Integrand size = 18, antiderivative size = 177 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx=\frac {b^2 (3 A b-10 a B) \sqrt {a+b x}}{64 a x^2}+\frac {b^3 (3 A b-10 a B) \sqrt {a+b x}}{128 a^2 x}+\frac {b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac {(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac {A (a+b x)^{7/2}}{5 a x^5}-\frac {b^4 (3 A b-10 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{5/2}} \]
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Time = 0.05 (sec) , antiderivative size = 177, 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, 43, 44, 65, 214} \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx=-\frac {b^4 (3 A b-10 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{5/2}}+\frac {b^3 \sqrt {a+b x} (3 A b-10 a B)}{128 a^2 x}+\frac {b^2 \sqrt {a+b x} (3 A b-10 a B)}{64 a x^2}+\frac {(a+b x)^{5/2} (3 A b-10 a B)}{40 a x^4}+\frac {b (a+b x)^{3/2} (3 A b-10 a B)}{48 a x^3}-\frac {A (a+b x)^{7/2}}{5 a x^5} \]
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Rule 43
Rule 44
Rule 65
Rule 79
Rule 214
Rubi steps \begin{align*} \text {integral}& = -\frac {A (a+b x)^{7/2}}{5 a x^5}+\frac {\left (-\frac {3 A b}{2}+5 a B\right ) \int \frac {(a+b x)^{5/2}}{x^5} \, dx}{5 a} \\ & = \frac {(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac {A (a+b x)^{7/2}}{5 a x^5}-\frac {(b (3 A b-10 a B)) \int \frac {(a+b x)^{3/2}}{x^4} \, dx}{16 a} \\ & = \frac {b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac {(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac {A (a+b x)^{7/2}}{5 a x^5}-\frac {\left (b^2 (3 A b-10 a B)\right ) \int \frac {\sqrt {a+b x}}{x^3} \, dx}{32 a} \\ & = \frac {b^2 (3 A b-10 a B) \sqrt {a+b x}}{64 a x^2}+\frac {b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac {(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac {A (a+b x)^{7/2}}{5 a x^5}-\frac {\left (b^3 (3 A b-10 a B)\right ) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{128 a} \\ & = \frac {b^2 (3 A b-10 a B) \sqrt {a+b x}}{64 a x^2}+\frac {b^3 (3 A b-10 a B) \sqrt {a+b x}}{128 a^2 x}+\frac {b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac {(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac {A (a+b x)^{7/2}}{5 a x^5}+\frac {\left (b^4 (3 A b-10 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{256 a^2} \\ & = \frac {b^2 (3 A b-10 a B) \sqrt {a+b x}}{64 a x^2}+\frac {b^3 (3 A b-10 a B) \sqrt {a+b x}}{128 a^2 x}+\frac {b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac {(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac {A (a+b x)^{7/2}}{5 a x^5}+\frac {\left (b^3 (3 A b-10 a B)\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{128 a^2} \\ & = \frac {b^2 (3 A b-10 a B) \sqrt {a+b x}}{64 a x^2}+\frac {b^3 (3 A b-10 a B) \sqrt {a+b x}}{128 a^2 x}+\frac {b (3 A b-10 a B) (a+b x)^{3/2}}{48 a x^3}+\frac {(3 A b-10 a B) (a+b x)^{5/2}}{40 a x^4}-\frac {A (a+b x)^{7/2}}{5 a x^5}-\frac {b^4 (3 A b-10 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{5/2}} \\ \end{align*}
Time = 0.31 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.73 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx=-\frac {\sqrt {a+b x} \left (-45 A b^4 x^4+30 a b^3 x^3 (A+5 B x)+96 a^4 (4 A+5 B x)+16 a^3 b x (63 A+85 B x)+4 a^2 b^2 x^2 (186 A+295 B x)\right )}{1920 a^2 x^5}+\frac {b^4 (-3 A b+10 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{128 a^{5/2}} \]
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Time = 0.55 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(-\frac {31 \left (\frac {15 x^{5} \left (A b -\frac {10 B a}{3}\right ) b^{4} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{248}+\sqrt {b x +a}\, \left (\frac {5 b^{3} x^{3} \left (5 B x +A \right ) a^{\frac {3}{2}}}{124}+b^{2} x^{2} \left (\frac {295 B x}{186}+A \right ) a^{\frac {5}{2}}+\frac {42 \left (\frac {85 B x}{63}+A \right ) x b \,a^{\frac {7}{2}}}{31}+\frac {4 \left (5 B x +4 A \right ) a^{\frac {9}{2}}}{31}-\frac {15 A \sqrt {a}\, b^{4} x^{4}}{248}\right )\right )}{80 a^{\frac {5}{2}} x^{5}}\) | \(118\) |
risch | \(-\frac {\sqrt {b x +a}\, \left (-45 A \,b^{4} x^{4}+150 B a \,b^{3} x^{4}+30 A a \,b^{3} x^{3}+1180 B \,a^{2} b^{2} x^{3}+744 A \,a^{2} b^{2} x^{2}+1360 B \,a^{3} b \,x^{2}+1008 A \,a^{3} b x +480 B \,a^{4} x +384 A \,a^{4}\right )}{1920 x^{5} a^{2}}-\frac {b^{4} \left (3 A b -10 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {5}{2}}}\) | \(131\) |
derivativedivides | \(2 b^{4} \left (-\frac {-\frac {\left (3 A b -10 B a \right ) \left (b x +a \right )^{\frac {9}{2}}}{256 a^{2}}+\frac {\left (21 A b +58 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{384 a}+\left (\frac {A b}{10}-\frac {B a}{3}\right ) \left (b x +a \right )^{\frac {5}{2}}-\frac {7 a \left (3 A b -10 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{384}+\frac {a^{2} \left (3 A b -10 B a \right ) \sqrt {b x +a}}{256}}{b^{5} x^{5}}-\frac {\left (3 A b -10 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{256 a^{\frac {5}{2}}}\right )\) | \(141\) |
