Integrand size = 24, antiderivative size = 263 \[ \int \frac {-d+c x}{x \sqrt [3]{-b+a x^3}} \, dx=-\frac {c \arctan \left (\frac {\frac {x}{\sqrt {3}}+\frac {2 \sqrt [3]{-b+a x^3}}{\sqrt {3} \sqrt [3]{a}}}{x}\right )}{\sqrt {3} \sqrt [3]{a}}+\frac {d \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-b+a x^3}}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {d \log \left (\sqrt [3]{b}+\sqrt [3]{-b+a x^3}\right )}{3 \sqrt [3]{b}}-\frac {c \log \left (-\sqrt [3]{a} x+\sqrt [3]{-b+a x^3}\right )}{3 \sqrt [3]{a}}-\frac {d \log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{-b+a x^3}+\left (-b+a x^3\right )^{2/3}\right )}{6 \sqrt [3]{b}}+\frac {c \log \left (a^{2/3} x^2+\sqrt [3]{a} x \sqrt [3]{-b+a x^3}+\left (-b+a x^3\right )^{2/3}\right )}{6 \sqrt [3]{a}} \]
-1/3*c*arctan((1/3*x*3^(1/2)+2/3*(a*x^3-b)^(1/3)*3^(1/2)/a^(1/3))/x)*3^(1/ 2)/a^(1/3)-1/3*d*arctan(-1/3*3^(1/2)+2/3*(a*x^3-b)^(1/3)*3^(1/2)/b^(1/3))* 3^(1/2)/b^(1/3)+1/3*d*ln(b^(1/3)+(a*x^3-b)^(1/3))/b^(1/3)-1/3*c*ln(-a^(1/3 )*x+(a*x^3-b)^(1/3))/a^(1/3)-1/6*d*ln(b^(2/3)-b^(1/3)*(a*x^3-b)^(1/3)+(a*x ^3-b)^(2/3))/b^(1/3)+1/6*c*ln(a^(2/3)*x^2+a^(1/3)*x*(a*x^3-b)^(1/3)+(a*x^3 -b)^(2/3))/a^(1/3)
Time = 5.20 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.95 \[ \int \frac {-d+c x}{x \sqrt [3]{-b+a x^3}} \, dx=\frac {1}{6} \left (\frac {2 \sqrt {3} d \arctan \left (\frac {1-\frac {2 \sqrt [3]{-b+a x^3}}{\sqrt [3]{b}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}-\frac {2 \sqrt {3} c \arctan \left (\frac {1+\frac {2 \sqrt [3]{-b+a x^3}}{\sqrt [3]{a} x}}{\sqrt {3}}\right )}{\sqrt [3]{a}}+\frac {2 d \log \left (\sqrt [3]{b}+\sqrt [3]{-b+a x^3}\right )}{\sqrt [3]{b}}-\frac {2 c \log \left (-\sqrt [3]{a} x+\sqrt [3]{-b+a x^3}\right )}{\sqrt [3]{a}}-\frac {d \log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{-b+a x^3}+\left (-b+a x^3\right )^{2/3}\right )}{\sqrt [3]{b}}+\frac {c \log \left (a^{2/3} x^2+\sqrt [3]{a} x \sqrt [3]{-b+a x^3}+\left (-b+a x^3\right )^{2/3}\right )}{\sqrt [3]{a}}\right ) \]
((2*Sqrt[3]*d*ArcTan[(1 - (2*(-b + a*x^3)^(1/3))/b^(1/3))/Sqrt[3]])/b^(1/3 ) - (2*Sqrt[3]*c*ArcTan[(1 + (2*(-b + a*x^3)^(1/3))/(a^(1/3)*x))/Sqrt[3]]) /a^(1/3) + (2*d*Log[b^(1/3) + (-b + a*x^3)^(1/3)])/b^(1/3) - (2*c*Log[-(a^ (1/3)*x) + (-b + a*x^3)^(1/3)])/a^(1/3) - (d*Log[b^(2/3) - b^(1/3)*(-b + a *x^3)^(1/3) + (-b + a*x^3)^(2/3)])/b^(1/3) + (c*Log[a^(2/3)*x^2 + a^(1/3)* x*(-b + a*x^3)^(1/3) + (-b + a*x^3)^(2/3)])/a^(1/3))/6
Time = 0.36 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.62, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2383, 2009}
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.
\(\displaystyle \int \frac {c x-d}{x \sqrt [3]{a x^3-b}} \, dx\) |
\(\Big \downarrow \) 2383 |
\(\displaystyle \int \left (\frac {c}{\sqrt [3]{a x^3-b}}-\frac {d}{x \sqrt [3]{a x^3-b}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c \arctan \left (\frac {\frac {2 \sqrt [3]{a} x}{\sqrt [3]{a x^3-b}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{a}}+\frac {d \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a x^3-b}}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {c \log \left (\sqrt [3]{a x^3-b}-\sqrt [3]{a} x\right )}{2 \sqrt [3]{a}}+\frac {d \log \left (\sqrt [3]{a x^3-b}+\sqrt [3]{b}\right )}{2 \sqrt [3]{b}}-\frac {d \log (x)}{2 \sqrt [3]{b}}\) |
(c*ArcTan[(1 + (2*a^(1/3)*x)/(-b + a*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*a^(1/3 )) + (d*ArcTan[(b^(1/3) - 2*(-b + a*x^3)^(1/3))/(Sqrt[3]*b^(1/3))])/(Sqrt[ 3]*b^(1/3)) - (d*Log[x])/(2*b^(1/3)) + (d*Log[b^(1/3) + (-b + a*x^3)^(1/3) ])/(2*b^(1/3)) - (c*Log[-(a^(1/3)*x) + (-b + a*x^3)^(1/3)])/(2*a^(1/3))
3.28.70.3.1 Defintions of rubi rules used
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> I nt[ExpandIntegrand[(c*x)^m*Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n , p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) && !IGtQ[m, 0]
\[\int \frac {c x -d}{x \left (a \,x^{3}-b \right )^{\frac {1}{3}}}d x\]
Timed out. \[ \int \frac {-d+c x}{x \sqrt [3]{-b+a x^3}} \, dx=\text {Timed out} \]
Result contains complex when optimal does not.
