Integrand size = 19, antiderivative size = 137 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=-\frac {b \sqrt {a x^2+b x^3}}{8 x^4}-\frac {b^2 \sqrt {a x^2+b x^3}}{32 a x^3}+\frac {3 b^3 \sqrt {a x^2+b x^3}}{64 a^2 x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}-\frac {3 b^4 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{64 a^{5/2}} \]
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Time = 0.12 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2045, 2050, 2033, 212} \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=-\frac {3 b^4 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{64 a^{5/2}}+\frac {3 b^3 \sqrt {a x^2+b x^3}}{64 a^2 x^2}-\frac {b^2 \sqrt {a x^2+b x^3}}{32 a x^3}-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}-\frac {b \sqrt {a x^2+b x^3}}{8 x^4} \]
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Rule 212
Rule 2033
Rule 2045
Rule 2050
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}+\frac {1}{8} (3 b) \int \frac {\sqrt {a x^2+b x^3}}{x^5} \, dx \\ & = -\frac {b \sqrt {a x^2+b x^3}}{8 x^4}-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}+\frac {1}{16} b^2 \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx \\ & = -\frac {b \sqrt {a x^2+b x^3}}{8 x^4}-\frac {b^2 \sqrt {a x^2+b x^3}}{32 a x^3}-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}-\frac {\left (3 b^3\right ) \int \frac {1}{x \sqrt {a x^2+b x^3}} \, dx}{64 a} \\ & = -\frac {b \sqrt {a x^2+b x^3}}{8 x^4}-\frac {b^2 \sqrt {a x^2+b x^3}}{32 a x^3}+\frac {3 b^3 \sqrt {a x^2+b x^3}}{64 a^2 x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}+\frac {\left (3 b^4\right ) \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx}{128 a^2} \\ & = -\frac {b \sqrt {a x^2+b x^3}}{8 x^4}-\frac {b^2 \sqrt {a x^2+b x^3}}{32 a x^3}+\frac {3 b^3 \sqrt {a x^2+b x^3}}{64 a^2 x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}-\frac {\left (3 b^4\right ) \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {x}{\sqrt {a x^2+b x^3}}\right )}{64 a^2} \\ & = -\frac {b \sqrt {a x^2+b x^3}}{8 x^4}-\frac {b^2 \sqrt {a x^2+b x^3}}{32 a x^3}+\frac {3 b^3 \sqrt {a x^2+b x^3}}{64 a^2 x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}-\frac {3 b^4 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{64 a^{5/2}} \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=-\frac {\sqrt {x^2 (a+b x)} \left (\sqrt {a} \sqrt {a+b x} \left (16 a^3+24 a^2 b x+2 a b^2 x^2-3 b^3 x^3\right )+3 b^4 x^4 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{64 a^{5/2} x^5 \sqrt {a+b x}} \]
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Time = 2.15 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.67
method | result | size |
risch | \(-\frac {\left (-3 b^{3} x^{3}+2 a \,b^{2} x^{2}+24 a^{2} b x +16 a^{3}\right ) \sqrt {x^{2} \left (b x +a \right )}}{64 x^{5} a^{2}}-\frac {3 b^{4} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{64 a^{\frac {5}{2}} x \sqrt {b x +a}}\) | \(92\) |
default | \(\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (3 \left (b x +a \right )^{\frac {7}{2}} a^{\frac {5}{2}}-11 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {7}{2}}-3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{2} x^{4} b^{4}-11 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {9}{2}}+3 \sqrt {b x +a}\, a^{\frac {11}{2}}\right )}{64 x^{7} \left (b x +a \right )^{\frac {3}{2}} a^{\frac {9}{2}}}\) | \(101\) |
pseudoelliptic | \(-\frac {5 \left (-\frac {63 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) x^{7} b^{7}}{1280}+\sqrt {b x +a}\, \left (\frac {63 \sqrt {a}\, b^{6} x^{6}}{1280}-\frac {21 a^{\frac {3}{2}} b^{5} x^{5}}{640}+\frac {21 a^{\frac {5}{2}} b^{4} x^{4}}{800}-\frac {9 a^{\frac {7}{2}} b^{3} x^{3}}{400}+\frac {a^{\frac {9}{2}} b^{2} x^{2}}{50}+a^{\frac {11}{2}} b x +\frac {4 a^{\frac {13}{2}}}{5}\right )\right )}{28 a^{\frac {11}{2}} x^{7}}\) | \(105\) |
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Time = 0.28 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\left [\frac {3 \, \sqrt {a} b^{4} x^{5} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (3 \, a b^{3} x^{3} - 2 \, a^{2} b^{2} x^{2} - 24 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt {b x^{3} + a x^{2}}}{128 \, a^{3} x^{5}}, \frac {3 \, \sqrt {-a} b^{4} x^{5} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + {\left (3 \, a b^{3} x^{3} - 2 \, a^{2} b^{2} x^{2} - 24 \, a^{3} b x - 16 \, a^{4}\right )} \sqrt {b x^{3} + a x^{2}}}{64 \, a^{3} x^{5}}\right ] \]
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\[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}{x^{8}}\, dx \]
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\[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}{x^{8}} \,d x } \]
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Time = 0.30 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\frac {\frac {3 \, b^{5} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{5} \mathrm {sgn}\left (x\right ) - 11 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{5} \mathrm {sgn}\left (x\right ) - 11 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{5} \mathrm {sgn}\left (x\right ) + 3 \, \sqrt {b x + a} a^{3} b^{5} \mathrm {sgn}\left (x\right )}{a^{2} b^{4} x^{4}}}{64 \, b} \]
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Timed out. \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{3/2}}{x^8} \,d x \]
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