Integrand size = 38, antiderivative size = 65 \[ \int \frac {-4 b+a x^3}{\left (-b+a x^3\right ) \sqrt [4]{b-a x^3+c x^4}} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b-a x^3+c x^4}}\right )}{\sqrt [4]{c}}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b-a x^3+c x^4}}\right )}{\sqrt [4]{c}} \]
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\[ \int \frac {-4 b+a x^3}{\left (-b+a x^3\right ) \sqrt [4]{b-a x^3+c x^4}} \, dx=\int \frac {-4 b+a x^3}{\left (-b+a x^3\right ) \sqrt [4]{b-a x^3+c x^4}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{\sqrt [4]{b-a x^3+c x^4}}-\frac {3 b}{\left (-b+a x^3\right ) \sqrt [4]{b-a x^3+c x^4}}\right ) \, dx \\ & = -\left ((3 b) \int \frac {1}{\left (-b+a x^3\right ) \sqrt [4]{b-a x^3+c x^4}} \, dx\right )+\int \frac {1}{\sqrt [4]{b-a x^3+c x^4}} \, dx \\ & = -\left ((3 b) \int \left (-\frac {1}{3 b^{2/3} \left (\sqrt [3]{b}-\sqrt [3]{a} x\right ) \sqrt [4]{b-a x^3+c x^4}}-\frac {1}{3 b^{2/3} \left (\sqrt [3]{b}+\sqrt [3]{-1} \sqrt [3]{a} x\right ) \sqrt [4]{b-a x^3+c x^4}}-\frac {1}{3 b^{2/3} \left (\sqrt [3]{b}-(-1)^{2/3} \sqrt [3]{a} x\right ) \sqrt [4]{b-a x^3+c x^4}}\right ) \, dx\right )+\int \frac {1}{\sqrt [4]{b-a x^3+c x^4}} \, dx \\ & = \sqrt [3]{b} \int \frac {1}{\left (\sqrt [3]{b}-\sqrt [3]{a} x\right ) \sqrt [4]{b-a x^3+c x^4}} \, dx+\sqrt [3]{b} \int \frac {1}{\left (\sqrt [3]{b}+\sqrt [3]{-1} \sqrt [3]{a} x\right ) \sqrt [4]{b-a x^3+c x^4}} \, dx+\sqrt [3]{b} \int \frac {1}{\left (\sqrt [3]{b}-(-1)^{2/3} \sqrt [3]{a} x\right ) \sqrt [4]{b-a x^3+c x^4}} \, dx+\int \frac {1}{\sqrt [4]{b-a x^3+c x^4}} \, dx \\ \end{align*}
Time = 0.60 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int \frac {-4 b+a x^3}{\left (-b+a x^3\right ) \sqrt [4]{b-a x^3+c x^4}} \, dx=\frac {2 \left (\arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b-a x^3+c x^4}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b-a x^3+c x^4}}\right )\right )}{\sqrt [4]{c}} \]
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Time = 2.71 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.20
method | result | size |
pseudoelliptic | \(\frac {-2 \arctan \left (\frac {\left (c \,x^{4}-a \,x^{3}+b \right )^{\frac {1}{4}}}{c^{\frac {1}{4}} x}\right )+\ln \left (\frac {c^{\frac {1}{4}} x +\left (c \,x^{4}-a \,x^{3}+b \right )^{\frac {1}{4}}}{-c^{\frac {1}{4}} x +\left (c \,x^{4}-a \,x^{3}+b \right )^{\frac {1}{4}}}\right )}{c^{\frac {1}{4}}}\) | \(78\) |
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Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.98 \[ \int \frac {-4 b+a x^3}{\left (-b+a x^3\right ) \sqrt [4]{b-a x^3+c x^4}} \, dx=\frac {\log \left (\frac {c^{\frac {1}{4}} x + {\left (c x^{4} - a x^{3} + b\right )}^{\frac {1}{4}}}{x}\right )}{c^{\frac {1}{4}}} - \frac {\log \left (-\frac {c^{\frac {1}{4}} x - {\left (c x^{4} - a x^{3} + b\right )}^{\frac {1}{4}}}{x}\right )}{c^{\frac {1}{4}}} - \frac {i \, \log \left (\frac {i \, c^{\frac {1}{4}} x + {\left (c x^{4} - a x^{3} + b\right )}^{\frac {1}{4}}}{x}\right )}{c^{\frac {1}{4}}} + \frac {i \, \log \left (\frac {-i \, c^{\frac {1}{4}} x + {\left (c x^{4} - a x^{3} + b\right )}^{\frac {1}{4}}}{x}\right )}{c^{\frac {1}{4}}} \]
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\[ \int \frac {-4 b+a x^3}{\left (-b+a x^3\right ) \sqrt [4]{b-a x^3+c x^4}} \, dx=\int \frac {a x^{3} - 4 b}{\left (a x^{3} - b\right ) \sqrt [4]{- a x^{3} + b + c x^{4}}}\, dx \]
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\[ \int \frac {-4 b+a x^3}{\left (-b+a x^3\right ) \sqrt [4]{b-a x^3+c x^4}} \, dx=\int { \frac {a x^{3} - 4 \, b}{{\left (c x^{4} - a x^{3} + b\right )}^{\frac {1}{4}} {\left (a x^{3} - b\right )}} \,d x } \]
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\[ \int \frac {-4 b+a x^3}{\left (-b+a x^3\right ) \sqrt [4]{b-a x^3+c x^4}} \, dx=\int { \frac {a x^{3} - 4 \, b}{{\left (c x^{4} - a x^{3} + b\right )}^{\frac {1}{4}} {\left (a x^{3} - b\right )}} \,d x } \]
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Timed out. \[ \int \frac {-4 b+a x^3}{\left (-b+a x^3\right ) \sqrt [4]{b-a x^3+c x^4}} \, dx=\int \frac {4\,b-a\,x^3}{\left (b-a\,x^3\right )\,{\left (c\,x^4-a\,x^3+b\right )}^{1/4}} \,d x \]
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