Integrand size = 19, antiderivative size = 119 \[ \int x^5 \sqrt {b x^2+c x^4} \, dx=\frac {5 b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^3}-\frac {5 b \left (b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^{7/2}} \]
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Time = 0.08 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2043, 684, 654, 626, 634, 212} \[ \int x^5 \sqrt {b x^2+c x^4} \, dx=-\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^{7/2}}+\frac {5 b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^3}-\frac {5 b \left (b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{8 c} \]
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Rule 212
Rule 626
Rule 634
Rule 654
Rule 684
Rule 2043
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x^2 \sqrt {b x+c x^2} \, dx,x,x^2\right ) \\ & = \frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {(5 b) \text {Subst}\left (\int x \sqrt {b x+c x^2} \, dx,x,x^2\right )}{16 c} \\ & = -\frac {5 b \left (b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{8 c}+\frac {\left (5 b^2\right ) \text {Subst}\left (\int \sqrt {b x+c x^2} \, dx,x,x^2\right )}{32 c^2} \\ & = \frac {5 b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^3}-\frac {5 b \left (b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {\left (5 b^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{256 c^3} \\ & = \frac {5 b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^3}-\frac {5 b \left (b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {\left (5 b^4\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^3} \\ & = \frac {5 b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^3}-\frac {5 b \left (b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac {x^2 \left (b x^2+c x^4\right )^{3/2}}{8 c}-\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^{7/2}} \\ \end{align*}
Time = 0.43 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.02 \[ \int x^5 \sqrt {b x^2+c x^4} \, dx=\frac {x \sqrt {b+c x^2} \left (\sqrt {c} x \sqrt {b+c x^2} \left (15 b^3-10 b^2 c x^2+8 b c^2 x^4+48 c^3 x^6\right )+30 b^4 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b}-\sqrt {b+c x^2}}\right )\right )}{384 c^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \]
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Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.84
method | result | size |
pseudoelliptic | \(-\frac {5 \left (\ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{4}+\left (-\frac {32 c^{\frac {7}{2}} x^{6}}{5}-\frac {16 c^{\frac {5}{2}} b \,x^{4}}{15}+\frac {4 c^{\frac {3}{2}} b^{2} x^{2}}{3}-2 \sqrt {c}\, b^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}-\ln \left (2\right ) b^{4}\right )}{256 c^{\frac {7}{2}}}\) | \(100\) |
risch | \(\frac {\left (48 c^{3} x^{6}+8 b \,c^{2} x^{4}-10 b^{2} c \,x^{2}+15 b^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{384 c^{3}}-\frac {5 b^{4} \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{128 c^{\frac {7}{2}} x \sqrt {c \,x^{2}+b}}\) | \(101\) |
default | \(-\frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (-48 x^{5} \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {5}{2}}+40 c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,x^{3}-30 \sqrt {c}\, \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} x +15 \sqrt {c}\, \sqrt {c \,x^{2}+b}\, b^{3} x +15 \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) b^{4}\right )}{384 x \sqrt {c \,x^{2}+b}\, c^{\frac {7}{2}}}\) | \(124\) |
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Time = 0.27 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.58 \[ \int x^5 \sqrt {b x^2+c x^4} \, dx=\left [\frac {15 \, b^{4} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (48 \, c^{4} x^{6} + 8 \, b c^{3} x^{4} - 10 \, b^{2} c^{2} x^{2} + 15 \, b^{3} c\right )} \sqrt {c x^{4} + b x^{2}}}{768 \, c^{4}}, \frac {15 \, b^{4} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (48 \, c^{4} x^{6} + 8 \, b c^{3} x^{4} - 10 \, b^{2} c^{2} x^{2} + 15 \, b^{3} c\right )} \sqrt {c x^{4} + b x^{2}}}{384 \, c^{4}}\right ] \]
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\[ \int x^5 \sqrt {b x^2+c x^4} \, dx=\int x^{5} \sqrt {x^{2} \left (b + c x^{2}\right )}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.02 \[ \int x^5 \sqrt {b x^2+c x^4} \, dx=\frac {5 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}}{64 \, c^{2}} + \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2}}{8 \, c} - \frac {5 \, b^{4} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{256 \, c^{\frac {7}{2}}} + \frac {5 \, \sqrt {c x^{4} + b x^{2}} b^{3}}{128 \, c^{3}} - \frac {5 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b}{48 \, c^{2}} \]
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Time = 0.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.85 \[ \int x^5 \sqrt {b x^2+c x^4} \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, x^{2} \mathrm {sgn}\left (x\right ) + \frac {b \mathrm {sgn}\left (x\right )}{c}\right )} x^{2} - \frac {5 \, b^{2} \mathrm {sgn}\left (x\right )}{c^{2}}\right )} x^{2} + \frac {15 \, b^{3} \mathrm {sgn}\left (x\right )}{c^{3}}\right )} \sqrt {c x^{2} + b} x + \frac {5 \, b^{4} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right ) \mathrm {sgn}\left (x\right )}{128 \, c^{\frac {7}{2}}} - \frac {5 \, b^{4} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{256 \, c^{\frac {7}{2}}} \]
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Time = 13.32 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.88 \[ \int x^5 \sqrt {b x^2+c x^4} \, dx=\frac {x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{8\,c}-\frac {5\,b\,\left (\frac {b^3\,\ln \left (\frac {2\,c\,x^2+b}{\sqrt {c}}+2\,\sqrt {c\,x^4+b\,x^2}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{24\,c^2}\right )}{16\,c} \]
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