\(\int \frac {x^{14}}{(1+x^4)^{3/2}} \, dx\) [953]

Optimal result

Integrand size = 13, antiderivative size = 156 \[ \int \frac {x^{14}}{\left (1+x^4\right )^{3/2}} \, dx=-\frac {x^{11}}{2 \sqrt {1+x^4}}-\frac {77}{90} x^3 \sqrt {1+x^4}+\frac {11}{18} x^7 \sqrt {1+x^4}+\frac {77 x \sqrt {1+x^4}}{30 \left (1+x^2\right )}-\frac {77 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{30 \sqrt {1+x^4}}+\frac {77 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{60 \sqrt {1+x^4}} \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {294, 327, 311, 226, 1210} \[ \int \frac {x^{14}}{\left (1+x^4\right )^{3/2}} \, dx=\frac {77 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{60 \sqrt {x^4+1}}-\frac {77 \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{30 \sqrt {x^4+1}}-\frac {x^{11}}{2 \sqrt {x^4+1}}+\frac {11}{18} \sqrt {x^4+1} x^7-\frac {77}{90} \sqrt {x^4+1} x^3+\frac {77 \sqrt {x^4+1} x}{30 \left (x^2+1\right )} \]

[In]

Int[x^14/(1 + x^4)^(3/2),x]

[Out]

-1/2*x^11/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 226

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

Rule 294

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 311

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 327

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 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps \begin{align*} \text {integral}& = -\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*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 7.06 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.35 \[ \int \frac {x^{14}}{\left (1+x^4\right )^{3/2}} \, dx=\frac {x^3 \left (77-11 x^4+5 x^8-77 \sqrt {1+x^4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-x^4\right )\right )}{45 \sqrt {1+x^4}} \]

[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])

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 4.56 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.11

method	result	size
meijerg	\(\frac {x^{15} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {3}{2},\frac {15}{4};\frac {19}{4};-x^{4}\right )}{15}\)	\(17\)
risch	\(\frac {x^{3} \left (10 x^{8}-22 x^{4}-77\right )}{90 \sqrt {x^{4}+1}}+\frac {77 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-E\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{30 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(107\)
default	\(-\frac {x^{3}}{2 \sqrt {x^{4}+1}}+\frac {x^{7} \sqrt {x^{4}+1}}{9}-\frac {16 x^{3} \sqrt {x^{4}+1}}{45}+\frac {77 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-E\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{30 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(119\)
elliptic	\(-\frac {x^{3}}{2 \sqrt {x^{4}+1}}+\frac {x^{7} \sqrt {x^{4}+1}}{9}-\frac {16 x^{3} \sqrt {x^{4}+1}}{45}+\frac {77 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-E\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{30 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(119\)

[In]

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

[Out]

1/15*x^15*hypergeom([3/2,15/4],[19/4],-x^4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.54 \[ \int \frac {x^{14}}{\left (1+x^4\right )^{3/2}} \, dx=-\frac {231 \, \sqrt {i} {\left (-i \, x^{5} - i \, x\right )} E(\arcsin \left (\frac {\sqrt {i}}{x}\right )\,|\,-1) + 231 \, \sqrt {i} {\left (i \, x^{5} + i \, x\right )} F(\arcsin \left (\frac {\sqrt {i}}{x}\right )\,|\,-1) - {\left (10 \, x^{12} - 22 \, x^{8} + 154 \, x^{4} + 231\right )} \sqrt {x^{4} + 1}}{90 \, {\left (x^{5} + x\right )}} \]

[In]

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

[Out]

-1/90*(231*sqrt(I)*(-I*x^5 - I*x)*elliptic_e(arcsin(sqrt(I)/x), -1) + 231*sqrt(I)*(I*x^5 + I*x)*elliptic_f(arc
sin(sqrt(I)/x), -1) - (10*x^12 - 22*x^8 + 154*x^4 + 231)*sqrt(x^4 + 1))/(x^5 + x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.69 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.19 \[ \int \frac {x^{14}}{\left (1+x^4\right )^{3/2}} \, dx=\frac {x^{15} \Gamma \left (\frac {15}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {15}{4} \\ \frac {19}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {19}{4}\right )} \]

[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))

Maxima [F]

\[ \int \frac {x^{14}}{\left (1+x^4\right )^{3/2}} \, dx=\int { \frac {x^{14}}{{\left (x^{4} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

integrate(x^14/(x^4 + 1)^(3/2), x)

Giac [F]

\[ \int \frac {x^{14}}{\left (1+x^4\right )^{3/2}} \, dx=\int { \frac {x^{14}}{{\left (x^{4} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

integrate(x^14/(x^4 + 1)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{14}}{\left (1+x^4\right )^{3/2}} \, dx=\int \frac {x^{14}}{{\left (x^4+1\right )}^{3/2}} \,d x \]

[In]

int(x^14/(x^4 + 1)^(3/2),x)

[Out]

int(x^14/(x^4 + 1)^(3/2), x)