Integrand size = 13, antiderivative size = 140 \[ \int \frac {x^{10}}{\left (1+x^4\right )^{3/2}} \, dx=-\frac {x^7}{2 \sqrt {1+x^4}}+\frac {7}{10} x^3 \sqrt {1+x^4}-\frac {21 x \sqrt {1+x^4}}{10 \left (1+x^2\right )}+\frac {21 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{10 \sqrt {1+x^4}}-\frac {21 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{20 \sqrt {1+x^4}} \] Output:
-1/2*x^7/(x^4+1)^(1/2)+7/10*x^3*(x^4+1)^(1/2)-21*x*(x^4+1)^(1/2)/(10*x^2+1 0)+21/10*(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)-21/20*(x^2+1)*((x^4+1)/(x^2+1)^2)^(1/2)*InverseJaco biAM(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 = 3.93 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.34 \[ \int \frac {x^{10}}{\left (1+x^4\right )^{3/2}} \, dx=\frac {x^3 \left (-7+x^4+7 \sqrt {1+x^4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-x^4\right )\right )}{5 \sqrt {1+x^4}} \] Input:
Integrate[x^10/(1 + x^4)^(3/2),x]
Output:
(x^3*(-7 + x^4 + 7*Sqrt[1 + x^4]*Hypergeometric2F1[3/4, 3/2, 7/4, -x^4]))/ (5*Sqrt[1 + x^4])
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 = {817, 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.
\(\displaystyle \int \frac {x^{10}}{\left (x^4+1\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 817 |
\(\displaystyle \frac {7}{2} \int \frac {x^6}{\sqrt {x^4+1}}dx-\frac {x^7}{2 \sqrt {x^4+1}}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {7}{2} \left (\frac {1}{5} x^3 \sqrt {x^4+1}-\frac {3}{5} \int \frac {x^2}{\sqrt {x^4+1}}dx\right )-\frac {x^7}{2 \sqrt {x^4+1}}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {7}{2} \left (\frac {1}{5} x^3 \sqrt {x^4+1}-\frac {3}{5} \left (\int \frac {1}{\sqrt {x^4+1}}dx-\int \frac {1-x^2}{\sqrt {x^4+1}}dx\right )\right )-\frac {x^7}{2 \sqrt {x^4+1}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {7}{2} \left (\frac {1}{5} x^3 \sqrt {x^4+1}-\frac {3}{5} \left (\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {x^4+1}}-\int \frac {1-x^2}{\sqrt {x^4+1}}dx\right )\right )-\frac {x^7}{2 \sqrt {x^4+1}}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {7}{2} \left (\frac {1}{5} x^3 \sqrt {x^4+1}-\frac {3}{5} \left (\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {x^4+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^4+1}}+\frac {\sqrt {x^4+1} x}{x^2+1}\right )\right )-\frac {x^7}{2 \sqrt {x^4+1}}\) |
Input:
Int[x^10/(1 + x^4)^(3/2),x]
Output:
-1/2*x^7/Sqrt[1 + x^4] + (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))/2
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[((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]
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 = 0.62 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.12
method | result | size |
meijerg | \(\frac {x^{11} \operatorname {hypergeom}\left (\left [\frac {3}{2}, \frac {11}{4}\right ], \left [\frac {15}{4}\right ], -x^{4}\right )}{11}\) | \(17\) |
risch | \(\frac {x^{3} \left (2 x^{4}+7\right )}{10 \sqrt {x^{4}+1}}-\frac {21 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{10 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\) | \(102\) |
default | \(\frac {x^{3}}{2 \sqrt {x^{4}+1}}+\frac {x^{3} \sqrt {x^{4}+1}}{5}-\frac {21 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{10 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\) | \(107\) |
elliptic | \(\frac {x^{3}}{2 \sqrt {x^{4}+1}}+\frac {x^{3} \sqrt {x^{4}+1}}{5}-\frac {21 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{10 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\) | \(107\) |
Input:
int(x^10/(x^4+1)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/11*x^11*hypergeom([3/2,11/4],[15/4],-x^4)
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.56 \[ \int \frac {x^{10}}{\left (1+x^4\right )^{3/2}} \, dx=-\frac {21 \, \sqrt {i} {\left (i \, x^{5} + i \, x\right )} E(\arcsin \left (\frac {\sqrt {i}}{x}\right )\,|\,-1) + 21 \, \sqrt {i} {\left (-i \, x^{5} - i \, x\right )} F(\arcsin \left (\frac {\sqrt {i}}{x}\right )\,|\,-1) - {\left (2 \, x^{8} - 14 \, x^{4} - 21\right )} \sqrt {x^{4} + 1}}{10 \, {\left (x^{5} + x\right )}} \] Input:
integrate(x^10/(x^4+1)^(3/2),x, algorithm="fricas")
Output:
-1/10*(21*sqrt(I)*(I*x^5 + I*x)*elliptic_e(arcsin(sqrt(I)/x), -1) + 21*sqr t(I)*(-I*x^5 - I*x)*elliptic_f(arcsin(sqrt(I)/x), -1) - (2*x^8 - 14*x^4 - 21)*sqrt(x^4 + 1))/(x^5 + x)
Result contains complex when optimal does not.
Time = 0.49 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.21 \[ \int \frac {x^{10}}{\left (1+x^4\right )^{3/2}} \, dx=\frac {x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {15}{4}\right )} \] Input:
integrate(x**10/(x**4+1)**(3/2),x)
Output:
x**11*gamma(11/4)*hyper((3/2, 11/4), (15/4,), x**4*exp_polar(I*pi))/(4*gam ma(15/4))
\[ \int \frac {x^{10}}{\left (1+x^4\right )^{3/2}} \, dx=\int { \frac {x^{10}}{{\left (x^{4} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^10/(x^4+1)^(3/2),x, algorithm="maxima")
Output:
integrate(x^10/(x^4 + 1)^(3/2), x)
\[ \int \frac {x^{10}}{\left (1+x^4\right )^{3/2}} \, dx=\int { \frac {x^{10}}{{\left (x^{4} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^10/(x^4+1)^(3/2),x, algorithm="giac")
Output:
integrate(x^10/(x^4 + 1)^(3/2), x)
Timed out. \[ \int \frac {x^{10}}{\left (1+x^4\right )^{3/2}} \, dx=\int \frac {x^{10}}{{\left (x^4+1\right )}^{3/2}} \,d x \] Input:
int(x^10/(x^4 + 1)^(3/2),x)
Output:
int(x^10/(x^4 + 1)^(3/2), x)
\[ \int \frac {x^{10}}{\left (1+x^4\right )^{3/2}} \, dx=\frac {\sqrt {x^{4}+1}\, x^{7}-7 \sqrt {x^{4}+1}\, x^{3}+21 \left (\int \frac {\sqrt {x^{4}+1}\, x^{2}}{x^{8}+2 x^{4}+1}d x \right ) x^{4}+21 \left (\int \frac {\sqrt {x^{4}+1}\, x^{2}}{x^{8}+2 x^{4}+1}d x \right )}{5 x^{4}+5} \] Input:
int(x^10/(x^4+1)^(3/2),x)
Output:
(sqrt(x**4 + 1)*x**7 - 7*sqrt(x**4 + 1)*x**3 + 21*int((sqrt(x**4 + 1)*x**2 )/(x**8 + 2*x**4 + 1),x)*x**4 + 21*int((sqrt(x**4 + 1)*x**2)/(x**8 + 2*x** 4 + 1),x))/(5*(x**4 + 1))