Integrand size = 46, antiderivative size = 210 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}+\frac {1}{4} \text {RootSum}\left [2 a^2-2 a b+a c-4 a \text {$\#$1}^4-c \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {-3 a \log (x)+c \log (x)+3 a \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )-c \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )+3 \log (x) \text {$\#$1}^4-3 \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{4 a \text {$\#$1}+c \text {$\#$1}-4 \text {$\#$1}^5}\&\right ] \]
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Leaf count is larger than twice the leaf count of optimal. \(611\) vs. \(2(210)=420\).
Time = 1.68 (sec) , antiderivative size = 611, normalized size of antiderivative = 2.91, number of steps used = 16, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6860, 246, 218, 212, 209, 385, 214, 211} \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=-\frac {\left (\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}+c\right ) \arctan \left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}{\sqrt [4]{\sqrt {16 a b+c^2}-c} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a} \left (\sqrt {16 a b+c^2}-c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}+\frac {\left (c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \arctan \left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}{\sqrt [4]{\sqrt {16 a b+c^2}+c} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a} \left (\sqrt {16 a b+c^2}+c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}+\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}-\frac {\left (\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}+c\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}{\sqrt [4]{\sqrt {16 a b+c^2}-c} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a} \left (\sqrt {16 a b+c^2}-c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}+\frac {\left (c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}{\sqrt [4]{\sqrt {16 a b+c^2}+c} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a} \left (\sqrt {16 a b+c^2}+c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}} \]
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Rule 209
Rule 211
Rule 212
Rule 214
Rule 218
Rule 246
Rule 385
Rule 6860
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{\sqrt [4]{-b+a x^4}}+\frac {3 b-c x^4}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )}\right ) \, dx \\ & = \int \frac {1}{\sqrt [4]{-b+a x^4}} \, dx+\int \frac {3 b-c x^4}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx \\ & = \int \left (\frac {-c+\frac {12 a b-c^2}{\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 {-c-\frac {12 a b-c^2}{\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+\text {Subst}\left (\int \frac {1}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )+\left (-c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\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 (-c+\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \int \frac {1}{\sqrt [4]{-b+a x^4} \left (-c-\sqrt {16 a b+c^2}+4 a x^4\right )} \, dx \\ & = \frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}+\left (-c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\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 (-c+\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\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 {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}-\frac {\left (c+\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\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 )}{2 \sqrt {-c+\sqrt {16 a b+c^2}}}-\frac {\left (c+\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\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 )}{2 \sqrt {-c+\sqrt {16 a b+c^2}}}+\frac {\left (c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\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 )}{2 \sqrt {c+\sqrt {16 a b+c^2}}}+\frac {\left (c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\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 )}{2 \sqrt {c+\sqrt {16 a b+c^2}}} \\ & = \frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}-\frac {\left (c+\frac {12 a b-c^2}{\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 )}{2 \sqrt [4]{a} \left (-c+\sqrt {16 a b+c^2}\right )^{3/4} \sqrt [4]{4 b-c+\sqrt {16 a b+c^2}}}+\frac {\left (c-\frac {12 a b-c^2}{\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 )}{2 \sqrt [4]{a} \left (c+\sqrt {16 a b+c^2}\right )^{3/4} \sqrt [4]{-4 b+c+\sqrt {16 a b+c^2}}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}-\frac {\left (c+\frac {12 a b-c^2}{\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 )}{2 \sqrt [4]{a} \left (-c+\sqrt {16 a b+c^2}\right )^{3/4} \sqrt [4]{4 b-c+\sqrt {16 a b+c^2}}}+\frac {\left (c-\frac {12 a b-c^2}{\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 )}{2 \sqrt [4]{a} \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 = 5.85 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.96 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\frac {1}{4} \left (\frac {2 \left (\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )\right )}{\sqrt [4]{a}}+\text {RootSum}\left [2 a^2-2 a b+a c-4 a \text {$\#$1}^4-c \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {3 a \log (x)-c \log (x)-3 a \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )+c \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )-3 \log (x) \text {$\#$1}^4+3 \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-4 a \text {$\#$1}-c \text {$\#$1}+4 \text {$\#$1}^5}\&\right ]\right ) \]
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Time = 0.38 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.75
method | result | size |
pseudoelliptic | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{8}+\left (-4 a -c \right ) \textit {\_Z}^{4}+2 a^{2}-2 a b +a c \right )}{\sum }\frac {\left (3 \textit {\_R}^{4}-3 a +c \right ) \ln \left (\frac {-\textit {\_R} x +\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R} \left (4 \textit {\_R}^{4}-4 a -c \right )}\right ) a^{\frac {1}{4}}-2 \arctan \left (\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+\ln \left (\frac {-x \,a^{\frac {1}{4}}-\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{x \,a^{\frac {1}{4}}-\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right )}{4 a^{\frac {1}{4}}}\) | \(157\) |
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Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 6.47 (sec) , antiderivative size = 16424, normalized size of antiderivative = 78.21 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\text {Timed out} \]
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Not integrable
Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.22 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\int { \frac {2 \, a x^{8} - 2 \, c x^{4} + b}{{\left (2 \, a x^{8} - c x^{4} - 2 \, b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}} \,d x } \]
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Not integrable
Time = 1.77 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.22 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\int { \frac {2 \, a x^{8} - 2 \, c x^{4} + b}{{\left (2 \, a x^{8} - c x^{4} - 2 \, b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}} \,d x } \]
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Not integrable
Time = 7.40 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.22 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\int -\frac {2\,a\,x^8-2\,c\,x^4+b}{{\left (a\,x^4-b\right )}^{1/4}\,\left (-2\,a\,x^8+c\,x^4+2\,b\right )} \,d x \]
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