Integrand size = 35, antiderivative size = 160 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=-\frac {2 (b d-a e) (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^{3/2}}+\frac {2 (2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt {d+e x}}+\frac {2 b B \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)} \]
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Time = 0.05 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {784, 78} \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (-a B e-A b e+2 b B d)}{e^3 (a+b x) \sqrt {d+e x}}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{3 e^3 (a+b x) (d+e x)^{3/2}}+\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x}}{e^3 (a+b x)} \]
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Rule 78
Rule 784
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right ) (A+B x)}{(d+e x)^{5/2}} \, dx}{a b+b^2 x} \\ & = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (-\frac {b (b d-a e) (-B d+A e)}{e^2 (d+e x)^{5/2}}+\frac {b (-2 b B d+A b e+a B e)}{e^2 (d+e x)^{3/2}}+\frac {b^2 B}{e^2 \sqrt {d+e x}}\right ) \, dx}{a b+b^2 x} \\ & = -\frac {2 (b d-a e) (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^{3/2}}+\frac {2 (2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt {d+e x}}+\frac {2 b B \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.54 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (A b e (2 d+3 e x)+a e (2 B d+A e+3 B e x)-b B \left (8 d^2+12 d e x+3 e^2 x^2\right )\right )}{3 e^3 (a+b x) (d+e x)^{3/2}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.49
method | result | size |
default | \(-\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (-3 B b \,e^{2} x^{2}+3 A b \,e^{2} x +3 B a \,e^{2} x -12 B b d e x +A a \,e^{2}+2 A b d e +2 B a d e -8 B b \,d^{2}\right )}{3 e^{3} \left (e x +d \right )^{\frac {3}{2}}}\) | \(78\) |
gosper | \(-\frac {2 \left (-3 B b \,e^{2} x^{2}+3 A b \,e^{2} x +3 B a \,e^{2} x -12 B b d e x +A a \,e^{2}+2 A b d e +2 B a d e -8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{3 \left (e x +d \right )^{\frac {3}{2}} e^{3} \left (b x +a \right )}\) | \(88\) |
risch | \(\frac {2 b B \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{e^{3} \left (b x +a \right )}-\frac {2 \left (3 A b \,e^{2} x +3 B a \,e^{2} x -6 B b d e x +A a \,e^{2}+2 A b d e +2 B a d e -5 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{3 e^{3} \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right )}\) | \(109\) |
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Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.57 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, B b e^{2} x^{2} + 8 \, B b d^{2} - A a e^{2} - 2 \, {\left (B a + A b\right )} d e + 3 \, {\left (4 \, B b d e - {\left (B a + A b\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]
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\[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=\int \frac {\left (A + B x\right ) \sqrt {\left (a + b x\right )^{2}}}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \]
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Time = 0.21 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.60 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=-\frac {2 \, {\left (3 \, b e x + 2 \, b d + a e\right )} A}{3 \, {\left (e^{3} x + d e^{2}\right )} \sqrt {e x + d}} + \frac {2 \, {\left (3 \, b e^{2} x^{2} + 8 \, b d^{2} - 2 \, a d e + 3 \, {\left (4 \, b d e - a e^{2}\right )} x\right )} B}{3 \, {\left (e^{4} x + d e^{3}\right )} \sqrt {e x + d}} \]
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Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.81 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, \sqrt {e x + d} B b \mathrm {sgn}\left (b x + a\right )}{e^{3}} + \frac {2 \, {\left (6 \, {\left (e x + d\right )} B b d \mathrm {sgn}\left (b x + a\right ) - B b d^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, {\left (e x + d\right )} B a e \mathrm {sgn}\left (b x + a\right ) - 3 \, {\left (e x + d\right )} A b e \mathrm {sgn}\left (b x + a\right ) + B a d e \mathrm {sgn}\left (b x + a\right ) + A b d e \mathrm {sgn}\left (b x + a\right ) - A a e^{2} \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, {\left (e x + d\right )}^{\frac {3}{2}} e^{3}} \]
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Time = 11.22 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,A\,a\,e^2-16\,B\,b\,d^2+4\,A\,b\,d\,e+4\,B\,a\,d\,e}{3\,b\,e^4}-\frac {2\,B\,x^2}{e^2}+\frac {x\,\left (6\,A\,b\,e^2+6\,B\,a\,e^2-24\,B\,b\,d\,e\right )}{3\,b\,e^4}\right )}{x^2\,\sqrt {d+e\,x}+\frac {a\,d\,\sqrt {d+e\,x}}{b\,e}+\frac {x\,\left (3\,a\,e^4+3\,b\,d\,e^3\right )\,\sqrt {d+e\,x}}{3\,b\,e^4}} \]
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