Integrand size = 15, antiderivative size = 138 \[ \int \frac {1}{x^5 \left (a+b x^2\right )^{2/3}} \, dx=-\frac {\sqrt [3]{a+b x^2}}{4 a x^4}+\frac {5 b \sqrt [3]{a+b x^2}}{12 a^2 x^2}-\frac {5 b^2 \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^2}}{\sqrt {3} \sqrt [3]{a}}\right )}{6 \sqrt {3} a^{8/3}}-\frac {5 b^2 \log (x)}{18 a^{8/3}}+\frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{12 a^{8/3}} \]
-1/4*(b*x^2+a)^(1/3)/a/x^4+5/12*b*(b*x^2+a)^(1/3)/a^2/x^2-5/18*b^2*ln(x)/a ^(8/3)+5/12*b^2*ln(a^(1/3)-(b*x^2+a)^(1/3))/a^(8/3)-5/18*b^2*arctan(1/3*(a ^(1/3)+2*(b*x^2+a)^(1/3))/a^(1/3)*3^(1/2))/a^(8/3)*3^(1/2)
Time = 0.15 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^5 \left (a+b x^2\right )^{2/3}} \, dx=\frac {\frac {3 a^{2/3} \sqrt [3]{a+b x^2} \left (-3 a+5 b x^2\right )}{x^4}-10 \sqrt {3} b^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+10 b^2 \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^2}\right )-5 b^2 \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )}{36 a^{8/3}} \]
((3*a^(2/3)*(a + b*x^2)^(1/3)*(-3*a + 5*b*x^2))/x^4 - 10*Sqrt[3]*b^2*ArcTa n[(1 + (2*(a + b*x^2)^(1/3))/a^(1/3))/Sqrt[3]] + 10*b^2*Log[-a^(1/3) + (a + b*x^2)^(1/3)] - 5*b^2*Log[a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x ^2)^(2/3)])/(36*a^(8/3))
Time = 0.23 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {243, 52, 52, 69, 16, 1082, 217}
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.
| \(\displaystyle \int \frac {1}{x^5 \left (a+b x^2\right )^{2/3}} \, dx\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{x^6 \left (b x^2+a\right )^{2/3}}dx^2\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {1}{2} \left (-\frac {5 b \int \frac {1}{x^4 \left (b x^2+a\right )^{2/3}}dx^2}{6 a}-\frac {\sqrt [3]{a+b x^2}}{2 a x^4}\right )\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {1}{2} \left (-\frac {5 b \left (-\frac {2 b \int \frac {1}{x^2 \left (b x^2+a\right )^{2/3}}dx^2}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x^2}\right )}{6 a}-\frac {\sqrt [3]{a+b x^2}}{2 a x^4}\right )\) |
| \(\Big \downarrow \) 69 |
| \(\displaystyle \frac {1}{2} \left (-\frac {5 b \left (-\frac {2 b \left (-\frac {3 \int \frac {1}{\sqrt [3]{a}-\sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 a^{2/3}}-\frac {3 \int \frac {1}{x^4+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 \sqrt [3]{a}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x^2}\right )}{6 a}-\frac {\sqrt [3]{a+b x^2}}{2 a x^4}\right )\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {1}{2} \left (-\frac {5 b \left (-\frac {2 b \left (-\frac {3 \int \frac {1}{x^4+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 \sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{2/3}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x^2}\right )}{6 a}-\frac {\sqrt [3]{a+b x^2}}{2 a x^4}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{2} \left (-\frac {5 b \left (-\frac {2 b \left (\frac {3 \int \frac {1}{-x^4-3}d\left (\frac {2 \sqrt [3]{b x^2+a}}{\sqrt [3]{a}}+1\right )}{a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{2/3}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x^2}\right )}{6 a}-\frac {\sqrt [3]{a+b x^2}}{2 a x^4}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{2} \left (-\frac {5 b \left (-\frac {2 b \left (-\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )}{a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{2/3}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x^2}\right )}{6 a}-\frac {\sqrt [3]{a+b x^2}}{2 a x^4}\right )\) |
