Integrand size = 20, antiderivative size = 149 \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )^3} \, dx=-\frac {A}{6 a^3 x^6}+\frac {3 A b-a B}{4 a^4 x^4}-\frac {3 b (2 A b-a B)}{2 a^5 x^2}-\frac {b^2 (A b-a B)}{4 a^4 \left (a+b x^2\right )^2}-\frac {b^2 (4 A b-3 a B)}{2 a^5 \left (a+b x^2\right )}-\frac {2 b^2 (5 A b-3 a B) \log (x)}{a^6}+\frac {b^2 (5 A b-3 a B) \log \left (a+b x^2\right )}{a^6} \]
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Time = 0.12 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {457, 78} \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )^3} \, dx=\frac {b^2 (5 A b-3 a B) \log \left (a+b x^2\right )}{a^6}-\frac {2 b^2 \log (x) (5 A b-3 a B)}{a^6}-\frac {b^2 (4 A b-3 a B)}{2 a^5 \left (a+b x^2\right )}-\frac {3 b (2 A b-a B)}{2 a^5 x^2}-\frac {b^2 (A b-a B)}{4 a^4 \left (a+b x^2\right )^2}+\frac {3 A b-a B}{4 a^4 x^4}-\frac {A}{6 a^3 x^6} \]
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Rule 78
Rule 457
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {A+B x}{x^4 (a+b x)^3} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {A}{a^3 x^4}+\frac {-3 A b+a B}{a^4 x^3}-\frac {3 b (-2 A b+a B)}{a^5 x^2}+\frac {2 b^2 (-5 A b+3 a B)}{a^6 x}-\frac {b^3 (-A b+a B)}{a^4 (a+b x)^3}-\frac {b^3 (-4 A b+3 a B)}{a^5 (a+b x)^2}-\frac {2 b^3 (-5 A b+3 a B)}{a^6 (a+b x)}\right ) \, dx,x,x^2\right ) \\ & = -\frac {A}{6 a^3 x^6}+\frac {3 A b-a B}{4 a^4 x^4}-\frac {3 b (2 A b-a B)}{2 a^5 x^2}-\frac {b^2 (A b-a B)}{4 a^4 \left (a+b x^2\right )^2}-\frac {b^2 (4 A b-3 a B)}{2 a^5 \left (a+b x^2\right )}-\frac {2 b^2 (5 A b-3 a B) \log (x)}{a^6}+\frac {b^2 (5 A b-3 a B) \log \left (a+b x^2\right )}{a^6} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )^3} \, dx=\frac {-\frac {2 a^3 A}{x^6}-\frac {3 a^2 (-3 A b+a B)}{x^4}+\frac {18 a b (-2 A b+a B)}{x^2}+\frac {3 a^2 b^2 (-A b+a B)}{\left (a+b x^2\right )^2}+\frac {6 a b^2 (-4 A b+3 a B)}{a+b x^2}+24 b^2 (-5 A b+3 a B) \log (x)+12 b^2 (5 A b-3 a B) \log \left (a+b x^2\right )}{12 a^6} \]
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Time = 2.50 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.96
method | result | size |
default | \(-\frac {A}{6 a^{3} x^{6}}-\frac {-3 A b +B a}{4 a^{4} x^{4}}-\frac {3 b \left (2 A b -B a \right )}{2 a^{5} x^{2}}-\frac {2 b^{2} \left (5 A b -3 B a \right ) \ln \left (x \right )}{a^{6}}+\frac {b^{3} \left (\frac {\left (10 A b -6 B a \right ) \ln \left (b \,x^{2}+a \right )}{b}-\frac {a^{2} \left (A b -B a \right )}{2 b \left (b \,x^{2}+a \right )^{2}}-\frac {a \left (4 A b -3 B a \right )}{b \left (b \,x^{2}+a \right )}\right )}{2 a^{6}}\) | \(143\) |
