Integrand size = 42, antiderivative size = 120 \[ \int \frac {-b+2 a x^4}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\frac {1}{4} \text {RootSum}\left [2 a^2-2 a b+a c-4 a \text {$\#$1}^4-c \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {-a \log (x)+a \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )-\log (x) \text {$\#$1}^4+\log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{4 a \text {$\#$1}+c \text {$\#$1}-4 \text {$\#$1}^5}\&\right ] \]
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Leaf count is larger than twice the leaf count of optimal. \(529\) vs. \(2(120)=240\).
Time = 1.04 (sec) , antiderivative size = 529, normalized size of antiderivative = 4.41, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {6860, 385, 218, 214, 211} \[ \int \frac {-b+2 a x^4}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\frac {a^{3/4} \left (\frac {2 b-c}{\sqrt {16 a b+c^2}}+1\right ) \arctan \left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}{\sqrt [4]{\sqrt {16 a b+c^2}-c} \sqrt [4]{a x^4-b}}\right )}{\left (\sqrt {16 a b+c^2}-c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}-\frac {a^{3/4} \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right ) \arctan \left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}{\sqrt [4]{\sqrt {16 a b+c^2}+c} \sqrt [4]{a x^4-b}}\right )}{\left (\sqrt {16 a b+c^2}+c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}+\frac {a^{3/4} \left (\frac {2 b-c}{\sqrt {16 a b+c^2}}+1\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}{\sqrt [4]{\sqrt {16 a b+c^2}-c} \sqrt [4]{a x^4-b}}\right )}{\left (\sqrt {16 a b+c^2}-c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}-\frac {a^{3/4} \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}{\sqrt [4]{\sqrt {16 a b+c^2}+c} \sqrt [4]{a x^4-b}}\right )}{\left (\sqrt {16 a b+c^2}+c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}} \]
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Rule 211
Rule 214
Rule 218
Rule 385
Rule 6860
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2 a-\frac {2 a (2 b-c)}{\sqrt {16 a b+c^2}}}{\sqrt [4]{-b+a x^4} \left (-c-\sqrt {16 a b+c^2}+4 a x^4\right )}+\frac {2 a+\frac {2 a (2 b-c)}{\sqrt {16 a b+c^2}}}{\sqrt [4]{-b+a x^4} \left (-c+\sqrt {16 a b+c^2}+4 a x^4\right )}\right ) \, dx \\ & = \left (2 a \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \int \frac {1}{\sqrt [4]{-b+a x^4} \left (-c-\sqrt {16 a b+c^2}+4 a x^4\right )} \, dx+\left (2 a \left (1+\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \int \frac {1}{\sqrt [4]{-b+a x^4} \left (-c+\sqrt {16 a b+c^2}+4 a x^4\right )} \, dx \\ & = \left (2 a \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \text {Subst}\left (\int \frac {1}{-c-\sqrt {16 a b+c^2}-\left (4 a b+a \left (-c-\sqrt {16 a b+c^2}\right )\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )+\left (2 a \left (1+\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \text {Subst}\left (\int \frac {1}{-c+\sqrt {16 a b+c^2}-\left (4 a b+a \left (-c+\sqrt {16 a b+c^2}\right )\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right ) \\ & = \frac {\left (a \left (1+\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-c+\sqrt {16 a b+c^2}}-\sqrt {a} \sqrt {4 b-c+\sqrt {16 a b+c^2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{\sqrt {-c+\sqrt {16 a b+c^2}}}+\frac {\left (a \left (1+\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-c+\sqrt {16 a b+c^2}}+\sqrt {a} \sqrt {4 b-c+\sqrt {16 a b+c^2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{\sqrt {-c+\sqrt {16 a b+c^2}}}-\frac {\left (a \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+\sqrt {16 a b+c^2}}-\sqrt {a} \sqrt {-4 b+c+\sqrt {16 a b+c^2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{\sqrt {c+\sqrt {16 a b+c^2}}}-\frac {\left (a \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+\sqrt {16 a b+c^2}}+\sqrt {a} \sqrt {-4 b+c+\sqrt {16 a b+c^2}} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{\sqrt {c+\sqrt {16 a b+c^2}}} \\ & = \frac {a^{3/4} \left (1+\frac {2 b-c}{\sqrt {16 a b+c^2}}\right ) \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{4 b-c+\sqrt {16 a b+c^2}} x}{\sqrt [4]{-c+\sqrt {16 a b+c^2}} \sqrt [4]{-b+a x^4}}\right )}{\left (-c+\sqrt {16 a b+c^2}\right )^{3/4} \sqrt [4]{4 b-c+\sqrt {16 a b+c^2}}}-\frac {a^{3/4} \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right ) \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{-4 b+c+\sqrt {16 a b+c^2}} x}{\sqrt [4]{c+\sqrt {16 a b+c^2}} \sqrt [4]{-b+a x^4}}\right )}{\left (c+\sqrt {16 a b+c^2}\right )^{3/4} \sqrt [4]{-4 b+c+\sqrt {16 a b+c^2}}}+\frac {a^{3/4} \left (1+\frac {2 b-c}{\sqrt {16 a b+c^2}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{4 b-c+\sqrt {16 a b+c^2}} x}{\sqrt [4]{-c+\sqrt {16 a b+c^2}} \sqrt [4]{-b+a x^4}}\right )}{\left (-c+\sqrt {16 a b+c^2}\right )^{3/4} \sqrt [4]{4 b-c+\sqrt {16 a b+c^2}}}-\frac {a^{3/4} \left (1-\frac {2 b-c}{\sqrt {16 a b+c^2}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{-4 b+c+\sqrt {16 a b+c^2}} x}{\sqrt [4]{c+\sqrt {16 a b+c^2}} \sqrt [4]{-b+a x^4}}\right )}{\left (c+\sqrt {16 a b+c^2}\right )^{3/4} \sqrt [4]{-4 b+c+\sqrt {16 a b+c^2}}} \\ \end{align*}
Time = 1.42 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.01 \[ \int \frac {-b+2 a x^4}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\frac {1}{4} \text {RootSum}\left [2 a^2-2 a b+a c-4 a \text {$\#$1}^4-c \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {a \log (x)-a \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-\log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-4 a \text {$\#$1}-c \text {$\#$1}+4 \text {$\#$1}^5}\&\right ] \]
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Time = 0.32 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{8}+\left (-4 a -c \right ) \textit {\_Z}^{4}+2 a^{2}-2 a b +a c \right )}{\sum }\left (-\frac {\left (\textit {\_R}^{4}+a \right ) \ln \left (\frac {-\textit {\_R} x +\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R} \left (-4 \textit {\_R}^{4}+4 a +c \right )}\right )\right )}{4}\) | \(79\) |
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Timed out. \[ \int \frac {-b+2 a x^4}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {-b+2 a x^4}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\text {Timed out} \]
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Not integrable
Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.35 \[ \int \frac {-b+2 a x^4}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\int { \frac {2 \, a x^{4} - b}{{\left (2 \, a x^{8} - c x^{4} - 2 \, b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}} \,d x } \]
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Not integrable
Time = 1.86 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.35 \[ \int \frac {-b+2 a x^4}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\int { \frac {2 \, a x^{4} - b}{{\left (2 \, a x^{8} - c x^{4} - 2 \, b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}} \,d x } \]
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Not integrable
Time = 6.56 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.32 \[ \int \frac {-b+2 a x^4}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\int \frac {b-2\,a\,x^4}{{\left (a\,x^4-b\right )}^{1/4}\,\left (-2\,a\,x^8+c\,x^4+2\,b\right )} \,d x \]
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