Integrand size = 13, antiderivative size = 66 \[ \int \frac {\left (a+b x^3\right )^5}{x^{10}} \, dx=-\frac {a^5}{9 x^9}-\frac {5 a^4 b}{6 x^6}-\frac {10 a^3 b^2}{3 x^3}+\frac {5}{3} a b^4 x^3+\frac {b^5 x^6}{6}+10 a^2 b^3 \log (x) \]
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Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int \frac {\left (a+b x^3\right )^5}{x^{10}} \, dx=-\frac {a^5}{9 x^9}-\frac {5 a^4 b}{6 x^6}-\frac {10 a^3 b^2}{3 x^3}+10 a^2 b^3 \log (x)+\frac {5}{3} a b^4 x^3+\frac {b^5 x^6}{6} \]
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Rule 45
Rule 272
Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {(a+b x)^5}{x^4} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (5 a b^4+\frac {a^5}{x^4}+\frac {5 a^4 b}{x^3}+\frac {10 a^3 b^2}{x^2}+\frac {10 a^2 b^3}{x}+b^5 x\right ) \, dx,x,x^3\right ) \\ & = -\frac {a^5}{9 x^9}-\frac {5 a^4 b}{6 x^6}-\frac {10 a^3 b^2}{3 x^3}+\frac {5}{3} a b^4 x^3+\frac {b^5 x^6}{6}+10 a^2 b^3 \log (x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^3\right )^5}{x^{10}} \, dx=-\frac {a^5}{9 x^9}-\frac {5 a^4 b}{6 x^6}-\frac {10 a^3 b^2}{3 x^3}+\frac {5}{3} a b^4 x^3+\frac {b^5 x^6}{6}+10 a^2 b^3 \log (x) \]
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Time = 3.64 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.86
method | result | size |
default | \(-\frac {a^{5}}{9 x^{9}}-\frac {5 a^{4} b}{6 x^{6}}-\frac {10 a^{3} b^{2}}{3 x^{3}}+\frac {5 a \,b^{4} x^{3}}{3}+\frac {b^{5} x^{6}}{6}+10 a^{2} b^{3} \ln \left (x \right )\) | \(57\) |
norman | \(\frac {-\frac {1}{9} a^{5}+\frac {1}{6} b^{5} x^{15}+\frac {5}{3} a \,b^{4} x^{12}-\frac {10}{3} a^{3} b^{2} x^{6}-\frac {5}{6} a^{4} b \,x^{3}}{x^{9}}+10 a^{2} b^{3} \ln \left (x \right )\) | \(59\) |
parallelrisch | \(\frac {3 b^{5} x^{15}+30 a \,b^{4} x^{12}+180 a^{2} b^{3} \ln \left (x \right ) x^{9}-60 a^{3} b^{2} x^{6}-15 a^{4} b \,x^{3}-2 a^{5}}{18 x^{9}}\) | \(62\) |
risch | \(\frac {b^{5} x^{6}}{6}+\frac {5 a \,b^{4} x^{3}}{3}+\frac {25 a^{2} b^{3}}{6}+\frac {-\frac {10}{3} a^{3} b^{2} x^{6}-\frac {5}{6} a^{4} b \,x^{3}-\frac {1}{9} a^{5}}{x^{9}}+10 a^{2} b^{3} \ln \left (x \right )\) | \(67\) |
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Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x^3\right )^5}{x^{10}} \, dx=\frac {3 \, b^{5} x^{15} + 30 \, a b^{4} x^{12} + 180 \, a^{2} b^{3} x^{9} \log \left (x\right ) - 60 \, a^{3} b^{2} x^{6} - 15 \, a^{4} b x^{3} - 2 \, a^{5}}{18 \, x^{9}} \]
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Time = 0.17 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^3\right )^5}{x^{10}} \, dx=10 a^{2} b^{3} \log {\left (x \right )} + \frac {5 a b^{4} x^{3}}{3} + \frac {b^{5} x^{6}}{6} + \frac {- 2 a^{5} - 15 a^{4} b x^{3} - 60 a^{3} b^{2} x^{6}}{18 x^{9}} \]
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Time = 0.23 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x^3\right )^5}{x^{10}} \, dx=\frac {1}{6} \, b^{5} x^{6} + \frac {5}{3} \, a b^{4} x^{3} + \frac {10}{3} \, a^{2} b^{3} \log \left (x^{3}\right ) - \frac {60 \, a^{3} b^{2} x^{6} + 15 \, a^{4} b x^{3} + 2 \, a^{5}}{18 \, x^{9}} \]
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Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^3\right )^5}{x^{10}} \, dx=\frac {1}{6} \, b^{5} x^{6} + \frac {5}{3} \, a b^{4} x^{3} + 10 \, a^{2} b^{3} \log \left ({\left | x \right |}\right ) - \frac {110 \, a^{2} b^{3} x^{9} + 60 \, a^{3} b^{2} x^{6} + 15 \, a^{4} b x^{3} + 2 \, a^{5}}{18 \, x^{9}} \]
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Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^3\right )^5}{x^{10}} \, dx=\frac {b^5\,x^6}{6}-\frac {\frac {a^5}{9}+\frac {5\,a^4\,b\,x^3}{6}+\frac {10\,a^3\,b^2\,x^6}{3}}{x^9}+\frac {5\,a\,b^4\,x^3}{3}+10\,a^2\,b^3\,\ln \left (x\right ) \]
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