Integrand size = 22, antiderivative size = 169 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=\frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}-\frac {7 (5 b B-9 A c)}{12 b^4 x^{3/2}}+\frac {7 c (5 b B-9 A c)}{4 b^5 \sqrt {x}}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}+\frac {7 c^{3/2} (5 b B-9 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{11/2}} \]
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Time = 0.05 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {795, 79, 44, 53, 65, 211} \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=\frac {7 c^{3/2} (5 b B-9 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{11/2}}+\frac {7 c (5 b B-9 A c)}{4 b^5 \sqrt {x}}-\frac {7 (5 b B-9 A c)}{12 b^4 x^{3/2}}+\frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2} \]
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Rule 44
Rule 53
Rule 65
Rule 79
Rule 211
Rule 795
Rubi steps \begin{align*} \text {integral}& = \int \frac {A+B x}{x^{7/2} (b+c x)^3} \, dx \\ & = -\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {\left (\frac {5 b B}{2}-\frac {9 A c}{2}\right ) \int \frac {1}{x^{7/2} (b+c x)^2} \, dx}{2 b c} \\ & = -\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}-\frac {(7 (5 b B-9 A c)) \int \frac {1}{x^{7/2} (b+c x)} \, dx}{8 b^2 c} \\ & = \frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}+\frac {(7 (5 b B-9 A c)) \int \frac {1}{x^{5/2} (b+c x)} \, dx}{8 b^3} \\ & = \frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}-\frac {7 (5 b B-9 A c)}{12 b^4 x^{3/2}}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}-\frac {(7 c (5 b B-9 A c)) \int \frac {1}{x^{3/2} (b+c x)} \, dx}{8 b^4} \\ & = \frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}-\frac {7 (5 b B-9 A c)}{12 b^4 x^{3/2}}+\frac {7 c (5 b B-9 A c)}{4 b^5 \sqrt {x}}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}+\frac {\left (7 c^2 (5 b B-9 A c)\right ) \int \frac {1}{\sqrt {x} (b+c x)} \, dx}{8 b^5} \\ & = \frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}-\frac {7 (5 b B-9 A c)}{12 b^4 x^{3/2}}+\frac {7 c (5 b B-9 A c)}{4 b^5 \sqrt {x}}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}+\frac {\left (7 c^2 (5 b B-9 A c)\right ) \text {Subst}\left (\int \frac {1}{b+c x^2} \, dx,x,\sqrt {x}\right )}{4 b^5} \\ & = \frac {7 (5 b B-9 A c)}{20 b^3 c x^{5/2}}-\frac {7 (5 b B-9 A c)}{12 b^4 x^{3/2}}+\frac {7 c (5 b B-9 A c)}{4 b^5 \sqrt {x}}-\frac {b B-A c}{2 b c x^{5/2} (b+c x)^2}-\frac {5 b B-9 A c}{4 b^2 c x^{5/2} (b+c x)}+\frac {7 c^{3/2} (5 b B-9 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{11/2}} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.83 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=\frac {5 b B x \left (-8 b^3+56 b^2 c x+175 b c^2 x^2+105 c^3 x^3\right )-3 A \left (8 b^4-24 b^3 c x+168 b^2 c^2 x^2+525 b c^3 x^3+315 c^4 x^4\right )}{60 b^5 x^{5/2} (b+c x)^2}+\frac {7 c^{3/2} (5 b B-9 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 b^{11/2}} \]
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Time = 0.14 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.72
method | result | size |
derivativedivides | \(-\frac {2 c^{2} \left (\frac {\left (\frac {15}{8} A \,c^{2}-\frac {11}{8} B b c \right ) x^{\frac {3}{2}}+\frac {b \left (17 A c -13 B b \right ) \sqrt {x}}{8}}{\left (c x +b \right )^{2}}+\frac {7 \left (9 A c -5 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{b^{5}}-\frac {2 A}{5 b^{3} x^{\frac {5}{2}}}-\frac {2 \left (-3 A c +B b \right )}{3 b^{4} x^{\frac {3}{2}}}-\frac {6 c \left (2 A c -B b \right )}{b^{5} \sqrt {x}}\) | \(121\) |
default | \(-\frac {2 c^{2} \left (\frac {\left (\frac {15}{8} A \,c^{2}-\frac {11}{8} B b c \right ) x^{\frac {3}{2}}+\frac {b \left (17 A c -13 B b \right ) \sqrt {x}}{8}}{\left (c x +b \right )^{2}}+\frac {7 \left (9 A c -5 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{b^{5}}-\frac {2 A}{5 b^{3} x^{\frac {5}{2}}}-\frac {2 \left (-3 A c +B b \right )}{3 b^{4} x^{\frac {3}{2}}}-\frac {6 c \left (2 A c -B b \right )}{b^{5} \sqrt {x}}\) | \(121\) |
