Integrand size = 11, antiderivative size = 304 \[ \int \frac {1}{\left (a+b x^6\right )^{5/3}} \, dx=\frac {x}{4 a \left (a+b x^6\right )^{2/3}}+\frac {3^{3/4} x \sqrt [3]{a+b x^6} \left (1-\frac {\sqrt [3]{b} x^2}{\sqrt [3]{a+b x^6}}\right ) \sqrt {\frac {1+\frac {b^{2/3} x^4}{\left (a+b x^6\right )^{2/3}}+\frac {\sqrt [3]{b} x^2}{\sqrt [3]{a+b x^6}}}{\left (1-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} x^2}{\sqrt [3]{a+b x^6}}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x^2}{\sqrt [3]{a+b x^6}}}{1-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} x^2}{\sqrt [3]{a+b x^6}}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{8 a^2 \sqrt {-\frac {\sqrt [3]{b} x^2 \left (1-\frac {\sqrt [3]{b} x^2}{\sqrt [3]{a+b x^6}}\right )}{\sqrt [3]{a+b x^6} \left (1-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} x^2}{\sqrt [3]{a+b x^6}}\right )^2}}} \] Output:
1/4*x/a/(b*x^6+a)^(2/3)+1/8*3^(3/4)*x*(b*x^6+a)^(1/3)*(1-b^(1/3)*x^2/(b*x^ 6+a)^(1/3))*((1+b^(2/3)*x^4/(b*x^6+a)^(2/3)+b^(1/3)*x^2/(b*x^6+a)^(1/3))/( 1-(1+3^(1/2))*b^(1/3)*x^2/(b*x^6+a)^(1/3))^2)^(1/2)*InverseJacobiAM(arccos ((1-(1-3^(1/2))*b^(1/3)*x^2/(b*x^6+a)^(1/3))/(1-(1+3^(1/2))*b^(1/3)*x^2/(b *x^6+a)^(1/3))),1/4*6^(1/2)+1/4*2^(1/2))/a^2/(-b^(1/3)*x^2*(1-b^(1/3)*x^2/ (b*x^6+a)^(1/3))/(b*x^6+a)^(1/3)/(1-(1+3^(1/2))*b^(1/3)*x^2/(b*x^6+a)^(1/3 ))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.18 \[ \int \frac {1}{\left (a+b x^6\right )^{5/3}} \, dx=\frac {x+3 x \left (1+\frac {b x^6}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {2}{3},\frac {7}{6},-\frac {b x^6}{a}\right )}{4 a \left (a+b x^6\right )^{2/3}} \] Input:
Integrate[(a + b*x^6)^(-5/3),x]
Output:
(x + 3*x*(1 + (b*x^6)/a)^(2/3)*Hypergeometric2F1[1/6, 2/3, 7/6, -((b*x^6)/ a)])/(4*a*(a + b*x^6)^(2/3))
Time = 0.42 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {749, 771, 766}
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.
\(\displaystyle \int \frac {1}{\left (a+b x^6\right )^{5/3}} \, dx\) |
\(\Big \downarrow \) 749 |
\(\displaystyle \frac {3 \int \frac {1}{\left (b x^6+a\right )^{2/3}}dx}{4 a}+\frac {x}{4 a \left (a+b x^6\right )^{2/3}}\) |
\(\Big \downarrow \) 771 |
\(\displaystyle \frac {3 \int \frac {1}{\sqrt {1-\frac {b x^6}{b x^6+a}}}d\frac {x}{\sqrt [6]{b x^6+a}}}{4 a \sqrt {\frac {a}{a+b x^6}} \sqrt {a+b x^6}}+\frac {x}{4 a \left (a+b x^6\right )^{2/3}}\) |
\(\Big \downarrow \) 766 |
\(\displaystyle \frac {3^{3/4} x \left (1-\frac {\sqrt [3]{b} x^2}{\sqrt [3]{a+b x^6}}\right ) \sqrt {\frac {\frac {b^{2/3} x^4}{\left (a+b x^6\right )^{2/3}}+\frac {\sqrt [3]{b} x^2}{\sqrt [3]{a+b x^6}}+1}{\left (1-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} x^2}{\sqrt [3]{a+b x^6}}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x^2}{\sqrt [3]{b x^6+a}}}{1-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} x^2}{\sqrt [3]{b x^6+a}}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{8 a \sqrt {\frac {a}{a+b x^6}} \left (a+b x^6\right )^{2/3} \sqrt {1-\frac {b x^6}{a+b x^6}} \sqrt {-\frac {\sqrt [3]{b} x^2 \left (1-\frac {\sqrt [3]{b} x^2}{\sqrt [3]{a+b x^6}}\right )}{\sqrt [3]{a+b x^6} \left (1-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} x^2}{\sqrt [3]{a+b x^6}}\right )^2}}}+\frac {x}{4 a \left (a+b x^6\right )^{2/3}}\) |
Input:
