Integrand size = 18, antiderivative size = 171 \[ \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx=-\frac {35 b^2 (3 A b-2 a B)}{24 a^4 (a+b x)^{3/2}}-\frac {A}{3 a x^3 (a+b x)^{3/2}}+\frac {3 A b-2 a B}{4 a^2 x^2 (a+b x)^{3/2}}-\frac {7 b (3 A b-2 a B)}{8 a^3 x (a+b x)^{3/2}}-\frac {35 b^2 (3 A b-2 a B)}{8 a^5 \sqrt {a+b x}}+\frac {35 b^2 (3 A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{11/2}} \]
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Time = 0.05 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {79, 44, 53, 65, 214} \[ \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx=\frac {35 b^2 (3 A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{11/2}}-\frac {35 b^2 (3 A b-2 a B)}{8 a^5 \sqrt {a+b x}}-\frac {35 b^2 (3 A b-2 a B)}{24 a^4 (a+b x)^{3/2}}-\frac {7 b (3 A b-2 a B)}{8 a^3 x (a+b x)^{3/2}}+\frac {3 A b-2 a B}{4 a^2 x^2 (a+b x)^{3/2}}-\frac {A}{3 a x^3 (a+b x)^{3/2}} \]
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Rule 44
Rule 53
Rule 65
Rule 79
Rule 214
Rubi steps \begin{align*} \text {integral}& = -\frac {A}{3 a x^3 (a+b x)^{3/2}}+\frac {\left (-\frac {9 A b}{2}+3 a B\right ) \int \frac {1}{x^3 (a+b x)^{5/2}} \, dx}{3 a} \\ & = -\frac {A}{3 a x^3 (a+b x)^{3/2}}+\frac {3 A b-2 a B}{4 a^2 x^2 (a+b x)^{3/2}}+\frac {(7 b (3 A b-2 a B)) \int \frac {1}{x^2 (a+b x)^{5/2}} \, dx}{8 a^2} \\ & = -\frac {A}{3 a x^3 (a+b x)^{3/2}}+\frac {3 A b-2 a B}{4 a^2 x^2 (a+b x)^{3/2}}-\frac {7 b (3 A b-2 a B)}{8 a^3 x (a+b x)^{3/2}}-\frac {\left (35 b^2 (3 A b-2 a B)\right ) \int \frac {1}{x (a+b x)^{5/2}} \, dx}{16 a^3} \\ & = -\frac {35 b^2 (3 A b-2 a B)}{24 a^4 (a+b x)^{3/2}}-\frac {A}{3 a x^3 (a+b x)^{3/2}}+\frac {3 A b-2 a B}{4 a^2 x^2 (a+b x)^{3/2}}-\frac {7 b (3 A b-2 a B)}{8 a^3 x (a+b x)^{3/2}}-\frac {\left (35 b^2 (3 A b-2 a B)\right ) \int \frac {1}{x (a+b x)^{3/2}} \, dx}{16 a^4} \\ & = -\frac {35 b^2 (3 A b-2 a B)}{24 a^4 (a+b x)^{3/2}}-\frac {A}{3 a x^3 (a+b x)^{3/2}}+\frac {3 A b-2 a B}{4 a^2 x^2 (a+b x)^{3/2}}-\frac {7 b (3 A b-2 a B)}{8 a^3 x (a+b x)^{3/2}}-\frac {35 b^2 (3 A b-2 a B)}{8 a^5 \sqrt {a+b x}}-\frac {\left (35 b^2 (3 A b-2 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{16 a^5} \\ & = -\frac {35 b^2 (3 A b-2 a B)}{24 a^4 (a+b x)^{3/2}}-\frac {A}{3 a x^3 (a+b x)^{3/2}}+\frac {3 A b-2 a B}{4 a^2 x^2 (a+b x)^{3/2}}-\frac {7 b (3 A b-2 a B)}{8 a^3 x (a+b x)^{3/2}}-\frac {35 b^2 (3 A b-2 a B)}{8 a^5 \sqrt {a+b x}}-\frac {(35 b (3 A b-2 a B)) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{8 a^5} \\ & = -\frac {35 b^2 (3 A b-2 a B)}{24 a^4 (a+b x)^{3/2}}-\frac {A}{3 a x^3 (a+b x)^{3/2}}+\frac {3 A b-2 a B}{4 a^2 x^2 (a+b x)^{3/2}}-\frac {7 b (3 A b-2 a B)}{8 a^3 x (a+b x)^{3/2}}-\frac {35 b^2 (3 A b-2 a B)}{8 a^5 \sqrt {a+b x}}+\frac {35 b^2 (3 A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{11/2}} \\ \end{align*}
Time = 0.26 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.76 \[ \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx=\frac {-315 A b^4 x^4+210 a b^3 x^3 (-2 A+B x)-4 a^4 (2 A+3 B x)+6 a^3 b x (3 A+7 B x)+7 a^2 b^2 x^2 (-9 A+40 B x)}{24 a^5 x^3 (a+b x)^{3/2}}+\frac {35 b^2 (3 A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{11/2}} \]
