Integrand size = 26, antiderivative size = 149 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=\frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right )^{5/2} \arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{64 c^{7/2} d} \]
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Time = 0.08 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {699, 702, 211} \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=-\frac {\left (b^2-4 a c\right )^{5/2} \arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{64 c^{7/2} d}+\frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d} \]
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Rule 211
Rule 699
Rule 702
Rubi steps \begin{align*} \text {integral}& = \frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right ) \int \frac {\left (a+b x+c x^2\right )^{3/2}}{b d+2 c d x} \, dx}{4 c} \\ & = -\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}+\frac {\left (b^2-4 a c\right )^2 \int \frac {\sqrt {a+b x+c x^2}}{b d+2 c d x} \, dx}{16 c^2} \\ & = \frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right )^3 \int \frac {1}{(b d+2 c d x) \sqrt {a+b x+c x^2}} \, dx}{64 c^3} \\ & = \frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right )^3 \text {Subst}\left (\int \frac {1}{2 b^2 c d-8 a c^2 d+8 c^2 d x^2} \, dx,x,\sqrt {a+b x+c x^2}\right )}{16 c^2} \\ & = \frac {\left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}{32 c^3 d}-\frac {\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}{24 c^2 d}+\frac {\left (a+b x+c x^2\right )^{5/2}}{10 c d}-\frac {\left (b^2-4 a c\right )^{5/2} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{64 c^{7/2} d} \\ \end{align*}
Time = 1.22 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=\frac {\sqrt {c} \sqrt {a+x (b+c x)} \left (15 b^4-20 b^3 c x+28 b^2 c \left (-5 a+c x^2\right )+16 b c^2 x \left (11 a+6 c x^2\right )+16 c^2 \left (23 a^2+11 a c x^2+3 c^2 x^4\right )\right )-15 \left (-b^2+4 a c\right )^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {-b^2+4 a c} x}{\sqrt {a} (b+2 c x)-b \sqrt {a+x (b+c x)}}\right )}{480 c^{7/2} d} \]
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Time = 2.87 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.06
method | result | size |
pseudoelliptic | \(-\frac {\left (-\frac {b^{2}}{4}+a c \right )^{3} \operatorname {arctanh}\left (\frac {2 c \sqrt {c \,x^{2}+b x +a}}{\sqrt {4 c^{2} a -b^{2} c}}\right )-\frac {23 \sqrt {4 c^{2} a -b^{2} c}\, \left (\frac {3 c^{4} x^{4}}{23}+\frac {11 x^{2} \left (\frac {6 b x}{11}+a \right ) c^{3}}{23}+\left (\frac {7}{92} b^{2} x^{2}+\frac {11}{23} a b x +a^{2}\right ) c^{2}-\frac {35 \left (\frac {b x}{7}+a \right ) b^{2} c}{92}+\frac {15 b^{4}}{368}\right ) \sqrt {c \,x^{2}+b x +a}}{30}}{\sqrt {4 c^{2} a -b^{2} c}\, c^{3} d}\) | \(158\) |
risch | \(\frac {\left (48 c^{4} x^{4}+96 b \,c^{3} x^{3}+176 x^{2} c^{3} a +28 b^{2} c^{2} x^{2}+176 a b \,c^{2} x -20 b^{3} c x +368 a^{2} c^{2}-140 a \,b^{2} c +15 b^{4}\right ) \sqrt {c \,x^{2}+b x +a}}{480 c^{3} d}-\frac {\left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{64 c^{4} \sqrt {\frac {4 a c -b^{2}}{c}}\, d}\) | \(227\) |
default | \(\frac {\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {5}{2}}}{5}+\frac {\left (4 a c -b^{2}\right ) \left (\frac {\left (\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}\right )^{\frac {3}{2}}}{3}+\frac {\left (4 a c -b^{2}\right ) \left (\frac {\sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{2 c \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{4 c}\right )}{4 c}}{2 d c}\) | \(245\) |
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Time = 0.34 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.50 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=\left [\frac {15 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-\frac {b^{2} - 4 \, a c}{c}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt {c x^{2} + b x + a} c \sqrt {-\frac {b^{2} - 4 \, a c}{c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) + 4 \, {\left (48 \, c^{4} x^{4} + 96 \, b c^{3} x^{3} + 15 \, b^{4} - 140 \, a b^{2} c + 368 \, a^{2} c^{2} + 4 \, {\left (7 \, b^{2} c^{2} + 44 \, a c^{3}\right )} x^{2} - 4 \, {\left (5 \, b^{3} c - 44 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{1920 \, c^{3} d}, \frac {15 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {\frac {b^{2} - 4 \, a c}{c}} \arctan \left (\frac {\sqrt {\frac {b^{2} - 4 \, a c}{c}}}{2 \, \sqrt {c x^{2} + b x + a}}\right ) + 2 \, {\left (48 \, c^{4} x^{4} + 96 \, b c^{3} x^{3} + 15 \, b^{4} - 140 \, a b^{2} c + 368 \, a^{2} c^{2} + 4 \, {\left (7 \, b^{2} c^{2} + 44 \, a c^{3}\right )} x^{2} - 4 \, {\left (5 \, b^{3} c - 44 \, a b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{960 \, c^{3} d}\right ] \]
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\[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=\frac {\int \frac {a^{2} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {b^{2} x^{2} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {c^{2} x^{4} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {2 a b x \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {2 a c x^{2} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx + \int \frac {2 b c x^{3} \sqrt {a + b x + c x^{2}}}{b + 2 c x}\, dx}{d} \]
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Exception generated. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.29 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=\frac {1}{480} \, \sqrt {c x^{2} + b x + a} {\left (4 \, {\left ({\left (12 \, {\left (\frac {c x}{d} + \frac {2 \, b}{d}\right )} x + \frac {7 \, b^{2} c^{9} d^{5} + 44 \, a c^{10} d^{5}}{c^{10} d^{6}}\right )} x - \frac {5 \, b^{3} c^{8} d^{5} - 44 \, a b c^{9} d^{5}}{c^{10} d^{6}}\right )} x + \frac {15 \, b^{4} c^{7} d^{5} - 140 \, a b^{2} c^{8} d^{5} + 368 \, a^{2} c^{9} d^{5}}{c^{10} d^{6}}\right )} - \frac {{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \arctan \left (-\frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} c + b \sqrt {c}}{\sqrt {b^{2} c - 4 \, a c^{2}}}\right )}{32 \, \sqrt {b^{2} c - 4 \, a c^{2}} c^{3} d} \]
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Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{5/2}}{b d+2 c d x} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{b\,d+2\,c\,d\,x} \,d x \]
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