Integrand size = 28, antiderivative size = 125 \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}}+\frac {4 \sqrt [4]{b^2-4 a c} d^{3/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{\sqrt {a+b x+c x^2}} \]
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Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {700, 705, 703, 227} \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {4 d^{3/2} \sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{\sqrt {a+b x+c x^2}}-\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}} \]
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Rule 227
Rule 700
Rule 703
Rule 705
Rubi steps \begin{align*} \text {integral}& = -\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}}+\left (2 c d^2\right ) \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx \\ & = -\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}}+\frac {\left (2 c d^2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {-\frac {a c}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {c^2 x^2}{b^2-4 a c}}} \, dx}{\sqrt {a+b x+c x^2}} \\ & = -\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}}+\frac {\left (4 d \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{\sqrt {a+b x+c x^2}} \\ & = -\frac {2 d \sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}}+\frac {4 \sqrt [4]{b^2-4 a c} d^{3/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{\sqrt {a+b x+c x^2}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.70 \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {2 d \sqrt {d (b+2 c x)} \left (-1+2 \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{\sqrt {a+x (b+c x)}} \]
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Time = 2.43 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.55
method | result | size |
default | \(\frac {2 \sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, d \left (\sqrt {-4 a c +b^{2}}\, \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, F\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right )-2 c x -b \right )}{2 c^{2} x^{3}+3 c b \,x^{2}+2 a c x +b^{2} x +a b}\) | \(194\) |
elliptic | \(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \sqrt {d \left (2 c x +b \right )}\, \left (-\frac {2 \left (2 c^{2} d x +b c d \right ) d}{c \sqrt {\left (\frac {a}{c}+\frac {b x}{c}+x^{2}\right ) \left (2 c^{2} d x +b c d \right )}}+\frac {4 c \,d^{2} \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, F\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +a b d}}\right )}{\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}\, d}\) | \(472\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.78 \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (\sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a} c d - \sqrt {2} {\left (c d x^{2} + b d x + a d\right )} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right )}}{c^{2} x^{2} + b c x + a c} \]
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\[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {\left (d \left (b + 2 c x\right )\right )^{\frac {3}{2}}}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]
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\[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]
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