Integrand size = 30, antiderivative size = 168 \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (a+b x)}{3 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (a+b x)}{(b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b^{3/2} (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
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Time = 0.05 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {660, 53, 65, 214} \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 b^{3/2} (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{5/2}}+\frac {2 b (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^2}+\frac {2 (a+b x)}{3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)} \]
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Rule 53
Rule 65
Rule 214
Rule 660
Rubi steps \begin{align*} \text {integral}& = \frac {\left (a b+b^2 x\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{5/2}} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {2 (a+b x)}{3 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (b \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{3/2}} \, dx}{(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {2 (a+b x)}{3 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (a+b x)}{(b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {2 (a+b x)}{3 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (a+b x)}{(b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 b^2 \left (a b+b^2 x\right )\right ) \text {Subst}\left (\int \frac {1}{a b-\frac {b^2 d}{e}+\frac {b^2 x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{e (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {2 (a+b x)}{3 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (a+b x)}{(b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b^{3/2} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (a+b x) \left (\sqrt {-b d+a e} (4 b d-a e+3 b e x)+3 b^{3/2} (d+e x)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )\right )}{3 (-b d+a e)^{5/2} \sqrt {(a+b x)^2} (d+e x)^{3/2}} \]
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Time = 2.58 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.77
method | result | size |
default | \(-\frac {2 \left (b x +a \right ) \left (-3 b^{2} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) \left (e x +d \right )^{\frac {3}{2}}-3 \sqrt {\left (a e -b d \right ) b}\, b e x +\sqrt {\left (a e -b d \right ) b}\, a e -4 \sqrt {\left (a e -b d \right ) b}\, b d \right )}{3 \sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}}\) | \(129\) |
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Time = 0.30 (sec) , antiderivative size = 398, normalized size of antiderivative = 2.37 \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [\frac {3 \, {\left (b e^{2} x^{2} + 2 \, b d e x + b d^{2}\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) + 2 \, {\left (3 \, b e x + 4 \, b d - a e\right )} \sqrt {e x + d}}{3 \, {\left (b^{2} d^{4} - 2 \, a b d^{3} e + a^{2} d^{2} e^{2} + {\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} x^{2} + 2 \, {\left (b^{2} d^{3} e - 2 \, a b d^{2} e^{2} + a^{2} d e^{3}\right )} x\right )}}, -\frac {2 \, {\left (3 \, {\left (b e^{2} x^{2} + 2 \, b d e x + b d^{2}\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (3 \, b e x + 4 \, b d - a e\right )} \sqrt {e x + d}\right )}}{3 \, {\left (b^{2} d^{4} - 2 \, a b d^{3} e + a^{2} d^{2} e^{2} + {\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} x^{2} + 2 \, {\left (b^{2} d^{3} e - 2 \, a b d^{2} e^{2} + a^{2} d e^{3}\right )} x\right )}}\right ] \]
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\[ \int \frac {1}{(d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {1}{\left (d + e x\right )^{\frac {5}{2}} \sqrt {\left (a + b x\right )^{2}}}\, dx \]
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\[ \int \frac {1}{(d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {1}{\sqrt {{\left (b x + a\right )}^{2}} {\left (e x + d\right )}^{\frac {5}{2}}} \,d x } \]
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Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2}{3} \, {\left (\frac {3 \, b^{2} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e}} + \frac {3 \, {\left (e x + d\right )} b + b d - a e}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} {\left (e x + d\right )}^{\frac {3}{2}}}\right )} \mathrm {sgn}\left (b x + a\right ) \]
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Timed out. \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {1}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^{5/2}} \,d x \]
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