Integrand size = 17, antiderivative size = 103 \[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2}{7 b x^3 \sqrt {b x+c x^2}}+\frac {16 c}{35 b^2 x^2 \sqrt {b x+c x^2}}-\frac {32 c^2}{35 b^3 x \sqrt {b x+c x^2}}+\frac {128 c^3 (b+2 c x)}{35 b^5 \sqrt {b x+c x^2}} \]
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Time = 0.03 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {672, 627} \[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {128 c^3 (b+2 c x)}{35 b^5 \sqrt {b x+c x^2}}-\frac {32 c^2}{35 b^3 x \sqrt {b x+c x^2}}+\frac {16 c}{35 b^2 x^2 \sqrt {b x+c x^2}}-\frac {2}{7 b x^3 \sqrt {b x+c x^2}} \]
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Rule 627
Rule 672
Rubi steps \begin{align*} \text {integral}& = -\frac {2}{7 b x^3 \sqrt {b x+c x^2}}-\frac {(8 c) \int \frac {1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx}{7 b} \\ & = -\frac {2}{7 b x^3 \sqrt {b x+c x^2}}+\frac {16 c}{35 b^2 x^2 \sqrt {b x+c x^2}}+\frac {\left (48 c^2\right ) \int \frac {1}{x \left (b x+c x^2\right )^{3/2}} \, dx}{35 b^2} \\ & = -\frac {2}{7 b x^3 \sqrt {b x+c x^2}}+\frac {16 c}{35 b^2 x^2 \sqrt {b x+c x^2}}-\frac {32 c^2}{35 b^3 x \sqrt {b x+c x^2}}-\frac {\left (64 c^3\right ) \int \frac {1}{\left (b x+c x^2\right )^{3/2}} \, dx}{35 b^3} \\ & = -\frac {2}{7 b x^3 \sqrt {b x+c x^2}}+\frac {16 c}{35 b^2 x^2 \sqrt {b x+c x^2}}-\frac {32 c^2}{35 b^3 x \sqrt {b x+c x^2}}+\frac {128 c^3 (b+2 c x)}{35 b^5 \sqrt {b x+c x^2}} \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \left (-5 b^4+8 b^3 c x-16 b^2 c^2 x^2+64 b c^3 x^3+128 c^4 x^4\right )}{35 b^5 x^3 \sqrt {x (b+c x)}} \]
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Time = 1.86 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.57
method | result | size |
pseudoelliptic | \(\frac {\frac {256}{35} c^{4} x^{4}+\frac {128}{35} b \,c^{3} x^{3}-\frac {32}{35} b^{2} c^{2} x^{2}+\frac {16}{35} b^{3} c x -\frac {2}{7} b^{4}}{x^{3} \sqrt {x \left (c x +b \right )}\, b^{5}}\) | \(59\) |
gosper | \(-\frac {2 \left (c x +b \right ) \left (-128 c^{4} x^{4}-64 b \,c^{3} x^{3}+16 b^{2} c^{2} x^{2}-8 b^{3} c x +5 b^{4}\right )}{35 x^{2} b^{5} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\) | \(66\) |
trager | \(-\frac {2 \left (-128 c^{4} x^{4}-64 b \,c^{3} x^{3}+16 b^{2} c^{2} x^{2}-8 b^{3} c x +5 b^{4}\right ) \sqrt {c \,x^{2}+b x}}{35 \left (c x +b \right ) b^{5} x^{4}}\) | \(68\) |
risch | \(-\frac {2 \left (c x +b \right ) \left (-93 c^{3} x^{3}+29 b \,c^{2} x^{2}-13 b^{2} c x +5 b^{3}\right )}{35 b^{5} x^{3} \sqrt {x \left (c x +b \right )}}+\frac {2 c^{4} x}{\sqrt {x \left (c x +b \right )}\, b^{5}}\) | \(72\) |
default | \(-\frac {2}{7 b \,x^{3} \sqrt {c \,x^{2}+b x}}-\frac {8 c \left (-\frac {2}{5 b \,x^{2} \sqrt {c \,x^{2}+b x}}-\frac {6 c \left (-\frac {2}{3 b x \sqrt {c \,x^{2}+b x}}+\frac {8 c \left (2 c x +b \right )}{3 b^{3} \sqrt {c \,x^{2}+b x}}\right )}{5 b}\right )}{7 b}\) | \(96\) |
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Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.70 \[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (128 \, c^{4} x^{4} + 64 \, b c^{3} x^{3} - 16 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x - 5 \, b^{4}\right )} \sqrt {c x^{2} + b x}}{35 \, {\left (b^{5} c x^{5} + b^{6} x^{4}\right )}} \]
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\[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{x^{3} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]
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Time = 0.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {256 \, c^{4} x}{35 \, \sqrt {c x^{2} + b x} b^{5}} + \frac {128 \, c^{3}}{35 \, \sqrt {c x^{2} + b x} b^{4}} - \frac {32 \, c^{2}}{35 \, \sqrt {c x^{2} + b x} b^{3} x} + \frac {16 \, c}{35 \, \sqrt {c x^{2} + b x} b^{2} x^{2}} - \frac {2}{7 \, \sqrt {c x^{2} + b x} b x^{3}} \]
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\[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{3}} \,d x } \]
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Time = 9.14 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c\,x^2+b\,x}\,\left (\frac {186\,c^3}{35\,b^4}+\frac {256\,c^4\,x}{35\,b^5}\right )}{x\,\left (b+c\,x\right )}-\frac {58\,c^2\,\sqrt {c\,x^2+b\,x}}{35\,b^4\,x^2}-\frac {2\,\sqrt {c\,x^2+b\,x}}{7\,b^2\,x^4}+\frac {26\,c\,\sqrt {c\,x^2+b\,x}}{35\,b^3\,x^3} \]
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