Integrand size = 21, antiderivative size = 132 \[ \int \tanh (x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \, dx=-\frac {(b+2 c) \text {arctanh}\left (\frac {b+2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{4 \sqrt {c}}+\frac {1}{2} \sqrt {a+b+c} \text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )-\frac {1}{2} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \]
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Time = 0.18 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3781, 1261, 748, 857, 635, 212, 738} \[ \int \tanh (x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \, dx=-\frac {(b+2 c) \text {arctanh}\left (\frac {b+2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{4 \sqrt {c}}+\frac {1}{2} \sqrt {a+b+c} \text {arctanh}\left (\frac {2 a+(b+2 c) \tanh ^2(x)+b}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )-\frac {1}{2} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \]
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Rule 212
Rule 635
Rule 738
Rule 748
Rule 857
Rule 1261
Rule 3781
Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {x \sqrt {a-b x^2+c x^4}}{1+x^2} \, dx,x,i \tanh (x)\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {a-b x+c x^2}}{1+x} \, dx,x,-\tanh ^2(x)\right )\right ) \\ & = -\frac {1}{2} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}+\frac {1}{4} \text {Subst}\left (\int \frac {-2 a-b+(b+2 c) x}{(1+x) \sqrt {a-b x+c x^2}} \, dx,x,-\tanh ^2(x)\right ) \\ & = -\frac {1}{2} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}+\frac {1}{2} (-a-b-c) \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a-b x+c x^2}} \, dx,x,-\tanh ^2(x)\right )+\frac {1}{4} (b+2 c) \text {Subst}\left (\int \frac {1}{\sqrt {a-b x+c x^2}} \, dx,x,-\tanh ^2(x)\right ) \\ & = -\frac {1}{2} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}+(a+b+c) \text {Subst}\left (\int \frac {1}{4 a+4 b+4 c-x^2} \, dx,x,\frac {2 a+b+(b+2 c) \tanh ^2(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )+\frac {1}{2} (b+2 c) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {-b-2 c \tanh ^2(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right ) \\ & = -\frac {(b+2 c) \text {arctanh}\left (\frac {b+2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{4 \sqrt {c}}+\frac {1}{2} \sqrt {a+b+c} \text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )-\frac {1}{2} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \\ \end{align*}
Time = 0.34 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.99 \[ \int \tanh (x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \, dx=\frac {1}{4} \left (\frac {(b+2 c) \text {arctanh}\left (\frac {-b-2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{\sqrt {c}}+2 \sqrt {a+b+c} \text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )-2 \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}\right ) \]
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Time = 0.75 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.25
method | result | size |
derivativedivides | \(-\frac {\sqrt {\left (\tanh \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+a +b +c}}{2}-\frac {\left (b +2 c \right ) \ln \left (\frac {\frac {b}{2}+c +c \left (\tanh \left (x \right )^{2}-1\right )}{\sqrt {c}}+\sqrt {\left (\tanh \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+a +b +c}\right )}{4 \sqrt {c}}+\frac {\sqrt {a +b +c}\, \ln \left (\frac {2 a +2 b +2 c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+2 \sqrt {a +b +c}\, \sqrt {\left (\tanh \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+a +b +c}}{\tanh \left (x \right )^{2}-1}\right )}{2}\) | \(165\) |
default | \(-\frac {\sqrt {\left (\tanh \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+a +b +c}}{2}-\frac {\left (b +2 c \right ) \ln \left (\frac {\frac {b}{2}+c +c \left (\tanh \left (x \right )^{2}-1\right )}{\sqrt {c}}+\sqrt {\left (\tanh \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+a +b +c}\right )}{4 \sqrt {c}}+\frac {\sqrt {a +b +c}\, \ln \left (\frac {2 a +2 b +2 c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+2 \sqrt {a +b +c}\, \sqrt {\left (\tanh \left (x \right )^{2}-1\right )^{2} c +\left (b +2 c \right ) \left (\tanh \left (x \right )^{2}-1\right )+a +b +c}}{\tanh \left (x \right )^{2}-1}\right )}{2}\) | \(165\) |
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Leaf count of result is larger than twice the leaf count of optimal. 1827 vs. \(2 (108) = 216\).
Time = 1.42 (sec) , antiderivative size = 7896, normalized size of antiderivative = 59.82 \[ \int \tanh (x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \, dx=\text {Too large to display} \]
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\[ \int \tanh (x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \, dx=\int \sqrt {a + b \tanh ^{2}{\left (x \right )} + c \tanh ^{4}{\left (x \right )}} \tanh {\left (x \right )}\, dx \]
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\[ \int \tanh (x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \, dx=\int { \sqrt {c \tanh \left (x\right )^{4} + b \tanh \left (x\right )^{2} + a} \tanh \left (x\right ) \,d x } \]
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\[ \int \tanh (x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \, dx=\int { \sqrt {c \tanh \left (x\right )^{4} + b \tanh \left (x\right )^{2} + a} \tanh \left (x\right ) \,d x } \]
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Timed out. \[ \int \tanh (x) \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)} \, dx=\int \mathrm {tanh}\left (x\right )\,\sqrt {c\,{\mathrm {tanh}\left (x\right )}^4+b\,{\mathrm {tanh}\left (x\right )}^2+a} \,d x \]
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