Int[(a + b*x^6)^(-5/3),x]
Output:
x/(4*a*(a + b*x^6)^(2/3)) + (3^(3/4)*x*(1 - (b^(1/3)*x^2)/(a + b*x^6)^(1/3 ))*Sqrt[(1 + (b^(2/3)*x^4)/(a + b*x^6)^(2/3) + (b^(1/3)*x^2)/(a + b*x^6)^( 1/3))/(1 - ((1 + Sqrt[3])*b^(1/3)*x^2)/(a + b*x^6)^(1/3))^2]*EllipticF[Arc Cos[(1 - ((1 - Sqrt[3])*b^(1/3)*x^2)/(a + b*x^6)^(1/3))/(1 - ((1 + Sqrt[3] )*b^(1/3)*x^2)/(a + b*x^6)^(1/3))], (2 + Sqrt[3])/4])/(8*a*Sqrt[a/(a + b*x ^6)]*(a + b*x^6)^(2/3)*Sqrt[1 - (b*x^6)/(a + b*x^6)]*Sqrt[-((b^(1/3)*x^2*( 1 - (b^(1/3)*x^2)/(a + b*x^6)^(1/3)))/((a + b*x^6)^(1/3)*(1 - ((1 + Sqrt[3 ])*b^(1/3)*x^2)/(a + b*x^6)^(1/3))^2))])
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1)) Int[(a + b*x^ n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/ (s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*((s + r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])* r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x ]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a/(a + b*x^n))^(p + 1 /n)*(a + b*x^n)^(p + 1/n) Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x /(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && LtQ[Denominator[p + 1/n], Denominator[p]]
\[\int \frac {1}{\left (b \,x^{6}+a \right )^{\frac {5}{3}}}d x\]
Input:
int(1/(b*x^6+a)^(5/3),x)
Output:
int(1/(b*x^6+a)^(5/3),x)
\[ \int \frac {1}{\left (a+b x^6\right )^{5/3}} \, dx=\int { \frac {1}{{\left (b x^{6} + a\right )}^{\frac {5}{3}}} \,d x } \] Input:
integrate(1/(b*x^6+a)^(5/3),x, algorithm="fricas")
Output:
integral((b*x^6 + a)^(1/3)/(b^2*x^12 + 2*a*b*x^6 + a^2), x)
Result contains complex when optimal does not.
Time = 0.48 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.12 \[ \int \frac {1}{\left (a+b x^6\right )^{5/3}} \, dx=\frac {x \Gamma \left (\frac {1}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{6}, \frac {5}{3} \\ \frac {7}{6} \end {matrix}\middle | {\frac {b x^{6} e^{i \pi }}{a}} \right )}}{6 a^{\frac {5}{3}} \Gamma \left (\frac {7}{6}\right )} \] Input:
integrate(1/(b*x**6+a)**(5/3),x)
Output:
x*gamma(1/6)*hyper((1/6, 5/3), (7/6,), b*x**6*exp_polar(I*pi)/a)/(6*a**(5/ 3)*gamma(7/6))
\[ \int \frac {1}{\left (a+b x^6\right )^{5/3}} \, dx=\int { \frac {1}{{\left (b x^{6} + a\right )}^{\frac {5}{3}}} \,d x } \] Input:
integrate(1/(b*x^6+a)^(5/3),x, algorithm="maxima")
Output:
integrate((b*x^6 + a)^(-5/3), x)
\[ \int \frac {1}{\left (a+b x^6\right )^{5/3}} \, dx=\int { \frac {1}{{\left (b x^{6} + a\right )}^{\frac {5}{3}}} \,d x } \] Input:
integrate(1/(b*x^6+a)^(5/3),x, algorithm="giac")
Output:
integrate((b*x^6 + a)^(-5/3), x)
Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.12 \[ \int \frac {1}{\left (a+b x^6\right )^{5/3}} \, dx=\frac {x\,{\left (\frac {b\,x^6}{a}+1\right )}^{5/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{6},\frac {5}{3};\ \frac {7}{6};\ -\frac {b\,x^6}{a}\right )}{{\left (b\,x^6+a\right )}^{5/3}} \] Input:
int(1/(a + b*x^6)^(5/3),x)
Output:
(x*((b*x^6)/a + 1)^(5/3)*hypergeom([1/6, 5/3], 7/6, -(b*x^6)/a))/(a + b*x^ 6)^(5/3)
\[ \int \frac {1}{\left (a+b x^6\right )^{5/3}} \, dx=\int \frac {1}{\left (b \,x^{6}+a \right )^{\frac {2}{3}} a +\left (b \,x^{6}+a \right )^{\frac {2}{3}} b \,x^{6}}d x \] Input:
int(1/(b*x^6+a)^(5/3),x)
Output:
int(1/((a + b*x**6)**(2/3)*a + (a + b*x**6)**(2/3)*b*x**6),x)