Integrand size = 18, antiderivative size = 86 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x} \, dx=2 a^2 A \sqrt {a+b x}+\frac {2}{3} a A (a+b x)^{3/2}+\frac {2}{5} A (a+b x)^{5/2}+\frac {2 B (a+b x)^{7/2}}{7 b}-2 a^{5/2} A \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
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Time = 0.02 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {81, 52, 65, 214} \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x} \, dx=-2 a^{5/2} A \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+2 a^2 A \sqrt {a+b x}+\frac {2}{5} A (a+b x)^{5/2}+\frac {2}{3} a A (a+b x)^{3/2}+\frac {2 B (a+b x)^{7/2}}{7 b} \]
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Rule 52
Rule 65
Rule 81
Rule 214
Rubi steps \begin{align*} \text {integral}& = \frac {2 B (a+b x)^{7/2}}{7 b}+A \int \frac {(a+b x)^{5/2}}{x} \, dx \\ & = \frac {2}{5} A (a+b x)^{5/2}+\frac {2 B (a+b x)^{7/2}}{7 b}+(a A) \int \frac {(a+b x)^{3/2}}{x} \, dx \\ & = \frac {2}{3} a A (a+b x)^{3/2}+\frac {2}{5} A (a+b x)^{5/2}+\frac {2 B (a+b x)^{7/2}}{7 b}+\left (a^2 A\right ) \int \frac {\sqrt {a+b x}}{x} \, dx \\ & = 2 a^2 A \sqrt {a+b x}+\frac {2}{3} a A (a+b x)^{3/2}+\frac {2}{5} A (a+b x)^{5/2}+\frac {2 B (a+b x)^{7/2}}{7 b}+\left (a^3 A\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx \\ & = 2 a^2 A \sqrt {a+b x}+\frac {2}{3} a A (a+b x)^{3/2}+\frac {2}{5} A (a+b x)^{5/2}+\frac {2 B (a+b x)^{7/2}}{7 b}+\frac {\left (2 a^3 A\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{b} \\ & = 2 a^2 A \sqrt {a+b x}+\frac {2}{3} a A (a+b x)^{3/2}+\frac {2}{5} A (a+b x)^{5/2}+\frac {2 B (a+b x)^{7/2}}{7 b}-2 a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x} \, dx=\frac {2 \left (105 a^2 A b \sqrt {a+b x}+35 a A b (a+b x)^{3/2}+21 A b (a+b x)^{5/2}+15 B (a+b x)^{7/2}\right )}{105 b}-2 a^{5/2} A \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]
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Time = 1.39 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.84
method | result | size |
derivativedivides | \(\frac {\frac {2 B \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 A b \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 A b a \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A \,a^{2} b \sqrt {b x +a}-2 A \,a^{\frac {5}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{b}\) | \(72\) |
default | \(\frac {\frac {2 B \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 A b \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 A b a \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A \,a^{2} b \sqrt {b x +a}-2 A \,a^{\frac {5}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{b}\) | \(72\) |
pseudoelliptic | \(\frac {-2 A \,a^{\frac {5}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {46 \left (\frac {3 x^{2} \left (\frac {5 B x}{7}+A \right ) b^{3}}{23}+\frac {11 \left (\frac {45 B x}{77}+A \right ) x a \,b^{2}}{23}+a^{2} \left (\frac {45 B x}{161}+A \right ) b +\frac {15 a^{3} B}{161}\right ) \sqrt {b x +a}}{15}}{b}\) | \(80\) |
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Time = 0.24 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.42 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x} \, dx=\left [\frac {105 \, A a^{\frac {5}{2}} b \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (15 \, B b^{3} x^{3} + 15 \, B a^{3} + 161 \, A a^{2} b + 3 \, {\left (15 \, B a b^{2} + 7 \, A b^{3}\right )} x^{2} + {\left (45 \, B a^{2} b + 77 \, A a b^{2}\right )} x\right )} \sqrt {b x + a}}{105 \, b}, \frac {2 \, {\left (105 \, A \sqrt {-a} a^{2} b \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, B b^{3} x^{3} + 15 \, B a^{3} + 161 \, A a^{2} b + 3 \, {\left (15 \, B a b^{2} + 7 \, A b^{3}\right )} x^{2} + {\left (45 \, B a^{2} b + 77 \, A a b^{2}\right )} x\right )} \sqrt {b x + a}\right )}}{105 \, b}\right ] \]
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Time = 1.58 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.21 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x} \, dx=\begin {cases} \frac {2 A a^{3} \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 A a^{2} \sqrt {a + b x} + \frac {2 A a \left (a + b x\right )^{\frac {3}{2}}}{3} + \frac {2 A \left (a + b x\right )^{\frac {5}{2}}}{5} + \frac {2 B \left (a + b x\right )^{\frac {7}{2}}}{7 b} & \text {for}\: b \neq 0 \\a^{\frac {5}{2}} \left (A \log {\left (B x \right )} + B x\right ) & \text {otherwise} \end {cases} \]
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Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x} \, dx=A a^{\frac {5}{2}} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2 \, {\left (15 \, {\left (b x + a\right )}^{\frac {7}{2}} B + 21 \, {\left (b x + a\right )}^{\frac {5}{2}} A b + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} A a b + 105 \, \sqrt {b x + a} A a^{2} b\right )}}{105 \, b} \]
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Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x} \, dx=\frac {2 \, A a^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (15 \, {\left (b x + a\right )}^{\frac {7}{2}} B b^{6} + 21 \, {\left (b x + a\right )}^{\frac {5}{2}} A b^{7} + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} A a b^{7} + 105 \, \sqrt {b x + a} A a^{2} b^{7}\right )}}{105 \, b^{7}} \]
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Time = 0.05 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.49 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x} \, dx=\left (\frac {2\,A\,b-2\,B\,a}{5\,b}+\frac {2\,B\,a}{5\,b}\right )\,{\left (a+b\,x\right )}^{5/2}+a^2\,\left (\frac {2\,A\,b-2\,B\,a}{b}+\frac {2\,B\,a}{b}\right )\,\sqrt {a+b\,x}+\frac {2\,B\,{\left (a+b\,x\right )}^{7/2}}{7\,b}+\frac {a\,\left (\frac {2\,A\,b-2\,B\,a}{b}+\frac {2\,B\,a}{b}\right )\,{\left (a+b\,x\right )}^{3/2}}{3}+A\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i} \]
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