Integrand size = 15, antiderivative size = 129 \[ \int \frac {x^{14}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {77 a^2 x^3}{120 b^3 \sqrt [4]{a+b x^4}}-\frac {11 a x^7}{60 b^2 \sqrt [4]{a+b x^4}}+\frac {x^{11}}{10 b \sqrt [4]{a+b x^4}}+\frac {77 a^{5/2} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{40 b^{7/2} \sqrt [4]{a+b x^4}} \]
77/120*a^2*x^3/b^3/(b*x^4+a)^(1/4)-11/60*a*x^7/b^2/(b*x^4+a)^(1/4)+1/10*x^ 11/b/(b*x^4+a)^(1/4)+77/40*a^(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*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))/b^(7/2)/(b*x^4+a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.64 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.62 \[ \int \frac {x^{14}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {x^3 \left (77 a^2-22 a b x^4+12 b^2 x^8-77 a^2 \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{120 b^3 \sqrt [4]{a+b x^4}} \]
(x^3*(77*a^2 - 22*a*b*x^4 + 12*b^2*x^8 - 77*a^2*(1 + (b*x^4)/a)^(1/4)*Hype rgeometric2F1[3/4, 5/4, 7/4, -((b*x^4)/a)]))/(120*b^3*(a + b*x^4)^(1/4))
Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {815, 815, 815, 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.
\(\displaystyle \int \frac {x^{14}}{\left (a+b x^4\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 815 |
\(\displaystyle \frac {x^{11}}{10 b \sqrt [4]{a+b x^4}}-\frac {11 a \int \frac {x^{10}}{\left (b x^4+a\right )^{5/4}}dx}{10 b}\) |
\(\Big \downarrow \) 815 |
\(\displaystyle \frac {x^{11}}{10 b \sqrt [4]{a+b x^4}}-\frac {11 a \left (\frac {x^7}{6 b \sqrt [4]{a+b x^4}}-\frac {7 a \int \frac {x^6}{\left (b x^4+a\right )^{5/4}}dx}{6 b}\right )}{10 b}\) |
\(\Big \downarrow \) 815 |
\(\displaystyle \frac {x^{11}}{10 b \sqrt [4]{a+b x^4}}-\frac {11 a \left (\frac {x^7}{6 b \sqrt [4]{a+b x^4}}-\frac {7 a \left (\frac {x^3}{2 b \sqrt [4]{a+b x^4}}-\frac {3 a \int \frac {x^2}{\left (b x^4+a\right )^{5/4}}dx}{2 b}\right )}{6 b}\right )}{10 b}\) |
\(\Big \downarrow \) 813 |
\(\displaystyle \frac {x^{11}}{10 b \sqrt [4]{a+b x^4}}-\frac {11 a \left (\frac {x^7}{6 b \sqrt [4]{a+b x^4}}-\frac {7 a \left (\frac {x^3}{2 b \sqrt [4]{a+b x^4}}-\frac {3 a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x^3}dx}{2 b^2 \sqrt [4]{a+b x^4}}\right )}{6 b}\right )}{10 b}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {x^{11}}{10 b \sqrt [4]{a+b x^4}}-\frac {11 a \left (\frac {x^7}{6 b \sqrt [4]{a+b x^4}}-\frac {7 a \left (\frac {3 a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x}d\frac {1}{x}}{2 b^2 \sqrt [4]{a+b x^4}}+\frac {x^3}{2 b \sqrt [4]{a+b x^4}}\right )}{6 b}\right )}{10 b}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {x^{11}}{10 b \sqrt [4]{a+b x^4}}-\frac {11 a \left (\frac {x^7}{6 b \sqrt [4]{a+b x^4}}-\frac {7 a \left (\frac {3 a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x^2}}{4 b^2 \sqrt [4]{a+b x^4}}+\frac {x^3}{2 b \sqrt [4]{a+b x^4}}\right )}{6 b}\right )}{10 b}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {x^{11}}{10 b \sqrt [4]{a+b x^4}}-\frac {11 a \left (\frac {x^7}{6 b \sqrt [4]{a+b x^4}}-\frac {7 a \left (\frac {3 \sqrt {a} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right )\right |2\right )}{2 b^{3/2} \sqrt [4]{a+b x^4}}+\frac {x^3}{2 b \sqrt [4]{a+b x^4}}\right )}{6 b}\right )}{10 b}\) |
x^11/(10*b*(a + b*x^4)^(1/4)) - (11*a*(x^7/(6*b*(a + b*x^4)^(1/4)) - (7*a* (x^3/(2*b*(a + b*x^4)^(1/4)) + (3*Sqrt[a]*(1 + a/(b*x^4))^(1/4)*x*Elliptic E[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(2*b^(3/2)*(a + b*x^4)^(1/4))))/(6* b)))/(10*b)
3.12.62.3.1 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 - 3)/(b*( m - 4)*(a + b*x^4)^(1/4)), x] - Simp[a*((m - 3)/(b*(m - 4))) Int[x^(m - 4 )/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a] && IGtQ[(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]
\[\int \frac {x^{14}}{\left (b \,x^{4}+a \right )^{\frac {5}{4}}}d x\]
\[ \int \frac {x^{14}}{\left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {x^{14}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}}} \,d x } \]
Result contains complex when optimal does not.
Time = 0.82 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.29 \[ \int \frac {x^{14}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {x^{15} \Gamma \left (\frac {15}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {15}{4} \\ \frac {19}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} \Gamma \left (\frac {19}{4}\right )} \]
x**15*gamma(15/4)*hyper((5/4, 15/4), (19/4,), b*x**4*exp_polar(I*pi)/a)/(4 *a**(5/4)*gamma(19/4))
\[ \int \frac {x^{14}}{\left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {x^{14}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}}} \,d x } \]
\[ \int \frac {x^{14}}{\left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {x^{14}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^{14}}{\left (a+b x^4\right )^{5/4}} \, dx=\int \frac {x^{14}}{{\left (b\,x^4+a\right )}^{5/4}} \,d x \]