Integrand size = 23, antiderivative size = 135 \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\frac {(b-2 c) \text {arctanh}\left (\frac {b+2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{4 c^{3/2}}+\frac {\text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a+b+c}}-\frac {\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{2 c} \]
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Time = 0.25 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3781, 1265, 1667, 857, 635, 212, 738} \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\frac {(b-2 c) \text {arctanh}\left (\frac {b+2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{4 c^{3/2}}+\frac {\text {arctanh}\left (\frac {2 a+(b+2 c) \tanh ^2(x)+b}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{2 \sqrt {a+b+c}}-\frac {\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{2 c} \]
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Rule 212
Rule 635
Rule 738
Rule 857
Rule 1265
Rule 1667
Rule 3781
Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {x^5}{\left (1+x^2\right ) \sqrt {a-b x^2+c x^4}} \, dx,x,i \tanh (x)\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {x^2}{(1+x) \sqrt {a-b x+c x^2}} \, dx,x,-\tanh ^2(x)\right )\right ) \\ & = -\frac {\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{2 c}-\frac {\text {Subst}\left (\int \frac {\frac {b}{2}+\frac {1}{2} (b-2 c) x}{(1+x) \sqrt {a-b x+c x^2}} \, dx,x,-\tanh ^2(x)\right )}{2 c} \\ & = -\frac {\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{2 c}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a-b x+c x^2}} \, dx,x,-\tanh ^2(x)\right )-\frac {(b-2 c) \text {Subst}\left (\int \frac {1}{\sqrt {a-b x+c x^2}} \, dx,x,-\tanh ^2(x)\right )}{4 c} \\ & = -\frac {\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{2 c}-\frac {(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 )}{2 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 {(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 c^{3/2}}+\frac {\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+c}}-\frac {\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{2 c} \\ \end{align*}
Time = 1.40 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01 \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\frac {1}{4} \left (\frac {(-b+2 c) \text {arctanh}\left (\frac {-b-2 c \tanh ^2(x)}{2 \sqrt {c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{c^{3/2}}+\frac {2 \text {arctanh}\left (\frac {2 a+b+(b+2 c) \tanh ^2(x)}{2 \sqrt {a+b+c} \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}\right )}{\sqrt {a+b+c}}-\frac {2 \sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}}{c}\right ) \]
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Time = 0.82 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.10
method | result | size |
derivativedivides | \(-\frac {\ln \left (\frac {\frac {b}{2}+c \tanh \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}\right )}{2 \sqrt {c}}-\frac {\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}}{2 c}+\frac {b \ln \left (\frac {\frac {b}{2}+c \tanh \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}\right )}{4 c^{\frac {3}{2}}}+\frac {\operatorname {arctanh}\left (\frac {b \tanh \left (x \right )^{2}+2 c \tanh \left (x \right )^{2}+2 a +b}{2 \sqrt {a +b +c}\, \sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b +c}}\) | \(149\) |
default | \(-\frac {\ln \left (\frac {\frac {b}{2}+c \tanh \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}\right )}{2 \sqrt {c}}-\frac {\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}}{2 c}+\frac {b \ln \left (\frac {\frac {b}{2}+c \tanh \left (x \right )^{2}}{\sqrt {c}}+\sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}\right )}{4 c^{\frac {3}{2}}}+\frac {\operatorname {arctanh}\left (\frac {b \tanh \left (x \right )^{2}+2 c \tanh \left (x \right )^{2}+2 a +b}{2 \sqrt {a +b +c}\, \sqrt {a +b \tanh \left (x \right )^{2}+c \tanh \left (x \right )^{4}}}\right )}{2 \sqrt {a +b +c}}\) | \(149\) |
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Leaf count of result is larger than twice the leaf count of optimal. 2075 vs. \(2 (111) = 222\).
Time = 1.14 (sec) , antiderivative size = 8891, normalized size of antiderivative = 65.86 \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\text {Too large to display} \]
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\[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\int \frac {\tanh ^{5}{\left (x \right )}}{\sqrt {a + b \tanh ^{2}{\left (x \right )} + c \tanh ^{4}{\left (x \right )}}}\, dx \]
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\[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\int { \frac {\tanh \left (x\right )^{5}}{\sqrt {c \tanh \left (x\right )^{4} + b \tanh \left (x\right )^{2} + a}} \,d x } \]
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\[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\int { \frac {\tanh \left (x\right )^{5}}{\sqrt {c \tanh \left (x\right )^{4} + b \tanh \left (x\right )^{2} + a}} \,d x } \]
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Timed out. \[ \int \frac {\tanh ^5(x)}{\sqrt {a+b \tanh ^2(x)+c \tanh ^4(x)}} \, dx=\int \frac {{\mathrm {tanh}\left (x\right )}^5}{\sqrt {c\,{\mathrm {tanh}\left (x\right )}^4+b\,{\mathrm {tanh}\left (x\right )}^2+a}} \,d x \]
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