risch | \(-\frac {2 \left (90 A \,c^{2} x^{2}-45 B b c \,x^{2}-15 A b c x +5 b^{2} B x +3 A \,b^{2}\right )}{15 b^{5} x^{\frac {5}{2}}}-\frac {c^{2} \left (\frac {2 \left (\frac {15}{8} A \,c^{2}-\frac {11}{8} B b c \right ) x^{\frac {3}{2}}+\frac {b \left (17 A c -13 B b \right ) \sqrt {x}}{4}}{\left (c x +b \right )^{2}}+\frac {7 \left (9 A c -5 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}}\right )}{b^{5}}\) | \(124\) |
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Time = 0.28 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.59 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=\left [-\frac {105 \, {\left ({\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{5} + 2 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{4} + {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{3}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x - 2 \, b \sqrt {x} \sqrt {-\frac {c}{b}} - b}{c x + b}\right ) + 2 \, {\left (24 \, A b^{4} - 105 \, {\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{4} - 175 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{3} - 56 \, {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2} + 8 \, {\left (5 \, B b^{4} - 9 \, A b^{3} c\right )} x\right )} \sqrt {x}}{120 \, {\left (b^{5} c^{2} x^{5} + 2 \, b^{6} c x^{4} + b^{7} x^{3}\right )}}, -\frac {105 \, {\left ({\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{5} + 2 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{4} + {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{3}\right )} \sqrt {\frac {c}{b}} \arctan \left (\frac {b \sqrt {\frac {c}{b}}}{c \sqrt {x}}\right ) + {\left (24 \, A b^{4} - 105 \, {\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{4} - 175 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{3} - 56 \, {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2} + 8 \, {\left (5 \, B b^{4} - 9 \, A b^{3} c\right )} x\right )} \sqrt {x}}{60 \, {\left (b^{5} c^{2} x^{5} + 2 \, b^{6} c x^{4} + b^{7} x^{3}\right )}}\right ] \]
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Leaf count of result is larger than twice the leaf count of optimal. 1882 vs. \(2 (163) = 326\).
Time = 122.82 (sec) , antiderivative size = 1882, normalized size of antiderivative = 11.14 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=\text {Too large to display} \]
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Time = 0.27 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=-\frac {24 \, A b^{4} - 105 \, {\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{4} - 175 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{3} - 56 \, {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2} + 8 \, {\left (5 \, B b^{4} - 9 \, A b^{3} c\right )} x}{60 \, {\left (b^{5} c^{2} x^{\frac {9}{2}} + 2 \, b^{6} c x^{\frac {7}{2}} + b^{7} x^{\frac {5}{2}}\right )}} + \frac {7 \, {\left (5 \, B b c^{2} - 9 \, A c^{3}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{5}} \]
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Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=\frac {7 \, {\left (5 \, B b c^{2} - 9 \, A c^{3}\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} b^{5}} + \frac {11 \, B b c^{3} x^{\frac {3}{2}} - 15 \, A c^{4} x^{\frac {3}{2}} + 13 \, B b^{2} c^{2} \sqrt {x} - 17 \, A b c^{3} \sqrt {x}}{4 \, {\left (c x + b\right )}^{2} b^{5}} + \frac {2 \, {\left (45 \, B b c x^{2} - 90 \, A c^{2} x^{2} - 5 \, B b^{2} x + 15 \, A b c x - 3 \, A b^{2}\right )}}{15 \, b^{5} x^{\frac {5}{2}}} \]
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Time = 10.29 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^3} \, dx=-\frac {\frac {2\,A}{5\,b}-\frac {2\,x\,\left (9\,A\,c-5\,B\,b\right )}{15\,b^2}+\frac {35\,c^2\,x^3\,\left (9\,A\,c-5\,B\,b\right )}{12\,b^4}+\frac {7\,c^3\,x^4\,\left (9\,A\,c-5\,B\,b\right )}{4\,b^5}+\frac {14\,c\,x^2\,\left (9\,A\,c-5\,B\,b\right )}{15\,b^3}}{b^2\,x^{5/2}+c^2\,x^{9/2}+2\,b\,c\,x^{7/2}}-\frac {7\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (9\,A\,c-5\,B\,b\right )}{4\,b^{11/2}} \]
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