Integrand size = 24, antiderivative size = 148 \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {3 b^2 (b B-2 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^3}-\frac {(b B-2 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac {3 b^4 (b B-2 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{7/2}} \]
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Time = 0.13 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2059, 654, 626, 634, 212} \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=-\frac {3 b^4 (b B-2 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{7/2}}+\frac {3 b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4} (b B-2 A c)}{256 c^3}-\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2} (b B-2 A c)}{32 c^2}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{10 c} \]
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Rule 212
Rule 626
Rule 634
Rule 654
Rule 2059
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int (A+B x) \left (b x+c x^2\right )^{3/2} \, dx,x,x^2\right ) \\ & = \frac {B \left (b x^2+c x^4\right )^{5/2}}{10 c}+\frac {(-b B+2 A c) \text {Subst}\left (\int \left (b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{4 c} \\ & = -\frac {(b B-2 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{10 c}+\frac {\left (3 b^2 (b B-2 A c)\right ) \text {Subst}\left (\int \sqrt {b x+c x^2} \, dx,x,x^2\right )}{64 c^2} \\ & = \frac {3 b^2 (b B-2 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^3}-\frac {(b B-2 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac {\left (3 b^4 (b B-2 A c)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{512 c^3} \\ & = \frac {3 b^2 (b B-2 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^3}-\frac {(b B-2 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac {\left (3 b^4 (b B-2 A c)\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^3} \\ & = \frac {3 b^2 (b B-2 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^3}-\frac {(b B-2 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac {3 b^4 (b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{7/2}} \\ \end{align*}
Time = 1.14 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.34 \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (15 b^4 B-30 A b^3 c-10 b^3 B c x^2+20 A b^2 c^2 x^2+8 b^2 B c^2 x^4+240 A b c^3 x^4+176 b B c^3 x^6+160 A c^4 x^6+128 B c^4 x^8\right )}{1280 c^3 x^2 \left (b+c x^2\right )}-\frac {3 b^4 (b B-2 A c) \left (x^2 \left (b+c x^2\right )\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )}{128 c^{7/2} x^3 \left (b+c x^2\right )^{3/2}} \]
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Time = 1.86 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.07
method | result | size |
pseudoelliptic | \(-\frac {3 \left (\left (-\frac {1}{2} A \,b^{4} c +\frac {1}{4} b^{5} B \right ) \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )+\left (b^{3} \left (\frac {x^{2} B}{3}+A \right ) c^{\frac {3}{2}}-\frac {2 x^{2} \left (\frac {2 x^{2} B}{5}+A \right ) b^{2} c^{\frac {5}{2}}}{3}-8 x^{4} \left (\frac {11 x^{2} B}{15}+A \right ) b \,c^{\frac {7}{2}}-\frac {16 x^{6} \left (\frac {4 x^{2} B}{5}+A \right ) c^{\frac {9}{2}}}{3}-\frac {B \sqrt {c}\, b^{4}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}+\frac {\ln \left (2\right ) \left (A c -\frac {B b}{2}\right ) b^{4}}{2}\right )}{128 c^{\frac {7}{2}}}\) | \(159\) |
risch | \(-\frac {\left (-128 B \,x^{8} c^{4}-160 A \,x^{6} c^{4}-176 B \,x^{6} b \,c^{3}-240 A \,x^{4} b \,c^{3}-8 B \,x^{4} b^{2} c^{2}-20 A \,x^{2} b^{2} c^{2}+10 B \,x^{2} b^{3} c +30 A \,b^{3} c -15 B \,b^{4}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{1280 c^{3}}+\frac {3 b^{4} \left (2 A c -B b \right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{256 c^{\frac {7}{2}} x \sqrt {c \,x^{2}+b}}\) | \(164\) |
default | \(\frac {\left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}} \left (128 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {5}{2}} x^{5}+160 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {5}{2}} x^{3}-80 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {3}{2}} b \,x^{3}-80 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {3}{2}} b x +20 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}} b^{2} x +40 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} \sqrt {c}\, b^{2} x +30 A \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} b^{3} x -10 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}\, b^{3} x -15 B \sqrt {c \,x^{2}+b}\, \sqrt {c}\, b^{4} x +30 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{4} c -15 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{5}\right )}{1280 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {7}{2}}}\) | \(244\) |
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Time = 0.31 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.14 \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\left [-\frac {15 \, {\left (B b^{5} - 2 \, A b^{4} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (128 \, B c^{5} x^{8} + 16 \, {\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{6} + 15 \, B b^{4} c - 30 \, A b^{3} c^{2} + 8 \, {\left (B b^{2} c^{3} + 30 \, A b c^{4}\right )} x^{4} - 10 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{2560 \, c^{4}}, \frac {15 \, {\left (B b^{5} - 2 \, A b^{4} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (128 \, B c^{5} x^{8} + 16 \, {\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{6} + 15 \, B b^{4} c - 30 \, A b^{3} c^{2} + 8 \, {\left (B b^{2} c^{3} + 30 \, A b c^{4}\right )} x^{4} - 10 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{1280 \, c^{4}}\right ] \]
