Integrand size = 15, antiderivative size = 104 \[ \int \frac {x^9}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {6 a x^2}{5 b^2 \sqrt [4]{a+b x^4}}+\frac {x^6}{5 b \sqrt [4]{a+b x^4}}+\frac {12 a^{3/2} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{5 b^{5/2} \sqrt [4]{a+b x^4}} \] Output:
-6/5*a*x^2/b^2/(b*x^4+a)^(1/4)+1/5*x^6/b/(b*x^4+a)^(1/4)+12/5*a^(3/2)*(1+b *x^4/a)^(1/4)*EllipticE(sin(1/2*arctan(b^(1/2)*x^2/a^(1/2))),2^(1/2))/b^(5 /2)/(b*x^4+a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.80 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.63 \[ \int \frac {x^9}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {x^2 \left (6 a+b x^4-6 a \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^4}{a}\right )\right )}{5 b^2 \sqrt [4]{a+b x^4}} \] Input:
Integrate[x^9/(a + b*x^4)^(5/4),x]
Output:
(x^2*(6*a + b*x^4 - 6*a*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^4)/a)]))/(5*b^2*(a + b*x^4)^(1/4))
Time = 0.35 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {807, 250, 250, 213, 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.
\(\displaystyle \int \frac {x^9}{\left (a+b x^4\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int \frac {x^8}{\left (b x^4+a\right )^{5/4}}dx^2\) |
\(\Big \downarrow \) 250 |
\(\displaystyle \frac {1}{2} \left (\frac {2 x^6}{5 b \sqrt [4]{a+b x^4}}-\frac {6 a \int \frac {x^4}{\left (b x^4+a\right )^{5/4}}dx^2}{5 b}\right )\) |
\(\Big \downarrow \) 250 |
\(\displaystyle \frac {1}{2} \left (\frac {2 x^6}{5 b \sqrt [4]{a+b x^4}}-\frac {6 a \left (\frac {2 x^2}{b \sqrt [4]{a+b x^4}}-\frac {2 a \int \frac {1}{\left (b x^4+a\right )^{5/4}}dx^2}{b}\right )}{5 b}\right )\) |
\(\Big \downarrow \) 213 |
\(\displaystyle \frac {1}{2} \left (\frac {2 x^6}{5 b \sqrt [4]{a+b x^4}}-\frac {6 a \left (\frac {2 x^2}{b \sqrt [4]{a+b x^4}}-\frac {2 \sqrt [4]{\frac {b x^4}{a}+1} \int \frac {1}{\left (\frac {b x^4}{a}+1\right )^{5/4}}dx^2}{b \sqrt [4]{a+b x^4}}\right )}{5 b}\right )\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {1}{2} \left (\frac {2 x^6}{5 b \sqrt [4]{a+b x^4}}-\frac {6 a \left (\frac {2 x^2}{b \sqrt [4]{a+b x^4}}-\frac {4 \sqrt {a} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{b^{3/2} \sqrt [4]{a+b x^4}}\right )}{5 b}\right )\) |
Input:
Int[x^9/(a + b*x^4)^(5/4),x]
Output:
((2*x^6)/(5*b*(a + b*x^4)^(1/4)) - (6*a*((2*x^2)/(b*(a + b*x^4)^(1/4)) - ( 4*Sqrt[a]*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(b^(3/2)*(a + b*x^4)^(1/4))))/(5*b))/2
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[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( a*(a + b*x^2)^(1/4)) Int[1/(1 + b*(x^2/a))^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a] && PosQ[b/a]
Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[2*c*(( c*x)^(m - 1)/(b*(2*m - 3)*(a + b*x^2)^(1/4))), x] - Simp[2*a*c^2*((m - 1)/( b*(2*m - 3))) Int[(c*x)^(m - 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a] && IntegerQ[2*m] && GtQ[m, 3/2]
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]
\[\int \frac {x^{9}}{\left (b \,x^{4}+a \right )^{\frac {5}{4}}}d x\]
Input:
int(x^9/(b*x^4+a)^(5/4),x)
Output:
int(x^9/(b*x^4+a)^(5/4),x)
\[ \int \frac {x^9}{\left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {x^{9}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(x^9/(b*x^4+a)^(5/4),x, algorithm="fricas")
Output:
integral((b*x^4 + a)^(3/4)*x^9/(b^2*x^8 + 2*a*b*x^4 + a^2), x)
Result contains complex when optimal does not.
Time = 0.55 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.26 \[ \int \frac {x^9}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {x^{10} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{10 a^{\frac {5}{4}}} \] Input:
integrate(x**9/(b*x**4+a)**(5/4),x)
Output:
x**10*hyper((5/4, 5/2), (7/2,), b*x**4*exp_polar(I*pi)/a)/(10*a**(5/4))
\[ \int \frac {x^9}{\left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {x^{9}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(x^9/(b*x^4+a)^(5/4),x, algorithm="maxima")
Output:
integrate(x^9/(b*x^4 + a)^(5/4), x)
\[ \int \frac {x^9}{\left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {x^{9}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(x^9/(b*x^4+a)^(5/4),x, algorithm="giac")
Output:
integrate(x^9/(b*x^4 + a)^(5/4), x)
Timed out. \[ \int \frac {x^9}{\left (a+b x^4\right )^{5/4}} \, dx=\int \frac {x^9}{{\left (b\,x^4+a\right )}^{5/4}} \,d x \] Input:
int(x^9/(a + b*x^4)^(5/4),x)
Output:
int(x^9/(a + b*x^4)^(5/4), x)
\[ \int \frac {x^9}{\left (a+b x^4\right )^{5/4}} \, dx=\int \frac {x^{9}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a +\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{4}}d x \] Input:
int(x^9/(b*x^4+a)^(5/4),x)
Output:
int(x**9/((a + b*x**4)**(1/4)*a + (a + b*x**4)**(1/4)*b*x**4),x)