Integrand size = 18, antiderivative size = 169 \[ \int \frac {x^{7/2} (A+B x)}{(a+b x)^3} \, dx=-\frac {7 a (5 A b-9 a B) \sqrt {x}}{4 b^5}+\frac {7 (5 A b-9 a B) x^{3/2}}{12 b^4}-\frac {7 (5 A b-9 a B) x^{5/2}}{20 a b^3}+\frac {(A b-a B) x^{9/2}}{2 a b (a+b x)^2}+\frac {(5 A b-9 a B) x^{7/2}}{4 a b^2 (a+b x)}+\frac {7 a^{3/2} (5 A b-9 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{11/2}} \]
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Time = 0.06 (sec) , antiderivative size = 169, 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, 52, 65, 211} \[ \int \frac {x^{7/2} (A+B x)}{(a+b x)^3} \, dx=\frac {7 a^{3/2} (5 A b-9 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{11/2}}-\frac {7 a \sqrt {x} (5 A b-9 a B)}{4 b^5}+\frac {7 x^{3/2} (5 A b-9 a B)}{12 b^4}-\frac {7 x^{5/2} (5 A b-9 a B)}{20 a b^3}+\frac {x^{7/2} (5 A b-9 a B)}{4 a b^2 (a+b x)}+\frac {x^{9/2} (A b-a B)}{2 a b (a+b x)^2} \]
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Rule 43
Rule 52
Rule 65
Rule 79
Rule 211
Rubi steps \begin{align*} \text {integral}& = \frac {(A b-a B) x^{9/2}}{2 a b (a+b x)^2}-\frac {\left (\frac {5 A b}{2}-\frac {9 a B}{2}\right ) \int \frac {x^{7/2}}{(a+b x)^2} \, dx}{2 a b} \\ & = \frac {(A b-a B) x^{9/2}}{2 a b (a+b x)^2}+\frac {(5 A b-9 a B) x^{7/2}}{4 a b^2 (a+b x)}-\frac {(7 (5 A b-9 a B)) \int \frac {x^{5/2}}{a+b x} \, dx}{8 a b^2} \\ & = -\frac {7 (5 A b-9 a B) x^{5/2}}{20 a b^3}+\frac {(A b-a B) x^{9/2}}{2 a b (a+b x)^2}+\frac {(5 A b-9 a B) x^{7/2}}{4 a b^2 (a+b x)}+\frac {(7 (5 A b-9 a B)) \int \frac {x^{3/2}}{a+b x} \, dx}{8 b^3} \\ & = \frac {7 (5 A b-9 a B) x^{3/2}}{12 b^4}-\frac {7 (5 A b-9 a B) x^{5/2}}{20 a b^3}+\frac {(A b-a B) x^{9/2}}{2 a b (a+b x)^2}+\frac {(5 A b-9 a B) x^{7/2}}{4 a b^2 (a+b x)}-\frac {(7 a (5 A b-9 a B)) \int \frac {\sqrt {x}}{a+b x} \, dx}{8 b^4} \\ & = -\frac {7 a (5 A b-9 a B) \sqrt {x}}{4 b^5}+\frac {7 (5 A b-9 a B) x^{3/2}}{12 b^4}-\frac {7 (5 A b-9 a B) x^{5/2}}{20 a b^3}+\frac {(A b-a B) x^{9/2}}{2 a b (a+b x)^2}+\frac {(5 A b-9 a B) x^{7/2}}{4 a b^2 (a+b x)}+\frac {\left (7 a^2 (5 A b-9 a B)\right ) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{8 b^5} \\ & = -\frac {7 a (5 A b-9 a B) \sqrt {x}}{4 b^5}+\frac {7 (5 A b-9 a B) x^{3/2}}{12 b^4}-\frac {7 (5 A b-9 a B) x^{5/2}}{20 a b^3}+\frac {(A b-a B) x^{9/2}}{2 a b (a+b x)^2}+\frac {(5 A b-9 a B) x^{7/2}}{4 a b^2 (a+b x)}+\frac {\left (7 a^2 (5 A b-9 a B)\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{4 b^5} \\ & = -\frac {7 a (5 A b-9 a B) \sqrt {x}}{4 b^5}+\frac {7 (5 A b-9 a B) x^{3/2}}{12 b^4}-\frac {7 (5 A b-9 a B) x^{5/2}}{20 a b^3}+\frac {(A b-a B) x^{9/2}}{2 a b (a+b x)^2}+\frac {(5 A b-9 a B) x^{7/2}}{4 a b^2 (a+b x)}+\frac {7 a^{3/2} (5 A b-9 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{11/2}} \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.76 \[ \int \frac {x^{7/2} (A+B x)}{(a+b x)^3} \, dx=\frac {\sqrt {x} \left (945 a^4 B-525 a^3 b (A-3 B x)+8 b^4 x^3 (5 A+3 B x)-8 a b^3 x^2 (35 A+9 B x)+7 a^2 b^2 x (-125 A+72 B x)\right )}{60 b^5 (a+b x)^2}-\frac {7 a^{3/2} (-5 A b+9 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{11/2}} \]
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Time = 0.51 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.71
method | result | size |
risch | \(-\frac {2 \left (-3 b^{2} B \,x^{2}-5 A \,b^{2} x +15 B a b x +45 a b A -90 a^{2} B \right ) \sqrt {x}}{15 b^{5}}+\frac {a^{2} \left (\frac {2 \left (-\frac {13}{8} b^{2} A +\frac {17}{8} a b B \right ) x^{\frac {3}{2}}-\frac {a \left (11 A b -15 B a \right ) \sqrt {x}}{4}}{\left (b x +a \right )^{2}}+\frac {7 \left (5 A b -9 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right )}{b^{5}}\) | \(120\) |
derivativedivides | \(-\frac {2 \left (-\frac {b^{2} B \,x^{\frac {5}{2}}}{5}-\frac {A \,b^{2} x^{\frac {3}{2}}}{3}+B a b \,x^{\frac {3}{2}}+3 a b A \sqrt {x}-6 a^{2} B \sqrt {x}\right )}{b^{5}}+\frac {2 a^{2} \left (\frac {\left (-\frac {13}{8} b^{2} A +\frac {17}{8} a b B \right ) x^{\frac {3}{2}}-\frac {a \left (11 A b -15 B a \right ) \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {7 \left (5 A b -9 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{5}}\) | \(126\) |
