Integrand size = 11, antiderivative size = 60 \[ \int \frac {(a+b x)^5}{x^3} \, dx=-\frac {a^5}{2 x^2}-\frac {5 a^4 b}{x}+10 a^2 b^3 x+\frac {5}{2} a b^4 x^2+\frac {b^5 x^3}{3}+10 a^3 b^2 \log (x) \]
[Out]
Time = 0.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int \frac {(a+b x)^5}{x^3} \, dx=-\frac {a^5}{2 x^2}-\frac {5 a^4 b}{x}+10 a^3 b^2 \log (x)+10 a^2 b^3 x+\frac {5}{2} a b^4 x^2+\frac {b^5 x^3}{3} \]
[In]
[Out]
Rule 45
Rubi steps \begin{align*} \text {integral}& = \int \left (10 a^2 b^3+\frac {a^5}{x^3}+\frac {5 a^4 b}{x^2}+\frac {10 a^3 b^2}{x}+5 a b^4 x+b^5 x^2\right ) \, dx \\ & = -\frac {a^5}{2 x^2}-\frac {5 a^4 b}{x}+10 a^2 b^3 x+\frac {5}{2} a b^4 x^2+\frac {b^5 x^3}{3}+10 a^3 b^2 \log (x) \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^5}{x^3} \, dx=-\frac {a^5}{2 x^2}-\frac {5 a^4 b}{x}+10 a^2 b^3 x+\frac {5}{2} a b^4 x^2+\frac {b^5 x^3}{3}+10 a^3 b^2 \log (x) \]
[In]
[Out]
Time = 0.17 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92
method | result | size |
default | \(-\frac {a^{5}}{2 x^{2}}-\frac {5 a^{4} b}{x}+10 a^{2} b^{3} x +\frac {5 a \,b^{4} x^{2}}{2}+\frac {b^{5} x^{3}}{3}+10 a^{3} b^{2} \ln \left (x \right )\) | \(55\) |
risch | \(\frac {b^{5} x^{3}}{3}+\frac {5 a \,b^{4} x^{2}}{2}+10 a^{2} b^{3} x +\frac {-5 a^{4} b x -\frac {1}{2} a^{5}}{x^{2}}+10 a^{3} b^{2} \ln \left (x \right )\) | \(55\) |
norman | \(\frac {-\frac {1}{2} a^{5}+\frac {1}{3} b^{5} x^{5}+\frac {5}{2} a \,b^{4} x^{4}+10 a^{2} b^{3} x^{3}-5 a^{4} b x}{x^{2}}+10 a^{3} b^{2} \ln \left (x \right )\) | \(57\) |
parallelrisch | \(\frac {2 b^{5} x^{5}+15 a \,b^{4} x^{4}+60 a^{3} b^{2} \ln \left (x \right ) x^{2}+60 a^{2} b^{3} x^{3}-30 a^{4} b x -3 a^{5}}{6 x^{2}}\) | \(60\) |
[In]
[Out]
none
Time = 0.22 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x)^5}{x^3} \, dx=\frac {2 \, b^{5} x^{5} + 15 \, a b^{4} x^{4} + 60 \, a^{2} b^{3} x^{3} + 60 \, a^{3} b^{2} x^{2} \log \left (x\right ) - 30 \, a^{4} b x - 3 \, a^{5}}{6 \, x^{2}} \]
[In]
[Out]
Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^5}{x^3} \, dx=10 a^{3} b^{2} \log {\left (x \right )} + 10 a^{2} b^{3} x + \frac {5 a b^{4} x^{2}}{2} + \frac {b^{5} x^{3}}{3} + \frac {- a^{5} - 10 a^{4} b x}{2 x^{2}} \]
[In]
[Out]
none
Time = 0.21 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b x)^5}{x^3} \, dx=\frac {1}{3} \, b^{5} x^{3} + \frac {5}{2} \, a b^{4} x^{2} + 10 \, a^{2} b^{3} x + 10 \, a^{3} b^{2} \log \left (x\right ) - \frac {10 \, a^{4} b x + a^{5}}{2 \, x^{2}} \]
[In]
[Out]
none
Time = 0.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b x)^5}{x^3} \, dx=\frac {1}{3} \, b^{5} x^{3} + \frac {5}{2} \, a b^{4} x^{2} + 10 \, a^{2} b^{3} x + 10 \, a^{3} b^{2} \log \left ({\left | x \right |}\right ) - \frac {10 \, a^{4} b x + a^{5}}{2 \, x^{2}} \]
[In]
[Out]
Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b x)^5}{x^3} \, dx=\frac {b^5\,x^3}{3}-\frac {\frac {a^5}{2}+5\,b\,x\,a^4}{x^2}+10\,a^2\,b^3\,x+\frac {5\,a\,b^4\,x^2}{2}+10\,a^3\,b^2\,\ln \left (x\right ) \]
[In]
[Out]