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Time = 0.57 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.71
method | result | size |
pseudoelliptic | \(\frac {\frac {105 x^{3} \left (b x +a \right )^{\frac {3}{2}} b^{2} \left (A b -\frac {2 B a}{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8}-\frac {35 x^{3} \left (-\frac {B x}{2}+A \right ) b^{3} a^{\frac {3}{2}}}{2}-\frac {21 x^{2} \left (-\frac {40 B x}{9}+A \right ) b^{2} a^{\frac {5}{2}}}{8}+\frac {3 x \left (\frac {7 B x}{3}+A \right ) b \,a^{\frac {7}{2}}}{4}+\frac {3 \left (-\frac {2 B x}{3}-\frac {4 A}{9}\right ) a^{\frac {9}{2}}}{4}-\frac {105 A \sqrt {a}\, b^{4} x^{4}}{8}}{a^{\frac {11}{2}} \left (b x +a \right )^{\frac {3}{2}} x^{3}}\) | \(122\) |
risch | \(-\frac {\sqrt {b x +a}\, \left (123 A \,b^{2} x^{2}-66 B a b \,x^{2}-34 a A b x +12 a^{2} B x +8 a^{2} A \right )}{24 a^{5} x^{3}}-\frac {b^{2} \left (-\frac {2 \left (-64 A b +48 B a \right )}{\sqrt {b x +a}}+\frac {32 a \left (A b -B a \right )}{3 \left (b x +a \right )^{\frac {3}{2}}}-\frac {2 \left (105 A b -70 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right )}{16 a^{5}}\) | \(125\) |
derivativedivides | \(2 b^{2} \left (\frac {-\frac {\left (\frac {41 A b}{16}-\frac {11 B a}{8}\right ) \left (b x +a \right )^{\frac {5}{2}}+\left (-\frac {35}{6} a b A +3 a^{2} B \right ) \left (b x +a \right )^{\frac {3}{2}}+\left (\frac {55}{16} a^{2} b A -\frac {13}{8} a^{3} B \right ) \sqrt {b x +a}}{b^{3} x^{3}}+\frac {35 \left (3 A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}}{a^{5}}-\frac {A b -B a}{3 a^{4} \left (b x +a \right )^{\frac {3}{2}}}-\frac {4 A b -3 B a}{a^{5} \sqrt {b x +a}}\right )\) | \(147\) |
default | \(2 b^{2} \left (\frac {-\frac {\left (\frac {41 A b}{16}-\frac {11 B a}{8}\right ) \left (b x +a \right )^{\frac {5}{2}}+\left (-\frac {35}{6} a b A +3 a^{2} B \right ) \left (b x +a \right )^{\frac {3}{2}}+\left (\frac {55}{16} a^{2} b A -\frac {13}{8} a^{3} B \right ) \sqrt {b x +a}}{b^{3} x^{3}}+\frac {35 \left (3 A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}}{a^{5}}-\frac {A b -B a}{3 a^{4} \left (b x +a \right )^{\frac {3}{2}}}-\frac {4 A b -3 B a}{a^{5} \sqrt {b x +a}}\right )\) | \(147\) |
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Time = 0.25 (sec) , antiderivative size = 448, normalized size of antiderivative = 2.62 \[ \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx=\left [-\frac {105 \, {\left ({\left (2 \, B a b^{4} - 3 \, A b^{5}\right )} x^{5} + 2 \, {\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} + {\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3}\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (8 \, A a^{5} - 105 \, {\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} - 140 \, {\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3} - 21 \, {\left (2 \, B a^{4} b - 3 \, A a^{3} b^{2}\right )} x^{2} + 6 \, {\left (2 \, B a^{5} - 3 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{48 \, {\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} x^{3}\right )}}, \frac {105 \, {\left ({\left (2 \, B a b^{4} - 3 \, A b^{5}\right )} x^{5} + 2 \, {\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} + {\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (8 \, A a^{5} - 105 \, {\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} - 140 \, {\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3} - 21 \, {\left (2 \, B a^{4} b - 3 \, A a^{3} b^{2}\right )} x^{2} + 6 \, {\left (2 \, B a^{5} - 3 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{24 \, {\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} x^{3}\right )}}\right ] \]
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Timed out. \[ \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx=\text {Timed out} \]
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Time = 0.36 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.19 \[ \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx=-\frac {1}{48} \, b^{3} {\left (\frac {2 \, {\left (16 \, B a^{5} - 16 \, A a^{4} b - 105 \, {\left (2 \, B a - 3 \, A b\right )} {\left (b x + a\right )}^{4} + 280 \, {\left (2 \, B a^{2} - 3 \, A a b\right )} {\left (b x + a\right )}^{3} - 231 \, {\left (2 \, B a^{3} - 3 \, A a^{2} b\right )} {\left (b x + a\right )}^{2} + 48 \, {\left (2 \, B a^{4} - 3 \, A a^{3} b\right )} {\left (b x + a\right )}\right )}}{{\left (b x + a\right )}^{\frac {9}{2}} a^{5} b - 3 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{6} b + 3 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{7} b - {\left (b x + a\right )}^{\frac {3}{2}} a^{8} b} - \frac {105 \, {\left (2 \, B a - 3 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {11}{2}} b}\right )} \]
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Time = 0.29 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.17 \[ \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx=\frac {35 \, {\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{5}} + \frac {210 \, {\left (b x + a\right )}^{4} B a b^{2} - 560 \, {\left (b x + a\right )}^{3} B a^{2} b^{2} + 462 \, {\left (b x + a\right )}^{2} B a^{3} b^{2} - 96 \, {\left (b x + a\right )} B a^{4} b^{2} - 16 \, B a^{5} b^{2} - 315 \, {\left (b x + a\right )}^{4} A b^{3} + 840 \, {\left (b x + a\right )}^{3} A a b^{3} - 693 \, {\left (b x + a\right )}^{2} A a^{2} b^{3} + 144 \, {\left (b x + a\right )} A a^{3} b^{3} + 16 \, A a^{4} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - \sqrt {b x + a} a\right )}^{3} a^{5}} \]
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Time = 0.14 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx=\frac {35\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (3\,A\,b-2\,B\,a\right )}{8\,a^{11/2}}-\frac {\frac {2\,\left (A\,b^3-B\,a\,b^2\right )}{3\,a}+\frac {2\,\left (3\,A\,b^3-2\,B\,a\,b^2\right )\,\left (a+b\,x\right )}{a^2}-\frac {77\,\left (3\,A\,b^3-2\,B\,a\,b^2\right )\,{\left (a+b\,x\right )}^2}{8\,a^3}+\frac {35\,\left (3\,A\,b^3-2\,B\,a\,b^2\right )\,{\left (a+b\,x\right )}^3}{3\,a^4}-\frac {35\,\left (3\,A\,b^3-2\,B\,a\,b^2\right )\,{\left (a+b\,x\right )}^4}{8\,a^5}}{3\,a\,{\left (a+b\,x\right )}^{7/2}-{\left (a+b\,x\right )}^{9/2}+a^3\,{\left (a+b\,x\right )}^{3/2}-3\,a^2\,{\left (a+b\,x\right )}^{5/2}} \]
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