Time = 1.48 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.30 \[ \int \frac {-d+c x}{x \sqrt [3]{-b+a x^3}} \, dx=\frac {c x e^{- \frac {i \pi }{3}} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {a x^{3}}{b}} \right )}}{3 \sqrt [3]{b} \Gamma \left (\frac {4}{3}\right )} + \frac {d \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{3}}} \right )}}{3 \sqrt [3]{a} x \Gamma \left (\frac {4}{3}\right )} \]
c*x*exp(-I*pi/3)*gamma(1/3)*hyper((1/3, 1/3), (4/3,), a*x**3/b)/(3*b**(1/3 )*gamma(4/3)) + d*gamma(1/3)*hyper((1/3, 1/3), (4/3,), b*exp_polar(2*I*pi) /(a*x**3))/(3*a**(1/3)*x*gamma(4/3))
Time = 0.27 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.79 \[ \int \frac {-d+c x}{x \sqrt [3]{-b+a x^3}} \, dx=-\frac {1}{6} \, {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (a^{\frac {1}{3}} + \frac {2 \, {\left (a x^{3} - b\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {1}{3}}} - \frac {\log \left (a^{\frac {2}{3}} + \frac {{\left (a x^{3} - b\right )}^{\frac {1}{3}} a^{\frac {1}{3}}}{x} + \frac {{\left (a x^{3} - b\right )}^{\frac {2}{3}}}{x^{2}}\right )}{a^{\frac {1}{3}}} + \frac {2 \, \log \left (-a^{\frac {1}{3}} + \frac {{\left (a x^{3} - b\right )}^{\frac {1}{3}}}{x}\right )}{a^{\frac {1}{3}}}\right )} c - \frac {1}{6} \, {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (a x^{3} - b\right )}^{\frac {1}{3}} - b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} + \frac {\log \left ({\left (a x^{3} - b\right )}^{\frac {2}{3}} - {\left (a x^{3} - b\right )}^{\frac {1}{3}} b^{\frac {1}{3}} + b^{\frac {2}{3}}\right )}{b^{\frac {1}{3}}} - \frac {2 \, \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{3}} + b^{\frac {1}{3}}\right )}{b^{\frac {1}{3}}}\right )} d \]
-1/6*(2*sqrt(3)*arctan(1/3*sqrt(3)*(a^(1/3) + 2*(a*x^3 - b)^(1/3)/x)/a^(1/ 3))/a^(1/3) - log(a^(2/3) + (a*x^3 - b)^(1/3)*a^(1/3)/x + (a*x^3 - b)^(2/3 )/x^2)/a^(1/3) + 2*log(-a^(1/3) + (a*x^3 - b)^(1/3)/x)/a^(1/3))*c - 1/6*(2 *sqrt(3)*arctan(1/3*sqrt(3)*(2*(a*x^3 - b)^(1/3) - b^(1/3))/b^(1/3))/b^(1/ 3) + log((a*x^3 - b)^(2/3) - (a*x^3 - b)^(1/3)*b^(1/3) + b^(2/3))/b^(1/3) - 2*log((a*x^3 - b)^(1/3) + b^(1/3))/b^(1/3))*d
\[ \int \frac {-d+c x}{x \sqrt [3]{-b+a x^3}} \, dx=\int { \frac {c x - d}{{\left (a x^{3} - b\right )}^{\frac {1}{3}} x} \,d x } \]
Time = 7.71 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.63 \[ \int \frac {-d+c x}{x \sqrt [3]{-b+a x^3}} \, dx=\frac {d\,\ln \left (d^2\,{\left (a\,x^3-b\right )}^{1/3}+b^{1/3}\,d^2\right )}{3\,b^{1/3}}-\frac {\ln \left (d^2\,{\left (a\,x^3-b\right )}^{1/3}+\frac {b^{1/3}\,{\left (d-\sqrt {3}\,d\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (d-\sqrt {3}\,d\,1{}\mathrm {i}\right )}{6\,b^{1/3}}-\frac {\ln \left (d^2\,{\left (a\,x^3-b\right )}^{1/3}+\frac {b^{1/3}\,{\left (d+\sqrt {3}\,d\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (d+\sqrt {3}\,d\,1{}\mathrm {i}\right )}{6\,b^{1/3}}+\frac {c\,x\,{\left (1-\frac {a\,x^3}{b}\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ \frac {a\,x^3}{b}\right )}{{\left (a\,x^3-b\right )}^{1/3}} \]
(d*log(d^2*(a*x^3 - b)^(1/3) + b^(1/3)*d^2))/(3*b^(1/3)) - (log(d^2*(a*x^3 - b)^(1/3) + (b^(1/3)*(d - 3^(1/2)*d*1i)^2)/4)*(d - 3^(1/2)*d*1i))/(6*b^( 1/3)) - (log(d^2*(a*x^3 - b)^(1/3) + (b^(1/3)*(d + 3^(1/2)*d*1i)^2)/4)*(d + 3^(1/2)*d*1i))/(6*b^(1/3)) + (c*x*(1 - (a*x^3)/b)^(1/3)*hypergeom([1/3, 1/3], 4/3, (a*x^3)/b))/(a*x^3 - b)^(1/3)