Integrand size = 15, antiderivative size = 104 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=-\frac {\left (a+b x^4\right )^{3/4}}{8 a x^8}+\frac {5 b \left (a+b x^4\right )^{3/4}}{32 a^2 x^4}+\frac {5 b^2 \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}-\frac {5 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}} \]
[Out]
Time = 0.05 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {272, 44, 65, 304, 209, 212} \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=\frac {5 b^2 \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}-\frac {5 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}+\frac {5 b \left (a+b x^4\right )^{3/4}}{32 a^2 x^4}-\frac {\left (a+b x^4\right )^{3/4}}{8 a x^8} \]
[In]
[Out]
Rule 44
Rule 65
Rule 209
Rule 212
Rule 272
Rule 304
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {1}{x^3 \sqrt [4]{a+b x}} \, dx,x,x^4\right ) \\ & = -\frac {\left (a+b x^4\right )^{3/4}}{8 a x^8}-\frac {(5 b) \text {Subst}\left (\int \frac {1}{x^2 \sqrt [4]{a+b x}} \, dx,x,x^4\right )}{32 a} \\ & = -\frac {\left (a+b x^4\right )^{3/4}}{8 a x^8}+\frac {5 b \left (a+b x^4\right )^{3/4}}{32 a^2 x^4}+\frac {\left (5 b^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt [4]{a+b x}} \, dx,x,x^4\right )}{128 a^2} \\ & = -\frac {\left (a+b x^4\right )^{3/4}}{8 a x^8}+\frac {5 b \left (a+b x^4\right )^{3/4}}{32 a^2 x^4}+\frac {(5 b) \text {Subst}\left (\int \frac {x^2}{-\frac {a}{b}+\frac {x^4}{b}} \, dx,x,\sqrt [4]{a+b x^4}\right )}{32 a^2} \\ & = -\frac {\left (a+b x^4\right )^{3/4}}{8 a x^8}+\frac {5 b \left (a+b x^4\right )^{3/4}}{32 a^2 x^4}-\frac {\left (5 b^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a}-x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )}{64 a^2}+\frac {\left (5 b^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a}+x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )}{64 a^2} \\ & = -\frac {\left (a+b x^4\right )^{3/4}}{8 a x^8}+\frac {5 b \left (a+b x^4\right )^{3/4}}{32 a^2 x^4}+\frac {5 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}-\frac {5 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}} \\ \end{align*}
Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=\frac {\left (a+b x^4\right )^{3/4} \left (-4 a+5 b x^4\right )}{32 a^2 x^8}+\frac {5 b^2 \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}}-\frac {5 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{9/4}} \]
[In]
[Out]
Time = 4.34 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.04
method | result | size |
pseudoelliptic | \(\frac {10 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) b^{2} x^{8}-5 \ln \left (\frac {-\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}{-\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}\right ) b^{2} x^{8}+20 b \,x^{4} a^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {3}{4}}-16 a^{\frac {5}{4}} \left (b \,x^{4}+a \right )^{\frac {3}{4}}}{128 a^{\frac {9}{4}} x^{8}}\) | \(108\) |
[In]
[Out]
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.10 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=-\frac {5 \, a^{2} x^{8} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} \log \left (125 \, a^{7} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{6}\right ) - 5 i \, a^{2} x^{8} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} \log \left (125 i \, a^{7} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{6}\right ) + 5 i \, a^{2} x^{8} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-125 i \, a^{7} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{6}\right ) - 5 \, a^{2} x^{8} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-125 \, a^{7} \left (\frac {b^{8}}{a^{9}}\right )^{\frac {3}{4}} + 125 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{6}\right ) - 4 \, {\left (5 \, b x^{4} - 4 \, a\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{128 \, a^{2} x^{8}} \]
[In]
[Out]
Result contains complex when optimal does not.
Time = 1.63 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.38 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=- \frac {\Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 \sqrt [4]{b} x^{9} \Gamma \left (\frac {13}{4}\right )} \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=\frac {5 \, b^{2} {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )}}{128 \, a^{2}} + \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} b^{2} - 9 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a b^{2}}{32 \, {\left ({\left (b x^{4} + a\right )}^{2} a^{2} - 2 \, {\left (b x^{4} + a\right )} a^{3} + a^{4}\right )}} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (80) = 160\).
Time = 0.28 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.35 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=\frac {\frac {10 \, \sqrt {2} b^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {10 \, \sqrt {2} b^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {5 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{3} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{a^{3}} + \frac {5 \, \sqrt {2} b^{3} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {1}{4}} a^{2}} + \frac {8 \, {\left (5 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} b^{3} - 9 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a b^{3}\right )}}{a^{2} b^{2} x^{8}}}{256 \, b} \]
[In]
[Out]
Time = 5.95 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^9 \sqrt [4]{a+b x^4}} \, dx=\frac {5\,b^2\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{64\,a^{9/4}}-\frac {9\,{\left (b\,x^4+a\right )}^{3/4}}{32\,a\,x^8}+\frac {5\,{\left (b\,x^4+a\right )}^{7/4}}{32\,a^2\,x^8}+\frac {b^2\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}\,1{}\mathrm {i}}{a^{1/4}}\right )\,5{}\mathrm {i}}{64\,a^{9/4}} \]
[In]
[Out]