\(\int \frac {x^{14}}{(1+x^4)^{3/2}} \, dx\) [432]

Optimal result

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

-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 
*x*(x^4+1)^(1/2)/(30*x^2+30)-77/30*(x^2+1)*((x^4+1)/(x^2+1)^2)^(1/2)*Ellip 
ticE(sin(2*arctan(x)),1/2*2^(1/2))/(x^4+1)^(1/2)+77/60*(x^2+1)*((x^4+1)/(x 
^2+1)^2)^(1/2)*InverseJacobiAM(2*arctan(x),1/2*2^(1/2))/(x^4+1)^(1/2)
 

Mathematica [C] (verified)

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

Time = 5.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.35 \[ \int \frac {x^{14}}{\left (1+x^4\right )^{3/2}} \, dx=\frac {x^3 \left (77-11 x^4+5 x^8-77 \sqrt {1+x^4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-x^4\right )\right )}{45 \sqrt {1+x^4}} \] Input:

Integrate[x^14/(1 + x^4)^(3/2),x]
 

Output:

(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])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {817, 843, 843, 834, 761, 1510}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 817

\(\displaystyle \frac {11}{2} \int \frac {x^{10}}{\sqrt {x^4+1}}dx-\frac {x^{11}}{2 \sqrt {x^4+1}}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {11}{2} \left (\frac {1}{9} x^7 \sqrt {x^4+1}-\frac {7}{9} \int \frac {x^6}{\sqrt {x^4+1}}dx\right )-\frac {x^{11}}{2 \sqrt {x^4+1}}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {11}{2} \left (\frac {1}{9} x^7 \sqrt {x^4+1}-\frac {7}{9} \left (\frac {1}{5} x^3 \sqrt {x^4+1}-\frac {3}{5} \int \frac {x^2}{\sqrt {x^4+1}}dx\right )\right )-\frac {x^{11}}{2 \sqrt {x^4+1}}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {11}{2} \left (\frac {1}{9} x^7 \sqrt {x^4+1}-\frac {7}{9} \left (\frac {1}{5} x^3 \sqrt {x^4+1}-\frac {3}{5} \left (\int \frac {1}{\sqrt {x^4+1}}dx-\int \frac {1-x^2}{\sqrt {x^4+1}}dx\right )\right )\right )-\frac {x^{11}}{2 \sqrt {x^4+1}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {11}{2} \left (\frac {1}{9} x^7 \sqrt {x^4+1}-\frac {7}{9} \left (\frac {1}{5} x^3 \sqrt {x^4+1}-\frac {3}{5} \left (\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {x^4+1}}-\int \frac {1-x^2}{\sqrt {x^4+1}}dx\right )\right )\right )-\frac {x^{11}}{2 \sqrt {x^4+1}}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {11}{2} \left (\frac {1}{9} x^7 \sqrt {x^4+1}-\frac {7}{9} \left (\frac {1}{5} x^3 \sqrt {x^4+1}-\frac {3}{5} \left (\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {x^4+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^4+1}}+\frac {\sqrt {x^4+1} x}{x^2+1}\right )\right )\right )-\frac {x^{11}}{2 \sqrt {x^4+1}}\)

Input:

Int[x^14/(1 + x^4)^(3/2),x]
 

Output:

-1/2*x^11/Sqrt[1 + x^4] + (11*((x^7*Sqrt[1 + x^4])/9 - (7*((x^3*Sqrt[1 + x 
^4])/5 - (3*((x*Sqrt[1 + x^4])/(1 + x^2) - ((1 + x^2)*Sqrt[(1 + x^4)/(1 + 
x^2)^2]*EllipticE[2*ArcTan[x], 1/2])/Sqrt[1 + x^4] + ((1 + x^2)*Sqrt[(1 + 
x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(2*Sqrt[1 + x^4])))/5))/9)) 
/2
 

Defintions of rubi rules used

rule 761

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 817

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] - Simp[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 834

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

rule 843

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] - Simp[ 
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 1510

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

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.11

method	result	size
meijerg	\(\frac {x^{15} \operatorname {hypergeom}\left (\left [\frac {3}{2}, \frac {15}{4}\right ], \left [\frac {19}{4}\right ], -x^{4}\right )}{15}\)	\(17\)
risch	\(\frac {x^{3} \left (10 x^{8}-22 x^{4}-77\right )}{90 \sqrt {x^{4}+1}}+\frac {77 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{30 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(107\)
default	\(-\frac {x^{3}}{2 \sqrt {x^{4}+1}}+\frac {x^{7} \sqrt {x^{4}+1}}{9}-\frac {16 x^{3} \sqrt {x^{4}+1}}{45}+\frac {77 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{30 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(119\)
elliptic	\(-\frac {x^{3}}{2 \sqrt {x^{4}+1}}+\frac {x^{7} \sqrt {x^{4}+1}}{9}-\frac {16 x^{3} \sqrt {x^{4}+1}}{45}+\frac {77 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{30 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(119\)

Input:

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

Output:

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 = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.54 \[ \int \frac {x^{14}}{\left (1+x^4\right )^{3/2}} \, dx=-\frac {231 \, \sqrt {i} {\left (-i \, x^{5} - i \, x\right )} E(\arcsin \left (\frac {\sqrt {i}}{x}\right )\,|\,-1) + 231 \, \sqrt {i} {\left (i \, x^{5} + i \, x\right )} F(\arcsin \left (\frac {\sqrt {i}}{x}\right )\,|\,-1) - {\left (10 \, x^{12} - 22 \, x^{8} + 154 \, x^{4} + 231\right )} \sqrt {x^{4} + 1}}{90 \, {\left (x^{5} + x\right )}} \] Input:

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

Output:

-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(arcsin(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 = 0.62 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.19 \[ \int \frac {x^{14}}{\left (1+x^4\right )^{3/2}} \, dx=\frac {x^{15} \Gamma \left (\frac {15}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {15}{4} \\ \frac {19}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {19}{4}\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^{14}}{\left (1+x^4\right )^{3/2}} \, dx=\frac {5 \sqrt {x^{4}+1}\, x^{11}-11 \sqrt {x^{4}+1}\, x^{7}+77 \sqrt {x^{4}+1}\, x^{3}-231 \left (\int \frac {\sqrt {x^{4}+1}\, x^{2}}{x^{8}+2 x^{4}+1}d x \right ) x^{4}-231 \left (\int \frac {\sqrt {x^{4}+1}\, x^{2}}{x^{8}+2 x^{4}+1}d x \right )}{45 x^{4}+45} \] Input:

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

Output:

(5*sqrt(x**4 + 1)*x**11 - 11*sqrt(x**4 + 1)*x**7 + 77*sqrt(x**4 + 1)*x**3 
- 231*int((sqrt(x**4 + 1)*x**2)/(x**8 + 2*x**4 + 1),x)*x**4 - 231*int((sqr 
t(x**4 + 1)*x**2)/(x**8 + 2*x**4 + 1),x))/(45*(x**4 + 1))