Integrand size = 12, antiderivative size = 102 \[ \int (b \text {sech}(c+d x))^{7/2} \, dx=\frac {6 i b^4 E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )}{5 d \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}+\frac {6 b^3 \sqrt {b \text {sech}(c+d x)} \sinh (c+d x)}{5 d}+\frac {2 b (b \text {sech}(c+d x))^{5/2} \sinh (c+d x)}{5 d} \]
2/5*b*(b*sech(d*x+c))^(5/2)*sinh(d*x+c)/d+6/5*I*b^4*(cosh(1/2*d*x+1/2*c)^2 )^(1/2)/cosh(1/2*d*x+1/2*c)*EllipticE(I*sinh(1/2*d*x+1/2*c),2^(1/2))/d/cos h(d*x+c)^(1/2)/(b*sech(d*x+c))^(1/2)+6/5*b^3*sinh(d*x+c)*(b*sech(d*x+c))^( 1/2)/d
Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67 \[ \int (b \text {sech}(c+d x))^{7/2} \, dx=\frac {b^2 (b \text {sech}(c+d x))^{3/2} \left (6 i \cosh ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )+3 \sinh (2 (c+d x))+2 \tanh (c+d x)\right )}{5 d} \]
(b^2*(b*Sech[c + d*x])^(3/2)*((6*I)*Cosh[c + d*x]^(3/2)*EllipticE[(I/2)*(c + d*x), 2] + 3*Sinh[2*(c + d*x)] + 2*Tanh[c + d*x]))/(5*d)
Time = 0.46 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4255, 3042, 4255, 3042, 4258, 3042, 3119}
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.
| \(\displaystyle \int (b \text {sech}(c+d x))^{7/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (b \csc \left (i c+i d x+\frac {\pi }{2}\right )\right )^{7/2}dx\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {3}{5} b^2 \int (b \text {sech}(c+d x))^{3/2}dx+\frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{5/2}}{5 d}+\frac {3}{5} b^2 \int \left (b \csc \left (i c+i d x+\frac {\pi }{2}\right )\right )^{3/2}dx\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {3}{5} b^2 \left (\frac {2 b \sinh (c+d x) \sqrt {b \text {sech}(c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \text {sech}(c+d x)}}dx\right )+\frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{5/2}}{5 d}+\frac {3}{5} b^2 \left (\frac {2 b \sinh (c+d x) \sqrt {b \text {sech}(c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \csc \left (i c+i d x+\frac {\pi }{2}\right )}}dx\right )\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {3}{5} b^2 \left (\frac {2 b \sinh (c+d x) \sqrt {b \text {sech}(c+d x)}}{d}-\frac {b^2 \int \sqrt {\cosh (c+d x)}dx}{\sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}\right )+\frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{5/2}}{5 d}+\frac {3}{5} b^2 \left (\frac {2 b \sinh (c+d x) \sqrt {b \text {sech}(c+d x)}}{d}-\frac {b^2 \int \sqrt {\sin \left (i c+i d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {2 b \sinh (c+d x) (b \text {sech}(c+d x))^{5/2}}{5 d}+\frac {3}{5} b^2 \left (\frac {2 b \sinh (c+d x) \sqrt {b \text {sech}(c+d x)}}{d}+\frac {2 i b^2 E\left (\left .\frac {1}{2} i (c+d x)\right |2\right )}{d \sqrt {\cosh (c+d x)} \sqrt {b \text {sech}(c+d x)}}\right )\) |
(2*b*(b*Sech[c + d*x])^(5/2)*Sinh[c + d*x])/(5*d) + (3*b^2*(((2*I)*b^2*Ell ipticE[(I/2)*(c + d*x), 2])/(d*Sqrt[Cosh[c + d*x]]*Sqrt[b*Sech[c + d*x]]) + (2*b*Sqrt[b*Sech[c + d*x]]*Sinh[c + d*x])/d))/5
3.1.15.3.1 Defintions of rubi rules used
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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]
