Optimal. Leaf size=270 \[ \frac {\log \left (x-\sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{2 b^{2/3}}+\frac {\log \left (\sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+x\right )}{2 b^{2/3}}-\frac {\log \left (-\sqrt [6]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+\sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+x^2\right )}{4 b^{2/3}}-\frac {\log \left (\sqrt [6]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+\sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+x^2\right )}{4 b^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x^2}{2 \sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+x^2}\right )}{2 b^{2/3}} \]
________________________________________________________________________________________
Rubi [F] time = 21.23, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {(-1+x) (-1+k x) \left (-2 x+(1+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {align*} \int \frac {(-1+x) (-1+k x) \left (-2 x+(1+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx &=\int \frac {(-1+x) x (-1+k x) (-2+(1+k) x)}{((1-x) x (1-k x))^{2/3} \left (b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx\\ &=\frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {(-1+x) \sqrt [3]{x} (-1+k x) (-2+(1+k) x)}{(1-x)^{2/3} (1-k x)^{2/3} \left (b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx}{((1-x) x (1-k x))^{2/3}}\\ &=-\frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {\sqrt [3]{1-x} \sqrt [3]{x} (-1+k x) (-2+(1+k) x)}{(1-k x)^{2/3} \left (b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} (-2+(1+k) x)}{b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4} \, dx}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^3 \sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-2+(1+k) x^3\right )}{b-2 b (1+k) x^3+\left (b+4 b k+b k^2\right ) x^6-2 b k (1+k) x^9+\left (-1+b k^2\right ) x^{12}} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^3 \sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (2-(1+k) x^3\right )}{x^{12}-b \left (-1+x^3\right )^2 \left (-1+k x^3\right )^2} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \left (\frac {(1+k) x^6 \sqrt [3]{1-x^3} \sqrt [3]{1-k x^3}}{b-2 b (1+k) x^3+b (1+k (4+k)) x^6-2 b k (1+k) x^9-\left (1-b k^2\right ) x^{12}}+\frac {2 x^3 \sqrt [3]{1-x^3} \sqrt [3]{1-k x^3}}{-b+2 b (1+k) x^3-b (1+k (4+k)) x^6+2 b k (1+k) x^9+\left (1-b k^2\right ) x^{12}}\right ) \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (6 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^3 \sqrt [3]{1-x^3} \sqrt [3]{1-k x^3}}{-b+2 b (1+k) x^3-b (1+k (4+k)) x^6+2 b k (1+k) x^9+\left (1-b k^2\right ) x^{12}} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}+\frac {\left (3 (1+k) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^6 \sqrt [3]{1-x^3} \sqrt [3]{1-k x^3}}{b-2 b (1+k) x^3+b (1+k (4+k)) x^6-2 b k (1+k) x^9-\left (1-b k^2\right ) x^{12}} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (6 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^3 \sqrt [3]{1-x^3} \sqrt [3]{1-k x^3}}{x^{12}-b \left (-1+x^3\right )^2 \left (-1+k x^3\right )^2} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}+\frac {\left (3 (1+k) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^6 \sqrt [3]{1-x^3} \sqrt [3]{1-k x^3}}{-x^{12}+b \left (-1+x^3\right )^2 \left (-1+k x^3\right )^2} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [F] time = 3.90, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-1+x) (-1+k x) \left (-2 x+(1+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
________________________________________________________________________________________
IntegrateAlgebraic [A] time = 0.63, size = 270, normalized size = 1.00 \begin {gather*} \frac {\log \left (x-\sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{2 b^{2/3}}+\frac {\log \left (\sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+x\right )}{2 b^{2/3}}-\frac {\log \left (-\sqrt [6]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+\sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+x^2\right )}{4 b^{2/3}}-\frac {\log \left (\sqrt [6]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+\sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+x^2\right )}{4 b^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x^2}{2 \sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+x^2}\right )}{2 b^{2/3}} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 2.63, size = 289, normalized size = 1.07 \begin {gather*} -\frac {{\left | b \right |} \log \left ({\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-\frac {1}{b}\right )^{\frac {1}{3}}\right )}{2 \, \left (-b^{5}\right )^{\frac {1}{3}}} + \frac {\sqrt {3} \left (-b^{5}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} \left (-\frac {1}{b}\right )^{\frac {1}{6}} + 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-\frac {1}{b}\right )^{\frac {1}{6}}}\right )}{2 \, b^{4}} - \frac {\sqrt {3} \left (-b^{5}\right )^{\frac {2}{3}} \arctan \left (-\frac {\sqrt {3} \left (-\frac {1}{b}\right )^{\frac {1}{6}} - 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-\frac {1}{b}\right )^{\frac {1}{6}}}\right )}{2 \, b^{4}} - \frac {\left (-b^{5}\right )^{\frac {2}{3}} \log \left (\sqrt {3} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} \left (-\frac {1}{b}\right )^{\frac {1}{6}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-\frac {1}{b}\right )^{\frac {1}{3}}\right )}{4 \, b^{4}} - \frac {\left (-b^{5}\right )^{\frac {2}{3}} \log \left (-\sqrt {3} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} \left (-\frac {1}{b}\right )^{\frac {1}{6}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-\frac {1}{b}\right )^{\frac {1}{3}}\right )}{4 \, b^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [F] time = 0.09, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-1+x \right ) \left (k x -1\right ) \left (-2 x +\left (1+k \right ) x^{2}\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (b -2 b \left (1+k \right ) x +\left (b \,k^{2}+4 b k +b \right ) x^{2}-2 b k \left (1+k \right ) x^{3}+\left (b \,k^{2}-1\right ) x^{4}\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left ({\left (k + 1\right )} x^{2} - 2 \, x\right )} {\left (k x - 1\right )} {\left (x - 1\right )}}{{\left (2 \, b {\left (k + 1\right )} k x^{3} - {\left (b k^{2} - 1\right )} x^{4} + 2 \, b {\left (k + 1\right )} x - {\left (b k^{2} + 4 \, b k + b\right )} x^{2} - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int -\frac {\left (2\,x-x^2\,\left (k+1\right )\right )\,\left (k\,x-1\right )\,\left (x-1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (\left (b\,k^2-1\right )\,x^4-2\,b\,k\,\left (k+1\right )\,x^3+\left (b\,k^2+4\,b\,k+b\right )\,x^2-2\,b\,\left (k+1\right )\,x+b\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________