Integrand size = 15, antiderivative size = 84 \[ \int \frac {1}{x \left (a+b x^3\right )^{2/3}} \, dx=-\frac {\arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3}}-\frac {\log (x)}{2 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 a^{2/3}} \]
-1/2*ln(x)/a^(2/3)+1/2*ln(a^(1/3)-(b*x^3+a)^(1/3))/a^(2/3)-1/3*arctan(1/3* (a^(1/3)+2*(b*x^3+a)^(1/3))/a^(1/3)*3^(1/2))/a^(2/3)*3^(1/2)
Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.20 \[ \int \frac {1}{x \left (a+b x^3\right )^{2/3}} \, dx=-\frac {2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-2 \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^3}\right )+\log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{6 a^{2/3}} \]
-1/6*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]] - 2*Lo g[-a^(1/3) + (a + b*x^3)^(1/3)] + Log[a^(2/3) + a^(1/3)*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/a^(2/3)
Time = 0.21 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {798, 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.
\(\displaystyle \int \frac {1}{x \left (a+b x^3\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{3} \int \frac {1}{x^3 \left (b x^3+a\right )^{2/3}}dx^3\) |
\(\Big \downarrow \) 69 |
\(\displaystyle \frac {1}{3} \left (-\frac {3 \int \frac {1}{\sqrt [3]{a}-\sqrt [3]{b x^3+a}}d\sqrt [3]{b x^3+a}}{2 a^{2/3}}-\frac {3 \int \frac {1}{x^6+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^3+a}}d\sqrt [3]{b x^3+a}}{2 \sqrt [3]{a}}-\frac {\log \left (x^3\right )}{2 a^{2/3}}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{3} \left (-\frac {3 \int \frac {1}{x^6+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^3+a}}d\sqrt [3]{b x^3+a}}{2 \sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 a^{2/3}}-\frac {\log \left (x^3\right )}{2 a^{2/3}}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{3} \left (\frac {3 \int \frac {1}{-x^6-3}d\left (\frac {2 \sqrt [3]{b x^3+a}}{\sqrt [3]{a}}+1\right )}{a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 a^{2/3}}-\frac {\log \left (x^3\right )}{2 a^{2/3}}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{3} \left (-\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )}{a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 a^{2/3}}-\frac {\log \left (x^3\right )}{2 a^{2/3}}\right )\) |
(-((Sqrt[3]*ArcTan[(1 + (2*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]])/a^(2/3)) - Log[x^3]/(2*a^(2/3)) + (3*Log[a^(1/3) - (a + b*x^3)^(1/3)])/(2*a^(2/3))) /3
3.6.65.3.1 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[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_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
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 = 4.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99
method | result | size |
pseudoelliptic | \(\frac {-2 \arctan \left (\frac {\left (a^{\frac {1}{3}}+2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) \sqrt {3}+2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )-\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{6 a^{\frac {2}{3}}}\) | \(83\) |
1/6*(-2*arctan(1/3*(a^(1/3)+2*(b*x^3+a)^(1/3))/a^(1/3)*3^(1/2))*3^(1/2)+2* ln((b*x^3+a)^(1/3)-a^(1/3))-ln((b*x^3+a)^(2/3)+a^(1/3)*(b*x^3+a)^(1/3)+a^( 2/3)))/a^(2/3)
Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (61) = 122\).
Time = 0.45 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.50 \[ \int \frac {1}{x \left (a+b x^3\right )^{2/3}} \, dx=-\frac {2 \, \sqrt {3} {\left (a^{2}\right )}^{\frac {1}{6}} a \arctan \left (\frac {{\left (a^{2}\right )}^{\frac {1}{6}} {\left (\sqrt {3} {\left (a^{2}\right )}^{\frac {1}{3}} a + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right )}}{3 \, a^{2}}\right ) + {\left (a^{2}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} a + {\left (a^{2}\right )}^{\frac {1}{3}} a + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right ) - 2 \, {\left (a^{2}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} a - {\left (a^{2}\right )}^{\frac {2}{3}}\right )}{6 \, a^{2}} \]
-1/6*(2*sqrt(3)*(a^2)^(1/6)*a*arctan(1/3*(a^2)^(1/6)*(sqrt(3)*(a^2)^(1/3)* a + 2*sqrt(3)*(b*x^3 + a)^(1/3)*(a^2)^(2/3))/a^2) + (a^2)^(2/3)*log((b*x^3 + a)^(2/3)*a + (a^2)^(1/3)*a + (b*x^3 + a)^(1/3)*(a^2)^(2/3)) - 2*(a^2)^( 2/3)*log((b*x^3 + a)^(1/3)*a - (a^2)^(2/3)))/a^2
Result contains complex when optimal does not.
Time = 0.54 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.46 \[ \int \frac {1}{x \left (a+b x^3\right )^{2/3}} \, dx=- \frac {\Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{3}}} \right )}}{3 b^{\frac {2}{3}} x^{2} \Gamma \left (\frac {5}{3}\right )} \]
-gamma(2/3)*hyper((2/3, 2/3), (5/3,), a*exp_polar(I*pi)/(b*x**3))/(3*b**(2 /3)*x**2*gamma(5/3))
Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x \left (a+b x^3\right )^{2/3}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{3 \, a^{\frac {2}{3}}} - \frac {\log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{6 \, a^{\frac {2}{3}}} + \frac {\log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{3 \, a^{\frac {2}{3}}} \]
-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^(1/3))/a ^(2/3) - 1/6*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3))/ a^(2/3) + 1/3*log((b*x^3 + a)^(1/3) - a^(1/3))/a^(2/3)
Time = 0.49 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x \left (a+b x^3\right )^{2/3}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{3 \, a^{\frac {2}{3}}} - \frac {\log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{6 \, a^{\frac {2}{3}}} + \frac {\log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{3 \, a^{\frac {2}{3}}} \]
-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^(1/3))/a ^(2/3) - 1/6*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3))/ a^(2/3) + 1/3*log(abs((b*x^3 + a)^(1/3) - a^(1/3)))/a^(2/3)
Time = 5.72 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x \left (a+b x^3\right )^{2/3}} \, dx=\frac {\ln \left (3\,{\left (b\,x^3+a\right )}^{1/3}-3\,a^{1/3}\right )}{3\,a^{2/3}}+\frac {\ln \left (\frac {3\,a^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}-3\,{\left (b\,x^3+a\right )}^{1/3}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,a^{2/3}}-\frac {\ln \left (\frac {3\,a^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}+3\,{\left (b\,x^3+a\right )}^{1/3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,a^{2/3}} \]