\[\int \left (b \,\operatorname {sech}\left (d x +c \right )\right )^{\frac {7}{2}}d x\]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 478, normalized size of antiderivative = 4.69 \[ \int (b \text {sech}(c+d x))^{7/2} \, dx=\frac {2 \, {\left (3 \, \sqrt {2} {\left (b^{3} \cosh \left (d x + c\right )^{4} + 4 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b^{3} \sinh \left (d x + c\right )^{4} + 2 \, b^{3} \cosh \left (d x + c\right )^{2} + b^{3} + 2 \, {\left (3 \, b^{3} \cosh \left (d x + c\right )^{2} + b^{3}\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (b^{3} \cosh \left (d x + c\right )^{3} + b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (3 \, b^{3} \cosh \left (d x + c\right )^{5} + 15 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 3 \, b^{3} \sinh \left (d x + c\right )^{5} + 8 \, b^{3} \cosh \left (d x + c\right )^{3} + b^{3} \cosh \left (d x + c\right ) + 2 \, {\left (15 \, b^{3} \cosh \left (d x + c\right )^{2} + 4 \, b^{3}\right )} \sinh \left (d x + c\right )^{3} + 6 \, {\left (5 \, b^{3} \cosh \left (d x + c\right )^{3} + 4 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + {\left (15 \, b^{3} \cosh \left (d x + c\right )^{4} + 24 \, b^{3} \cosh \left (d x + c\right )^{2} + b^{3}\right )} \sinh \left (d x + c\right )\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}}\right )}}{5 \, {\left (d \cosh \left (d x + c\right )^{4} + 4 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + d \sinh \left (d x + c\right )^{4} + 2 \, d \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{2} + d\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (d \cosh \left (d x + c\right )^{3} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + d\right )}} \]
2/5*(3*sqrt(2)*(b^3*cosh(d*x + c)^4 + 4*b^3*cosh(d*x + c)*sinh(d*x + c)^3 + b^3*sinh(d*x + c)^4 + 2*b^3*cosh(d*x + c)^2 + b^3 + 2*(3*b^3*cosh(d*x + c)^2 + b^3)*sinh(d*x + c)^2 + 4*(b^3*cosh(d*x + c)^3 + b^3*cosh(d*x + c))* sinh(d*x + c))*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c osh(d*x + c) + sinh(d*x + c))) + sqrt(2)*(3*b^3*cosh(d*x + c)^5 + 15*b^3*c osh(d*x + c)*sinh(d*x + c)^4 + 3*b^3*sinh(d*x + c)^5 + 8*b^3*cosh(d*x + c) ^3 + b^3*cosh(d*x + c) + 2*(15*b^3*cosh(d*x + c)^2 + 4*b^3)*sinh(d*x + c)^ 3 + 6*(5*b^3*cosh(d*x + c)^3 + 4*b^3*cosh(d*x + c))*sinh(d*x + c)^2 + (15* b^3*cosh(d*x + c)^4 + 24*b^3*cosh(d*x + c)^2 + b^3)*sinh(d*x + c))*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)))/(d*cosh(d*x + c)^4 + 4*d*cosh(d*x + c)*s inh(d*x + c)^3 + d*sinh(d*x + c)^4 + 2*d*cosh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^2 + 4*(d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sin h(d*x + c) + d)
Timed out. \[ \int (b \text {sech}(c+d x))^{7/2} \, dx=\text {Timed out} \]
\[ \int (b \text {sech}(c+d x))^{7/2} \, dx=\int { \left (b \operatorname {sech}\left (d x + c\right )\right )^{\frac {7}{2}} \,d x } \]
\[ \int (b \text {sech}(c+d x))^{7/2} \, dx=\int { \left (b \operatorname {sech}\left (d x + c\right )\right )^{\frac {7}{2}} \,d x } \]
Timed out. \[ \int (b \text {sech}(c+d x))^{7/2} \, dx=\int {\left (\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}\right )}^{7/2} \,d x \]