Integrand size = 15, antiderivative size = 135 \[ \int \frac {\sqrt [3]{a+b x^2}}{x^5} \, dx=-\frac {\sqrt [3]{a+b x^2}}{4 x^4}-\frac {b \sqrt [3]{a+b x^2}}{12 a x^2}+\frac {b^2 \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^2}}{\sqrt {3} \sqrt [3]{a}}\right )}{6 \sqrt {3} a^{5/3}}+\frac {b^2 \log (x)}{18 a^{5/3}}-\frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{12 a^{5/3}} \]
-1/4*(b*x^2+a)^(1/3)/x^4-1/12*b*(b*x^2+a)^(1/3)/a/x^2+1/18*b^2*ln(x)/a^(5/ 3)-1/12*b^2*ln(a^(1/3)-(b*x^2+a)^(1/3))/a^(5/3)+1/18*b^2*arctan(1/3*(a^(1/ 3)+2*(b*x^2+a)^(1/3))/a^(1/3)*3^(1/2))/a^(5/3)*3^(1/2)
Time = 0.16 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.17 \[ \int \frac {\sqrt [3]{a+b x^2}}{x^5} \, dx=\frac {\left (-3 a-b x^2\right ) \sqrt [3]{a+b x^2}}{12 a x^4}+\frac {b^2 \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt {3} \sqrt [3]{a}}\right )}{6 \sqrt {3} a^{5/3}}-\frac {b^2 \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^2}\right )}{18 a^{5/3}}+\frac {b^2 \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )}{36 a^{5/3}} \]
((-3*a - b*x^2)*(a + b*x^2)^(1/3))/(12*a*x^4) + (b^2*ArcTan[1/Sqrt[3] + (2 *(a + b*x^2)^(1/3))/(Sqrt[3]*a^(1/3))])/(6*Sqrt[3]*a^(5/3)) - (b^2*Log[-a^ (1/3) + (a + b*x^2)^(1/3)])/(18*a^(5/3)) + (b^2*Log[a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3)])/(36*a^(5/3))
Time = 0.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {243, 51, 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.
| \(\displaystyle \int \frac {\sqrt [3]{a+b x^2}}{x^5} \, dx\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {1}{2} \int \frac {\sqrt [3]{b x^2+a}}{x^6}dx^2\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{6} b \int \frac {1}{x^4 \left (b x^2+a\right )^{2/3}}dx^2-\frac {\sqrt [3]{a+b x^2}}{2 x^4}\right )\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{6} b \left (-\frac {2 b \int \frac {1}{x^2 \left (b x^2+a\right )^{2/3}}dx^2}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x^2}\right )-\frac {\sqrt [3]{a+b x^2}}{2 x^4}\right )\) |
| \(\Big \downarrow \) 69 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{6} b \left (-\frac {2 b \left (-\frac {3 \int \frac {1}{\sqrt [3]{a}-\sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 a^{2/3}}-\frac {3 \int \frac {1}{x^4+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 \sqrt [3]{a}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x^2}\right )-\frac {\sqrt [3]{a+b x^2}}{2 x^4}\right )\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{6} b \left (-\frac {2 b \left (-\frac {3 \int \frac {1}{x^4+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 \sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{2/3}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x^2}\right )-\frac {\sqrt [3]{a+b x^2}}{2 x^4}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{6} b \left (-\frac {2 b \left (\frac {3 \int \frac {1}{-x^4-3}d\left (\frac {2 \sqrt [3]{b x^2+a}}{\sqrt [3]{a}}+1\right )}{a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{2/3}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x^2}\right )-\frac {\sqrt [3]{a+b x^2}}{2 x^4}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{6} b \left (-\frac {2 b \left (-\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )}{a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{2/3}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x^2}\right )-\frac {\sqrt [3]{a+b x^2}}{2 x^4}\right )\) |
(-1/2*(a + b*x^2)^(1/3)/x^4 + (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)) - Lo g[x^2]/(2*a^(2/3)) + (3*Log[a^(1/3) - (a + b*x^2)^(1/3)])/(2*a^(2/3))))/(3 *a)))/6)/2
3.7.71.3.1 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/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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 = 2.07 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00
| method | result | size |
| pseudoelliptic | \(\frac {2 b^{2} \arctan \left (\frac {\left (a^{\frac {1}{3}}+2 \left (b \,x^{2}+a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) \sqrt {3}\, x^{4}-2 b^{2} \ln \left (\left (b \,x^{2}+a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right ) x^{4}+b^{2} \ln \left (a^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b \,x^{2}+a \right )^{\frac {1}{3}}+\left (b \,x^{2}+a \right )^{\frac {2}{3}}\right ) x^{4}-3 b \,x^{2} \left (b \,x^{2}+a \right )^{\frac {1}{3}} a^{\frac {2}{3}}-9 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a^{\frac {5}{3}}}{36 a^{\frac {5}{3}} x^{4}}\) | \(135\) |
1/36/a^(5/3)*(2*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^4-2*b^2*ln((b*x^2+a)^(1/3)-a^(1/3))*x^4+b^2*ln(a^(2/3)+a^(1/3) *(b*x^2+a)^(1/3)+(b*x^2+a)^(2/3))*x^4-3*b*x^2*(b*x^2+a)^(1/3)*a^(2/3)-9*(b *x^2+a)^(1/3)*a^(5/3))/x^4
Time = 0.26 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.47 \[ \int \frac {\sqrt [3]{a+b x^2}}{x^5} \, dx=\frac {2 \, \sqrt {3} a b^{2} x^{4} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {{\left (\sqrt {3} \left (-a^{2}\right )^{\frac {1}{3}} a - 2 \, \sqrt {3} {\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {2}{3}}\right )} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}}}{3 \, a^{2}}\right ) + \left (-a^{2}\right )^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} a - \left (-a^{2}\right )^{\frac {1}{3}} a + {\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {2}{3}}\right ) - 2 \, \left (-a^{2}\right )^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} a - \left (-a^{2}\right )^{\frac {2}{3}}\right ) - 3 \, {\left (a^{2} b x^{2} + 3 \, a^{3}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}}}{36 \, a^{3} x^{4}} \]
1/36*(2*sqrt(3)*a*b^2*x^4*sqrt(-(-a^2)^(1/3))*arctan(-1/3*(sqrt(3)*(-a^2)^ (1/3)*a - 2*sqrt(3)*(b*x^2 + a)^(1/3)*(-a^2)^(2/3))*sqrt(-(-a^2)^(1/3))/a^ 2) + (-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)) - 2*(-a^2)^(2/3)*b^2*x^4*log((b*x^2 + a)^(1/3)* a - (-a^2)^(2/3)) - 3*(a^2*b*x^2 + 3*a^3)*(b*x^2 + a)^(1/3))/(a^3*x^4)
Result contains complex when optimal does not.
