Integrand size = 35, antiderivative size = 164 \[ \int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 (b d-a e) (B d-A e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x)}-\frac {2 (2 b B d-A b e-a B e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)}+\frac {2 b B (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^3 (a+b x)} \]
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Time = 0.05 (sec) , antiderivative size = 164, 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} \[ \int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (-a B e-A b e+2 b B d)}{5 e^3 (a+b x)}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e) (B d-A e)}{3 e^3 (a+b x)}+\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^3 (a+b x)} \]
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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 \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) \sqrt {d+e x}}{e^2}+\frac {b (-2 b B d+A b e+a B e) (d+e x)^{3/2}}{e^2}+\frac {b^2 B (d+e x)^{5/2}}{e^2}\right ) \, dx}{a b+b^2 x} \\ & = \frac {2 (b d-a e) (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 (2 b B d-A b e-a B e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)}+\frac {2 b B (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^3 (a+b x)} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.54 \[ \int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \sqrt {(a+b x)^2} (d+e x)^{3/2} \left (7 A b e (-2 d+3 e x)+7 a e (-2 B d+5 A e+3 B e x)+b B \left (8 d^2-12 d e x+15 e^2 x^2\right )\right )}{105 e^3 (a+b x)} \]
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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 = 0.48
method | result | size |
default | \(\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (e x +d \right )^{\frac {3}{2}} \left (15 B b \,e^{2} x^{2}+21 A b \,e^{2} x +21 B a \,e^{2} x -12 B b d e x +35 A a \,e^{2}-14 A b d e -14 B a d e +8 B b \,d^{2}\right )}{105 e^{3}}\) | \(79\) |
gosper | \(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (15 B b \,e^{2} x^{2}+21 A b \,e^{2} x +21 B a \,e^{2} x -12 B b d e x +35 A a \,e^{2}-14 A b d e -14 B a d e +8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{105 e^{3} \left (b x +a \right )}\) | \(89\) |
risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (15 x^{3} B b \,e^{3}+21 A b \,e^{3} x^{2}+21 B \,x^{2} a \,e^{3}+3 B b d \,e^{2} x^{2}+35 A a \,e^{3} x +7 A b d \,e^{2} x +7 B a d \,e^{2} x -4 B b \,d^{2} e x +35 A a d \,e^{2}-14 A b \,d^{2} e -14 B a \,d^{2} e +8 B b \,d^{3}\right ) \sqrt {e x +d}}{105 \left (b x +a \right ) e^{3}}\) | \(137\) |
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Time = 0.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.66 \[ \int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (15 \, B b e^{3} x^{3} + 8 \, B b d^{3} + 35 \, A a d e^{2} - 14 \, {\left (B a + A b\right )} d^{2} e + 3 \, {\left (B b d e^{2} + 7 \, {\left (B a + A b\right )} e^{3}\right )} x^{2} - {\left (4 \, B b d^{2} e - 35 \, A a e^{3} - 7 \, {\left (B a + A b\right )} d e^{2}\right )} x\right )} \sqrt {e x + d}}{105 \, e^{3}} \]
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\[ \int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\int \left (A + B x\right ) \sqrt {d + e x} \sqrt {\left (a + b x\right )^{2}}\, dx \]
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Time = 0.21 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.73 \[ \int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (3 \, b e^{2} x^{2} - 2 \, b d^{2} + 5 \, a d e + {\left (b d e + 5 \, a e^{2}\right )} x\right )} \sqrt {e x + d} A}{15 \, e^{2}} + \frac {2 \, {\left (15 \, b e^{3} x^{3} + 8 \, b d^{3} - 14 \, a d^{2} e + 3 \, {\left (b d e^{2} + 7 \, a e^{3}\right )} x^{2} - {\left (4 \, b d^{2} e - 7 \, a d e^{2}\right )} x\right )} \sqrt {e x + d} B}{105 \, e^{3}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (119) = 238\).
Time = 0.28 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.88 \[ \int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (105 \, \sqrt {e x + d} A a d \mathrm {sgn}\left (b x + a\right ) + 35 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A a \mathrm {sgn}\left (b x + a\right ) + \frac {35 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} B a d \mathrm {sgn}\left (b x + a\right )}{e} + \frac {35 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A b d \mathrm {sgn}\left (b x + a\right )}{e} + \frac {7 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} B b d \mathrm {sgn}\left (b x + a\right )}{e^{2}} + \frac {7 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} B a \mathrm {sgn}\left (b x + a\right )}{e} + \frac {7 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} A b \mathrm {sgn}\left (b x + a\right )}{e} + \frac {3 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} B b \mathrm {sgn}\left (b x + a\right )}{e^{2}}\right )}}{105 \, e} \]
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Timed out. \[ \int (A+B x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\int \sqrt {{\left (a+b\,x\right )}^2}\,\left (A+B\,x\right )\,\sqrt {d+e\,x} \,d x \]
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