Integrand size = 35, antiderivative size = 150 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)}-\frac {4 b (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x)}+\frac {2 b^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)} \]
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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} \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^3 (a+b x)}-\frac {4 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}{3 e^3 (a+b x)}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^2}{e^3 (a+b x)} \]
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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 )}{\sqrt {d+e x}} \, 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}{\sqrt {d+e x}} \, 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 \sqrt {d+e x}}-\frac {2 b (b d-a e) \sqrt {d+e x}}{e^2}+\frac {b^2 (d+e x)^{3/2}}{e^2}\right ) \, dx}{a b+b^2 x} \\ & = \frac {2 (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)}-\frac {4 b (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x)}+\frac {2 b^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.52 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {(a+b x)^2} \sqrt {d+e x} \left (15 a^2 e^2+10 a b e (-2 d+e x)+b^2 \left (8 d^2-4 d e x+3 e^2 x^2\right )\right )}{15 e^3 (a+b x)} \]
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Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.46
method | result | size |
default | \(\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \sqrt {e x +d}\, \left (3 b^{2} e^{2} x^{2}+10 a b \,e^{2} x -4 b^{2} d e x +15 e^{2} a^{2}-20 a b d e +8 b^{2} d^{2}\right )}{15 e^{3}}\) | \(69\) |
gosper | \(\frac {2 \sqrt {e x +d}\, \left (3 b^{2} e^{2} x^{2}+10 a b \,e^{2} x -4 b^{2} d e x +15 e^{2} a^{2}-20 a b d e +8 b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{3} \left (b x +a \right )}\) | \(79\) |
risch | \(\frac {2 \sqrt {e x +d}\, \left (3 b^{2} e^{2} x^{2}+10 a b \,e^{2} x -4 b^{2} d e x +15 e^{2} a^{2}-20 a b d e +8 b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{3} \left (b x +a \right )}\) | \(79\) |
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Time = 0.33 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.43 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (3 \, b^{2} e^{2} x^{2} + 8 \, b^{2} d^{2} - 20 \, a b d e + 15 \, a^{2} e^{2} - 2 \, {\left (2 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, e^{3}} \]
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\[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\int \frac {\left (a + b x\right ) \sqrt {\left (a + b x\right )^{2}}}{\sqrt {d + e x}}\, dx \]
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Time = 0.21 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.79 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (b e^{2} x^{2} - 2 \, b d^{2} + 3 \, a d e - {\left (b d e - 3 \, a e^{2}\right )} x\right )} a}{3 \, \sqrt {e x + d} e^{2}} + \frac {2 \, {\left (3 \, b e^{3} x^{3} + 8 \, b d^{3} - 10 \, a d^{2} e - {\left (b d e^{2} - 5 \, a e^{3}\right )} x^{2} + {\left (4 \, b d^{2} e - 5 \, a d e^{2}\right )} x\right )} b}{15 \, \sqrt {e x + d} e^{3}} \]
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Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.67 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15 \, \sqrt {e x + d} a^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {10 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a b \mathrm {sgn}\left (b x + a\right )}{e} + \frac {{\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} b^{2} \mathrm {sgn}\left (b x + a\right )}{e^{2}}\right )}}{15 \, e} \]
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Time = 11.37 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,b\,x^3}{5}+\frac {2\,x^2\,\left (10\,a\,e-b\,d\right )}{15\,e}+\frac {30\,a^2\,d\,e^2-40\,a\,b\,d^2\,e+16\,b^2\,d^3}{15\,b\,e^3}+\frac {x\,\left (30\,a^2\,e^3-20\,a\,b\,d\,e^2+8\,b^2\,d^2\,e\right )}{15\,b\,e^3}\right )}{x\,\sqrt {d+e\,x}+\frac {a\,\sqrt {d+e\,x}}{b}} \]
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