Integrand size = 19, antiderivative size = 130 \[ \int \frac {x^{25/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {x^{3/2}}{b^4 \sqrt {a x+b x^3}}+\frac {\text {arctanh}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x+b x^3}}\right )}{b^{9/2}} \]
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Time = 0.13 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2047, 2054, 212} \[ \int \frac {x^{25/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x+b x^3}}\right )}{b^{9/2}}-\frac {x^{3/2}}{b^4 \sqrt {a x+b x^3}}-\frac {x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]
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Rule 212
Rule 2047
Rule 2054
Rubi steps \begin{align*} \text {integral}& = -\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}+\frac {\int \frac {x^{19/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{b} \\ & = -\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}+\frac {\int \frac {x^{13/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{b^2} \\ & = -\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}+\frac {\int \frac {x^{7/2}}{\left (a x+b x^3\right )^{3/2}} \, dx}{b^3} \\ & = -\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {x^{3/2}}{b^4 \sqrt {a x+b x^3}}+\frac {\int \frac {\sqrt {x}}{\sqrt {a x+b x^3}} \, dx}{b^4} \\ & = -\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {x^{3/2}}{b^4 \sqrt {a x+b x^3}}+\frac {\text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^{3/2}}{\sqrt {a x+b x^3}}\right )}{b^4} \\ & = -\frac {x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {x^{3/2}}{b^4 \sqrt {a x+b x^3}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x+b x^3}}\right )}{b^{9/2}} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.90 \[ \int \frac {x^{25/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {x^{9/2} \left (-\sqrt {b} x \left (a+b x^2\right ) \left (105 a^3+350 a^2 b x^2+406 a b^2 x^4+176 b^3 x^6\right )+210 \left (a+b x^2\right )^{9/2} \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )\right )}{105 b^{9/2} \left (x \left (a+b x^2\right )\right )^{9/2}} \]
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Time = 2.08 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.52
method | result | size |
default | \(\frac {\sqrt {x \left (b \,x^{2}+a \right )}\, \left (105 \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) b^{3} x^{6} \sqrt {b \,x^{2}+a}-176 x^{7} b^{\frac {7}{2}}+315 \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) a \,b^{2} x^{4} \sqrt {b \,x^{2}+a}-406 b^{\frac {5}{2}} a \,x^{5}+315 \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) a^{2} b \,x^{2} \sqrt {b \,x^{2}+a}-350 b^{\frac {3}{2}} a^{2} x^{3}+105 \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) a^{3} \sqrt {b \,x^{2}+a}-105 \sqrt {b}\, a^{3} x \right )}{105 b^{\frac {9}{2}} \sqrt {x}\, \left (b \,x^{2}+a \right )^{4}}\) | \(198\) |
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Time = 0.42 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.68 \[ \int \frac {x^{25/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\left [\frac {105 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {b} \log \left (2 \, b x^{2} + 2 \, \sqrt {b x^{3} + a x} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (176 \, b^{4} x^{6} + 406 \, a b^{3} x^{4} + 350 \, a^{2} b^{2} x^{2} + 105 \, a^{3} b\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{210 \, {\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}, -\frac {105 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{3} + a x} \sqrt {-b}}{b x^{\frac {3}{2}}}\right ) + {\left (176 \, b^{4} x^{6} + 406 \, a b^{3} x^{4} + 350 \, a^{2} b^{2} x^{2} + 105 \, a^{3} b\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{105 \, {\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}\right ] \]
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Timed out. \[ \int \frac {x^{25/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\text {Timed out} \]
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\[ \int \frac {x^{25/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int { \frac {x^{\frac {25}{2}}}{{\left (b x^{3} + a x\right )}^{\frac {9}{2}}} \,d x } \]
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Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.66 \[ \int \frac {x^{25/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {{\left (2 \, {\left (x^{2} {\left (\frac {88 \, x^{2}}{b} + \frac {203 \, a}{b^{2}}\right )} + \frac {175 \, a^{2}}{b^{3}}\right )} x^{2} + \frac {105 \, a^{3}}{b^{4}}\right )} x}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} - \frac {\log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {9}{2}}} + \frac {\log \left ({\left | a \right |}\right )}{2 \, b^{\frac {9}{2}}} \]
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Timed out. \[ \int \frac {x^{25/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int \frac {x^{25/2}}{{\left (b\,x^3+a\,x\right )}^{9/2}} \,d x \]
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