default | \(2 b^{4} \left (-\frac {-\frac {\left (3 A b -10 B a \right ) \left (b x +a \right )^{\frac {9}{2}}}{256 a^{2}}+\frac {\left (21 A b +58 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{384 a}+\left (\frac {A b}{10}-\frac {B a}{3}\right ) \left (b x +a \right )^{\frac {5}{2}}-\frac {7 a \left (3 A b -10 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{384}+\frac {a^{2} \left (3 A b -10 B a \right ) \sqrt {b x +a}}{256}}{b^{5} x^{5}}-\frac {\left (3 A b -10 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{256 a^{\frac {5}{2}}}\right )\) | \(141\) |
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Time = 0.24 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.73 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx=\left [-\frac {15 \, {\left (10 \, B a b^{4} - 3 \, A b^{5}\right )} \sqrt {a} x^{5} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (384 \, A a^{5} + 15 \, {\left (10 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} + 10 \, {\left (118 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{3} + 8 \, {\left (170 \, B a^{4} b + 93 \, A a^{3} b^{2}\right )} x^{2} + 48 \, {\left (10 \, B a^{5} + 21 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{3840 \, a^{3} x^{5}}, -\frac {15 \, {\left (10 \, B a b^{4} - 3 \, A b^{5}\right )} \sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (384 \, A a^{5} + 15 \, {\left (10 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} + 10 \, {\left (118 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{3} + 8 \, {\left (170 \, B a^{4} b + 93 \, A a^{3} b^{2}\right )} x^{2} + 48 \, {\left (10 \, B a^{5} + 21 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{1920 \, a^{3} x^{5}}\right ] \]
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Timed out. \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx=\text {Timed out} \]
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Time = 0.29 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx=-\frac {1}{3840} \, b^{5} {\left (\frac {2 \, {\left (15 \, {\left (10 \, B a - 3 \, A b\right )} {\left (b x + a\right )}^{\frac {9}{2}} + 10 \, {\left (58 \, B a^{2} + 21 \, A a b\right )} {\left (b x + a\right )}^{\frac {7}{2}} - 128 \, {\left (10 \, B a^{3} - 3 \, A a^{2} b\right )} {\left (b x + a\right )}^{\frac {5}{2}} + 70 \, {\left (10 \, B a^{4} - 3 \, A a^{3} b\right )} {\left (b x + a\right )}^{\frac {3}{2}} - 15 \, {\left (10 \, B a^{5} - 3 \, A a^{4} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{5} a^{2} b - 5 \, {\left (b x + a\right )}^{4} a^{3} b + 10 \, {\left (b x + a\right )}^{3} a^{4} b - 10 \, {\left (b x + a\right )}^{2} a^{5} b + 5 \, {\left (b x + a\right )} a^{6} b - a^{7} b} + \frac {15 \, {\left (10 \, B a - 3 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}} b}\right )} \]
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Time = 0.30 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.18 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx=-\frac {\frac {15 \, {\left (10 \, B a b^{5} - 3 \, A b^{6}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {150 \, {\left (b x + a\right )}^{\frac {9}{2}} B a b^{5} + 580 \, {\left (b x + a\right )}^{\frac {7}{2}} B a^{2} b^{5} - 1280 \, {\left (b x + a\right )}^{\frac {5}{2}} B a^{3} b^{5} + 700 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{4} b^{5} - 150 \, \sqrt {b x + a} B a^{5} b^{5} - 45 \, {\left (b x + a\right )}^{\frac {9}{2}} A b^{6} + 210 \, {\left (b x + a\right )}^{\frac {7}{2}} A a b^{6} + 384 \, {\left (b x + a\right )}^{\frac {5}{2}} A a^{2} b^{6} - 210 \, {\left (b x + a\right )}^{\frac {3}{2}} A a^{3} b^{6} + 45 \, \sqrt {b x + a} A a^{4} b^{6}}{a^{2} b^{5} x^{5}}}{1920 \, b} \]
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Time = 0.21 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx=\frac {\left (\frac {A\,b^5}{5}-\frac {2\,B\,a\,b^4}{3}\right )\,{\left (a+b\,x\right )}^{5/2}+\left (\frac {3\,A\,a^2\,b^5}{128}-\frac {5\,B\,a^3\,b^4}{64}\right )\,\sqrt {a+b\,x}+\left (\frac {35\,B\,a^2\,b^4}{96}-\frac {7\,A\,a\,b^5}{64}\right )\,{\left (a+b\,x\right )}^{3/2}-\frac {\left (3\,A\,b^5-10\,B\,a\,b^4\right )\,{\left (a+b\,x\right )}^{9/2}}{128\,a^2}+\frac {\left (21\,A\,b^5+58\,B\,a\,b^4\right )\,{\left (a+b\,x\right )}^{7/2}}{192\,a}}{5\,a\,{\left (a+b\,x\right )}^4-5\,a^4\,\left (a+b\,x\right )-{\left (a+b\,x\right )}^5-10\,a^2\,{\left (a+b\,x\right )}^3+10\,a^3\,{\left (a+b\,x\right )}^2+a^5}-\frac {b^4\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (3\,A\,b-10\,B\,a\right )}{128\,a^{5/2}} \]
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