Integrand size = 17, antiderivative size = 81 \[ \int \frac {\left (-b+a x^4\right )^{3/4}}{x^4} \, dx=-\frac {\left (-b+a x^4\right )^{3/4}}{3 x^3}+\frac {1}{2} a^{3/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\frac {1}{2} a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right ) \]
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Time = 0.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {283, 246, 218, 212, 209} \[ \int \frac {\left (-b+a x^4\right )^{3/4}}{x^4} \, dx=\frac {1}{2} a^{3/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+\frac {1}{2} a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )-\frac {\left (a x^4-b\right )^{3/4}}{3 x^3} \]
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Rule 209
Rule 212
Rule 218
Rule 246
Rule 283
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (-b+a x^4\right )^{3/4}}{3 x^3}+a \int \frac {1}{\sqrt [4]{-b+a x^4}} \, dx \\ & = -\frac {\left (-b+a x^4\right )^{3/4}}{3 x^3}+a \text {Subst}\left (\int \frac {1}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right ) \\ & = -\frac {\left (-b+a x^4\right )^{3/4}}{3 x^3}+\frac {1}{2} a \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} a \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right ) \\ & = -\frac {\left (-b+a x^4\right )^{3/4}}{3 x^3}+\frac {1}{2} a^{3/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\frac {1}{2} a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right ) \\ \end{align*}
Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-b+a x^4\right )^{3/4}}{x^4} \, dx=-\frac {\left (-b+a x^4\right )^{3/4}}{3 x^3}+\frac {1}{2} a^{3/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\frac {1}{2} a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right ) \]
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Time = 0.88 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.21
method | result | size |
pseudoelliptic | \(\frac {-6 \arctan \left (\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) a^{\frac {3}{4}} x^{3}+3 \ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right ) a^{\frac {3}{4}} x^{3}-4 \left (a \,x^{4}-b \right )^{\frac {3}{4}}}{12 x^{3}}\) | \(98\) |
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Timed out. \[ \int \frac {\left (-b+a x^4\right )^{3/4}}{x^4} \, dx=\text {Timed out} \]
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Result contains complex when optimal does not.
Time = 0.71 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.57 \[ \int \frac {\left (-b+a x^4\right )^{3/4}}{x^4} \, dx=\frac {b^{\frac {3}{4}} e^{\frac {3 i \pi }{4}} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {a x^{4}}{b}} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} \]
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none
Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.15 \[ \int \frac {\left (-b+a x^4\right )^{3/4}}{x^4} \, dx=-\frac {1}{4} \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {1}{4}}}\right )} - \frac {{\left (a x^{4} - b\right )}^{\frac {3}{4}}}{3 \, x^{3}} \]
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\[ \int \frac {\left (-b+a x^4\right )^{3/4}}{x^4} \, dx=\int { \frac {{\left (a x^{4} - b\right )}^{\frac {3}{4}}}{x^{4}} \,d x } \]
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Timed out. \[ \int \frac {\left (-b+a x^4\right )^{3/4}}{x^4} \, dx=\int \frac {{\left (a\,x^4-b\right )}^{3/4}}{x^4} \,d x \]
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