Integrand size = 17, antiderivative size = 100 \[ \int \frac {1}{x^4 \sqrt {b x+c x^2}} \, dx=-\frac {2 \sqrt {b x+c x^2}}{7 b x^4}+\frac {12 c \sqrt {b x+c x^2}}{35 b^2 x^3}-\frac {16 c^2 \sqrt {b x+c x^2}}{35 b^3 x^2}+\frac {32 c^3 \sqrt {b x+c x^2}}{35 b^4 x} \]
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Time = 0.03 (sec) , antiderivative size = 100, 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, 664} \[ \int \frac {1}{x^4 \sqrt {b x+c x^2}} \, dx=\frac {32 c^3 \sqrt {b x+c x^2}}{35 b^4 x}-\frac {16 c^2 \sqrt {b x+c x^2}}{35 b^3 x^2}+\frac {12 c \sqrt {b x+c x^2}}{35 b^2 x^3}-\frac {2 \sqrt {b x+c x^2}}{7 b x^4} \]
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Rule 664
Rule 672
Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt {b x+c x^2}}{7 b x^4}-\frac {(6 c) \int \frac {1}{x^3 \sqrt {b x+c x^2}} \, dx}{7 b} \\ & = -\frac {2 \sqrt {b x+c x^2}}{7 b x^4}+\frac {12 c \sqrt {b x+c x^2}}{35 b^2 x^3}+\frac {\left (24 c^2\right ) \int \frac {1}{x^2 \sqrt {b x+c x^2}} \, dx}{35 b^2} \\ & = -\frac {2 \sqrt {b x+c x^2}}{7 b x^4}+\frac {12 c \sqrt {b x+c x^2}}{35 b^2 x^3}-\frac {16 c^2 \sqrt {b x+c x^2}}{35 b^3 x^2}-\frac {\left (16 c^3\right ) \int \frac {1}{x \sqrt {b x+c x^2}} \, dx}{35 b^3} \\ & = -\frac {2 \sqrt {b x+c x^2}}{7 b x^4}+\frac {12 c \sqrt {b x+c x^2}}{35 b^2 x^3}-\frac {16 c^2 \sqrt {b x+c x^2}}{35 b^3 x^2}+\frac {32 c^3 \sqrt {b x+c x^2}}{35 b^4 x} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^4 \sqrt {b x+c x^2}} \, dx=\frac {2 \sqrt {x (b+c x)} \left (-5 b^3+6 b^2 c x-8 b c^2 x^2+16 c^3 x^3\right )}{35 b^4 x^4} \]
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Time = 1.97 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.48
method | result | size |
pseudoelliptic | \(-\frac {2 \left (-16 c^{3} x^{3}+8 b \,c^{2} x^{2}-6 b^{2} c x +5 b^{3}\right ) \sqrt {x \left (c x +b \right )}}{35 b^{4} x^{4}}\) | \(48\) |
trager | \(-\frac {2 \left (-16 c^{3} x^{3}+8 b \,c^{2} x^{2}-6 b^{2} c x +5 b^{3}\right ) \sqrt {c \,x^{2}+b x}}{35 b^{4} x^{4}}\) | \(50\) |
risch | \(-\frac {2 \left (c x +b \right ) \left (-16 c^{3} x^{3}+8 b \,c^{2} x^{2}-6 b^{2} c x +5 b^{3}\right )}{35 b^{4} x^{3} \sqrt {x \left (c x +b \right )}}\) | \(53\) |
gosper | \(-\frac {2 \left (c x +b \right ) \left (-16 c^{3} x^{3}+8 b \,c^{2} x^{2}-6 b^{2} c x +5 b^{3}\right )}{35 x^{3} b^{4} \sqrt {c \,x^{2}+b x}}\) | \(55\) |
default | \(-\frac {2 \sqrt {c \,x^{2}+b x}}{7 b \,x^{4}}-\frac {6 c \left (-\frac {2 \sqrt {c \,x^{2}+b x}}{5 b \,x^{3}}-\frac {4 c \left (-\frac {2 \sqrt {c \,x^{2}+b x}}{3 b \,x^{2}}+\frac {4 c \sqrt {c \,x^{2}+b x}}{3 b^{2} x}\right )}{5 b}\right )}{7 b}\) | \(93\) |
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Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x^4 \sqrt {b x+c x^2}} \, dx=\frac {2 \, {\left (16 \, c^{3} x^{3} - 8 \, b c^{2} x^{2} + 6 \, b^{2} c x - 5 \, b^{3}\right )} \sqrt {c x^{2} + b x}}{35 \, b^{4} x^{4}} \]
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\[ \int \frac {1}{x^4 \sqrt {b x+c x^2}} \, dx=\int \frac {1}{x^{4} \sqrt {x \left (b + c x\right )}}\, dx \]
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Time = 0.19 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^4 \sqrt {b x+c x^2}} \, dx=\frac {32 \, \sqrt {c x^{2} + b x} c^{3}}{35 \, b^{4} x} - \frac {16 \, \sqrt {c x^{2} + b x} c^{2}}{35 \, b^{3} x^{2}} + \frac {12 \, \sqrt {c x^{2} + b x} c}{35 \, b^{2} x^{3}} - \frac {2 \, \sqrt {c x^{2} + b x}}{7 \, b x^{4}} \]
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Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x^4 \sqrt {b x+c x^2}} \, dx=\frac {2 \, {\left (70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} c^{\frac {3}{2}} + 84 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b c + 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{2} \sqrt {c} + 5 \, b^{3}\right )}}{35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7}} \]
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Time = 9.05 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^4 \sqrt {b x+c x^2}} \, dx=\frac {32\,c^3\,\sqrt {c\,x^2+b\,x}}{35\,b^4\,x}-\frac {16\,c^2\,\sqrt {c\,x^2+b\,x}}{35\,b^3\,x^2}-\frac {2\,\sqrt {c\,x^2+b\,x}}{7\,b\,x^4}+\frac {12\,c\,\sqrt {c\,x^2+b\,x}}{35\,b^2\,x^3} \]
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