Integrand size = 23, antiderivative size = 316 \[ \int \frac {\coth ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}-\frac {(2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{2 (a-b)^{5/2} d}+\frac {b \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{5/2} d}-\frac {b \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{5/2} d}-\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{5/2} d}-\frac {2 b^4}{a \left (a^2-b^2\right )^2 d \sqrt {a+b \text {sech}(c+d x)}}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 (a+b)^2 d (1-\text {sech}(c+d x))}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 (a-b)^2 d (1+\text {sech}(c+d x))} \]
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Time = 0.31 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3970, 912, 1349, 212, 205} \[ \int \frac {\coth ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}-\frac {2 b^4}{a d \left (a^2-b^2\right )^2 \sqrt {a+b \text {sech}(c+d x)}}+\frac {b \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{4 d (a-b)^{5/2}}-\frac {b \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{4 d (a+b)^{5/2}}-\frac {(2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{2 d (a-b)^{5/2}}-\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{2 d (a+b)^{5/2}}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 d (a+b)^2 (1-\text {sech}(c+d x))}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 d (a-b)^2 (\text {sech}(c+d x)+1)} \]
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Rule 205
Rule 212
Rule 912
Rule 1349
Rule 3970
Rubi steps \begin{align*} \text {integral}& = -\frac {b^4 \text {Subst}\left (\int \frac {1}{x (a+x)^{3/2} \left (b^2-x^2\right )^2} \, dx,x,b \text {sech}(c+d x)\right )}{d} \\ & = -\frac {\left (2 b^4\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (-a+x^2\right ) \left (-a^2+b^2+2 a x^2-x^4\right )^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{d} \\ & = -\frac {\left (2 b^4\right ) \text {Subst}\left (\int \left (-\frac {1}{a (a-b)^2 (a+b)^2 x^2}-\frac {1}{a b^4 \left (a-x^2\right )}-\frac {1}{4 (a-b) b^3 \left (a-b-x^2\right )^2}+\frac {2 a-3 b}{4 (a-b)^2 b^4 \left (a-b-x^2\right )}+\frac {1}{4 b^3 (a+b) \left (a+b-x^2\right )^2}+\frac {2 a+3 b}{4 b^4 (a+b)^2 \left (a+b-x^2\right )}\right ) \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{d} \\ & = -\frac {2 b^4}{a \left (a^2-b^2\right )^2 d \sqrt {a+b \text {sech}(c+d x)}}+\frac {2 \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{a d}-\frac {(2 a-3 b) \text {Subst}\left (\int \frac {1}{a-b-x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{2 (a-b)^2 d}+\frac {b \text {Subst}\left (\int \frac {1}{\left (a-b-x^2\right )^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{2 (a-b) d}-\frac {b \text {Subst}\left (\int \frac {1}{\left (a+b-x^2\right )^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{2 (a+b) d}-\frac {(2 a+3 b) \text {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{2 (a+b)^2 d} \\ & = \frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}-\frac {(2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{2 (a-b)^{5/2} d}-\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{5/2} d}-\frac {2 b^4}{a \left (a^2-b^2\right )^2 d \sqrt {a+b \text {sech}(c+d x)}}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 (a+b)^2 d (1-\text {sech}(c+d x))}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 (a-b)^2 d (1+\text {sech}(c+d x))}+\frac {b \text {Subst}\left (\int \frac {1}{a-b-x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{4 (a-b)^2 d}-\frac {b \text {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{4 (a+b)^2 d} \\ & = \frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}-\frac {(2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{2 (a-b)^{5/2} d}+\frac {b \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{5/2} d}-\frac {b \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{5/2} d}-\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{5/2} d}-\frac {2 b^4}{a \left (a^2-b^2\right )^2 d \sqrt {a+b \text {sech}(c+d x)}}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 (a+b)^2 d (1-\text {sech}(c+d x))}-\frac {\sqrt {a+b \text {sech}(c+d x)}}{4 (a-b)^2 d (1+\text {sech}(c+d x))} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.69 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.00 \[ \int \frac {\coth ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=-\frac {-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}-\frac {2 a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \text {sech}(c+d x)}{a-b}\right )}{(a-b) \sqrt {a+b \text {sech}(c+d x)}}+\frac {2 a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \text {sech}(c+d x)}{a+b}\right )}{(a+b) \sqrt {a+b \text {sech}(c+d x)}}+\frac {4 b \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \text {sech}(c+d x)}{a}\right )}{a \sqrt {a+b \text {sech}(c+d x)}}+\frac {b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\frac {a+b \text {sech}(c+d x)}{a-b}\right )}{(a-b)^2 \sqrt {a+b \text {sech}(c+d x)}}-\frac {b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\frac {a+b \text {sech}(c+d x)}{a+b}\right )}{(a+b)^2 \sqrt {a+b \text {sech}(c+d x)}}}{2 b d} \]
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\[\int \frac {\coth \left (d x +c \right )^{3}}{\left (a +b \,\operatorname {sech}\left (d x +c \right )\right )^{\frac {3}{2}}}d x\]
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Leaf count of result is larger than twice the leaf count of optimal. 5984 vs. \(2 (266) = 532\).
Time = 6.62 (sec) , antiderivative size = 53212, normalized size of antiderivative = 168.39 \[ \int \frac {\coth ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\text {Too large to display} \]
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\[ \int \frac {\coth ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\int \frac {\coth ^{3}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
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\[ \int \frac {\coth ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\int { \frac {\coth \left (d x + c\right )^{3}}{{\left (b \operatorname {sech}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {\coth ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\int { \frac {\coth \left (d x + c\right )^{3}}{{\left (b \operatorname {sech}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int \frac {\coth ^3(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {coth}\left (c+d\,x\right )}^3}{{\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}\right )}^{3/2}} \,d x \]
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