Integrand size = 12, antiderivative size = 118 \[ \int \frac {1}{(b \sinh (c+d x))^{7/2}} \, dx=-\frac {2 \cosh (c+d x)}{5 b d (b \sinh (c+d x))^{5/2}}+\frac {6 \cosh (c+d x)}{5 b^3 d \sqrt {b \sinh (c+d x)}}+\frac {6 i E\left (\left .\frac {1}{2} \left (i c-\frac {\pi }{2}+i d x\right )\right |2\right ) \sqrt {b \sinh (c+d x)}}{5 b^4 d \sqrt {i \sinh (c+d x)}} \] Output:
-2/5*cosh(d*x+c)/b/d/(b*sinh(d*x+c))^(5/2)+6/5*cosh(d*x+c)/b^3/d/(b*sinh(d *x+c))^(1/2)-6/5*I*EllipticE(cos(1/2*I*c+1/4*Pi+1/2*I*d*x),2^(1/2))*(b*sin h(d*x+c))^(1/2)/b^4/d/(I*sinh(d*x+c))^(1/2)
Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(b \sinh (c+d x))^{7/2}} \, dx=-\frac {2 \left (-3 \cosh (c+d x)+\coth (c+d x) \text {csch}(c+d x)+3 E\left (\left .\frac {1}{4} (-2 i c+\pi -2 i d x)\right |2\right ) \sqrt {i \sinh (c+d x)}\right )}{5 b^3 d \sqrt {b \sinh (c+d x)}} \] Input:
Integrate[(b*Sinh[c + d*x])^(-7/2),x]
Output:
(-2*(-3*Cosh[c + d*x] + Coth[c + d*x]*Csch[c + d*x] + 3*EllipticE[((-2*I)* c + Pi - (2*I)*d*x)/4, 2]*Sqrt[I*Sinh[c + d*x]]))/(5*b^3*d*Sqrt[b*Sinh[c + d*x]])
Time = 0.48 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3116, 3042, 3116, 3042, 3121, 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.
\(\displaystyle \int \frac {1}{(b \sinh (c+d x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(-i b \sin (i c+i d x))^{7/2}}dx\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle -\frac {3 \int \frac {1}{(b \sinh (c+d x))^{3/2}}dx}{5 b^2}-\frac {2 \cosh (c+d x)}{5 b d (b \sinh (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 \cosh (c+d x)}{5 b d (b \sinh (c+d x))^{5/2}}-\frac {3 \int \frac {1}{(-i b \sin (i c+i d x))^{3/2}}dx}{5 b^2}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle -\frac {3 \left (\frac {\int \sqrt {b \sinh (c+d x)}dx}{b^2}-\frac {2 \cosh (c+d x)}{b d \sqrt {b \sinh (c+d x)}}\right )}{5 b^2}-\frac {2 \cosh (c+d x)}{5 b d (b \sinh (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 \cosh (c+d x)}{5 b d (b \sinh (c+d x))^{5/2}}-\frac {3 \left (-\frac {2 \cosh (c+d x)}{b d \sqrt {b \sinh (c+d x)}}+\frac {\int \sqrt {-i b \sin (i c+i d x)}dx}{b^2}\right )}{5 b^2}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle -\frac {2 \cosh (c+d x)}{5 b d (b \sinh (c+d x))^{5/2}}-\frac {3 \left (-\frac {2 \cosh (c+d x)}{b d \sqrt {b \sinh (c+d x)}}+\frac {\sqrt {b \sinh (c+d x)} \int \sqrt {i \sinh (c+d x)}dx}{b^2 \sqrt {i \sinh (c+d x)}}\right )}{5 b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 \cosh (c+d x)}{5 b d (b \sinh (c+d x))^{5/2}}-\frac {3 \left (-\frac {2 \cosh (c+d x)}{b d \sqrt {b \sinh (c+d x)}}+\frac {\sqrt {b \sinh (c+d x)} \int \sqrt {\sin (i c+i d x)}dx}{b^2 \sqrt {i \sinh (c+d x)}}\right )}{5 b^2}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle -\frac {2 \cosh (c+d x)}{5 b d (b \sinh (c+d x))^{5/2}}-\frac {3 \left (-\frac {2 \cosh (c+d x)}{b d \sqrt {b \sinh (c+d x)}}-\frac {2 i E\left (\left .\frac {1}{2} \left (i c+i d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \sinh (c+d x)}}{b^2 d \sqrt {i \sinh (c+d x)}}\right )}{5 b^2}\) |
Input:
Int[(b*Sinh[c + d*x])^(-7/2),x]
Output:
(-2*Cosh[c + d*x])/(5*b*d*(b*Sinh[c + d*x])^(5/2)) - (3*((-2*Cosh[c + d*x] )/(b*d*Sqrt[b*Sinh[c + d*x]]) - ((2*I)*EllipticE[(I*c - Pi/2 + I*d*x)/2, 2 ]*Sqrt[b*Sinh[c + d*x]])/(b^2*d*Sqrt[I*Sinh[c + d*x]])))/(5*b^2)
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
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[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (97 ) = 194\).
