Integrand size = 15, antiderivative size = 145 \[ \int \frac {\left (a+b x^2\right )^{3/4}}{x^6} \, dx=-\frac {3 b^3 x}{20 a^2 \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{5 x^5}-\frac {b \left (a+b x^2\right )^{3/4}}{10 a x^3}+\frac {3 b^2 \left (a+b x^2\right )^{3/4}}{20 a^2 x}+\frac {3 b^{5/2} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{20 a^{3/2} \sqrt [4]{a+b x^2}} \]
-3/20*b^3*x/a^2/(b*x^2+a)^(1/4)-1/5*(b*x^2+a)^(3/4)/x^5-1/10*b*(b*x^2+a)^( 3/4)/a/x^3+3/20*b^2*(b*x^2+a)^(3/4)/a^2/x+3/20*b^(5/2)*(1+b*x^2/a)^(1/4)*( cos(1/2*arctan(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1/2)/a^(1/ 2)))*EllipticE(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))/a^(3/2)/(b*x^2+ a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.35 \[ \int \frac {\left (a+b x^2\right )^{3/4}}{x^6} \, dx=-\frac {\left (a+b x^2\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {3}{4},-\frac {3}{2},-\frac {b x^2}{a}\right )}{5 x^5 \left (1+\frac {b x^2}{a}\right )^{3/4}} \]
-1/5*((a + b*x^2)^(3/4)*Hypergeometric2F1[-5/2, -3/4, -3/2, -((b*x^2)/a)]) /(x^5*(1 + (b*x^2)/a)^(3/4))
Time = 0.22 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {247, 264, 264, 227, 225, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/4}}{x^6} \, dx\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {3}{10} b \int \frac {1}{x^4 \sqrt [4]{b x^2+a}}dx-\frac {\left (a+b x^2\right )^{3/4}}{5 x^5}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {3}{10} b \left (-\frac {b \int \frac {1}{x^2 \sqrt [4]{b x^2+a}}dx}{2 a}-\frac {\left (a+b x^2\right )^{3/4}}{3 a x^3}\right )-\frac {\left (a+b x^2\right )^{3/4}}{5 x^5}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {3}{10} b \left (-\frac {b \left (\frac {b \int \frac {1}{\sqrt [4]{b x^2+a}}dx}{2 a}-\frac {\left (a+b x^2\right )^{3/4}}{a x}\right )}{2 a}-\frac {\left (a+b x^2\right )^{3/4}}{3 a x^3}\right )-\frac {\left (a+b x^2\right )^{3/4}}{5 x^5}\) |
\(\Big \downarrow \) 227 |
\(\displaystyle \frac {3}{10} b \left (-\frac {b \left (\frac {b \sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1}}dx}{2 a \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{a x}\right )}{2 a}-\frac {\left (a+b x^2\right )^{3/4}}{3 a x^3}\right )-\frac {\left (a+b x^2\right )^{3/4}}{5 x^5}\) |
\(\Big \downarrow \) 225 |
\(\displaystyle \frac {3}{10} b \left (-\frac {b \left (\frac {b \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx\right )}{2 a \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{a x}\right )}{2 a}-\frac {\left (a+b x^2\right )^{3/4}}{3 a x^3}\right )-\frac {\left (a+b x^2\right )^{3/4}}{5 x^5}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {3}{10} b \left (-\frac {b \left (\frac {b \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{2 a \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{a x}\right )}{2 a}-\frac {\left (a+b x^2\right )^{3/4}}{3 a x^3}\right )-\frac {\left (a+b x^2\right )^{3/4}}{5 x^5}\) |
-1/5*(a + b*x^2)^(3/4)/x^5 + (3*b*(-1/3*(a + b*x^2)^(3/4)/(a*x^3) - (b*(-( (a + b*x^2)^(3/4)/(a*x)) + (b*(1 + (b*x^2)/a)^(1/4)*((2*x)/(1 + (b*x^2)/a) ^(1/4) - (2*Sqrt[a]*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/Sqrt[b])) /(2*a*(a + b*x^2)^(1/4))))/(2*a)))/10
3.9.4.3.1 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[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) , x] - Simp[a Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ a, 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( a + b*x^2)^(1/4) Int[1/(1 + b*(x^2/a))^(1/4), 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]
\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {3}{4}}}{x^{6}}d x\]
\[ \int \frac {\left (a+b x^2\right )^{3/4}}{x^6} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}}}{x^{6}} \,d x } \]
Result contains complex when optimal does not.
Time = 0.68 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a+b x^2\right )^{3/4}}{x^6} \, dx=- \frac {a^{\frac {3}{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, - \frac {3}{4} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 x^{5}} \]
\[ \int \frac {\left (a+b x^2\right )^{3/4}}{x^6} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}}}{x^{6}} \,d x } \]
\[ \int \frac {\left (a+b x^2\right )^{3/4}}{x^6} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}}}{x^{6}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/4}}{x^6} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{3/4}}{x^6} \,d x \]