Integrand size = 17, antiderivative size = 71 \[ \int \frac {1}{x \sqrt [3]{a^3+b^3 x}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {a+2 \sqrt [3]{a^3+b^3 x}}{\sqrt {3} a}\right )}{a}-\frac {\log (x)}{2 a}+\frac {3 \log \left (a-\sqrt [3]{a^3+b^3 x}\right )}{2 a} \]
-1/2*ln(x)/a+3/2*ln(a-(b^3*x+a^3)^(1/3))/a+arctan(1/3*(a+2*(b^3*x+a^3)^(1/ 3))/a*3^(1/2))*3^(1/2)/a
Time = 0.05 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.37 \[ \int \frac {1}{x \sqrt [3]{a^3+b^3 x}} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {a+2 \sqrt [3]{a^3+b^3 x}}{\sqrt {3} a}\right )+2 \log \left (a-\sqrt [3]{a^3+b^3 x}\right )-\log \left (a^2+a \sqrt [3]{a^3+b^3 x}+\left (a^3+b^3 x\right )^{2/3}\right )}{2 a} \]
(2*Sqrt[3]*ArcTan[(a + 2*(a^3 + b^3*x)^(1/3))/(Sqrt[3]*a)] + 2*Log[a - (a^ 3 + b^3*x)^(1/3)] - Log[a^2 + a*(a^3 + b^3*x)^(1/3) + (a^3 + b^3*x)^(2/3)] )/(2*a)
Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {67, 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.
\(\displaystyle \int \frac {1}{x \sqrt [3]{a^3+b^3 x}} \, dx\) |
\(\Big \downarrow \) 67 |
\(\displaystyle -\frac {3 \int \frac {1}{a-\sqrt [3]{a^3+b^3 x}}d\sqrt [3]{a^3+b^3 x}}{2 a}+\frac {3}{2} \int \frac {1}{a^2+\sqrt [3]{a^3+b^3 x} a+\left (a^3+b^3 x\right )^{2/3}}d\sqrt [3]{a^3+b^3 x}-\frac {\log (x)}{2 a}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {3}{2} \int \frac {1}{a^2+\sqrt [3]{a^3+b^3 x} a+\left (a^3+b^3 x\right )^{2/3}}d\sqrt [3]{a^3+b^3 x}+\frac {3 \log \left (a-\sqrt [3]{a^3+b^3 x}\right )}{2 a}-\frac {\log (x)}{2 a}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle -\frac {3 \int \frac {1}{-\left (a^3+b^3 x\right )^{2/3}-3}d\left (\frac {2 \sqrt [3]{a^3+b^3 x}}{a}+1\right )}{a}+\frac {3 \log \left (a-\sqrt [3]{a^3+b^3 x}\right )}{2 a}-\frac {\log (x)}{2 a}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{a^3+b^3 x}}{a}+1}{\sqrt {3}}\right )}{a}+\frac {3 \log \left (a-\sqrt [3]{a^3+b^3 x}\right )}{2 a}-\frac {\log (x)}{2 a}\) |
(Sqrt[3]*ArcTan[(1 + (2*(a^3 + b^3*x)^(1/3))/a)/Sqrt[3]])/a - Log[x]/(2*a) + (3*Log[a - (a^3 + b^3*x)^(1/3)])/(2*a)
3.5.20.3.1 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[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x ] + (Simp[3/(2*b) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], x] - Simp[3/(2*b*q) 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[((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 = 0.40 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.20
method | result | size |
pseudoelliptic | \(\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (a +2 \left (b^{3} x +a^{3}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a}\right )+2 \ln \left (-a +\left (b^{3} x +a^{3}\right )^{\frac {1}{3}}\right )-\ln \left (a^{2}+a \left (b^{3} x +a^{3}\right )^{\frac {1}{3}}+\left (b^{3} x +a^{3}\right )^{\frac {2}{3}}\right )}{2 a}\) | \(85\) |
derivativedivides | \(\frac {-\frac {\ln \left (a^{2}+a \left (b^{3} x +a^{3}\right )^{\frac {1}{3}}+\left (b^{3} x +a^{3}\right )^{\frac {2}{3}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\left (a +2 \left (b^{3} x +a^{3}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a}\right )}{a}+\frac {\ln \left (a -\left (b^{3} x +a^{3}\right )^{\frac {1}{3}}\right )}{a}\) | \(86\) |
default | \(\frac {-\frac {\ln \left (a^{2}+a \left (b^{3} x +a^{3}\right )^{\frac {1}{3}}+\left (b^{3} x +a^{3}\right )^{\frac {2}{3}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\left (a +2 \left (b^{3} x +a^{3}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a}\right )}{a}+\frac {\ln \left (a -\left (b^{3} x +a^{3}\right )^{\frac {1}{3}}\right )}{a}\) | \(86\) |
1/2*(2*3^(1/2)*arctan(1/3*(a+2*(b^3*x+a^3)^(1/3))/a*3^(1/2))+2*ln(-a+(b^3* x+a^3)^(1/3))-ln(a^2+a*(b^3*x+a^3)^(1/3)+(b^3*x+a^3)^(2/3)))/a
Time = 0.22 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.24 \[ \int \frac {1}{x \sqrt [3]{a^3+b^3 x}} \, dx=\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} a + 2 \, \sqrt {3} {\left (b^{3} x + a^{3}\right )}^{\frac {1}{3}}}{3 \, a}\right ) - \log \left (a^{2} + {\left (b^{3} x + a^{3}\right )}^{\frac {1}{3}} a + {\left (b^{3} x + a^{3}\right )}^{\frac {2}{3}}\right ) + 2 \, \log \left (-a + {\left (b^{3} x + a^{3}\right )}^{\frac {1}{3}}\right )}{2 \, a} \]
1/2*(2*sqrt(3)*arctan(1/3*(sqrt(3)*a + 2*sqrt(3)*(b^3*x + a^3)^(1/3))/a) - log(a^2 + (b^3*x + a^3)^(1/3)*a + (b^3*x + a^3)^(2/3)) + 2*log(-a + (b^3* x + a^3)^(1/3)))/a
Result contains complex when optimal does not.