(-1/2*(a + b*x^2)^(1/3)/(a*x^4) - (5*b*(-((a + b*x^2)^(1/3)/(a*x^2)) - (2* b*(-((Sqrt[3]*ArcTan[(1 + (2*(a + b*x^2)^(1/3))/a^(1/3))/Sqrt[3]])/a^(2/3) ) - Log[x^2]/(2*a^(2/3)) + (3*Log[a^(1/3) - (a + b*x^2)^(1/3)])/(2*a^(2/3) )))/(3*a)))/(6*a))/2
3.8.20.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Simp[3/(2*b*q) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 /3)], x] - Simp[3/(2*b*q^2) Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Time = 2.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.99
| method | result | size |
| pseudoelliptic | \(\frac {-10 b^{2} \arctan \left (\frac {\left (a^{\frac {1}{3}}+2 \left (b \,x^{2}+a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) \sqrt {3}\, x^{4}+10 b^{2} \ln \left (\left (b \,x^{2}+a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right ) x^{4}-5 b^{2} \ln \left (a^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b \,x^{2}+a \right )^{\frac {1}{3}}+\left (b \,x^{2}+a \right )^{\frac {2}{3}}\right ) x^{4}-9 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a^{\frac {5}{3}}+15 b \,x^{2} \left (b \,x^{2}+a \right )^{\frac {1}{3}} a^{\frac {2}{3}}}{36 a^{\frac {8}{3}} x^{4}}\) | \(136\) |
1/36*(-10*b^2*arctan(1/3*(a^(1/3)+2*(b*x^2+a)^(1/3))/a^(1/3)*3^(1/2))*3^(1 /2)*x^4+10*b^2*ln((b*x^2+a)^(1/3)-a^(1/3))*x^4-5*b^2*ln(a^(2/3)+a^(1/3)*(b *x^2+a)^(1/3)+(b*x^2+a)^(2/3))*x^4-9*(b*x^2+a)^(1/3)*a^(5/3)+15*b*x^2*(b*x ^2+a)^(1/3)*a^(2/3))/a^(8/3)/x^4
Time = 0.26 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.26 \[ \int \frac {1}{x^5 \left (a+b x^2\right )^{2/3}} \, dx=-\frac {10 \, \sqrt {3} {\left (a^{2}\right )}^{\frac {1}{6}} a b^{2} x^{4} \arctan \left (\frac {{\left (a^{2}\right )}^{\frac {1}{6}} {\left (\sqrt {3} {\left (a^{2}\right )}^{\frac {1}{3}} a + 2 \, \sqrt {3} {\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right )}}{3 \, a^{2}}\right ) + 5 \, {\left (a^{2}\right )}^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} a + {\left (a^{2}\right )}^{\frac {1}{3}} a + {\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right ) - 10 \, {\left (a^{2}\right )}^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} a - {\left (a^{2}\right )}^{\frac {2}{3}}\right ) - 3 \, {\left (5 \, a^{2} b x^{2} - 3 \, a^{3}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}}}{36 \, a^{4} x^{4}} \]
-1/36*(10*sqrt(3)*(a^2)^(1/6)*a*b^2*x^4*arctan(1/3*(a^2)^(1/6)*(sqrt(3)*(a ^2)^(1/3)*a + 2*sqrt(3)*(b*x^2 + a)^(1/3)*(a^2)^(2/3))/a^2) + 5*(a^2)^(2/3 )*b^2*x^4*log((b*x^2 + a)^(2/3)*a + (a^2)^(1/3)*a + (b*x^2 + a)^(1/3)*(a^2 )^(2/3)) - 10*(a^2)^(2/3)*b^2*x^4*log((b*x^2 + a)^(1/3)*a - (a^2)^(2/3)) - 3*(5*a^2*b*x^2 - 3*a^3)*(b*x^2 + a)^(1/3))/(a^4*x^4)
Result contains complex when optimal does not.
