac {1}{10} \sqrt {x^8+1} x^6-\frac {3 \sqrt {x^8+1} x^2}{10 \left (x^4+1\right )}-\frac {3 \left (x^4+1\right ) \sqrt {\frac {x^8+1}{\left (x^4+1\right )^2}} F\left (2 \tan ^{-1}\left (x^2\right )|\frac {1}{2}\right )}{20 \sqrt {x^8+1}}+\frac {3 \left (x^4+1\right ) \sqrt {\frac {x^8+1}{\left (x^4+1\right )^2}} E\left (2 \tan ^{-1}\left (x^2\right )|\frac {1}{2}\right )}{10 \sqrt {x^8+1}} \]

[Out]

1/10*x^6*(x^8+1)^(1/2)-3/10*x^2*(x^8+1)^(1/2)/(x^4+1)+3/10*(x^4+1)*(cos(2*arctan(x^2))^2)^(1/2)/cos(2*arctan(x
^2))*EllipticE(sin(2*arctan(x^2)),1/2*2^(1/2))*((x^8+1)/(x^4+1)^2)^(1/2)/(x^8+1)^(1/2)-3/20*(x^4+1)*(cos(2*arc
tan(x^2))^2)^(1/2)/cos(2*arctan(x^2))*EllipticF(sin(2*arctan(x^2)),1/2*2^(1/2))*((x^8+1)/(x^4+1)^2)^(1/2)/(x^8
+1)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {275, 321, 305, 220, 1196} \[ \frac {1}{10} \sqrt {x^8+1} x^6-\frac {3 \sqrt {x^8+1} x^2}{10 \left (x^4+1\right )}-\frac {3 \left (x^4+1\right ) \sqrt {\frac {x^8+1}{\left (x^4+1\right )^2}} F\left (2 \tan ^{-1}\left (x^2\right )|\frac {1}{2}\right )}{20 \sqrt {x^8+1}}+\frac {3 \left (x^4+1\right ) \sqrt {\frac {x^8+1}{\left (x^4+1\right )^2}} E\left (2 \tan ^{-1}\left (x^2\right )|\frac {1}{2}\right )}{10 \sqrt {x^8+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^13/Sqrt[1 + x^8],x]

[Out]

(x^6*Sqrt[1 + x^8])/10 - (3*x^2*Sqrt[1 + x^8])/(10*(1 + x^4)) + (3*(1 + x^4)*Sqrt[(1 + x^8)/(1 + x^4)^2]*Ellip
ticE[2*ArcTan[x^2], 1/2])/(10*Sqrt[1 + x^8]) - (3*(1 + x^4)*Sqrt[(1 + x^8)/(1 + x^4)^2]*EllipticF[2*ArcTan[x^2
], 1/2])/(20*Sqrt[1 + x^8])

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 275

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

Rule 305

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

Rule 321

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

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {x^{13}}{\sqrt {1+x^8}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^6}{\sqrt {1+x^4}} \, dx,x,x^2\right )\\ &=\frac {1}{10} x^6 \sqrt {1+x^8}-\frac {3}{10} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^4}} \, dx,x,x^2\right )\\ &=\frac {1}{10} x^6 \sqrt {1+x^8}-\frac {3}{10} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,x^2\right )+\frac {3}{10} \operatorname {Subst}\left (\int \frac {1-x^2}{\sqrt {1+x^4}} \, dx,x,x^2\right )\\ &=\frac {1}{10} x^6 \sqrt {1+x^8}-\frac {3 x^2 \sqrt {1+x^8}}{10 \left (1+x^4\right )}+\frac {3 \left (1+x^4\right ) \sqrt {\frac {1+x^8}{\left (1+x^4\right )^2}} E\left (2 \tan ^{-1}\left (x^2\right )|\frac {1}{2}\right )}{10 \sqrt {1+x^8}}-\frac {3 \left (1+x^4\right ) \sqrt {\frac {1+x^8}{\left (1+x^4\right )^2}} F\left (2 \tan ^{-1}\left (x^2\right )|\frac {1}{2}\right )}{20 \sqrt {1+x^8}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 34, normalized size = 0.26 \[ \frac {1}{10} x^6 \left (\sqrt {x^8+1}-\, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-x^8\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^13/Sqrt[1 + x^8],x]

[Out]

(x^6*(Sqrt[1 + x^8] - Hypergeometric2F1[1/2, 3/4, 7/4, -x^8]))/10

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fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{13}}{\sqrt {x^{8} + 1}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(x^8+1)^(1/2),x, algorithm="fricas")

[Out]

integral(x^13/sqrt(x^8 + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{13}}{\sqrt {x^{8} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(x^8+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^13/sqrt(x^8 + 1), x)

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maple [C]  time = 0.15, size = 30, normalized size = 0.23 \[ -\frac {x^{6} \hypergeom \left (\left [\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -x^{8}\right )}{10}+\frac {\sqrt {x^{8}+1}\, x^{6}}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^13/(x^8+1)^(1/2),x)

[Out]

1/10*x^6*(x^8+1)^(1/2)-1/10*x^6*hypergeom([1/2,3/4],[7/4],-x^8)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{13}}{\sqrt {x^{8} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(x^8+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^13/sqrt(x^8 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{13}}{\sqrt {x^8+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^13/(x^8 + 1)^(1/2),x)

[Out]

int(x^13/(x^8 + 1)^(1/2), x)

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sympy [C]  time = 1.29, size = 29, normalized size = 0.22 \[ \frac {x^{14} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {x^{8} e^{i \pi }} \right )}}{8 \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**13/(x**8+1)**(1/2),x)

[Out]

x**14*gamma(7/4)*hyper((1/2, 7/4), (11/4,), x**8*exp_polar(I*pi))/(8*gamma(11/4))

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