Time = 1.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.31 \[ \int \frac {\sqrt [3]{a+b x^2}}{x^5} \, dx=- \frac {\sqrt [3]{b} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 x^{\frac {10}{3}} \Gamma \left (\frac {8}{3}\right )} \]
-b**(1/3)*gamma(5/3)*hyper((-1/3, 5/3), (8/3,), a*exp_polar(I*pi)/(b*x**2) )/(2*x**(10/3)*gamma(8/3))
Time = 0.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt [3]{a+b x^2}}{x^5} \, dx=\frac {\sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{18 \, a^{\frac {5}{3}}} + \frac {b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{36 \, a^{\frac {5}{3}}} - \frac {b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{18 \, a^{\frac {5}{3}}} - \frac {{\left (b x^{2} + a\right )}^{\frac {4}{3}} b^{2} + 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a b^{2}}{12 \, {\left ({\left (b x^{2} + a\right )}^{2} a - 2 \, {\left (b x^{2} + a\right )} a^{2} + a^{3}\right )}} \]
1/18*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3 ))/a^(5/3) + 1/36*b^2*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(5/3) - 1/18*b^2*log((b*x^2 + a)^(1/3) - a^(1/3))/a^(5/3) - 1/1 2*((b*x^2 + a)^(4/3)*b^2 + 2*(b*x^2 + a)^(1/3)*a*b^2)/((b*x^2 + a)^2*a - 2 *(b*x^2 + a)*a^2 + a^3)
Time = 0.54 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.04 \[ \int \frac {\sqrt [3]{a+b x^2}}{x^5} \, dx=\frac {\frac {2 \, \sqrt {3} b^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {5}{3}}} + \frac {b^{3} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {5}{3}}} - \frac {2 \, b^{3} \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {5}{3}}} - \frac {3 \, {\left ({\left (b x^{2} + a\right )}^{\frac {4}{3}} b^{3} + 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a b^{3}\right )}}{a b^{2} x^{4}}}{36 \, b} \]
1/36*(2*sqrt(3)*b^3*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^( 1/3))/a^(5/3) + b^3*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^ (2/3))/a^(5/3) - 2*b^3*log(abs((b*x^2 + a)^(1/3) - a^(1/3)))/a^(5/3) - 3*( (b*x^2 + a)^(4/3)*b^3 + 2*(b*x^2 + a)^(1/3)*a*b^3)/(a*b^2*x^4))/b
Time = 5.37 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.61 \[ \int \frac {\sqrt [3]{a+b x^2}}{x^5} \, dx=\frac {b^2\,\ln \left (\frac {b^2}{2\,{\left (-a\right )}^{2/3}}-\frac {b^2\,{\left (b\,x^2+a\right )}^{1/3}}{2\,a}\right )}{18\,{\left (-a\right )}^{5/3}}-\frac {\ln \left (\frac {b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{2/3}}+\frac {b^2\,{\left (b\,x^2+a\right )}^{1/3}}{2\,a}\right )\,\left (b^2+\sqrt {3}\,b^2\,1{}\mathrm {i}\right )}{36\,{\left (-a\right )}^{5/3}}-\frac {\frac {b^2\,{\left (b\,x^2+a\right )}^{1/3}}{3}+\frac {b^2\,{\left (b\,x^2+a\right )}^{4/3}}{6\,a}}{2\,{\left (b\,x^2+a\right )}^2-4\,a\,\left (b\,x^2+a\right )+2\,a^2}+\frac {b^2\,\ln \left (\frac {b^2\,{\left (b\,x^2+a\right )}^{1/3}}{2\,a}-\frac {b^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,{\left (-a\right )}^{2/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{18\,{\left (-a\right )}^{5/3}} \]
(b^2*log(b^2/(2*(-a)^(2/3)) - (b^2*(a + b*x^2)^(1/3))/(2*a)))/(18*(-a)^(5/ 3)) - (log((3^(1/2)*b^2*1i + b^2)/(4*(-a)^(2/3)) + (b^2*(a + b*x^2)^(1/3)) /(2*a))*(3^(1/2)*b^2*1i + b^2))/(36*(-a)^(5/3)) - ((b^2*(a + b*x^2)^(1/3)) /3 + (b^2*(a + b*x^2)^(4/3))/(6*a))/(2*(a + b*x^2)^2 - 4*a*(a + b*x^2) + 2 *a^2) + (b^2*log((b^2*(a + b*x^2)^(1/3))/(2*a) - (b^2*((3^(1/2)*1i)/2 - 1/ 2))/(2*(-a)^(2/3)))*((3^(1/2)*1i)/2 - 1/2))/(18*(-a)^(5/3))