3.953 \(\int \frac {x^{14}}{(1+x^4)^{3/2}} \, dx\)

Optimal. Leaf size=156 \[ -\frac {x^{11}}{2 \sqrt {x^4+1}}+\frac {11}{18} \sqrt {x^4+1} x^7-\frac {77}{90} \sqrt {x^4+1} x^3+\frac {77 \sqrt {x^4+1} x}{30 \left (x^2+1\right )}+\frac {77 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{60 \sqrt {x^4+1}}-\frac {77 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{30 \sqrt {x^4+1}} \]

[Out]

-1/2*x^11/(x^4+1)^(1/2)-77/90*x^3*(x^4+1)^(1/2)+11/18*x^7*(x^4+1)^(1/2)+77/30*x*(x^4+1)^(1/2)/(x^2+1)-77/30*(x
^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*EllipticE(sin(2*arctan(x)),1/2*2^(1/2))*((x^4+1)/(x^2+1)^2)^
(1/2)/(x^4+1)^(1/2)+77/60*(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*EllipticF(sin(2*arctan(x)),1/2*2
^(1/2))*((x^4+1)/(x^2+1)^2)^(1/2)/(x^4+1)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {288, 321, 305, 220, 1196} \[ -\frac {x^{11}}{2 \sqrt {x^4+1}}+\frac {11}{18} \sqrt {x^4+1} x^7-\frac {77}{90} \sqrt {x^4+1} x^3+\frac {77 \sqrt {x^4+1} x}{30 \left (x^2+1\right )}+\frac {77 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{60 \sqrt {x^4+1}}-\frac {77 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{30 \sqrt {x^4+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^14/(1 + x^4)^(3/2),x]

[Out]

-x^11/(2*Sqrt[1 + x^4]) - (77*x^3*Sqrt[1 + x^4])/90 + (11*x^7*Sqrt[1 + x^4])/18 + (77*x*Sqrt[1 + x^4])/(30*(1
+ x^2)) - (77*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/2])/(30*Sqrt[1 + x^4]) + (77*(1 +
x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(60*Sqrt[1 + x^4])

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 288

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

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^{14}}{\left (1+x^4\right )^{3/2}} \, dx &=-\frac {x^{11}}{2 \sqrt {1+x^4}}+\frac {11}{2} \int \frac {x^{10}}{\sqrt {1+x^4}} \, dx\\ &=-\frac {x^{11}}{2 \sqrt {1+x^4}}+\frac {11}{18} x^7 \sqrt {1+x^4}-\frac {77}{18} \int \frac {x^6}{\sqrt {1+x^4}} \, dx\\ &=-\frac {x^{11}}{2 \sqrt {1+x^4}}-\frac {77}{90} x^3 \sqrt {1+x^4}+\frac {11}{18} x^7 \sqrt {1+x^4}+\frac {77}{30} \int \frac {x^2}{\sqrt {1+x^4}} \, dx\\ &=-\frac {x^{11}}{2 \sqrt {1+x^4}}-\frac {77}{90} x^3 \sqrt {1+x^4}+\frac {11}{18} x^7 \sqrt {1+x^4}+\frac {77}{30} \int \frac {1}{\sqrt {1+x^4}} \, dx-\frac {77}{30} \int \frac {1-x^2}{\sqrt {1+x^4}} \, dx\\ &=-\frac {x^{11}}{2 \sqrt {1+x^4}}-\frac {77}{90} x^3 \sqrt {1+x^4}+\frac {11}{18} x^7 \sqrt {1+x^4}+\frac {77 x \sqrt {1+x^4}}{30 \left (1+x^2\right )}-\frac {77 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{30 \sqrt {1+x^4}}+\frac {77 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{60 \sqrt {1+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 54, normalized size = 0.35 \[ \frac {x^3 \left (-77 \sqrt {x^4+1} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-x^4\right )+5 x^8-11 x^4+77\right )}{45 \sqrt {x^4+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^14/(1 + x^4)^(3/2),x]

[Out]

(x^3*(77 - 11*x^4 + 5*x^8 - 77*Sqrt[1 + x^4]*Hypergeometric2F1[3/4, 3/2, 7/4, -x^4]))/(45*Sqrt[1 + x^4])

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 1} x^{14}}{x^{8} + 2 \, x^{4} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(x^4+1)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^4 + 1)*x^14/(x^8 + 2*x^4 + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{14}}{{\left (x^{4} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(x^4+1)^(3/2),x, algorithm="giac")

[Out]

integrate(x^14/(x^4 + 1)^(3/2), x)

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maple [C]  time = 0.01, size = 119, normalized size = 0.76 \[ \frac {\sqrt {x^{4}+1}\, x^{7}}{9}-\frac {x^{3}}{2 \sqrt {x^{4}+1}}-\frac {16 \sqrt {x^{4}+1}\, x^{3}}{45}+\frac {77 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (-\EllipticE \left (\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) x , i\right )+\EllipticF \left (\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) x , i\right )\right )}{30 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/(x^4+1)^(3/2),x)

[Out]

-1/2*x^3/(x^4+1)^(1/2)+1/9*(x^4+1)^(1/2)*x^7-16/45*(x^4+1)^(1/2)*x^3+77/30*I/(1/2*2^(1/2)+1/2*I*2^(1/2))*(-I*x
^2+1)^(1/2)*(I*x^2+1)^(1/2)/(x^4+1)^(1/2)*(EllipticF((1/2*2^(1/2)+1/2*I*2^(1/2))*x,I)-EllipticE((1/2*2^(1/2)+1
/2*I*2^(1/2))*x,I))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{14}}{{\left (x^{4} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^14/(x^4+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^14/(x^4 + 1)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{14}}{{\left (x^4+1\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^14/(x^4 + 1)^(3/2),x)

[Out]

int(x^14/(x^4 + 1)^(3/2), x)

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sympy [C]  time = 2.24, size = 29, normalized size = 0.19 \[ \frac {x^{15} \Gamma \left (\frac {15}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {15}{4} \\ \frac {19}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {19}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**14/(x**4+1)**(3/2),x)

[Out]

x**15*gamma(15/4)*hyper((3/2, 15/4), (19/4,), x**4*exp_polar(I*pi))/(4*gamma(19/4))

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