Integrand size = 35, antiderivative size = 162 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 (b d-a e) (B d-A e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)}-\frac {2 (2 b B d-A b e-a B e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x)}+\frac {2 b B (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)} \]
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Time = 0.05 (sec) , antiderivative size = 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} \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (-a B e-A b e+2 b B d)}{3 e^3 (a+b x)}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e) (B d-A e)}{e^3 (a+b x)}+\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^3 (a+b x)} \]
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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)}{\sqrt {d+e x}} \, 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 \sqrt {d+e x}}+\frac {b (-2 b B d+A b e+a B e) \sqrt {d+e x}}{e^2}+\frac {b^2 B (d+e x)^{3/2}}{e^2}\right ) \, dx}{a b+b^2 x} \\ & = \frac {2 (b d-a e) (B d-A e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)}-\frac {2 (2 b B d-A b e-a B 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 B (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 = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.53 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {(a+b x)^2} \sqrt {d+e x} \left (5 A b e (-2 d+e x)+5 a e (-2 B d+3 A e+B e x)+b B \left (8 d^2-4 d e x+3 e^2 x^2\right )\right )}{15 e^3 (a+b x)} \]
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Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.22 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.49
method | result | size |
default | \(\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \sqrt {e x +d}\, \left (3 B b \,e^{2} x^{2}+5 A b \,e^{2} x +5 B a \,e^{2} x -4 B b d e x +15 A a \,e^{2}-10 A b d e -10 B a d e +8 B b \,d^{2}\right )}{15 e^{3}}\) | \(79\) |
gosper | \(\frac {2 \sqrt {e x +d}\, \left (3 B b \,e^{2} x^{2}+5 A b \,e^{2} x +5 B a \,e^{2} x -4 B b d e x +15 A a \,e^{2}-10 A b d e -10 B a d e +8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{3} \left (b x +a \right )}\) | \(89\) |
risch | \(\frac {2 \sqrt {e x +d}\, \left (3 B b \,e^{2} x^{2}+5 A b \,e^{2} x +5 B a \,e^{2} x -4 B b d e x +15 A a \,e^{2}-10 A b d e -10 B a d e +8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{3} \left (b x +a \right )}\) | \(89\) |
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Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.43 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (3 \, B b e^{2} x^{2} + 8 \, B b d^{2} + 15 \, A a e^{2} - 10 \, {\left (B a + A b\right )} d e - {\left (4 \, B b d e - 5 \, {\left (B a + A b\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, e^{3}} \]
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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 = 0.73 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (b e^{2} x^{2} - 2 \, b d^{2} + 3 \, a d e - {\left (b d e - 3 \, a e^{2}\right )} x\right )} A}{3 \, \sqrt {e x + d} e^{2}} + \frac {2 \, {\left (3 \, b e^{3} x^{3} + 8 \, b d^{3} - 10 \, a d^{2} e - {\left (b d e^{2} - 5 \, a e^{3}\right )} x^{2} + {\left (4 \, b d^{2} e - 5 \, a d e^{2}\right )} x\right )} B}{15 \, \sqrt {e x + d} e^{3}} \]
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Time = 0.28 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.80 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15 \, \sqrt {e x + d} A a \mathrm {sgn}\left (b x + a\right ) + \frac {5 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} B a \mathrm {sgn}\left (b x + a\right )}{e} + \frac {5 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A b \mathrm {sgn}\left (b x + a\right )}{e} + \frac {{\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} B b \mathrm {sgn}\left (b x + a\right )}{e^{2}}\right )}}{15 \, e} \]
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Time = 11.07 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx=\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,B\,x^3}{5}+\frac {16\,B\,b\,d^3+30\,A\,a\,d\,e^2-20\,A\,b\,d^2\,e-20\,B\,a\,d^2\,e}{15\,b\,e^3}+\frac {x\,\left (30\,A\,a\,e^3-10\,A\,b\,d\,e^2-10\,B\,a\,d\,e^2+8\,B\,b\,d^2\,e\right )}{15\,b\,e^3}+\frac {x^2\,\left (10\,A\,b\,e^3+10\,B\,a\,e^3-2\,B\,b\,d\,e^2\right )}{15\,b\,e^3}\right )}{x\,\sqrt {d+e\,x}+\frac {a\,\sqrt {d+e\,x}}{b}} \]
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