norman | \(\frac {-\frac {A}{6 a}+\frac {\left (5 A b -3 B a \right ) x^{2}}{12 a^{2}}-\frac {b \left (5 A b -3 B a \right ) x^{4}}{3 a^{3}}+\frac {2 b \left (5 b^{3} A -3 a \,b^{2} B \right ) x^{8}}{a^{5}}+\frac {b^{2} \left (15 b^{3} A -9 a \,b^{2} B \right ) x^{10}}{2 a^{6}}}{x^{6} \left (b \,x^{2}+a \right )^{2}}+\frac {b^{2} \left (5 A b -3 B a \right ) \ln \left (b \,x^{2}+a \right )}{a^{6}}-\frac {2 b^{2} \left (5 A b -3 B a \right ) \ln \left (x \right )}{a^{6}}\) | \(148\) |
risch | \(\frac {-\frac {b^{3} \left (5 A b -3 B a \right ) x^{8}}{a^{5}}-\frac {3 b^{2} \left (5 A b -3 B a \right ) x^{6}}{2 a^{4}}-\frac {b \left (5 A b -3 B a \right ) x^{4}}{3 a^{3}}+\frac {\left (5 A b -3 B a \right ) x^{2}}{12 a^{2}}-\frac {A}{6 a}}{x^{6} \left (b \,x^{2}+a \right )^{2}}-\frac {10 b^{3} \ln \left (x \right ) A}{a^{6}}+\frac {6 b^{2} \ln \left (x \right ) B}{a^{5}}+\frac {5 b^{3} \ln \left (-b \,x^{2}-a \right ) A}{a^{6}}-\frac {3 b^{2} \ln \left (-b \,x^{2}-a \right ) B}{a^{5}}\) | \(159\) |
parallelrisch | \(-\frac {120 A \ln \left (x \right ) x^{10} b^{5}-60 A \ln \left (b \,x^{2}+a \right ) x^{10} b^{5}-72 B \ln \left (x \right ) x^{10} a \,b^{4}+36 B \ln \left (b \,x^{2}+a \right ) x^{10} a \,b^{4}-90 A \,b^{5} x^{10}+54 B a \,b^{4} x^{10}+240 A \ln \left (x \right ) x^{8} a \,b^{4}-120 A \ln \left (b \,x^{2}+a \right ) x^{8} a \,b^{4}-144 B \ln \left (x \right ) x^{8} a^{2} b^{3}+72 B \ln \left (b \,x^{2}+a \right ) x^{8} a^{2} b^{3}-120 a A \,b^{4} x^{8}+72 B \,a^{2} b^{3} x^{8}+120 A \ln \left (x \right ) x^{6} a^{2} b^{3}-60 A \ln \left (b \,x^{2}+a \right ) x^{6} a^{2} b^{3}-72 B \ln \left (x \right ) x^{6} a^{3} b^{2}+36 B \ln \left (b \,x^{2}+a \right ) x^{6} a^{3} b^{2}+20 a^{3} A \,b^{2} x^{4}-12 B \,a^{4} b \,x^{4}-5 a^{4} A b \,x^{2}+3 a^{5} B \,x^{2}+2 a^{5} A}{12 a^{6} x^{6} \left (b \,x^{2}+a \right )^{2}}\) | \(297\) |
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Time = 0.28 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.79 \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )^3} \, dx=\frac {12 \, {\left (3 \, B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{8} + 18 \, {\left (3 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{6} - 2 \, A a^{5} + 4 \, {\left (3 \, B a^{4} b - 5 \, A a^{3} b^{2}\right )} x^{4} - {\left (3 \, B a^{5} - 5 \, A a^{4} b\right )} x^{2} - 12 \, {\left ({\left (3 \, B a b^{4} - 5 \, A b^{5}\right )} x^{10} + 2 \, {\left (3 \, B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{8} + {\left (3 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{6}\right )} \log \left (b x^{2} + a\right ) + 24 \, {\left ({\left (3 \, B a b^{4} - 5 \, A b^{5}\right )} x^{10} + 2 \, {\left (3 \, B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{8} + {\left (3 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{6}\right )} \log \left (x\right )}{12 \, {\left (a^{6} b^{2} x^{10} + 2 \, a^{7} b x^{8} + a^{8} x^{6}\right )}} \]
