Integrand size = 15, antiderivative size = 70 \[ \int \frac {\left (b+a x^3\right )^{3/4}}{x} \, dx=\frac {4}{9} \left (b+a x^3\right )^{3/4}+\frac {2}{3} b^{3/4} \arctan \left (\frac {\sqrt [4]{b+a x^3}}{\sqrt [4]{b}}\right )-\frac {2}{3} b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{b+a x^3}}{\sqrt [4]{b}}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {272, 52, 65, 304, 209, 212} \[ \int \frac {\left (b+a x^3\right )^{3/4}}{x} \, dx=\frac {2}{3} b^{3/4} \arctan \left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )-\frac {2}{3} b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )+\frac {4}{9} \left (a x^3+b\right )^{3/4} \]
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Rule 52
Rule 65
Rule 209
Rule 212
Rule 272
Rule 304
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {(b+a x)^{3/4}}{x} \, dx,x,x^3\right ) \\ & = \frac {4}{9} \left (b+a x^3\right )^{3/4}+\frac {1}{3} b \text {Subst}\left (\int \frac {1}{x \sqrt [4]{b+a x}} \, dx,x,x^3\right ) \\ & = \frac {4}{9} \left (b+a x^3\right )^{3/4}+\frac {(4 b) \text {Subst}\left (\int \frac {x^2}{-\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{b+a x^3}\right )}{3 a} \\ & = \frac {4}{9} \left (b+a x^3\right )^{3/4}-\frac {1}{3} (2 b) \text {Subst}\left (\int \frac {1}{\sqrt {b}-x^2} \, dx,x,\sqrt [4]{b+a x^3}\right )+\frac {1}{3} (2 b) \text {Subst}\left (\int \frac {1}{\sqrt {b}+x^2} \, dx,x,\sqrt [4]{b+a x^3}\right ) \\ & = \frac {4}{9} \left (b+a x^3\right )^{3/4}+\frac {2}{3} b^{3/4} \arctan \left (\frac {\sqrt [4]{b+a x^3}}{\sqrt [4]{b}}\right )-\frac {2}{3} b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{b+a x^3}}{\sqrt [4]{b}}\right ) \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b+a x^3\right )^{3/4}}{x} \, dx=\frac {4}{9} \left (b+a x^3\right )^{3/4}+\frac {2}{3} b^{3/4} \arctan \left (\frac {\sqrt [4]{b+a x^3}}{\sqrt [4]{b}}\right )-\frac {2}{3} b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{b+a x^3}}{\sqrt [4]{b}}\right ) \]
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Time = 0.95 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.99
method | result | size |
pseudoelliptic | \(\frac {4 \left (a \,x^{3}+b \right )^{\frac {3}{4}}}{9}+\frac {2 b^{\frac {3}{4}} \arctan \left (\frac {\left (a \,x^{3}+b \right )^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{3}-\frac {\ln \left (\frac {\left (a \,x^{3}+b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{\left (a \,x^{3}+b \right )^{\frac {1}{4}}-b^{\frac {1}{4}}}\right ) b^{\frac {3}{4}}}{3}\) | \(69\) |
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Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.80 \[ \int \frac {\left (b+a x^3\right )^{3/4}}{x} \, dx=-\frac {1}{3} \, {\left (b^{3}\right )}^{\frac {1}{4}} \log \left ({\left (a x^{3} + b\right )}^{\frac {1}{4}} b^{2} + {\left (b^{3}\right )}^{\frac {3}{4}}\right ) + \frac {1}{3} i \, {\left (b^{3}\right )}^{\frac {1}{4}} \log \left ({\left (a x^{3} + b\right )}^{\frac {1}{4}} b^{2} + i \, {\left (b^{3}\right )}^{\frac {3}{4}}\right ) - \frac {1}{3} i \, {\left (b^{3}\right )}^{\frac {1}{4}} \log \left ({\left (a x^{3} + b\right )}^{\frac {1}{4}} b^{2} - i \, {\left (b^{3}\right )}^{\frac {3}{4}}\right ) + \frac {1}{3} \, {\left (b^{3}\right )}^{\frac {1}{4}} \log \left ({\left (a x^{3} + b\right )}^{\frac {1}{4}} b^{2} - {\left (b^{3}\right )}^{\frac {3}{4}}\right ) + \frac {4}{9} \, {\left (a x^{3} + b\right )}^{\frac {3}{4}} \]
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Result contains complex when optimal does not.
Time = 0.73 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.66 \[ \int \frac {\left (b+a x^3\right )^{3/4}}{x} \, dx=- \frac {a^{\frac {3}{4}} x^{\frac {9}{4}} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 \Gamma \left (\frac {1}{4}\right )} \]
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none
Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.01 \[ \int \frac {\left (b+a x^3\right )^{3/4}}{x} \, dx=\frac {1}{3} \, b {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{3} + b\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (a x^{3} + b\right )}^{\frac {1}{4}} - b^{\frac {1}{4}}}{{\left (a x^{3} + b\right )}^{\frac {1}{4}} + b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}}\right )} + \frac {4}{9} \, {\left (a x^{3} + b\right )}^{\frac {3}{4}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (50) = 100\).
Time = 0.26 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.64 \[ \int \frac {\left (b+a x^3\right )^{3/4}}{x} \, dx=-\frac {1}{3} \, \sqrt {2} \left (-b\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} + 2 \, {\left (a x^{3} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right ) - \frac {1}{3} \, \sqrt {2} \left (-b\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} - 2 \, {\left (a x^{3} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right ) + \frac {1}{6} \, \sqrt {2} \left (-b\right )^{\frac {3}{4}} \log \left (\sqrt {2} {\left (a x^{3} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{3} + b} + \sqrt {-b}\right ) - \frac {1}{6} \, \sqrt {2} \left (-b\right )^{\frac {3}{4}} \log \left (-\sqrt {2} {\left (a x^{3} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{3} + b} + \sqrt {-b}\right ) + \frac {4}{9} \, {\left (a x^{3} + b\right )}^{\frac {3}{4}} \]
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Time = 5.65 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.71 \[ \int \frac {\left (b+a x^3\right )^{3/4}}{x} \, dx=\frac {2\,b^{3/4}\,\mathrm {atan}\left (\frac {{\left (a\,x^3+b\right )}^{1/4}}{b^{1/4}}\right )}{3}-\frac {2\,b^{3/4}\,\mathrm {atanh}\left (\frac {{\left (a\,x^3+b\right )}^{1/4}}{b^{1/4}}\right )}{3}+\frac {4\,{\left (a\,x^3+b\right )}^{3/4}}{9} \]
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