Integrand size = 12, antiderivative size = 90 \[ \int \frac {1}{(b \text {csch}(c+d x))^{3/2}} \, dx=\frac {2 \cosh (c+d x)}{3 b d \sqrt {b \text {csch}(c+d x)}}+\frac {2 i \sqrt {b \text {csch}(c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right ),2\right ) \sqrt {i \sinh (c+d x)}}{3 b^2 d} \]
2/3*cosh(d*x+c)/b/d/(b*csch(d*x+c))^(1/2)-2/3*I*(sin(1/2*I*c+1/4*Pi+1/2*I* d*x)^2)^(1/2)/sin(1/2*I*c+1/4*Pi+1/2*I*d*x)*EllipticF(cos(1/2*I*c+1/4*Pi+1 /2*I*d*x),2^(1/2))*(b*csch(d*x+c))^(1/2)*(I*sinh(d*x+c))^(1/2)/b^2/d
Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(b \text {csch}(c+d x))^{3/2}} \, dx=\frac {\text {csch}^2(c+d x) \left (-2 i \operatorname {EllipticF}\left (\frac {1}{4} (-2 i c+\pi -2 i d x),2\right ) \sqrt {i \sinh (c+d x)}+\sinh (2 (c+d x))\right )}{3 d (b \text {csch}(c+d x))^{3/2}} \]
(Csch[c + d*x]^2*((-2*I)*EllipticF[((-2*I)*c + Pi - (2*I)*d*x)/4, 2]*Sqrt[ I*Sinh[c + d*x]] + Sinh[2*(c + d*x)]))/(3*d*(b*Csch[c + d*x])^(3/2))
Time = 0.35 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4256, 3042, 4258, 3042, 3120}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{(b \text {csch}(c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(i b \csc (i c+i d x))^{3/2}}dx\) |
\(\Big \downarrow \) 4256 |
\(\displaystyle \frac {2 \cosh (c+d x)}{3 b d \sqrt {b \text {csch}(c+d x)}}-\frac {\int \sqrt {b \text {csch}(c+d x)}dx}{3 b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \cosh (c+d x)}{3 b d \sqrt {b \text {csch}(c+d x)}}-\frac {\int \sqrt {i b \csc (i c+i d x)}dx}{3 b^2}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {2 \cosh (c+d x)}{3 b d \sqrt {b \text {csch}(c+d x)}}-\frac {\sqrt {i \sinh (c+d x)} \sqrt {b \text {csch}(c+d x)} \int \frac {1}{\sqrt {i \sinh (c+d x)}}dx}{3 b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \cosh (c+d x)}{3 b d \sqrt {b \text {csch}(c+d x)}}-\frac {\sqrt {i \sinh (c+d x)} \sqrt {b \text {csch}(c+d x)} \int \frac {1}{\sqrt {\sin (i c+i d x)}}dx}{3 b^2}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 \cosh (c+d x)}{3 b d \sqrt {b \text {csch}(c+d x)}}+\frac {2 i \sqrt {i \sinh (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right ),2\right ) \sqrt {b \text {csch}(c+d x)}}{3 b^2 d}\) |
(2*Cosh[c + d*x])/(3*b*d*Sqrt[b*Csch[c + d*x]]) + (((2*I)/3)*Sqrt[b*Csch[c + d*x]]*EllipticF[(I*c - Pi/2 + I*d*x)/2, 2]*Sqrt[I*Sinh[c + d*x]])/(b^2* d)
3.1.18.3.1 Defintions of rubi rules used
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n) Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
\[\int \frac {1}{\left (b \,\operatorname {csch}\left (d x +c \right )\right )^{\frac {3}{2}}}d x\]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.58 \[ \int \frac {1}{(b \text {csch}(c+d x))^{3/2}} \, dx=-\frac {4 \, \sqrt {2} {\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) - \sqrt {2} {\left (\cosh \left (d x + c\right )^{4} + 4 \, \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 6 \, \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 4 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + \sinh \left (d x + c\right )^{4} - 1\right )} \sqrt {\frac {b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1}}}{6 \, {\left (b^{2} d \cosh \left (d x + c\right )^{2} + 2 \, b^{2} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} d \sinh \left (d x + c\right )^{2}\right )}} \]
-1/6*(4*sqrt(2)*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d* x + c)^2)*sqrt(b)*weierstrassPInverse(4, 0, cosh(d*x + c) + sinh(d*x + c)) - sqrt(2)*(cosh(d*x + c)^4 + 4*cosh(d*x + c)^3*sinh(d*x + c) + 6*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^ 4 - 1)*sqrt((b*cosh(d*x + c) + b*sinh(d*x + c))/(cosh(d*x + c)^2 + 2*cosh( d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)))/(b^2*d*cosh(d*x + c)^2 + 2 *b^2*d*cosh(d*x + c)*sinh(d*x + c) + b^2*d*sinh(d*x + c)^2)
\[ \int \frac {1}{(b \text {csch}(c+d x))^{3/2}} \, dx=\int \frac {1}{\left (b \operatorname {csch}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{(b \text {csch}(c+d x))^{3/2}} \, dx=\int { \frac {1}{\left (b \operatorname {csch}\left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {1}{(b \text {csch}(c+d x))^{3/2}} \, dx=\int { \frac {1}{\left (b \operatorname {csch}\left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(b \text {csch}(c+d x))^{3/2}} \, dx=\int \frac {1}{{\left (\frac {b}{\mathrm {sinh}\left (c+d\,x\right )}\right )}^{3/2}} \,d x \]