Integrand size = 46, antiderivative size = 200 \[ \int \frac {\sqrt {d+e x} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=-\frac {8 (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^3 d^3 e \sqrt {d+e x}}+\frac {8 g (c d f-a e g) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^2 d^2 e}+\frac {2 (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d \sqrt {d+e x}} \]
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Time = 0.14 (sec) , antiderivative size = 200, 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 = {884, 808, 662} \[ \int \frac {\sqrt {d+e x} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=-\frac {8 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right )}{15 c^3 d^3 e \sqrt {d+e x}}+\frac {8 g \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g)}{15 c^2 d^2 e}+\frac {2 (f+g x)^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 c d \sqrt {d+e x}} \]
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Rule 662
Rule 808
Rule 884
Rubi steps \begin{align*} \text {integral}& = \frac {2 (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d \sqrt {d+e x}}+\frac {\left (4 \left (c d e^2 f+c d^2 e g-e \left (c d^2+a e^2\right ) g\right )\right ) \int \frac {\sqrt {d+e x} (f+g x)}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{5 c d e^2} \\ & = \frac {8 g (c d f-a e g) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^2 d^2 e}+\frac {2 (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d \sqrt {d+e x}}-\frac {\left (4 (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right )\right ) \int \frac {\sqrt {d+e x}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{15 c^2 d^2 e} \\ & = -\frac {8 (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^3 d^3 e \sqrt {d+e x}}+\frac {8 g (c d f-a e g) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{15 c^2 d^2 e}+\frac {2 (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c d \sqrt {d+e x}} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.44 \[ \int \frac {\sqrt {d+e x} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \sqrt {(a e+c d x) (d+e x)} \left (8 a^2 e^2 g^2-4 a c d e g (5 f+g x)+c^2 d^2 \left (15 f^2+10 f g x+3 g^2 x^2\right )\right )}{15 c^3 d^3 \sqrt {d+e x}} \]
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Time = 0.53 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.49
method | result | size |
default | \(\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (3 g^{2} x^{2} c^{2} d^{2}-4 a c d e \,g^{2} x +10 c^{2} d^{2} f g x +8 a^{2} e^{2} g^{2}-20 a c d e f g +15 c^{2} d^{2} f^{2}\right )}{15 \sqrt {e x +d}\, c^{3} d^{3}}\) | \(98\) |
gosper | \(\frac {2 \left (c d x +a e \right ) \left (3 g^{2} x^{2} c^{2} d^{2}-4 a c d e \,g^{2} x +10 c^{2} d^{2} f g x +8 a^{2} e^{2} g^{2}-20 a c d e f g +15 c^{2} d^{2} f^{2}\right ) \sqrt {e x +d}}{15 c^{3} d^{3} \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}\) | \(116\) |
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Time = 0.31 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {d+e x} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, {\left (3 \, c^{2} d^{2} g^{2} x^{2} + 15 \, c^{2} d^{2} f^{2} - 20 \, a c d e f g + 8 \, a^{2} e^{2} g^{2} + 2 \, {\left (5 \, c^{2} d^{2} f g - 2 \, a c d e g^{2}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{15 \, {\left (c^{3} d^{3} e x + c^{3} d^{4}\right )}} \]
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\[ \int \frac {\sqrt {d+e x} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\int \frac {\sqrt {d + e x} \left (f + g x\right )^{2}}{\sqrt {\left (d + e x\right ) \left (a e + c d x\right )}}\, dx \]
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Time = 0.24 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt {d+e x} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, \sqrt {c d x + a e} f^{2}}{c d} + \frac {4 \, {\left (c^{2} d^{2} x^{2} - a c d e x - 2 \, a^{2} e^{2}\right )} f g}{3 \, \sqrt {c d x + a e} c^{2} d^{2}} + \frac {2 \, {\left (3 \, c^{3} d^{3} x^{3} - a c^{2} d^{2} e x^{2} + 4 \, a^{2} c d e^{2} x + 8 \, a^{3} e^{3}\right )} g^{2}}{15 \, \sqrt {c d x + a e} c^{3} d^{3}} \]
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Time = 0.29 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.75 \[ \int \frac {\sqrt {d+e x} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, e {\left (\frac {15 \, {\left (c^{2} d^{2} f^{2} - 2 \, a c d e f g + a^{2} e^{2} g^{2}\right )} \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}}}{c^{3} d^{3} e} - \frac {15 \, \sqrt {-c d^{2} e + a e^{3}} c^{2} d^{2} e^{2} f^{2} - 10 \, \sqrt {-c d^{2} e + a e^{3}} c^{2} d^{3} e f g - 20 \, \sqrt {-c d^{2} e + a e^{3}} a c d e^{3} f g + 3 \, \sqrt {-c d^{2} e + a e^{3}} c^{2} d^{4} g^{2} + 4 \, \sqrt {-c d^{2} e + a e^{3}} a c d^{2} e^{2} g^{2} + 8 \, \sqrt {-c d^{2} e + a e^{3}} a^{2} e^{4} g^{2}}{c^{3} d^{3} e^{3}} + \frac {10 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c d e^{2} f g - 10 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a e^{3} g^{2} + 3 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}} g^{2}}{c^{3} d^{3} e^{5}}\right )}}{15 \, {\left | e \right |}} \]
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Time = 12.50 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {d+e x} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {\sqrt {d+e\,x}\,\left (16\,a^2\,e^2\,g^2-40\,a\,c\,d\,e\,f\,g+30\,c^2\,d^2\,f^2\right )}{15\,c^3\,d^3\,e}+\frac {2\,g^2\,x^2\,\sqrt {d+e\,x}}{5\,c\,d\,e}-\frac {4\,g\,x\,\left (2\,a\,e\,g-5\,c\,d\,f\right )\,\sqrt {d+e\,x}}{15\,c^2\,d^2\,e}\right )}{x+\frac {d}{e}} \]
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