Integrand size = 46, antiderivative size = 193 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {4 (2 c d-b e) (5 c e f+3 c d g-4 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{15 c^3 e^2 \sqrt {d+e x}}-\frac {2 (5 c e f+3 c d g-4 b e g) \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{15 c^2 e^2}-\frac {2 g (d+e x)^{3/2} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{5 c e^2} \]
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Time = 0.20 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {808, 670, 662} \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {4 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-4 b e g+3 c d g+5 c e f)}{15 c^3 e^2 \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-4 b e g+3 c d g+5 c e f)}{15 c^2 e^2}-\frac {2 g (d+e x)^{3/2} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{5 c e^2} \]
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Rule 662
Rule 670
Rule 808
Rubi steps \begin{align*} \text {integral}& = -\frac {2 g (d+e x)^{3/2} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{5 c e^2}-\frac {\left (2 \left (\frac {1}{2} e \left (-2 c e^2 f+b e^2 g\right )+\frac {3}{2} \left (-c e^3 f+\left (-c d e^2+b e^3\right ) g\right )\right )\right ) \int \frac {(d+e x)^{3/2}}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{5 c e^3} \\ & = -\frac {2 (5 c e f+3 c d g-4 b e g) \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{15 c^2 e^2}-\frac {2 g (d+e x)^{3/2} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{5 c e^2}+\frac {(2 (2 c d-b e) (5 c e f+3 c d g-4 b e g)) \int \frac {\sqrt {d+e x}}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{15 c^2 e} \\ & = -\frac {4 (2 c d-b e) (5 c e f+3 c d g-4 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{15 c^3 e^2 \sqrt {d+e x}}-\frac {2 (5 c e f+3 c d g-4 b e g) \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{15 c^2 e^2}-\frac {2 g (d+e x)^{3/2} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{5 c e^2} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.61 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {2 \sqrt {d+e x} (-c d+b e+c e x) \left (8 b^2 e^2 g-2 b c e (5 e f+13 d g+2 e g x)+c^2 \left (18 d^2 g+e^2 x (5 f+3 g x)+d e (25 f+9 g x)\right )\right )}{15 c^3 e^2 \sqrt {(d+e x) (-b e+c (d-e x))}} \]
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Time = 0.37 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.62
method | result | size |
default | \(-\frac {2 \sqrt {-\left (e x +d \right ) \left (x c e +b e -c d \right )}\, \left (3 g \,x^{2} c^{2} e^{2}-4 b c \,e^{2} g x +9 c^{2} d e g x +5 c^{2} e^{2} f x +8 b^{2} e^{2} g -26 b c d e g -10 b c \,e^{2} f +18 c^{2} d^{2} g +25 c^{2} d e f \right )}{15 \sqrt {e x +d}\, c^{3} e^{2}}\) | \(119\) |
gosper | \(\frac {2 \left (x c e +b e -c d \right ) \left (3 g \,x^{2} c^{2} e^{2}-4 b c \,e^{2} g x +9 c^{2} d e g x +5 c^{2} e^{2} f x +8 b^{2} e^{2} g -26 b c d e g -10 b c \,e^{2} f +18 c^{2} d^{2} g +25 c^{2} d e f \right ) \sqrt {e x +d}}{15 c^{3} e^{2} \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}}\) | \(139\) |
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Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.74 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {2 \, {\left (3 \, c^{2} e^{2} g x^{2} + 5 \, {\left (5 \, c^{2} d e - 2 \, b c e^{2}\right )} f + 2 \, {\left (9 \, c^{2} d^{2} - 13 \, b c d e + 4 \, b^{2} e^{2}\right )} g + {\left (5 \, c^{2} e^{2} f + {\left (9 \, c^{2} d e - 4 \, b c e^{2}\right )} g\right )} x\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {e x + d}}{15 \, {\left (c^{3} e^{3} x + c^{3} d e^{2}\right )}} \]
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\[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\int \frac {\left (d + e x\right )^{\frac {3}{2}} \left (f + g x\right )}{\sqrt {- \left (d + e x\right ) \left (b e - c d + c e x\right )}}\, dx \]
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Time = 0.24 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.04 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {2 \, {\left (c^{2} e^{2} x^{2} - 5 \, c^{2} d^{2} + 7 \, b c d e - 2 \, b^{2} e^{2} + {\left (4 \, c^{2} d e - b c e^{2}\right )} x\right )} f}{3 \, \sqrt {-c e x + c d - b e} c^{2} e} + \frac {2 \, {\left (3 \, c^{3} e^{3} x^{3} - 18 \, c^{3} d^{3} + 44 \, b c^{2} d^{2} e - 34 \, b^{2} c d e^{2} + 8 \, b^{3} e^{3} + {\left (6 \, c^{3} d e^{2} - b c^{2} e^{3}\right )} x^{2} + {\left (9 \, c^{3} d^{2} e - 13 \, b c^{2} d e^{2} + 4 \, b^{2} c e^{3}\right )} x\right )} g}{15 \, \sqrt {-c e x + c d - b e} c^{3} e^{2}} \]
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Time = 0.31 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.55 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {2 \, {\left (\frac {15 \, {\left (2 \, c^{2} d e f - b c e^{2} f + 2 \, c^{2} d^{2} g - 3 \, b c d e g + b^{2} e^{2} g\right )} \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e}}{c^{3} e} - \frac {2 \, {\left (10 \, \sqrt {2 \, c d - b e} c^{2} d e f - 5 \, \sqrt {2 \, c d - b e} b c e^{2} f + 6 \, \sqrt {2 \, c d - b e} c^{2} d^{2} g - 11 \, \sqrt {2 \, c d - b e} b c d e g + 4 \, \sqrt {2 \, c d - b e} b^{2} e^{2} g\right )}}{c^{3} e} - \frac {5 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} c e f + 15 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} c d g - 10 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} b e g - 3 \, {\left ({\left (e x + d\right )} c - 2 \, c d + b e\right )}^{2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} g}{c^{3} e}\right )}}{15 \, e} \]
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Time = 11.40 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.77 \[ \int \frac {(d+e x)^{3/2} (f+g x)}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {\left (\frac {\sqrt {d+e\,x}\,\left (16\,g\,b^2\,e^2-52\,g\,b\,c\,d\,e-20\,f\,b\,c\,e^2+36\,g\,c^2\,d^2+50\,f\,c^2\,d\,e\right )}{15\,c^3\,e^3}+\frac {2\,g\,x^2\,\sqrt {d+e\,x}}{5\,c\,e}+\frac {2\,x\,\sqrt {d+e\,x}\,\left (9\,c\,d\,g-4\,b\,e\,g+5\,c\,e\,f\right )}{15\,c^2\,e^2}\right )\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}}{x+\frac {d}{e}} \]
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