Integrand size = 15, antiderivative size = 120 \[ \int \frac {\left (a+b x^2\right )^{5/4}}{x^6} \, dx=-\frac {b \sqrt [4]{a+b x^2}}{6 x^3}-\frac {b^2 \sqrt [4]{a+b x^2}}{12 a x}-\frac {\left (a+b x^2\right )^{5/4}}{5 x^5}-\frac {b^{5/2} \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{12 \sqrt {a} \left (a+b x^2\right )^{3/4}} \] Output:
-1/6*b*(b*x^2+a)^(1/4)/x^3-1/12*b^2*(b*x^2+a)^(1/4)/a/x-1/5*(b*x^2+a)^(5/4 )/x^5-1/12*b^(5/2)*(1+b*x^2/a)^(3/4)*InverseJacobiAM(1/2*arctan(b^(1/2)*x/ a^(1/2)),2^(1/2))/a^(1/2)/(b*x^2+a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.43 \[ \int \frac {\left (a+b x^2\right )^{5/4}}{x^6} \, dx=-\frac {a \sqrt [4]{a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {5}{4},-\frac {3}{2},-\frac {b x^2}{a}\right )}{5 x^5 \sqrt [4]{1+\frac {b x^2}{a}}} \] Input:
Integrate[(a + b*x^2)^(5/4)/x^6,x]
Output:
-1/5*(a*(a + b*x^2)^(1/4)*Hypergeometric2F1[-5/2, -5/4, -3/2, -((b*x^2)/a) ])/(x^5*(1 + (b*x^2)/a)^(1/4))
Time = 0.21 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {247, 247, 264, 231, 229}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{5/4}}{x^6} \, dx\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {1}{2} b \int \frac {\sqrt [4]{b x^2+a}}{x^4}dx-\frac {\left (a+b x^2\right )^{5/4}}{5 x^5}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {1}{2} b \left (\frac {1}{6} b \int \frac {1}{x^2 \left (b x^2+a\right )^{3/4}}dx-\frac {\sqrt [4]{a+b x^2}}{3 x^3}\right )-\frac {\left (a+b x^2\right )^{5/4}}{5 x^5}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {1}{2} b \left (\frac {1}{6} b \left (-\frac {b \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx}{2 a}-\frac {\sqrt [4]{a+b x^2}}{a x}\right )-\frac {\sqrt [4]{a+b x^2}}{3 x^3}\right )-\frac {\left (a+b x^2\right )^{5/4}}{5 x^5}\) |
\(\Big \downarrow \) 231 |
\(\displaystyle \frac {1}{2} b \left (\frac {1}{6} b \left (-\frac {b \left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{2 a \left (a+b x^2\right )^{3/4}}-\frac {\sqrt [4]{a+b x^2}}{a x}\right )-\frac {\sqrt [4]{a+b x^2}}{3 x^3}\right )-\frac {\left (a+b x^2\right )^{5/4}}{5 x^5}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle \frac {1}{2} b \left (\frac {1}{6} b \left (-\frac {\sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\sqrt {a} \left (a+b x^2\right )^{3/4}}-\frac {\sqrt [4]{a+b x^2}}{a x}\right )-\frac {\sqrt [4]{a+b x^2}}{3 x^3}\right )-\frac {\left (a+b x^2\right )^{5/4}}{5 x^5}\) |
Input:
Int[(a + b*x^2)^(5/4)/x^6,x]
Output:
-1/5*(a + b*x^2)^(5/4)/x^5 + (b*(-1/3*(a + b*x^2)^(1/4)/x^3 + (b*(-((a + b *x^2)^(1/4)/(a*x)) - (Sqrt[b]*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt [b]*x)/Sqrt[a]]/2, 2])/(Sqrt[a]*(a + b*x^2)^(3/4))))/6))/2
Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) )*EllipticF[(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)^(-3/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(3/4)/( a + b*x^2)^(3/4) Int[1/(1 + b*(x^2/a))^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {5}{4}}}{x^{6}}d x\]
Input:
int((b*x^2+a)^(5/4)/x^6,x)
Output:
int((b*x^2+a)^(5/4)/x^6,x)
\[ \int \frac {\left (a+b x^2\right )^{5/4}}{x^6} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {5}{4}}}{x^{6}} \,d x } \] Input:
integrate((b*x^2+a)^(5/4)/x^6,x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(5/4)/x^6, x)
Result contains complex when optimal does not.
Time = 0.69 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.28 \[ \int \frac {\left (a+b x^2\right )^{5/4}}{x^6} \, dx=- \frac {a^{\frac {5}{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, - \frac {5}{4} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 x^{5}} \] Input:
integrate((b*x**2+a)**(5/4)/x**6,x)
Output:
-a**(5/4)*hyper((-5/2, -5/4), (-3/2,), b*x**2*exp_polar(I*pi)/a)/(5*x**5)
\[ \int \frac {\left (a+b x^2\right )^{5/4}}{x^6} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {5}{4}}}{x^{6}} \,d x } \] Input:
integrate((b*x^2+a)^(5/4)/x^6,x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(5/4)/x^6, x)
\[ \int \frac {\left (a+b x^2\right )^{5/4}}{x^6} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {5}{4}}}{x^{6}} \,d x } \] Input:
integrate((b*x^2+a)^(5/4)/x^6,x, algorithm="giac")
Output:
integrate((b*x^2 + a)^(5/4)/x^6, x)
Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/4}}{x^6} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{5/4}}{x^6} \,d x \] Input:
int((a + b*x^2)^(5/4)/x^6,x)
Output:
int((a + b*x^2)^(5/4)/x^6, x)
\[ \int \frac {\left (a+b x^2\right )^{5/4}}{x^6} \, dx=\frac {-8 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a -18 \left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}+5 \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{b \,x^{8}+a \,x^{6}}d x \right ) a^{2} x^{5}}{45 x^{5}} \] Input:
int((b*x^2+a)^(5/4)/x^6,x)
Output:
( - 8*(a + b*x**2)**(1/4)*a - 18*(a + b*x**2)**(1/4)*b*x**2 + 5*int((a + b *x**2)**(1/4)/(a*x**6 + b*x**8),x)*a**2*x**5)/(45*x**5)