Integrand size = 15, antiderivative size = 151 \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{3/2}} \, dx=-\frac {x^9}{2 b \sqrt {a+b x^4}}-\frac {15 a x \sqrt {a+b x^4}}{14 b^3}+\frac {9 x^5 \sqrt {a+b x^4}}{14 b^2}+\frac {15 a^{7/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{28 b^{13/4} \sqrt {a+b x^4}} \]
-1/2*x^9/b/(b*x^4+a)^(1/2)-15/14*a*x*(b*x^4+a)^(1/2)/b^3+9/14*x^5*(b*x^4+a )^(1/2)/b^2+15/28*a^(7/4)*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2 *arctan(b^(1/4)*x/a^(1/4)))*EllipticF(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^(13/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.52 \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {-15 a^2 x-6 a b x^5+2 b^2 x^9+15 a^2 x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )}{14 b^3 \sqrt {a+b x^4}} \]
(-15*a^2*x - 6*a*b*x^5 + 2*b^2*x^9 + 15*a^2*x*Sqrt[1 + (b*x^4)/a]*Hypergeo metric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)])/(14*b^3*Sqrt[a + b*x^4])
Time = 0.25 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {817, 843, 843, 761}
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.
\(\displaystyle \int \frac {x^{12}}{\left (a+b x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 817 |
\(\displaystyle \frac {9 \int \frac {x^8}{\sqrt {b x^4+a}}dx}{2 b}-\frac {x^9}{2 b \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {9 \left (\frac {x^5 \sqrt {a+b x^4}}{7 b}-\frac {5 a \int \frac {x^4}{\sqrt {b x^4+a}}dx}{7 b}\right )}{2 b}-\frac {x^9}{2 b \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {9 \left (\frac {x^5 \sqrt {a+b x^4}}{7 b}-\frac {5 a \left (\frac {x \sqrt {a+b x^4}}{3 b}-\frac {a \int \frac {1}{\sqrt {b x^4+a}}dx}{3 b}\right )}{7 b}\right )}{2 b}-\frac {x^9}{2 b \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {9 \left (\frac {x^5 \sqrt {a+b x^4}}{7 b}-\frac {5 a \left (\frac {x \sqrt {a+b x^4}}{3 b}-\frac {a^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 b^{5/4} \sqrt {a+b x^4}}\right )}{7 b}\right )}{2 b}-\frac {x^9}{2 b \sqrt {a+b x^4}}\) |
-1/2*x^9/(b*Sqrt[a + b*x^4]) + (9*((x^5*Sqrt[a + b*x^4])/(7*b) - (5*a*((x* Sqrt[a + b*x^4])/(3*b) - (a^(3/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4) /(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2]) /(6*b^(5/4)*Sqrt[a + b*x^4])))/(7*b)))/(2*b)
3.9.61.3.1 Defintions of rubi rules used
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[((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]
Result contains complex when optimal does not.
Time = 5.32 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.88
method | result | size |
default | \(-\frac {a^{2} x}{2 b^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x^{5} \sqrt {b \,x^{4}+a}}{7 b^{2}}-\frac {4 a x \sqrt {b \,x^{4}+a}}{7 b^{3}}+\frac {15 a^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{14 b^{3} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(133\) |
elliptic | \(-\frac {a^{2} x}{2 b^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x^{5} \sqrt {b \,x^{4}+a}}{7 b^{2}}-\frac {4 a x \sqrt {b \,x^{4}+a}}{7 b^{3}}+\frac {15 a^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{14 b^{3} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(133\) |
risch | \(-\frac {x \left (-b \,x^{4}+4 a \right ) \sqrt {b \,x^{4}+a}}{7 b^{3}}+\frac {a^{2} \left (4 a \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+11 b \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\right )}{7 b^{3}}\) | \(228\) |
-1/2/b^3*a^2*x/((x^4+a/b)*b)^(1/2)+1/7*x^5*(b*x^4+a)^(1/2)/b^2-4/7*a*x*(b* x^4+a)^(1/2)/b^3+15/14*a^2/b^3/(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)
Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.58 \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {15 \, {\left (a b x^{4} + a^{2}\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (2 \, b^{2} x^{9} - 6 \, a b x^{5} - 15 \, a^{2} x\right )} \sqrt {b x^{4} + a}}{14 \, {\left (b^{4} x^{4} + a b^{3}\right )}} \]
1/14*(15*(a*b*x^4 + a^2)*sqrt(b)*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/ 4)/x), -1) + (2*b^2*x^9 - 6*a*b*x^5 - 15*a^2*x)*sqrt(b*x^4 + a))/(b^4*x^4 + a*b^3)
Result contains complex when optimal does not.
Time = 0.68 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.25 \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {17}{4}\right )} \]
x**13*gamma(13/4)*hyper((3/2, 13/4), (17/4,), b*x**4*exp_polar(I*pi)/a)/(4 *a**(3/2)*gamma(17/4))
\[ \int \frac {x^{12}}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {x^{12}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x^{12}}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {x^{12}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{3/2}} \, dx=\int \frac {x^{12}}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \]