Integrand size = 35, antiderivative size = 150 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x) (d+e x)^{5/2}}+\frac {4 b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^{3/2}}-\frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt {d+e x}} \]
[Out]
Time = 0.05 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {784, 21, 45} \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt {d+e x}}+\frac {4 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{3 e^3 (a+b x) (d+e x)^{3/2}}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{5 e^3 (a+b x) (d+e x)^{5/2}} \]
[In]
[Out]
Rule 21
Rule 45
Rule 784
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x) \left (a b+b^2 x\right )}{(d+e x)^{7/2}} \, dx}{a b+b^2 x} \\ & = \frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {(a+b x)^2}{(d+e x)^{7/2}} \, dx}{a b+b^2 x} \\ & = \frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac {(-b d+a e)^2}{e^2 (d+e x)^{7/2}}-\frac {2 b (b d-a e)}{e^2 (d+e x)^{5/2}}+\frac {b^2}{e^2 (d+e x)^{3/2}}\right ) \, dx}{a b+b^2 x} \\ & = -\frac {2 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x) (d+e x)^{5/2}}+\frac {4 b (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 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt {d+e x}} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.53 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (3 a^2 e^2+2 a b e (2 d+5 e x)+b^2 \left (8 d^2+20 d e x+15 e^2 x^2\right )\right )}{15 e^3 (a+b x) (d+e x)^{5/2}} \]
[In]
[Out]
Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.46
method | result | size |
default | \(-\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (15 b^{2} e^{2} x^{2}+10 a b \,e^{2} x +20 b^{2} d e x +3 e^{2} a^{2}+4 a b d e +8 b^{2} d^{2}\right )}{15 e^{3} \left (e x +d \right )^{\frac {5}{2}}}\) | \(69\) |
gosper | \(-\frac {2 \left (15 b^{2} e^{2} x^{2}+10 a b \,e^{2} x +20 b^{2} d e x +3 e^{2} a^{2}+4 a b d e +8 b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 \left (e x +d \right )^{\frac {5}{2}} e^{3} \left (b x +a \right )}\) | \(79\) |
[In]
[Out]
none
Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.63 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (15 \, b^{2} e^{2} x^{2} + 8 \, b^{2} d^{2} + 4 \, a b d e + 3 \, a^{2} e^{2} + 10 \, {\left (2 \, b^{2} d e + a b e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \]
[In]
[Out]
Timed out. \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=\text {Timed out} \]
[In]
[Out]
none
Time = 0.21 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.79 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (5 \, b e x + 2 \, b d + 3 \, a e\right )} a}{15 \, {\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )} \sqrt {e x + d}} - \frac {2 \, {\left (15 \, b e^{2} x^{2} + 8 \, b d^{2} + 2 \, a d e + 5 \, {\left (4 \, b d e + a e^{2}\right )} x\right )} b}{15 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )} \sqrt {e x + d}} \]
[In]
[Out]
none
Time = 0.35 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.69 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} b^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, {\left (e x + d\right )} b^{2} d \mathrm {sgn}\left (b x + a\right ) + 3 \, b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, {\left (e x + d\right )} a b e \mathrm {sgn}\left (b x + a\right ) - 6 \, a b d e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{3}} \]
[In]
[Out]
Time = 11.56 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{7/2}} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {4\,x\,\left (a\,e+2\,b\,d\right )}{3\,e^4}+\frac {2\,b\,x^2}{e^3}+\frac {\frac {2\,a^2\,e^2}{5}+\frac {8\,a\,b\,d\,e}{15}+\frac {16\,b^2\,d^2}{15}}{b\,e^5}\right )}{x^3\,\sqrt {d+e\,x}+\frac {a\,d^2\,\sqrt {d+e\,x}}{b\,e^2}+\frac {x^2\,\left (a\,e^5+2\,b\,d\,e^4\right )\,\sqrt {d+e\,x}}{b\,e^5}+\frac {d\,x\,\left (2\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}} \]
[In]
[Out]