Integrand size = 29, antiderivative size = 188 \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {3 a B}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^4}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 B}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a^2 B}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]
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Time = 0.07 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {784, 79, 45} \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {x^4 (A b-a B)}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 a B}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a^2 B}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 B}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]
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Rule 45
Rule 79
Rule 784
Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {x^3 (A+B x)}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {(A b-a B) x^4}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (b^2 B \left (a b+b^2 x\right )\right ) \int \frac {x^3}{\left (a b+b^2 x\right )^4} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {(A b-a B) x^4}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (b^2 B \left (a b+b^2 x\right )\right ) \int \left (-\frac {a^3}{b^7 (a+b x)^4}+\frac {3 a^2}{b^7 (a+b x)^3}-\frac {3 a}{b^7 (a+b x)^2}+\frac {1}{b^7 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {3 a B}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^4}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 B}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a^2 B}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}
Time = 1.04 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.55 \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {25 a^4 B-12 A b^4 x^3-12 a^2 b^2 x (A-9 B x)+6 a b^3 x^2 (-3 A+8 B x)+a^3 (-3 A b+88 b B x)+12 B (a+b x)^4 \log (a+b x)}{12 b^5 (a+b x)^3 \sqrt {(a+b x)^2}} \]
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Time = 0.34 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.61
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {\left (A b -4 B a \right ) x^{3}}{b^{2}}-\frac {3 a \left (A b -6 B a \right ) x^{2}}{2 b^{3}}-\frac {a^{2} \left (3 A b -22 B a \right ) x}{3 b^{4}}-\frac {a^{3} \left (3 A b -25 B a \right )}{12 b^{5}}\right )}{\left (b x +a \right )^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, B \ln \left (b x +a \right )}{\left (b x +a \right ) b^{5}}\) | \(115\) |
default | \(-\frac {\left (-12 B \ln \left (b x +a \right ) b^{4} x^{4}-48 B \ln \left (b x +a \right ) x^{3} a \,b^{3}+12 A \,b^{4} x^{3}-72 B \ln \left (b x +a \right ) x^{2} a^{2} b^{2}-48 B a \,b^{3} x^{3}+18 A a \,b^{3} x^{2}-48 B \ln \left (b x +a \right ) x \,a^{3} b -108 B \,a^{2} b^{2} x^{2}+12 A \,a^{2} b^{2} x -12 B \ln \left (b x +a \right ) a^{4}-88 B \,a^{3} b x +3 A \,a^{3} b -25 B \,a^{4}\right ) \left (b x +a \right )}{12 b^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(168\) |
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Time = 0.43 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.93 \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {25 \, B a^{4} - 3 \, A a^{3} b + 12 \, {\left (4 \, B a b^{3} - A b^{4}\right )} x^{3} + 18 \, {\left (6 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 4 \, {\left (22 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x + 12 \, {\left (B b^{4} x^{4} + 4 \, B a b^{3} x^{3} + 6 \, B a^{2} b^{2} x^{2} + 4 \, B a^{3} b x + B a^{4}\right )} \log \left (b x + a\right )}{12 \, {\left (b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}\right )}} \]
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\[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x^{3} \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]
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Time = 0.26 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.07 \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {1}{12} \, B {\left (\frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4}}{b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}} + \frac {12 \, \log \left (b x + a\right )}{b^{5}}\right )} - \frac {1}{12} \, A {\left (\frac {12 \, x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} + \frac {6 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{3}}{b^{8} {\left (x + \frac {a}{b}\right )}^{4}}\right )} \]
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Time = 0.29 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.62 \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {B \log \left ({\left | b x + a \right |}\right )}{b^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {12 \, {\left (4 \, B a b^{2} - A b^{3}\right )} x^{3} + 18 \, {\left (6 \, B a^{2} b - A a b^{2}\right )} x^{2} + 4 \, {\left (22 \, B a^{3} - 3 \, A a^{2} b\right )} x + \frac {25 \, B a^{4} - 3 \, A a^{3} b}{b}}{12 \, {\left (b x + a\right )}^{4} b^{4} \mathrm {sgn}\left (b x + a\right )} \]
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Timed out. \[ \int \frac {x^3 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x^3\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]
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