Integrand size = 12, antiderivative size = 57 \[ \int x^3 \text {sech}^{-1}\left (a+b x^4\right ) \, dx=\frac {\left (a+b x^4\right ) \text {sech}^{-1}\left (a+b x^4\right )}{4 b}-\frac {\arctan \left (\sqrt {\frac {1-a-b x^4}{1+a+b x^4}}\right )}{2 b} \]
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Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6847, 6448, 1983, 12, 209} \[ \int x^3 \text {sech}^{-1}\left (a+b x^4\right ) \, dx=\frac {\left (a+b x^4\right ) \text {sech}^{-1}\left (a+b x^4\right )}{4 b}-\frac {\arctan \left (\sqrt {\frac {-a-b x^4+1}{a+b x^4+1}}\right )}{2 b} \]
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Rule 12
Rule 209
Rule 1983
Rule 6448
Rule 6847
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \text {sech}^{-1}(a+b x) \, dx,x,x^4\right ) \\ & = \frac {\left (a+b x^4\right ) \text {sech}^{-1}\left (a+b x^4\right )}{4 b}+\frac {1}{4} \text {Subst}\left (\int \frac {\sqrt {\frac {1-a-b x}{1+a+b x}}}{1-a-b x} \, dx,x,x^4\right ) \\ & = \frac {\left (a+b x^4\right ) \text {sech}^{-1}\left (a+b x^4\right )}{4 b}-b \text {Subst}\left (\int \frac {1}{2 b^2 \left (1+x^2\right )} \, dx,x,\sqrt {\frac {1-a-b x^4}{1+a+b x^4}}\right ) \\ & = \frac {\left (a+b x^4\right ) \text {sech}^{-1}\left (a+b x^4\right )}{4 b}-\frac {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {\frac {1-a-b x^4}{1+a+b x^4}}\right )}{2 b} \\ & = \frac {\left (a+b x^4\right ) \text {sech}^{-1}\left (a+b x^4\right )}{4 b}-\frac {\arctan \left (\sqrt {\frac {1-a-b x^4}{1+a+b x^4}}\right )}{2 b} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.86 \[ \int x^3 \text {sech}^{-1}\left (a+b x^4\right ) \, dx=\frac {\left (a+b x^4\right ) \text {sech}^{-1}\left (a+b x^4\right )+\frac {2 \sqrt {-\frac {-1+a+b x^4}{1+a+b x^4}} \sqrt {1-\left (a+b x^4\right )^2} \arctan \left (\frac {\sqrt {1-\left (a+b x^4\right )^2}}{-1+a+b x^4}\right )}{-1+a+b x^4}}{4 b} \]
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Time = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {\left (b \,x^{4}+a \right ) \operatorname {arcsech}\left (b \,x^{4}+a \right )-\arctan \left (\sqrt {\frac {1}{b \,x^{4}+a}-1}\, \sqrt {\frac {1}{b \,x^{4}+a}+1}\right )}{4 b}\) | \(53\) |
default | \(\frac {\left (b \,x^{4}+a \right ) \operatorname {arcsech}\left (b \,x^{4}+a \right )-\arctan \left (\sqrt {\frac {1}{b \,x^{4}+a}-1}\, \sqrt {\frac {1}{b \,x^{4}+a}+1}\right )}{4 b}\) | \(53\) |
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Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (49) = 98\).
Time = 0.27 (sec) , antiderivative size = 283, normalized size of antiderivative = 4.96 \[ \int x^3 \text {sech}^{-1}\left (a+b x^4\right ) \, dx=\frac {2 \, b x^{4} \log \left (\frac {{\left (b x^{4} + a\right )} \sqrt {-\frac {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}} + 1}{b x^{4} + a}\right ) + a \log \left (\frac {{\left (b x^{4} + a\right )} \sqrt {-\frac {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}} + 1}{x^{4}}\right ) - a \log \left (\frac {{\left (b x^{4} + a\right )} \sqrt {-\frac {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}} - 1}{x^{4}}\right ) - 2 \, \arctan \left (\frac {{\left (b^{2} x^{8} + 2 \, a b x^{4} + a^{2}\right )} \sqrt {-\frac {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2} - 1}\right )}{8 \, b} \]
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\[ \int x^3 \text {sech}^{-1}\left (a+b x^4\right ) \, dx=\int x^{3} \operatorname {asech}{\left (a + b x^{4} \right )}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.67 \[ \int x^3 \text {sech}^{-1}\left (a+b x^4\right ) \, dx=\frac {{\left (b x^{4} + a\right )} \operatorname {arsech}\left (b x^{4} + a\right ) - \arctan \left (\sqrt {\frac {1}{{\left (b x^{4} + a\right )}^{2}} - 1}\right )}{4 \, b} \]
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\[ \int x^3 \text {sech}^{-1}\left (a+b x^4\right ) \, dx=\int { x^{3} \operatorname {arsech}\left (b x^{4} + a\right ) \,d x } \]
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Time = 5.58 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98 \[ \int x^3 \text {sech}^{-1}\left (a+b x^4\right ) \, dx=\frac {\mathrm {atan}\left (\frac {1}{\sqrt {\frac {1}{b\,x^4+a}-1}\,\sqrt {\frac {1}{b\,x^4+a}+1}}\right )}{4\,b}+\frac {\mathrm {acosh}\left (\frac {1}{b\,x^4+a}\right )\,\left (b\,x^4+a\right )}{4\,b} \]
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