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Time = 1.01 (sec) , antiderivative size = 614, normalized size of antiderivative = 4.15 \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {A b \left (\begin {cases} \frac {b^{3} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{16 c^{2}} + \sqrt {b x^{2} + c x^{4}} \left (- \frac {b^{2}}{8 c^{2}} + \frac {b x^{2}}{12 c} + \frac {x^{4}}{3}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x^{2}\right )^{\frac {5}{2}}}{5 b^{2}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{2} + \frac {A c \left (\begin {cases} - \frac {5 b^{4} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{128 c^{3}} + \sqrt {b x^{2} + c x^{4}} \cdot \left (\frac {5 b^{3}}{64 c^{3}} - \frac {5 b^{2} x^{2}}{96 c^{2}} + \frac {b x^{4}}{24 c} + \frac {x^{6}}{4}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x^{2}\right )^{\frac {7}{2}}}{7 b^{3}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{2} + \frac {B b \left (\begin {cases} - \frac {5 b^{4} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{128 c^{3}} + \sqrt {b x^{2} + c x^{4}} \cdot \left (\frac {5 b^{3}}{64 c^{3}} - \frac {5 b^{2} x^{2}}{96 c^{2}} + \frac {b x^{4}}{24 c} + \frac {x^{6}}{4}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x^{2}\right )^{\frac {7}{2}}}{7 b^{3}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{2} + \frac {B c \left (\begin {cases} \frac {7 b^{5} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x^{2} + c x^{4}} + 2 c x^{2} \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x^{2}\right ) \log {\left (\frac {b}{2 c} + x^{2} \right )}}{\sqrt {c \left (\frac {b}{2 c} + x^{2}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{256 c^{4}} + \sqrt {b x^{2} + c x^{4}} \left (- \frac {7 b^{4}}{128 c^{4}} + \frac {7 b^{3} x^{2}}{192 c^{3}} - \frac {7 b^{2} x^{4}}{240 c^{2}} + \frac {b x^{6}}{40 c} + \frac {x^{8}}{5}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (b x^{2}\right )^{\frac {9}{2}}}{9 b^{4}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (128) = 256\).
Time = 0.22 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.80 \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {1}{256} \, {\left (32 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2} - \frac {12 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}}{c} + \frac {3 \, b^{4} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {5}{2}}} - \frac {6 \, \sqrt {c x^{4} + b x^{2}} b^{3}}{c^{2}} + \frac {16 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b}{c}\right )} A + \frac {1}{2560} \, {\left (\frac {60 \, \sqrt {c x^{4} + b x^{2}} b^{3} x^{2}}{c^{2}} - \frac {160 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b x^{2}}{c} - \frac {15 \, b^{5} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}} + \frac {30 \, \sqrt {c x^{4} + b x^{2}} b^{4}}{c^{3}} - \frac {80 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2}}{c^{2}} + \frac {256 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {5}{2}}}{c}\right )} B \]
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Time = 0.29 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.40 \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {1}{1280} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, B c x^{2} \mathrm {sgn}\left (x\right ) + \frac {11 \, B b c^{8} \mathrm {sgn}\left (x\right ) + 10 \, A c^{9} \mathrm {sgn}\left (x\right )}{c^{8}}\right )} x^{2} + \frac {B b^{2} c^{7} \mathrm {sgn}\left (x\right ) + 30 \, A b c^{8} \mathrm {sgn}\left (x\right )}{c^{8}}\right )} x^{2} - \frac {5 \, {\left (B b^{3} c^{6} \mathrm {sgn}\left (x\right ) - 2 \, A b^{2} c^{7} \mathrm {sgn}\left (x\right )\right )}}{c^{8}}\right )} x^{2} + \frac {15 \, {\left (B b^{4} c^{5} \mathrm {sgn}\left (x\right ) - 2 \, A b^{3} c^{6} \mathrm {sgn}\left (x\right )\right )}}{c^{8}}\right )} \sqrt {c x^{2} + b} x + \frac {3 \, {\left (B b^{5} \mathrm {sgn}\left (x\right ) - 2 \, A b^{4} c \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{256 \, c^{\frac {7}{2}}} - \frac {3 \, {\left (B b^{5} \log \left ({\left | b \right |}\right ) - 2 \, A b^{4} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{512 \, c^{\frac {7}{2}}} \]
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Time = 9.87 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.59 \[ \int x \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {B\,{\left (c\,x^4+b\,x^2\right )}^{5/2}}{10\,c}+\frac {A\,{\left (c\,x^4+b\,x^2\right )}^{3/2}\,\left (c\,x^2+\frac {b}{2}\right )}{8\,c}-\frac {3\,A\,b^2\,\left (\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2}-\frac {b^2\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{8\,c^{3/2}}\right )}{32\,c}-\frac {B\,b\,\left (\frac {x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{4}-\frac {3\,b^2\,\left (\frac {\left (2\,c\,x^2+b\right )\,\sqrt {c\,x^4+b\,x^2}}{4\,c}-\frac {b^2\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{8\,c^{3/2}}\right )}{16\,c}+\frac {b\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{8\,c}\right )}{4\,c} \]
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