Integrand size = 15, antiderivative size = 129 \[ \int \frac {1}{x^{10} \left (a+b x^4\right )^{5/4}} \, dx=-\frac {1}{9 a x^9 \sqrt [4]{a+b x^4}}+\frac {2 b}{9 a^2 x^5 \sqrt [4]{a+b x^4}}-\frac {4 b^2}{3 a^3 x \sqrt [4]{a+b x^4}}+\frac {8 b^{5/2} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{3 a^{7/2} \sqrt [4]{a+b x^4}} \]
-1/9/a/x^9/(b*x^4+a)^(1/4)+2/9*b/a^2/x^5/(b*x^4+a)^(1/4)-4/3*b^2/a^3/x/(b* x^4+a)^(1/4)+8/3*b^(5/2)*(1+a/b/x^4)^(1/4)*x*(cos(1/2*arccot(x^2*b^(1/2)/a ^(1/2)))^2)^(1/2)/cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*a rccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))/a^(7/2)/(b*x^4+a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.42 \[ \int \frac {1}{x^{10} \left (a+b x^4\right )^{5/4}} \, dx=-\frac {\sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {9}{4},\frac {5}{4},-\frac {5}{4},-\frac {b x^4}{a}\right )}{9 a x^9 \sqrt [4]{a+b x^4}} \]
-1/9*((1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[-9/4, 5/4, -5/4, -((b*x^4)/a )])/(a*x^9*(a + b*x^4)^(1/4))
Time = 0.30 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {816, 816, 816, 813, 858, 807, 212}
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.
\(\displaystyle \int \frac {1}{x^{10} \left (a+b x^4\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 816 |
\(\displaystyle -\frac {10 b \int \frac {1}{x^6 \left (b x^4+a\right )^{5/4}}dx}{9 a}-\frac {1}{9 a x^9 \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 816 |
\(\displaystyle -\frac {10 b \left (-\frac {6 b \int \frac {1}{x^2 \left (b x^4+a\right )^{5/4}}dx}{5 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}\right )}{9 a}-\frac {1}{9 a x^9 \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 816 |
\(\displaystyle -\frac {10 b \left (-\frac {6 b \left (-\frac {2 b \int \frac {x^2}{\left (b x^4+a\right )^{5/4}}dx}{a}-\frac {1}{a x \sqrt [4]{a+b x^4}}\right )}{5 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}\right )}{9 a}-\frac {1}{9 a x^9 \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 813 |
\(\displaystyle -\frac {10 b \left (-\frac {6 b \left (-\frac {2 x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x^3}dx}{a \sqrt [4]{a+b x^4}}-\frac {1}{a x \sqrt [4]{a+b x^4}}\right )}{5 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}\right )}{9 a}-\frac {1}{9 a x^9 \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {10 b \left (-\frac {6 b \left (\frac {2 x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x}d\frac {1}{x}}{a \sqrt [4]{a+b x^4}}-\frac {1}{a x \sqrt [4]{a+b x^4}}\right )}{5 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}\right )}{9 a}-\frac {1}{9 a x^9 \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {10 b \left (-\frac {6 b \left (\frac {x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x^2}}{a \sqrt [4]{a+b x^4}}-\frac {1}{a x \sqrt [4]{a+b x^4}}\right )}{5 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}\right )}{9 a}-\frac {1}{9 a x^9 \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle -\frac {10 b \left (-\frac {6 b \left (\frac {2 \sqrt {b} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right )\right |2\right )}{a^{3/2} \sqrt [4]{a+b x^4}}-\frac {1}{a x \sqrt [4]{a+b x^4}}\right )}{5 a}-\frac {1}{5 a x^5 \sqrt [4]{a+b x^4}}\right )}{9 a}-\frac {1}{9 a x^9 \sqrt [4]{a+b x^4}}\) |
-1/9*1/(a*x^9*(a + b*x^4)^(1/4)) - (10*b*(-1/5*1/(a*x^5*(a + b*x^4)^(1/4)) - (6*b*(-(1/(a*x*(a + b*x^4)^(1/4))) + (2*Sqrt[b]*(1 + a/(b*x^4))^(1/4)*x *EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(a^(3/2)*(a + b*x^4)^(1/4) )))/(5*a)))/(9*a)
3.12.68.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) )*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x*((1 + a/(b*x^4) )^(1/4)/(b*(a + b*x^4)^(1/4))) Int[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x^(m + 1)/(a*( m + 1)*(a + b*x^4)^(1/4)), x] - Simp[b*(m/(a*(m + 1))) Int[x^(m + 4)/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a] && ILtQ[(m - 2)/4, 0 ]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
\[\int \frac {1}{x^{10} \left (b \,x^{4}+a \right )^{\frac {5}{4}}}d x\]
\[ \int \frac {1}{x^{10} \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{10}} \,d x } \]
Result contains complex when optimal does not.
Time = 0.90 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.34 \[ \int \frac {1}{x^{10} \left (a+b x^4\right )^{5/4}} \, dx=\frac {\Gamma \left (- \frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {9}{4}, \frac {5}{4} \\ - \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} x^{9} \Gamma \left (- \frac {5}{4}\right )} \]
gamma(-9/4)*hyper((-9/4, 5/4), (-5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(5 /4)*x**9*gamma(-5/4))
\[ \int \frac {1}{x^{10} \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{10}} \,d x } \]
\[ \int \frac {1}{x^{10} \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{10}} \,d x } \]
Timed out. \[ \int \frac {1}{x^{10} \left (a+b x^4\right )^{5/4}} \, dx=\int \frac {1}{x^{10}\,{\left (b\,x^4+a\right )}^{5/4}} \,d x \]