Integrand size = 37, antiderivative size = 116 \[ \int \frac {b+2 a x^3}{\left (-b+x+a x^3\right ) \sqrt [4]{-b x^3+a x^6}} \, dx=-\sqrt {2} \arctan \left (\frac {\sqrt {2} x \sqrt [4]{-b x^3+a x^6}}{-x^2+\sqrt {-b x^3+a x^6}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\frac {x^2}{\sqrt {2}}+\frac {\sqrt {-b x^3+a x^6}}{\sqrt {2}}}{x \sqrt [4]{-b x^3+a x^6}}\right ) \]
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\[ \int \frac {b+2 a x^3}{\left (-b+x+a x^3\right ) \sqrt [4]{-b x^3+a x^6}} \, dx=\int \frac {b+2 a x^3}{\left (-b+x+a x^3\right ) \sqrt [4]{-b x^3+a x^6}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^{3/4} \sqrt [4]{-b+a x^3}\right ) \int \frac {b+2 a x^3}{x^{3/4} \sqrt [4]{-b+a x^3} \left (-b+x+a x^3\right )} \, dx}{\sqrt [4]{-b x^3+a x^6}} \\ & = \frac {\left (4 x^{3/4} \sqrt [4]{-b+a x^3}\right ) \text {Subst}\left (\int \frac {b+2 a x^{12}}{\sqrt [4]{-b+a x^{12}} \left (-b+x^4+a x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^6}} \\ & = \frac {\left (4 x^{3/4} \sqrt [4]{-b+a x^3}\right ) \text {Subst}\left (\int \left (\frac {2}{\sqrt [4]{-b+a x^{12}}}+\frac {3 b-2 x^4}{\sqrt [4]{-b+a x^{12}} \left (-b+x^4+a x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^6}} \\ & = \frac {\left (4 x^{3/4} \sqrt [4]{-b+a x^3}\right ) \text {Subst}\left (\int \frac {3 b-2 x^4}{\sqrt [4]{-b+a x^{12}} \left (-b+x^4+a x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^6}}+\frac {\left (8 x^{3/4} \sqrt [4]{-b+a x^3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{-b+a x^{12}}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^6}} \\ & = \frac {\left (4 x^{3/4} \sqrt [4]{-b+a x^3}\right ) \text {Subst}\left (\int \left (-\frac {3 b}{\left (b-x^4-a x^{12}\right ) \sqrt [4]{-b+a x^{12}}}-\frac {2 x^4}{\sqrt [4]{-b+a x^{12}} \left (-b+x^4+a x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^6}}+\frac {\left (8 x^{3/4} \sqrt [4]{1-\frac {a x^3}{b}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1-\frac {a x^{12}}{b}}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^6}} \\ & = \frac {8 x \sqrt [4]{1-\frac {a x^3}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{12},\frac {1}{4},\frac {13}{12},\frac {a x^3}{b}\right )}{\sqrt [4]{-b x^3+a x^6}}-\frac {\left (8 x^{3/4} \sqrt [4]{-b+a x^3}\right ) \text {Subst}\left (\int \frac {x^4}{\sqrt [4]{-b+a x^{12}} \left (-b+x^4+a x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^6}}-\frac {\left (12 b x^{3/4} \sqrt [4]{-b+a x^3}\right ) \text {Subst}\left (\int \frac {1}{\left (b-x^4-a x^{12}\right ) \sqrt [4]{-b+a x^{12}}} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-b x^3+a x^6}} \\ \end{align*}
\[ \int \frac {b+2 a x^3}{\left (-b+x+a x^3\right ) \sqrt [4]{-b x^3+a x^6}} \, dx=\int \frac {b+2 a x^3}{\left (-b+x+a x^3\right ) \sqrt [4]{-b x^3+a x^6}} \, dx \]
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Time = 1.01 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.28
method | result | size |
pseudoelliptic | \(\frac {\sqrt {2}\, \left (\ln \left (\frac {-\left (x^{3} \left (a \,x^{3}-b \right )\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{3} \left (a \,x^{3}-b \right )}}{\left (x^{3} \left (a \,x^{3}-b \right )\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{3} \left (a \,x^{3}-b \right )}}\right )+2 \arctan \left (\frac {\left (x^{3} \left (a \,x^{3}-b \right )\right )^{\frac {1}{4}} \sqrt {2}+x}{x}\right )+2 \arctan \left (\frac {\left (x^{3} \left (a \,x^{3}-b \right )\right )^{\frac {1}{4}} \sqrt {2}-x}{x}\right )\right )}{2}\) | \(148\) |
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Timed out. \[ \int \frac {b+2 a x^3}{\left (-b+x+a x^3\right ) \sqrt [4]{-b x^3+a x^6}} \, dx=\text {Timed out} \]
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\[ \int \frac {b+2 a x^3}{\left (-b+x+a x^3\right ) \sqrt [4]{-b x^3+a x^6}} \, dx=\int \frac {2 a x^{3} + b}{\sqrt [4]{x^{3} \left (a x^{3} - b\right )} \left (a x^{3} - b + x\right )}\, dx \]
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\[ \int \frac {b+2 a x^3}{\left (-b+x+a x^3\right ) \sqrt [4]{-b x^3+a x^6}} \, dx=\int { \frac {2 \, a x^{3} + b}{{\left (a x^{6} - b x^{3}\right )}^{\frac {1}{4}} {\left (a x^{3} - b + x\right )}} \,d x } \]
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\[ \int \frac {b+2 a x^3}{\left (-b+x+a x^3\right ) \sqrt [4]{-b x^3+a x^6}} \, dx=\int { \frac {2 \, a x^{3} + b}{{\left (a x^{6} - b x^{3}\right )}^{\frac {1}{4}} {\left (a x^{3} - b + x\right )}} \,d x } \]
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Timed out. \[ \int \frac {b+2 a x^3}{\left (-b+x+a x^3\right ) \sqrt [4]{-b x^3+a x^6}} \, dx=\int \frac {2\,a\,x^3+b}{{\left (a\,x^6-b\,x^3\right )}^{1/4}\,\left (a\,x^3+x-b\right )} \,d x \]
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