Integrand size = 33, antiderivative size = 158 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=-\frac {(b d-a e) (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x) (d+e x)^5}+\frac {(2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^3 (a+b x) (d+e x)^4}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^3} \]
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Time = 0.06 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {784, 78} \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=\frac {\sqrt {a^2+2 a b x+b^2 x^2} (-a B e-A b e+2 b B d)}{4 e^3 (a+b x) (d+e x)^4}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{5 e^3 (a+b x) (d+e x)^5}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^3} \]
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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)^6} \, 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)^6}+\frac {b (-2 b B d+A b e+a B e)}{e^2 (d+e x)^5}+\frac {b^2 B}{e^2 (d+e x)^4}\right ) \, dx}{a b+b^2 x} \\ & = -\frac {(b d-a e) (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x) (d+e x)^5}+\frac {(2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^3 (a+b x) (d+e x)^4}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^3} \\ \end{align*}
Time = 1.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.53 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=-\frac {\sqrt {(a+b x)^2} \left (3 a e (4 A e+B (d+5 e x))+b \left (3 A e (d+5 e x)+2 B \left (d^2+5 d e x+10 e^2 x^2\right )\right )\right )}{60 e^3 (a+b x) (d+e x)^5} \]
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Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.86 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.50
method | result | size |
default | \(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (20 B b \,e^{2} x^{2}+15 A b \,e^{2} x +15 B a \,e^{2} x +10 B b d e x +12 A a \,e^{2}+3 A b d e +3 B a d e +2 B b \,d^{2}\right )}{60 e^{3} \left (e x +d \right )^{5}}\) | \(79\) |
gosper | \(-\frac {\left (20 B b \,e^{2} x^{2}+15 A b \,e^{2} x +15 B a \,e^{2} x +10 B b d e x +12 A a \,e^{2}+3 A b d e +3 B a d e +2 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{60 e^{3} \left (e x +d \right )^{5} \left (b x +a \right )}\) | \(89\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B b \,x^{2}}{3 e}-\frac {\left (3 A b e +3 B a e +2 B b d \right ) x}{12 e^{2}}-\frac {12 A a \,e^{2}+3 A b d e +3 B a d e +2 B b \,d^{2}}{60 e^{3}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{5}}\) | \(90\) |
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Time = 0.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.74 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=-\frac {20 \, B b e^{2} x^{2} + 2 \, B b d^{2} + 12 \, A a e^{2} + 3 \, {\left (B a + A b\right )} d e + 5 \, {\left (2 \, B b d e + 3 \, {\left (B a + A b\right )} e^{2}\right )} x}{60 \, {\left (e^{8} x^{5} + 5 \, d e^{7} x^{4} + 10 \, d^{2} e^{6} x^{3} + 10 \, d^{3} e^{5} x^{2} + 5 \, d^{4} e^{4} x + d^{5} e^{3}\right )}} \]
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Timed out. \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.28 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.32 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=\frac {{\left (2 \, B b^{5} d - 5 \, B a b^{4} e + 3 \, A b^{5} e\right )} \mathrm {sgn}\left (b x + a\right )}{60 \, {\left (b^{4} d^{4} e^{3} - 4 \, a b^{3} d^{3} e^{4} + 6 \, a^{2} b^{2} d^{2} e^{5} - 4 \, a^{3} b d e^{6} + a^{4} e^{7}\right )}} - \frac {20 \, B b e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, B b d e x \mathrm {sgn}\left (b x + a\right ) + 15 \, B a e^{2} x \mathrm {sgn}\left (b x + a\right ) + 15 \, A b e^{2} x \mathrm {sgn}\left (b x + a\right ) + 2 \, B b d^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, B a d e \mathrm {sgn}\left (b x + a\right ) + 3 \, A b d e \mathrm {sgn}\left (b x + a\right ) + 12 \, A a e^{2} \mathrm {sgn}\left (b x + a\right )}{60 \, {\left (e x + d\right )}^{5} e^{3}} \]
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Time = 11.02 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.56 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^6} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (12\,A\,a\,e^2+2\,B\,b\,d^2+15\,A\,b\,e^2\,x+15\,B\,a\,e^2\,x+20\,B\,b\,e^2\,x^2+3\,A\,b\,d\,e+3\,B\,a\,d\,e+10\,B\,b\,d\,e\,x\right )}{60\,e^3\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5} \]
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