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Time = 0.75 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.11 \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )^3} \, dx=\frac {- 2 A a^{4} + x^{8} \left (- 60 A b^{4} + 36 B a b^{3}\right ) + x^{6} \left (- 90 A a b^{3} + 54 B a^{2} b^{2}\right ) + x^{4} \left (- 20 A a^{2} b^{2} + 12 B a^{3} b\right ) + x^{2} \cdot \left (5 A a^{3} b - 3 B a^{4}\right )}{12 a^{7} x^{6} + 24 a^{6} b x^{8} + 12 a^{5} b^{2} x^{10}} + \frac {2 b^{2} \left (- 5 A b + 3 B a\right ) \log {\left (x \right )}}{a^{6}} - \frac {b^{2} \left (- 5 A b + 3 B a\right ) \log {\left (\frac {a}{b} + x^{2} \right )}}{a^{6}} \]
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Time = 0.20 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.14 \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )^3} \, dx=\frac {12 \, {\left (3 \, B a b^{3} - 5 \, A b^{4}\right )} x^{8} + 18 \, {\left (3 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{6} - 2 \, A a^{4} + 4 \, {\left (3 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{4} - {\left (3 \, B a^{4} - 5 \, A a^{3} b\right )} x^{2}}{12 \, {\left (a^{5} b^{2} x^{10} + 2 \, a^{6} b x^{8} + a^{7} x^{6}\right )}} - \frac {{\left (3 \, B a b^{2} - 5 \, A b^{3}\right )} \log \left (b x^{2} + a\right )}{a^{6}} + \frac {{\left (3 \, B a b^{2} - 5 \, A b^{3}\right )} \log \left (x^{2}\right )}{a^{6}} \]
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Time = 0.30 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.35 \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )^3} \, dx=\frac {{\left (3 \, B a b^{2} - 5 \, A b^{3}\right )} \log \left (x^{2}\right )}{a^{6}} - \frac {{\left (3 \, B a b^{3} - 5 \, A b^{4}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{a^{6} b} + \frac {18 \, B a b^{4} x^{4} - 30 \, A b^{5} x^{4} + 42 \, B a^{2} b^{3} x^{2} - 68 \, A a b^{4} x^{2} + 25 \, B a^{3} b^{2} - 39 \, A a^{2} b^{3}}{4 \, {\left (b x^{2} + a\right )}^{2} a^{6}} - \frac {66 \, B a b^{2} x^{6} - 110 \, A b^{3} x^{6} - 18 \, B a^{2} b x^{4} + 36 \, A a b^{2} x^{4} + 3 \, B a^{3} x^{2} - 9 \, A a^{2} b x^{2} + 2 \, A a^{3}}{12 \, a^{6} x^{6}} \]
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Time = 4.99 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.04 \[ \int \frac {A+B x^2}{x^7 \left (a+b x^2\right )^3} \, dx=\frac {\ln \left (b\,x^2+a\right )\,\left (5\,A\,b^3-3\,B\,a\,b^2\right )}{a^6}-\frac {\frac {A}{6\,a}-\frac {x^2\,\left (5\,A\,b-3\,B\,a\right )}{12\,a^2}+\frac {3\,b^2\,x^6\,\left (5\,A\,b-3\,B\,a\right )}{2\,a^4}+\frac {b^3\,x^8\,\left (5\,A\,b-3\,B\,a\right )}{a^5}+\frac {b\,x^4\,\left (5\,A\,b-3\,B\,a\right )}{3\,a^3}}{a^2\,x^6+2\,a\,b\,x^8+b^2\,x^{10}}-\frac {\ln \left (x\right )\,\left (10\,A\,b^3-6\,B\,a\,b^2\right )}{a^6} \]
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