Time = 1.63 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.30 \[ \int \frac {1}{x^5 \left (a+b x^2\right )^{2/3}} \, dx=- \frac {\Gamma \left (\frac {8}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {8}{3} \\ \frac {11}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 b^{\frac {2}{3}} x^{\frac {16}{3}} \Gamma \left (\frac {11}{3}\right )} \]
-gamma(8/3)*hyper((2/3, 8/3), (11/3,), a*exp_polar(I*pi)/(b*x**2))/(2*b**( 2/3)*x**(16/3)*gamma(11/3))
Time = 0.28 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.14 \[ \int \frac {1}{x^5 \left (a+b x^2\right )^{2/3}} \, dx=-\frac {5 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{18 \, a^{\frac {8}{3}}} - \frac {5 \, b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{36 \, a^{\frac {8}{3}}} + \frac {5 \, b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{18 \, a^{\frac {8}{3}}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} b^{2} - 8 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a b^{2}}{12 \, {\left ({\left (b x^{2} + a\right )}^{2} a^{2} - 2 \, {\left (b x^{2} + a\right )} a^{3} + a^{4}\right )}} \]
-5/18*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/ 3))/a^(8/3) - 5/36*b^2*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(8/3) + 5/18*b^2*log((b*x^2 + a)^(1/3) - a^(1/3))/a^(8/3) + 1/ 12*(5*(b*x^2 + a)^(4/3)*b^2 - 8*(b*x^2 + a)^(1/3)*a*b^2)/((b*x^2 + a)^2*a^ 2 - 2*(b*x^2 + a)*a^3 + a^4)
Time = 0.51 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^5 \left (a+b x^2\right )^{2/3}} \, dx=-\frac {\frac {10 \, \sqrt {3} b^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {8}{3}}} + \frac {5 \, b^{3} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {8}{3}}} - \frac {10 \, b^{3} \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {8}{3}}} - \frac {3 \, {\left (5 \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} b^{3} - 8 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a b^{3}\right )}}{a^{2} b^{2} x^{4}}}{36 \, b} \]
-1/36*(10*sqrt(3)*b^3*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a ^(1/3))/a^(8/3) + 5*b^3*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(8/3) - 10*b^3*log(abs((b*x^2 + a)^(1/3) - a^(1/3)))/a^(8/3) - 3*(5*(b*x^2 + a)^(4/3)*b^3 - 8*(b*x^2 + a)^(1/3)*a*b^3)/(a^2*b^2*x^4))/b
Time = 4.98 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.40 \[ \int \frac {1}{x^5 \left (a+b x^2\right )^{2/3}} \, dx=\frac {5\,b^2\,\ln \left ({\left (b\,x^2+a\right )}^{1/3}-a^{1/3}\right )}{18\,a^{8/3}}-\frac {\frac {4\,b^2\,{\left (b\,x^2+a\right )}^{1/3}}{3\,a}-\frac {5\,b^2\,{\left (b\,x^2+a\right )}^{4/3}}{6\,a^2}}{2\,{\left (b\,x^2+a\right )}^2-4\,a\,\left (b\,x^2+a\right )+2\,a^2}+\frac {5\,b^2\,\ln \left (\frac {5\,b^2\,{\left (b\,x^2+a\right )}^{1/3}}{2\,a^2}-\frac {5\,b^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,a^{5/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{18\,a^{8/3}}-\frac {5\,b^2\,\ln \left (\frac {5\,b^2\,{\left (b\,x^2+a\right )}^{1/3}}{2\,a^2}+\frac {5\,b^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,a^{5/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{18\,a^{8/3}} \]
(5*b^2*log((a + b*x^2)^(1/3) - a^(1/3)))/(18*a^(8/3)) - ((4*b^2*(a + b*x^2 )^(1/3))/(3*a) - (5*b^2*(a + b*x^2)^(4/3))/(6*a^2))/(2*(a + b*x^2)^2 - 4*a *(a + b*x^2) + 2*a^2) + (5*b^2*log((5*b^2*(a + b*x^2)^(1/3))/(2*a^2) - (5* b^2*((3^(1/2)*1i)/2 - 1/2))/(2*a^(5/3)))*((3^(1/2)*1i)/2 - 1/2))/(18*a^(8/ 3)) - (5*b^2*log((5*b^2*(a + b*x^2)^(1/3))/(2*a^2) + (5*b^2*((3^(1/2)*1i)/ 2 + 1/2))/(2*a^(5/3)))*((3^(1/2)*1i)/2 + 1/2))/(18*a^(8/3))