Integrand size = 19, antiderivative size = 101 \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {16 \sqrt {x}}{35 b^4 \sqrt {a x+b x^3}} \]
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Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2040, 2039} \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {16 \sqrt {x}}{35 b^4 \sqrt {a x+b x^3}}-\frac {8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]
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Rule 2039
Rule 2040
Rubi steps \begin{align*} \text {integral}& = -\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}+\frac {6 \int \frac {x^{17/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 b} \\ & = -\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}+\frac {24 \int \frac {x^{11/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{35 b^2} \\ & = -\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}+\frac {16 \int \frac {x^{5/2}}{\left (a x+b x^3\right )^{3/2}} \, dx}{35 b^3} \\ & = -\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {16 \sqrt {x}}{35 b^4 \sqrt {a x+b x^3}} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {x^{9/2} \left (a+b x^2\right ) \left (-16 a^3-56 a^2 b x^2-70 a b^2 x^4-35 b^3 x^6\right )}{35 b^4 \left (x \left (a+b x^2\right )\right )^{9/2}} \]
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Time = 2.00 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.58
method | result | size |
gosper | \(-\frac {\left (b \,x^{2}+a \right ) \left (35 b^{3} x^{6}+70 a \,b^{2} x^{4}+56 a^{2} b \,x^{2}+16 a^{3}\right ) x^{\frac {9}{2}}}{35 b^{4} \left (b \,x^{3}+a x \right )^{\frac {9}{2}}}\) | \(59\) |
default | \(-\frac {\sqrt {x \left (b \,x^{2}+a \right )}\, \left (35 b^{3} x^{6}+70 a \,b^{2} x^{4}+56 a^{2} b \,x^{2}+16 a^{3}\right )}{35 \sqrt {x}\, \left (b \,x^{2}+a \right )^{4} b^{4}}\) | \(61\) |
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Time = 0.42 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.96 \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=-\frac {{\left (35 \, b^{3} x^{6} + 70 \, a b^{2} x^{4} + 56 \, a^{2} b x^{2} + 16 \, a^{3}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{35 \, {\left (b^{8} x^{9} + 4 \, a b^{7} x^{7} + 6 \, a^{2} b^{6} x^{5} + 4 \, a^{3} b^{5} x^{3} + a^{4} b^{4} x\right )}} \]
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Timed out. \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\text {Timed out} \]
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\[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int { \frac {x^{\frac {23}{2}}}{{\left (b x^{3} + a x\right )}^{\frac {9}{2}}} \,d x } \]
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Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\frac {16}{35 \, \sqrt {a} b^{4}} - \frac {35 \, {\left (b x^{2} + a\right )}^{3} - 35 \, {\left (b x^{2} + a\right )}^{2} a + 21 \, {\left (b x^{2} + a\right )} a^{2} - 5 \, a^{3}}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{4}} \]
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Timed out. \[ \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx=\int \frac {x^{23/2}}{{\left (b\,x^3+a\,x\right )}^{9/2}} \,d x \]
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