Integrand size = 15, antiderivative size = 300 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx=-\frac {\sqrt [6]{a+b x^2}}{3 a x^3}+\frac {8 b \sqrt [6]{a+b x^2}}{9 a^2 x}+\frac {16 \sqrt {2-\sqrt {3}} b \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{\frac {a}{a+b x^2}}\right ) \sqrt {\frac {1+\sqrt [3]{\frac {a}{a+b x^2}}+\left (\frac {a}{a+b x^2}\right )^{2/3}}{\left (1-\sqrt {3}-\sqrt [3]{\frac {a}{a+b x^2}}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-\sqrt [3]{\frac {a}{a+b x^2}}}{1-\sqrt {3}-\sqrt [3]{\frac {a}{a+b x^2}}}\right ),-7+4 \sqrt {3}\right )}{9 \sqrt [4]{3} a^2 x \sqrt [3]{\frac {a}{a+b x^2}} \sqrt {-\frac {1-\sqrt [3]{\frac {a}{a+b x^2}}}{\left (1-\sqrt {3}-\sqrt [3]{\frac {a}{a+b x^2}}\right )^2}}} \]
-1/3*(b*x^2+a)^(1/6)/a/x^3+8/9*b*(b*x^2+a)^(1/6)/a^2/x+16/27*b*(b*x^2+a)^( 1/6)*(1-(a/(b*x^2+a))^(1/3))*EllipticF((1-(a/(b*x^2+a))^(1/3)+3^(1/2))/(1- (a/(b*x^2+a))^(1/3)-3^(1/2)),2*I-I*3^(1/2))*(1/2*6^(1/2)-1/2*2^(1/2))*((1+ (a/(b*x^2+a))^(1/3)+(a/(b*x^2+a))^(2/3))/(1-(a/(b*x^2+a))^(1/3)-3^(1/2))^2 )^(1/2)*3^(3/4)/a^2/x/(a/(b*x^2+a))^(1/3)/((-1+(a/(b*x^2+a))^(1/3))/(1-(a/ (b*x^2+a))^(1/3)-3^(1/2))^2)^(1/2)
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 = 0.17 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx=-\frac {\left (1+\frac {b x^2}{a}\right )^{5/6} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{6},-\frac {1}{2},-\frac {b x^2}{a}\right )}{3 x^3 \left (a+b x^2\right )^{5/6}} \]
-1/3*((1 + (b*x^2)/a)^(5/6)*Hypergeometric2F1[-3/2, 5/6, -1/2, -((b*x^2)/a )])/(x^3*(a + b*x^2)^(5/6))
Time = 0.29 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.28, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {264, 264, 236, 234, 760}
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.
\(\displaystyle \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {8 b \int \frac {1}{x^2 \left (b x^2+a\right )^{5/6}}dx}{9 a}-\frac {\sqrt [6]{a+b x^2}}{3 a x^3}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {8 b \left (-\frac {2 b \int \frac {1}{\left (b x^2+a\right )^{5/6}}dx}{3 a}-\frac {\sqrt [6]{a+b x^2}}{a x}\right )}{9 a}-\frac {\sqrt [6]{a+b x^2}}{3 a x^3}\) |
\(\Big \downarrow \) 236 |
\(\displaystyle -\frac {8 b \left (-\frac {2 b \int \frac {1}{\left (1-\frac {b x^2}{b x^2+a}\right )^{2/3}}d\frac {x}{\sqrt {b x^2+a}}}{3 a \sqrt [3]{\frac {a}{a+b x^2}} \sqrt [3]{a+b x^2}}-\frac {\sqrt [6]{a+b x^2}}{a x}\right )}{9 a}-\frac {\sqrt [6]{a+b x^2}}{3 a x^3}\) |
\(\Big \downarrow \) 234 |
\(\displaystyle -\frac {8 b \left (\frac {\sqrt {-\frac {b x^2}{a+b x^2}} \sqrt [6]{a+b x^2} \int \frac {1}{\sqrt {\frac {x^3}{\left (b x^2+a\right )^{3/2}}-1}}d\sqrt [3]{1-\frac {b x^2}{b x^2+a}}}{a x \sqrt [3]{\frac {a}{a+b x^2}}}-\frac {\sqrt [6]{a+b x^2}}{a x}\right )}{9 a}-\frac {\sqrt [6]{a+b x^2}}{3 a x^3}\) |
\(\Big \downarrow \) 760 |
\(\displaystyle -\frac {8 b \left (-\frac {2 \sqrt {2-\sqrt {3}} \sqrt {-\frac {b x^2}{a+b x^2}} \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}\right ) \sqrt {\frac {\frac {x^2}{a+b x^2}+\sqrt [3]{1-\frac {b x^2}{a+b x^2}}+1}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}+\sqrt {3}+1}{-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} a x \sqrt [3]{\frac {a}{a+b x^2}} \sqrt {\frac {x^3}{\left (a+b x^2\right )^{3/2}}-1} \sqrt {-\frac {1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}}}-\frac {\sqrt [6]{a+b x^2}}{a x}\right )}{9 a}-\frac {\sqrt [6]{a+b x^2}}{3 a x^3}\) |
-1/3*(a + b*x^2)^(1/6)/(a*x^3) - (8*b*(-((a + b*x^2)^(1/6)/(a*x)) - (2*Sqr t[2 - Sqrt[3]]*Sqrt[-((b*x^2)/(a + b*x^2))]*(a + b*x^2)^(1/6)*(1 - (1 - (b *x^2)/(a + b*x^2))^(1/3))*Sqrt[(1 + x^2/(a + b*x^2) + (1 - (b*x^2)/(a + b* x^2))^(1/3))/(1 - Sqrt[3] - (1 - (b*x^2)/(a + b*x^2))^(1/3))^2]*EllipticF[ ArcSin[(1 + Sqrt[3] - (1 - (b*x^2)/(a + b*x^2))^(1/3))/(1 - Sqrt[3] - (1 - (b*x^2)/(a + b*x^2))^(1/3))], -7 + 4*Sqrt[3]])/(3^(1/4)*a*x*(a/(a + b*x^2 ))^(1/3)*Sqrt[-1 + x^3/(a + b*x^2)^(3/2)]*Sqrt[-((1 - (1 - (b*x^2)/(a + b* x^2))^(1/3))/(1 - Sqrt[3] - (1 - (b*x^2)/(a + b*x^2))^(1/3))^2)])))/(9*a)
3.11.30.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b }, x]
Int[((a_) + (b_.)*(x_)^2)^(-5/6), x_Symbol] :> Simp[1/((a/(a + b*x^2))^(1/3 )*(a + b*x^2)^(1/3)) Subst[Int[1/(1 - b*x^2)^(2/3), x], x, x/Sqrt[a + b*x ^2]], x] /; FreeQ[{a, b}, 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[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) *s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x ] && NegQ[a]
\[\int \frac {1}{x^{4} \left (b \,x^{2}+a \right )^{\frac {5}{6}}}d x\]
\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{6}} x^{4}} \,d x } \]
Time = 0.65 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.11 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx=- \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {5}{6} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {5}{6}} x^{3}} \]
\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{6}} x^{4}} \,d x } \]
\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{6}} x^{4}} \,d x } \]
Timed out. \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx=\int \frac {1}{x^4\,{\left (b\,x^2+a\right )}^{5/6}} \,d x \]