Integrand size = 19, antiderivative size = 108 \[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {32 b^3 \sqrt {x}}{5 c^4 \sqrt {b x+c x^2}}+\frac {16 b^2 x^{3/2}}{5 c^3 \sqrt {b x+c x^2}}-\frac {4 b x^{5/2}}{5 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{7/2}}{5 c \sqrt {b x+c x^2}} \]
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Time = 0.03 (sec) , antiderivative size = 108, 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 = {670, 662} \[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {32 b^3 \sqrt {x}}{5 c^4 \sqrt {b x+c x^2}}+\frac {16 b^2 x^{3/2}}{5 c^3 \sqrt {b x+c x^2}}-\frac {4 b x^{5/2}}{5 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{7/2}}{5 c \sqrt {b x+c x^2}} \]
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Rule 662
Rule 670
Rubi steps \begin{align*} \text {integral}& = \frac {2 x^{7/2}}{5 c \sqrt {b x+c x^2}}-\frac {(6 b) \int \frac {x^{7/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{5 c} \\ & = -\frac {4 b x^{5/2}}{5 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{7/2}}{5 c \sqrt {b x+c x^2}}+\frac {\left (8 b^2\right ) \int \frac {x^{5/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{5 c^2} \\ & = \frac {16 b^2 x^{3/2}}{5 c^3 \sqrt {b x+c x^2}}-\frac {4 b x^{5/2}}{5 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{7/2}}{5 c \sqrt {b x+c x^2}}-\frac {\left (16 b^3\right ) \int \frac {x^{3/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{5 c^3} \\ & = \frac {32 b^3 \sqrt {x}}{5 c^4 \sqrt {b x+c x^2}}+\frac {16 b^2 x^{3/2}}{5 c^3 \sqrt {b x+c x^2}}-\frac {4 b x^{5/2}}{5 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{7/2}}{5 c \sqrt {b x+c x^2}} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.48 \[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {x} \left (16 b^3+8 b^2 c x-2 b c^2 x^2+c^3 x^3\right )}{5 c^4 \sqrt {x (b+c x)}} \]
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Time = 2.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.50
method | result | size |
gosper | \(\frac {2 \left (c x +b \right ) \left (c^{3} x^{3}-2 b \,c^{2} x^{2}+8 b^{2} c x +16 b^{3}\right ) x^{\frac {3}{2}}}{5 c^{4} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\) | \(54\) |
default | \(\frac {2 \sqrt {x \left (c x +b \right )}\, \left (c^{3} x^{3}-2 b \,c^{2} x^{2}+8 b^{2} c x +16 b^{3}\right )}{5 \sqrt {x}\, \left (c x +b \right ) c^{4}}\) | \(54\) |
risch | \(\frac {2 \left (c^{2} x^{2}-3 b c x +11 b^{2}\right ) \left (c x +b \right ) \sqrt {x}}{5 c^{4} \sqrt {x \left (c x +b \right )}}+\frac {2 b^{3} \sqrt {x}}{c^{4} \sqrt {x \left (c x +b \right )}}\) | \(62\) |
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Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.56 \[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (c^{3} x^{3} - 2 \, b c^{2} x^{2} + 8 \, b^{2} c x + 16 \, b^{3}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{5 \, {\left (c^{5} x^{2} + b c^{4} x\right )}} \]
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\[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^{\frac {9}{2}}}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]
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\[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {x^{\frac {9}{2}}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}}} \,d x } \]
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Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.64 \[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {32 \, b^{\frac {5}{2}}}{5 \, c^{4}} + \frac {2 \, b^{3}}{\sqrt {c x + b} c^{4}} + \frac {2 \, {\left ({\left (c x + b\right )}^{\frac {5}{2}} c^{16} - 5 \, {\left (c x + b\right )}^{\frac {3}{2}} b c^{16} + 15 \, \sqrt {c x + b} b^{2} c^{16}\right )}}{5 \, c^{20}} \]
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Timed out. \[ \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^{9/2}}{{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \]
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