Integrand size = 28, antiderivative size = 231 \[ \int \frac {(b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}} \, dx=\frac {4}{5} d (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}+\frac {6 \left (b^2-4 a c\right )^{7/4} d^{5/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{5 c \sqrt {a+b x+c x^2}}-\frac {6 \left (b^2-4 a c\right )^{7/4} d^{5/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{5 c \sqrt {a+b x+c x^2}} \]
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Time = 0.15 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {706, 705, 704, 313, 227, 1213, 435} \[ \int \frac {(b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}} \, dx=-\frac {6 d^{5/2} \left (b^2-4 a c\right )^{7/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {6 d^{5/2} \left (b^2-4 a c\right )^{7/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{5 c \sqrt {a+b x+c x^2}}+\frac {4}{5} d \sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2} \]
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Rule 227
Rule 313
Rule 435
Rule 704
Rule 705
Rule 706
Rule 1213
Rubi steps \begin{align*} \text {integral}& = \frac {4}{5} d (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}+\frac {1}{5} \left (3 \left (b^2-4 a c\right ) d^2\right ) \int \frac {\sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}} \, dx \\ & = \frac {4}{5} d (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}+\frac {\left (3 \left (b^2-4 a c\right ) d^2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \int \frac {\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}{5 \sqrt {a+b x+c x^2}} \\ & = \frac {4}{5} d (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}+\frac {\left (6 \left (b^2-4 a c\right ) d \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-\frac {x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{5 c \sqrt {a+b x+c x^2}} \\ & = \frac {4}{5} d (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}-\frac {\left (6 \left (b^2-4 a c\right )^{3/2} d^2 \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 )}{5 c \sqrt {a+b x+c x^2}}+\frac {\left (6 \left (b^2-4 a c\right )^{3/2} d^2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \text {Subst}\left (\int \frac {1+\frac {x^2}{\sqrt {b^2-4 a c} d}}{\sqrt {1-\frac {x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{5 c \sqrt {a+b x+c x^2}} \\ & = \frac {4}{5} d (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}-\frac {6 \left (b^2-4 a c\right )^{7/4} d^{5/2} \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 )}{5 c \sqrt {a+b x+c x^2}}+\frac {\left (6 \left (b^2-4 a c\right )^{3/2} d^2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {x^2}{\sqrt {b^2-4 a c} d}}}{\sqrt {1-\frac {x^2}{\sqrt {b^2-4 a c} d}}} \, dx,x,\sqrt {b d+2 c d x}\right )}{5 c \sqrt {a+b x+c x^2}} \\ & = \frac {4}{5} d (b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}+\frac {6 \left (b^2-4 a c\right )^{7/4} d^{5/2} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\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 )}{5 c \sqrt {a+b x+c x^2}}-\frac {6 \left (b^2-4 a c\right )^{7/4} d^{5/2} \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 )}{5 c \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 = 10.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.48 \[ \int \frac {(b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}} \, dx=\frac {2 d (d (b+2 c x))^{3/2} \left (2 c (a+x (b+c x))+\left (b^2-4 a c\right ) \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{5 c \sqrt {a+x (b+c x)}} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(495\) vs. \(2(193)=386\).
Time = 3.40 (sec) , antiderivative size = 496, normalized size of antiderivative = 2.15
method | result | size |
default | \(-\frac {\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, d^{2} \left (48 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) a^{2} c^{2}-24 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) a \,b^{2} c +3 \sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-b -2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, E\left (\frac {\sqrt {\frac {b +2 c x +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {2}}{2}, \sqrt {2}\right ) b^{4}-16 c^{4} x^{4}-32 b \,c^{3} x^{3}-16 x^{2} c^{3} a -20 b^{2} c^{2} x^{2}-16 a b \,c^{2} x -4 b^{3} c x -4 a \,b^{2} c \right )}{5 c \left (2 c^{2} x^{3}+3 c b \,x^{2}+2 a c x +b^{2} x +a b \right )}\) | \(496\) |
elliptic | \(\text {Expression too large to display}\) | \(1090\) |
risch | \(\text {Expression too large to display}\) | \(1355\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.25 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.47 \[ \int \frac {(b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}} \, dx=-\frac {2 \, {\left (3 \, \sqrt {2} \sqrt {c^{2} d} {\left (b^{2} - 4 \, a c\right )} d^{2} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right ) - 2 \, {\left (2 \, c^{2} d^{2} x + b c d^{2}\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}\right )}}{5 \, c} \]
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\[ \int \frac {(b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}} \, dx=\int \frac {\left (d \left (b + 2 c x\right )\right )^{\frac {5}{2}}}{\sqrt {a + b x + c x^{2}}}\, dx \]
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\[ \int \frac {(b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {5}{2}}}{\sqrt {c x^{2} + b x + a}} \,d x } \]
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\[ \int \frac {(b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}} \, dx=\int { \frac {{\left (2 \, c d x + b d\right )}^{\frac {5}{2}}}{\sqrt {c x^{2} + b x + a}} \,d x } \]
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Timed out. \[ \int \frac {(b d+2 c d x)^{5/2}}{\sqrt {a+b x+c x^2}} \, dx=\int \frac {{\left (b\,d+2\,c\,d\,x\right )}^{5/2}}{\sqrt {c\,x^2+b\,x+a}} \,d x \]
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