Time = 1.34 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.94 \[ \int \frac {1}{x \sqrt [3]{a^3+b^3 x}} \, dx=\frac {e^{\frac {i \pi }{3}} \log {\left (- \frac {a e^{\frac {2 i \pi }{3}}}{b \sqrt [3]{\frac {a^{3}}{b^{3}} + x}} + 1 \right )} \Gamma \left (- \frac {1}{3}\right )}{3 a \Gamma \left (\frac {2}{3}\right )} + \frac {e^{- \frac {i \pi }{3}} \log {\left (- \frac {a e^{\frac {4 i \pi }{3}}}{b \sqrt [3]{\frac {a^{3}}{b^{3}} + x}} + 1 \right )} \Gamma \left (- \frac {1}{3}\right )}{3 a \Gamma \left (\frac {2}{3}\right )} - \frac {\log {\left (- \frac {a e^{2 i \pi }}{b \sqrt [3]{\frac {a^{3}}{b^{3}} + x}} + 1 \right )} \Gamma \left (- \frac {1}{3}\right )}{3 a \Gamma \left (\frac {2}{3}\right )} \]
exp(I*pi/3)*log(-a*exp_polar(2*I*pi/3)/(b*(a**3/b**3 + x)**(1/3)) + 1)*gam ma(-1/3)/(3*a*gamma(2/3)) + exp(-I*pi/3)*log(-a*exp_polar(4*I*pi/3)/(b*(a* *3/b**3 + x)**(1/3)) + 1)*gamma(-1/3)/(3*a*gamma(2/3)) - log(-a*exp_polar( 2*I*pi)/(b*(a**3/b**3 + x)**(1/3)) + 1)*gamma(-1/3)/(3*a*gamma(2/3))
Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x \sqrt [3]{a^3+b^3 x}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (a + 2 \, {\left (b^{3} x + a^{3}\right )}^{\frac {1}{3}}\right )}}{3 \, a}\right )}{a} - \frac {\log \left (a^{2} + {\left (b^{3} x + a^{3}\right )}^{\frac {1}{3}} a + {\left (b^{3} x + a^{3}\right )}^{\frac {2}{3}}\right )}{2 \, a} + \frac {\log \left (-a + {\left (b^{3} x + a^{3}\right )}^{\frac {1}{3}}\right )}{a} \]
sqrt(3)*arctan(1/3*sqrt(3)*(a + 2*(b^3*x + a^3)^(1/3))/a)/a - 1/2*log(a^2 + (b^3*x + a^3)^(1/3)*a + (b^3*x + a^3)^(2/3))/a + log(-a + (b^3*x + a^3)^ (1/3))/a
Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x \sqrt [3]{a^3+b^3 x}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (a + 2 \, {\left (b^{3} x + a^{3}\right )}^{\frac {1}{3}}\right )}}{3 \, a}\right )}{a} - \frac {\log \left (a^{2} + {\left (b^{3} x + a^{3}\right )}^{\frac {1}{3}} a + {\left (b^{3} x + a^{3}\right )}^{\frac {2}{3}}\right )}{2 \, a} + \frac {\log \left ({\left | -a + {\left (b^{3} x + a^{3}\right )}^{\frac {1}{3}} \right |}\right )}{a} \]
sqrt(3)*arctan(1/3*sqrt(3)*(a + 2*(b^3*x + a^3)^(1/3))/a)/a - 1/2*log(a^2 + (b^3*x + a^3)^(1/3)*a + (b^3*x + a^3)^(2/3))/a + log(abs(-a + (b^3*x + a ^3)^(1/3)))/a
Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.48 \[ \int \frac {1}{x \sqrt [3]{a^3+b^3 x}} \, dx=\frac {\ln \left (9\,{\left (a^3+x\,b^3\right )}^{1/3}-9\,a\right )}{a}+\frac {\ln \left (9\,{\left (a^3+x\,b^3\right )}^{1/3}-\frac {9\,a\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,a}-\frac {\ln \left (9\,{\left (a^3+x\,b^3\right )}^{1/3}-\frac {9\,a\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,a} \]