Integrand size = 35, antiderivative size = 162 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 (b d-a e) (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x) (d+e x)^{5/2}}+\frac {2 (2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^{3/2}}-\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt {d+e x}} \]
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Time = 0.05 (sec) , antiderivative size = 162, 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} \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (-a B e-A b e+2 b B d)}{3 e^3 (a+b x) (d+e x)^{3/2}}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{5 e^3 (a+b x) (d+e x)^{5/2}}-\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt {d+e x}} \]
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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)^{7/2}} \, 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)^{7/2}}+\frac {b (-2 b B d+A b e+a B e)}{e^2 (d+e x)^{5/2}}+\frac {b^2 B}{e^2 (d+e x)^{3/2}}\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}}{5 e^3 (a+b x) (d+e x)^{5/2}}+\frac {2 (2 b B d-A b e-a B 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 b B \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt {d+e x}} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.53 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (A b e (2 d+5 e x)+a e (2 B d+3 A e+5 B e x)+b B \left (8 d^2+20 d e x+15 e^2 x^2\right )\right )}{15 e^3 (a+b x) (d+e x)^{5/2}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.49
method | result | size |
default | \(-\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (15 B b \,e^{2} x^{2}+5 A b \,e^{2} x +5 B a \,e^{2} x +20 B b d e x +3 A a \,e^{2}+2 A b d e +2 B a d e +8 B b \,d^{2}\right )}{15 e^{3} \left (e x +d \right )^{\frac {5}{2}}}\) | \(79\) |
gosper | \(-\frac {2 \left (15 B b \,e^{2} x^{2}+5 A b \,e^{2} x +5 B a \,e^{2} x +20 B b d e x +3 A a \,e^{2}+2 A b d e +2 B a d e +8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 \left (e x +d \right )^{\frac {5}{2}} e^{3} \left (b x +a \right )}\) | \(89\) |
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Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.62 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (15 \, B b e^{2} x^{2} + 8 \, B b d^{2} + 3 \, A a e^{2} + 2 \, {\left (B a + A b\right )} d e + 5 \, {\left (4 \, B b d e + {\left (B a + A b\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \]
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Timed out. \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=\text {Timed out} \]
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Time = 0.22 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.73 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (5 \, b e x + 2 \, b d + 3 \, a e\right )} A}{15 \, {\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )} \sqrt {e x + d}} - \frac {2 \, {\left (15 \, b e^{2} x^{2} + 8 \, b d^{2} + 2 \, a d e + 5 \, {\left (4 \, b d e + a e^{2}\right )} x\right )} B}{15 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )} \sqrt {e x + d}} \]
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Time = 0.29 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.79 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} B b \mathrm {sgn}\left (b x + a\right ) - 10 \, {\left (e x + d\right )} B b d \mathrm {sgn}\left (b x + a\right ) + 3 \, B b d^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (e x + d\right )} B a e \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (e x + d\right )} A b e \mathrm {sgn}\left (b x + a\right ) - 3 \, B a d e \mathrm {sgn}\left (b x + a\right ) - 3 \, A b d e \mathrm {sgn}\left (b x + a\right ) + 3 \, A a e^{2} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{3}} \]
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Time = 11.24 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,B\,x^2}{e^3}+\frac {6\,A\,a\,e^2+16\,B\,b\,d^2+4\,A\,b\,d\,e+4\,B\,a\,d\,e}{15\,b\,e^5}+\frac {x\,\left (10\,A\,b\,e^2+10\,B\,a\,e^2+40\,B\,b\,d\,e\right )}{15\,b\,e^5}\right )}{x^3\,\sqrt {d+e\,x}+\frac {a\,d^2\,\sqrt {d+e\,x}}{b\,e^2}+\frac {x^2\,\left (15\,a\,e^5+30\,b\,d\,e^4\right )\,\sqrt {d+e\,x}}{15\,b\,e^5}+\frac {d\,x\,\left (2\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}} \]
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