Integrand size = 19, antiderivative size = 124 \[ \int x^3 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {3 b^3 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^3}-\frac {b \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {\left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac {3 b^5 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{7/2}} \]
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Time = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2043, 654, 626, 634, 212} \[ \int x^3 \left (b x^2+c x^4\right )^{3/2} \, dx=-\frac {3 b^5 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{7/2}}+\frac {3 b^3 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^3}-\frac {b \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {\left (b x^2+c x^4\right )^{5/2}}{10 c} \]
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Rule 212
Rule 626
Rule 634
Rule 654
Rule 2043
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x \left (b x+c x^2\right )^{3/2} \, dx,x,x^2\right ) \\ & = \frac {\left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac {b \text {Subst}\left (\int \left (b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{4 c} \\ & = -\frac {b \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {\left (b x^2+c x^4\right )^{5/2}}{10 c}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \sqrt {b x+c x^2} \, dx,x,x^2\right )}{64 c^2} \\ & = \frac {3 b^3 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^3}-\frac {b \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {\left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac {\left (3 b^5\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{512 c^3} \\ & = \frac {3 b^3 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^3}-\frac {b \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {\left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac {\left (3 b^5\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^3 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^3}-\frac {b \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {\left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac {3 b^5 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{7/2}} \\ \end{align*}
Time = 0.45 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.06 \[ \int x^3 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {x \sqrt {b+c x^2} \left (\sqrt {c} x \sqrt {b+c x^2} \left (15 b^4-10 b^3 c x^2+8 b^2 c^2 x^4+176 b c^3 x^6+128 c^4 x^8\right )+30 b^5 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b}-\sqrt {b+c x^2}}\right )\right )}{1280 c^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \]
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Time = 0.14 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.90
method | result | size |
pseudoelliptic | \(-\frac {3 \left (\ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{5}+\left (-\frac {256 c^{\frac {9}{2}} x^{8}}{15}-\frac {352 c^{\frac {7}{2}} b \,x^{6}}{15}-\frac {16 c^{\frac {5}{2}} b^{2} x^{4}}{15}+\frac {4 c^{\frac {3}{2}} b^{3} x^{2}}{3}-2 \sqrt {c}\, b^{4}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}-\ln \left (2\right ) b^{5}\right )}{512 c^{\frac {7}{2}}}\) | \(111\) |
risch | \(\frac {\left (128 c^{4} x^{8}+176 c^{3} x^{6} b +8 b^{2} c^{2} x^{4}-10 x^{2} c \,b^{3}+15 b^{4}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{1280 c^{3}}-\frac {3 b^{5} \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{256 c^{\frac {7}{2}} x \sqrt {c \,x^{2}+b}}\) | \(112\) |
default | \(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (128 x^{5} \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {5}{2}}-80 c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \,x^{3}+40 \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2} x -10 \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{3} x -15 \sqrt {c}\, \sqrt {c \,x^{2}+b}\, b^{4} x -15 \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) b^{5}\right )}{1280 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {7}{2}}}\) | \(142\) |
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Time = 0.28 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.69 \[ \int x^3 \left (b x^2+c x^4\right )^{3/2} \, dx=\left [\frac {15 \, b^{5} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (128 \, c^{5} x^{8} + 176 \, b c^{4} x^{6} + 8 \, b^{2} c^{3} x^{4} - 10 \, b^{3} c^{2} x^{2} + 15 \, b^{4} c\right )} \sqrt {c x^{4} + b x^{2}}}{2560 \, c^{4}}, \frac {15 \, b^{5} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (128 \, c^{5} x^{8} + 176 \, b c^{4} x^{6} + 8 \, b^{2} c^{3} x^{4} - 10 \, b^{3} c^{2} x^{2} + 15 \, b^{4} c\right )} \sqrt {c x^{4} + b x^{2}}}{1280 \, c^{4}}\right ] \]
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\[ \int x^3 \left (b x^2+c x^4\right )^{3/2} \, dx=\int x^{3} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.15 \[ \int x^3 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {3 \, \sqrt {c x^{4} + b x^{2}} b^{3} x^{2}}{128 \, c^{2}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b x^{2}}{16 \, c} - \frac {3 \, b^{5} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{512 \, c^{\frac {7}{2}}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}} b^{4}}{256 \, c^{3}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2}}{32 \, c^{2}} + \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {5}{2}}}{10 \, c} \]
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Time = 0.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.93 \[ \int x^3 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {3 \, b^{5} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right ) \mathrm {sgn}\left (x\right )}{256 \, c^{\frac {7}{2}}} - \frac {3 \, b^{5} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{512 \, c^{\frac {7}{2}}} + \frac {1}{1280} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, c x^{2} \mathrm {sgn}\left (x\right ) + 11 \, b \mathrm {sgn}\left (x\right )\right )} x^{2} + \frac {b^{2} \mathrm {sgn}\left (x\right )}{c}\right )} x^{2} - \frac {5 \, b^{3} \mathrm {sgn}\left (x\right )}{c^{2}}\right )} x^{2} + \frac {15 \, b^{4} \mathrm {sgn}\left (x\right )}{c^{3}}\right )} \sqrt {c x^{2} + b} x \]
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Time = 13.12 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.08 \[ \int x^3 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {{\left (c\,x^4+b\,x^2\right )}^{5/2}}{10\,c}-\frac {b\,\left (\frac {x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{4}-\frac {3\,b^2\,\left (\frac {\left (2\,c\,x^2+b\right )\,\sqrt {c\,x^4+b\,x^2}}{4\,c}-\frac {b^2\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{8\,c^{3/2}}\right )}{16\,c}+\frac {b\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{8\,c}\right )}{4\,c} \]
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