default | \(-\frac {2 \left (-\frac {b^{2} B \,x^{\frac {5}{2}}}{5}-\frac {A \,b^{2} x^{\frac {3}{2}}}{3}+B a b \,x^{\frac {3}{2}}+3 a b A \sqrt {x}-6 a^{2} B \sqrt {x}\right )}{b^{5}}+\frac {2 a^{2} \left (\frac {\left (-\frac {13}{8} b^{2} A +\frac {17}{8} a b B \right ) x^{\frac {3}{2}}-\frac {a \left (11 A b -15 B a \right ) \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {7 \left (5 A b -9 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{5}}\) | \(126\) |
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Time = 0.24 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.41 \[ \int \frac {x^{7/2} (A+B x)}{(a+b x)^3} \, dx=\left [-\frac {105 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b + {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 2 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (24 \, B b^{4} x^{4} + 945 \, B a^{4} - 525 \, A a^{3} b - 8 \, {\left (9 \, B a b^{3} - 5 \, A b^{4}\right )} x^{3} + 56 \, {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 175 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{120 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}, -\frac {105 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b + {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 2 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (24 \, B b^{4} x^{4} + 945 \, B a^{4} - 525 \, A a^{3} b - 8 \, {\left (9 \, B a b^{3} - 5 \, A b^{4}\right )} x^{3} + 56 \, {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} + 175 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{60 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}\right ] \]
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Leaf count of result is larger than twice the leaf count of optimal. 1652 vs. \(2 (163) = 326\).
Time = 58.20 (sec) , antiderivative size = 1652, normalized size of antiderivative = 9.78 \[ \int \frac {x^{7/2} (A+B x)}{(a+b x)^3} \, dx=\text {Too large to display} \]
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Time = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.89 \[ \int \frac {x^{7/2} (A+B x)}{(a+b x)^3} \, dx=\frac {{\left (17 \, B a^{3} b - 13 \, A a^{2} b^{2}\right )} x^{\frac {3}{2}} + {\left (15 \, B a^{4} - 11 \, A a^{3} b\right )} \sqrt {x}}{4 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} - \frac {7 \, {\left (9 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{5}} + \frac {2 \, {\left (3 \, B b^{2} x^{\frac {5}{2}} - 5 \, {\left (3 \, B a b - A b^{2}\right )} x^{\frac {3}{2}} + 45 \, {\left (2 \, B a^{2} - A a b\right )} \sqrt {x}\right )}}{15 \, b^{5}} \]
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Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.86 \[ \int \frac {x^{7/2} (A+B x)}{(a+b x)^3} \, dx=-\frac {7 \, {\left (9 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{5}} + \frac {17 \, B a^{3} b x^{\frac {3}{2}} - 13 \, A a^{2} b^{2} x^{\frac {3}{2}} + 15 \, B a^{4} \sqrt {x} - 11 \, A a^{3} b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{5}} + \frac {2 \, {\left (3 \, B b^{12} x^{\frac {5}{2}} - 15 \, B a b^{11} x^{\frac {3}{2}} + 5 \, A b^{12} x^{\frac {3}{2}} + 90 \, B a^{2} b^{10} \sqrt {x} - 45 \, A a b^{11} \sqrt {x}\right )}}{15 \, b^{15}} \]
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Time = 0.49 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.08 \[ \int \frac {x^{7/2} (A+B x)}{(a+b x)^3} \, dx=x^{3/2}\,\left (\frac {2\,A}{3\,b^3}-\frac {2\,B\,a}{b^4}\right )-\frac {x^{3/2}\,\left (\frac {13\,A\,a^2\,b^2}{4}-\frac {17\,B\,a^3\,b}{4}\right )-\sqrt {x}\,\left (\frac {15\,B\,a^4}{4}-\frac {11\,A\,a^3\,b}{4}\right )}{a^2\,b^5+2\,a\,b^6\,x+b^7\,x^2}-\sqrt {x}\,\left (\frac {3\,a\,\left (\frac {2\,A}{b^3}-\frac {6\,B\,a}{b^4}\right )}{b}+\frac {6\,B\,a^2}{b^5}\right )+\frac {2\,B\,x^{5/2}}{5\,b^3}-\frac {7\,a^{3/2}\,\mathrm {atan}\left (\frac {a^{3/2}\,\sqrt {b}\,\sqrt {x}\,\left (5\,A\,b-9\,B\,a\right )}{9\,B\,a^3-5\,A\,a^2\,b}\right )\,\left (5\,A\,b-9\,B\,a\right )}{4\,b^{11/2}} \]
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