Integrand size = 13, antiderivative size = 140 \[ \int \frac {x^{10}}{\sqrt {1+x^4}} \, dx=-\frac {7}{45} x^3 \sqrt {1+x^4}+\frac {1}{9} x^7 \sqrt {1+x^4}+\frac {7 x \sqrt {1+x^4}}{15 \left (1+x^2\right )}-\frac {7 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{15 \sqrt {1+x^4}}+\frac {7 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{30 \sqrt {1+x^4}} \] Output:
-7/45*x^3*(x^4+1)^(1/2)+1/9*x^7*(x^4+1)^(1/2)+7*x*(x^4+1)^(1/2)/(15*x^2+15 )-7/15*(x^2+1)*((x^4+1)/(x^2+1)^2)^(1/2)*EllipticE(sin(2*arctan(x)),1/2*2^ (1/2))/(x^4+1)^(1/2)+7/30*(x^2+1)*((x^4+1)/(x^2+1)^2)^(1/2)*InverseJacobiA M(2*arctan(x),1/2*2^(1/2))/(x^4+1)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.30 \[ \int \frac {x^{10}}{\sqrt {1+x^4}} \, dx=\frac {1}{45} x^3 \left (\sqrt {1+x^4} \left (-7+5 x^4\right )+7 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-x^4\right )\right ) \] Input:
Integrate[x^10/Sqrt[1 + x^4],x]
Output:
(x^3*(Sqrt[1 + x^4]*(-7 + 5*x^4) + 7*Hypergeometric2F1[1/2, 3/4, 7/4, -x^4 ]))/45
Time = 0.42 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {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.
\(\displaystyle \int \frac {x^{10}}{\sqrt {x^4+1}} \, dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {1}{9} x^7 \sqrt {x^4+1}-\frac {7}{9} \int \frac {x^6}{\sqrt {x^4+1}}dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {1}{9} x^7 \sqrt {x^4+1}-\frac {7}{9} \left (\frac {1}{5} x^3 \sqrt {x^4+1}-\frac {3}{5} \int \frac {x^2}{\sqrt {x^4+1}}dx\right )\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {1}{9} x^7 \sqrt {x^4+1}-\frac {7}{9} \left (\frac {1}{5} x^3 \sqrt {x^4+1}-\frac {3}{5} \left (\int \frac {1}{\sqrt {x^4+1}}dx-\int \frac {1-x^2}{\sqrt {x^4+1}}dx\right )\right )\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {1}{9} x^7 \sqrt {x^4+1}-\frac {7}{9} \left (\frac {1}{5} x^3 \sqrt {x^4+1}-\frac {3}{5} \left (\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {x^4+1}}-\int \frac {1-x^2}{\sqrt {x^4+1}}dx\right )\right )\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {1}{9} x^7 \sqrt {x^4+1}-\frac {7}{9} \left (\frac {1}{5} x^3 \sqrt {x^4+1}-\frac {3}{5} \left (\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {x^4+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^4+1}}+\frac {\sqrt {x^4+1} x}{x^2+1}\right )\right )\) |
Input:
Int[x^10/Sqrt[1 + x^4],x]
Output:
(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*A rcTan[x], 1/2])/(2*Sqrt[1 + x^4])))/5))/9
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]
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]
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]
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]
Time = 0.65 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.12
method | result | size |
meijerg | \(\frac {x^{11} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {11}{4}\right ], \left [\frac {15}{4}\right ], -x^{4}\right )}{11}\) | \(17\) |
risch | \(\frac {x^{3} \left (5 x^{4}-7\right ) \sqrt {x^{4}+1}}{45}+\frac {7 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{15 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\) | \(102\) |
default | \(\frac {x^{7} \sqrt {x^{4}+1}}{9}-\frac {7 x^{3} \sqrt {x^{4}+1}}{45}+\frac {7 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{15 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\) | \(107\) |
elliptic | \(\frac {x^{7} \sqrt {x^{4}+1}}{9}-\frac {7 x^{3} \sqrt {x^{4}+1}}{45}+\frac {7 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{15 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\) | \(107\) |
Input:
int(x^10/(x^4+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/11*x^11*hypergeom([1/2,11/4],[15/4],-x^4)
Result contains complex when optimal does not.
Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.41 \[ \int \frac {x^{10}}{\sqrt {1+x^4}} \, dx=\frac {21 i \, \sqrt {i} x E(\arcsin \left (\frac {\sqrt {i}}{x}\right )\,|\,-1) - 21 i \, \sqrt {i} x F(\arcsin \left (\frac {\sqrt {i}}{x}\right )\,|\,-1) + {\left (5 \, x^{8} - 7 \, x^{4} + 21\right )} \sqrt {x^{4} + 1}}{45 \, x} \] Input:
integrate(x^10/(x^4+1)^(1/2),x, algorithm="fricas")
Output:
1/45*(21*I*sqrt(I)*x*elliptic_e(arcsin(sqrt(I)/x), -1) - 21*I*sqrt(I)*x*el liptic_f(arcsin(sqrt(I)/x), -1) + (5*x^8 - 7*x^4 + 21)*sqrt(x^4 + 1))/x
Result contains complex when optimal does not.
Time = 0.53 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.21 \[ \int \frac {x^{10}}{\sqrt {1+x^4}} \, dx=\frac {x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {15}{4}\right )} \] Input:
integrate(x**10/(x**4+1)**(1/2),x)
Output:
x**11*gamma(11/4)*hyper((1/2, 11/4), (15/4,), x**4*exp_polar(I*pi))/(4*gam ma(15/4))
\[ \int \frac {x^{10}}{\sqrt {1+x^4}} \, dx=\int { \frac {x^{10}}{\sqrt {x^{4} + 1}} \,d x } \] Input:
integrate(x^10/(x^4+1)^(1/2),x, algorithm="maxima")
Output:
integrate(x^10/sqrt(x^4 + 1), x)
\[ \int \frac {x^{10}}{\sqrt {1+x^4}} \, dx=\int { \frac {x^{10}}{\sqrt {x^{4} + 1}} \,d x } \] Input:
integrate(x^10/(x^4+1)^(1/2),x, algorithm="giac")
Output:
integrate(x^10/sqrt(x^4 + 1), x)
Timed out. \[ \int \frac {x^{10}}{\sqrt {1+x^4}} \, dx=\int \frac {x^{10}}{\sqrt {x^4+1}} \,d x \] Input:
int(x^10/(x^4 + 1)^(1/2),x)
Output:
int(x^10/(x^4 + 1)^(1/2), x)
\[ \int \frac {x^{10}}{\sqrt {1+x^4}} \, dx=\frac {\sqrt {x^{4}+1}\, x^{7}}{9}-\frac {7 \sqrt {x^{4}+1}\, x^{3}}{45}+\frac {7 \left (\int \frac {\sqrt {x^{4}+1}\, x^{2}}{x^{4}+1}d x \right )}{15} \] Input:
int(x^10/(x^4+1)^(1/2),x)
Output:
(5*sqrt(x**4 + 1)*x**7 - 7*sqrt(x**4 + 1)*x**3 + 21*int((sqrt(x**4 + 1)*x* *2)/(x**4 + 1),x))/45