Integrand size = 15, antiderivative size = 258 \[ \int \frac {x^{10}}{\left (a+b x^4\right )^{3/2}} \, dx=-\frac {x^7}{2 b \sqrt {a+b x^4}}+\frac {7 x^3 \sqrt {a+b x^4}}{10 b^2}-\frac {21 a x \sqrt {a+b x^4}}{10 b^{5/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {21 a^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{10 b^{11/4} \sqrt {a+b x^4}}-\frac {21 a^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{20 b^{11/4} \sqrt {a+b x^4}} \]
-1/2*x^7/b/(b*x^4+a)^(1/2)+7/10*x^3*(b*x^4+a)^(1/2)/b^2-21/10*a*x*(b*x^4+a )^(1/2)/b^(5/2)/(a^(1/2)+x^2*b^(1/2))+21/10*a^(5/4)*(cos(2*arctan(b^(1/4)* x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arct an(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1 /2)+x^2*b^(1/2))^2)^(1/2)/b^(11/4)/(b*x^4+a)^(1/2)-21/20*a^(5/4)*(cos(2*ar ctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*Ellipti cF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b *x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/b^(11/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.26 \[ \int \frac {x^{10}}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {x^3 \left (-7 a+b x^4+7 a \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{5 b^2 \sqrt {a+b x^4}} \]
(x^3*(-7*a + b*x^4 + 7*a*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[3/4, 3/2, 7 /4, -((b*x^4)/a)]))/(5*b^2*Sqrt[a + b*x^4])
Time = 0.33 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {817, 843, 834, 27, 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.
| \(\displaystyle \int \frac {x^{10}}{\left (a+b x^4\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {7 \int \frac {x^6}{\sqrt {b x^4+a}}dx}{2 b}-\frac {x^7}{2 b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 843 |
| \(\displaystyle \frac {7 \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^4+a}}dx}{5 b}\right )}{2 b}-\frac {x^7}{2 b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 834 |
| \(\displaystyle \frac {7 \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^4+a}}dx}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {a} \sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{5 b}\right )}{2 b}-\frac {x^7}{2 b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {7 \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^4+a}}dx}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{5 b}\right )}{2 b}-\frac {x^7}{2 b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {7 \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{5 b}\right )}{2 b}-\frac {x^7}{2 b \sqrt {a+b x^4}}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {7 \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^4}}-\frac {x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}}{\sqrt {b}}\right )}{5 b}\right )}{2 b}-\frac {x^7}{2 b \sqrt {a+b x^4}}\) |
-1/2*x^7/(b*Sqrt[a + b*x^4]) + (7*((x^3*Sqrt[a + b*x^4])/(5*b) - (3*a*(-(( -((x*Sqrt[a + b*x^4])/(Sqrt[a] + Sqrt[b]*x^2)) + (a^(1/4)*(Sqrt[a] + Sqrt[ b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^ (1/4)*x)/a^(1/4)], 1/2])/(b^(1/4)*Sqrt[a + b*x^4]))/Sqrt[b]) + (a^(1/4)*(S qrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Elliptic F[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(2*b^(3/4)*Sqrt[a + b*x^4])))/(5*b) ))/(2*b)
3.9.68.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Result contains complex when optimal does not.
Time = 5.58 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.53
| method | result | size |
| default | \(\frac {a \,x^{3}}{2 b^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x^{3} \sqrt {b \,x^{4}+a}}{5 b^{2}}-\frac {21 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{10 b^{\frac {5}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(137\) |
| elliptic | \(\frac {a \,x^{3}}{2 b^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x^{3} \sqrt {b \,x^{4}+a}}{5 b^{2}}-\frac {21 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{10 b^{\frac {5}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(137\) |
| risch | \(\frac {x^{3} \sqrt {b \,x^{4}+a}}{5 b^{2}}-\frac {a \left (8 b \left (-\frac {x^{3}}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+3 a \left (\frac {x^{3}}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {i \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )\right )}{5 b^{2}}\) | \(268\) |
1/2/b^2*a*x^3/((x^4+a/b)*b)^(1/2)+1/5*x^3*(b*x^4+a)^(1/2)/b^2-21/10*I*a^(3 /2)/b^(5/2)/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2)^(1/2)*(1+I /a^(1/2)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*(EllipticF(x*(I/a^(1/2)*b^(1/2 ))^(1/2),I)-EllipticE(x*(I/a^(1/2)*b^(1/2))^(1/2),I))
Time = 0.10 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.51 \[ \int \frac {x^{10}}{\left (a+b x^4\right )^{3/2}} \, dx=-\frac {21 \, {\left (a b x^{5} + a^{2} x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 21 \, {\left (a b x^{5} + a^{2} x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - {\left (2 \, b^{2} x^{8} - 14 \, a b x^{4} - 21 \, a^{2}\right )} \sqrt {b x^{4} + a}}{10 \, {\left (b^{4} x^{5} + a b^{3} x\right )}} \]
-1/10*(21*(a*b*x^5 + a^2*x)*sqrt(b)*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^ (1/4)/x), -1) - 21*(a*b*x^5 + a^2*x)*sqrt(b)*(-a/b)^(3/4)*elliptic_f(arcsi n((-a/b)^(1/4)/x), -1) - (2*b^2*x^8 - 14*a*b*x^4 - 21*a^2)*sqrt(b*x^4 + a) )/(b^4*x^5 + a*b^3*x)
Result contains complex when optimal does not.
Time = 0.61 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.14 \[ \int \frac {x^{10}}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {15}{4}\right )} \]
x**11*gamma(11/4)*hyper((3/2, 11/4), (15/4,), b*x**4*exp_polar(I*pi)/a)/(4 *a**(3/2)*gamma(15/4))
\[ \int \frac {x^{10}}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {x^{10}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x^{10}}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {x^{10}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^{10}}{\left (a+b x^4\right )^{3/2}} \, dx=\int \frac {x^{10}}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \]