Integrand size = 15, antiderivative size = 91 \[ \int \sqrt [4]{b x^5+a x^8} \, dx=\frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {b \arctan \left (\frac {\sqrt [4]{b x^5+a x^8}}{\sqrt [4]{a} x^2}\right )}{6 a^{3/4}}+\frac {b \text {arctanh}\left (\frac {\sqrt [4]{b x^5+a x^8}}{\sqrt [4]{a} x^2}\right )}{6 a^{3/4}} \]
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Time = 0.08 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.64, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {2029, 2057, 335, 281, 338, 304, 209, 212} \[ \int \sqrt [4]{b x^5+a x^8} \, dx=-\frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{6 a^{3/4} \left (a x^8+b x^5\right )^{3/4}}+\frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{6 a^{3/4} \left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5} \]
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Rule 209
Rule 212
Rule 281
Rule 304
Rule 335
Rule 338
Rule 2029
Rule 2057
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {1}{4} b \int \frac {x^5}{\left (b x^5+a x^8\right )^{3/4}} \, dx \\ & = \frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {\left (b x^{15/4} \left (b+a x^3\right )^{3/4}\right ) \int \frac {x^{5/4}}{\left (b+a x^3\right )^{3/4}} \, dx}{4 \left (b x^5+a x^8\right )^{3/4}} \\ & = \frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {\left (b x^{15/4} \left (b+a x^3\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^8}{\left (b+a x^{12}\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{\left (b x^5+a x^8\right )^{3/4}} \\ & = \frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {\left (b x^{15/4} \left (b+a x^3\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx,x,x^{3/4}\right )}{3 \left (b x^5+a x^8\right )^{3/4}} \\ & = \frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {\left (b x^{15/4} \left (b+a x^3\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{3 \left (b x^5+a x^8\right )^{3/4}} \\ & = \frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {\left (b x^{15/4} \left (b+a x^3\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{6 \sqrt {a} \left (b x^5+a x^8\right )^{3/4}}-\frac {\left (b x^{15/4} \left (b+a x^3\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{6 \sqrt {a} \left (b x^5+a x^8\right )^{3/4}} \\ & = \frac {1}{3} x \sqrt [4]{b x^5+a x^8}-\frac {b x^{15/4} \left (b+a x^3\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{6 a^{3/4} \left (b x^5+a x^8\right )^{3/4}}+\frac {b x^{15/4} \left (b+a x^3\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )}{6 a^{3/4} \left (b x^5+a x^8\right )^{3/4}} \\ \end{align*}
Time = 6.70 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.26 \[ \int \sqrt [4]{b x^5+a x^8} \, dx=\frac {\sqrt [4]{x^5 \left (b+a x^3\right )} \left (2 a^{3/4} x^{9/4} \sqrt [4]{b+a x^3}-b \arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )+b \text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )\right )}{6 a^{3/4} x^{5/4} \sqrt [4]{b+a x^3}} \]
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Time = 3.52 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.13
method | result | size |
pseudoelliptic | \(\frac {4 \left (x^{5} \left (a \,x^{3}+b \right )\right )^{\frac {1}{4}} x \,a^{\frac {3}{4}}+\ln \left (\frac {-a^{\frac {1}{4}} x^{2}-\left (x^{5} \left (a \,x^{3}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x^{2}-\left (x^{5} \left (a \,x^{3}+b \right )\right )^{\frac {1}{4}}}\right ) b +2 \arctan \left (\frac {\left (x^{5} \left (a \,x^{3}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x^{2}}\right ) b}{12 a^{\frac {3}{4}}}\) | \(103\) |
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Timed out. \[ \int \sqrt [4]{b x^5+a x^8} \, dx=\text {Timed out} \]
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\[ \int \sqrt [4]{b x^5+a x^8} \, dx=\int \sqrt [4]{a x^{8} + b x^{5}}\, dx \]
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\[ \int \sqrt [4]{b x^5+a x^8} \, dx=\int { {\left (a x^{8} + b x^{5}\right )}^{\frac {1}{4}} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (71) = 142\).
Time = 0.28 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.34 \[ \int \sqrt [4]{b x^5+a x^8} \, dx=\frac {8 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} b x^{3} + \frac {2 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {2 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {\sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right )}{a} + \frac {\sqrt {2} b^{2} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right )}{\left (-a\right )^{\frac {3}{4}}}}{24 \, b} \]
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Time = 5.94 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.46 \[ \int \sqrt [4]{b x^5+a x^8} \, dx=\frac {4\,x\,{\left (a\,x^8+b\,x^5\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {3}{4};\ \frac {7}{4};\ -\frac {a\,x^3}{b}\right )}{9\,{\left (\frac {a\,x^3}{b}+1\right )}^{1/4}} \]
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