Time = 0.24 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.74
method | result | size |
default | \(-\frac {6 \sqrt {-i \left (\sinh \left (d x +c \right )+i\right )}\, \sqrt {2}\, \sqrt {-i \left (i-\sinh \left (d x +c \right )\right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \sinh \left (d x +c \right )^{2} \operatorname {EllipticE}\left (\sqrt {-i \left (\sinh \left (d x +c \right )+i\right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-i \left (\sinh \left (d x +c \right )+i\right )}\, \sqrt {2}\, \sqrt {-i \left (i-\sinh \left (d x +c \right )\right )}\, \sqrt {i \sinh \left (d x +c \right )}\, \sinh \left (d x +c \right )^{2} \operatorname {EllipticF}\left (\sqrt {-i \left (\sinh \left (d x +c \right )+i\right )}, \frac {\sqrt {2}}{2}\right )-6 \sinh \left (d x +c \right )^{4}-4 \sinh \left (d x +c \right )^{2}+2}{5 b^{3} \sinh \left (d x +c \right )^{2} \cosh \left (d x +c \right ) \sqrt {b \sinh \left (d x +c \right )}\, d}\) | \(205\) |
Input:
int(1/(b*sinh(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
Output:
-1/5/b^3/sinh(d*x+c)^2*(6*(-I*(sinh(d*x+c)+I))^(1/2)*2^(1/2)*(-I*(I-sinh(d *x+c)))^(1/2)*(I*sinh(d*x+c))^(1/2)*sinh(d*x+c)^2*EllipticE((-I*(sinh(d*x+ c)+I))^(1/2),1/2*2^(1/2))-3*(-I*(sinh(d*x+c)+I))^(1/2)*2^(1/2)*(-I*(I-sinh (d*x+c)))^(1/2)*(I*sinh(d*x+c))^(1/2)*sinh(d*x+c)^2*EllipticF((-I*(sinh(d* x+c)+I))^(1/2),1/2*2^(1/2))-6*sinh(d*x+c)^4-4*sinh(d*x+c)^2+2)/cosh(d*x+c) /(b*sinh(d*x+c))^(1/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 624 vs. \(2 (93) = 186\).
Time = 0.09 (sec) , antiderivative size = 624, normalized size of antiderivative = 5.29 \[ \int \frac {1}{(b \sinh (c+d x))^{7/2}} \, dx =\text {Too large to display} \] Input:
integrate(1/(b*sinh(d*x+c))^(7/2),x, algorithm="fricas")
Output:
4/5*(3*sqrt(1/2)*(cosh(d*x + c)^6 + 6*cosh(d*x + c)*sinh(d*x + c)^5 + sinh (d*x + c)^6 + 3*(5*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^4 - 3*cosh(d*x + c)^ 4 + 4*(5*cosh(d*x + c)^3 - 3*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*cosh(d* x + c)^4 - 6*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^2 + 3*cosh(d*x + c)^2 + 6* (cosh(d*x + c)^5 - 2*cosh(d*x + c)^3 + cosh(d*x + c))*sinh(d*x + c) - 1)*s qrt(b)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cosh(d*x + c) + sin h(d*x + c))) + (3*cosh(d*x + c)^6 + 18*cosh(d*x + c)*sinh(d*x + c)^5 + 3*s inh(d*x + c)^6 + (45*cosh(d*x + c)^2 - 8)*sinh(d*x + c)^4 - 8*cosh(d*x + c )^4 + 4*(15*cosh(d*x + c)^3 - 8*cosh(d*x + c))*sinh(d*x + c)^3 + (45*cosh( d*x + c)^4 - 48*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^2 + cosh(d*x + c)^2 + 2 *(9*cosh(d*x + c)^5 - 16*cosh(d*x + c)^3 + cosh(d*x + c))*sinh(d*x + c))*s qrt(b*sinh(d*x + c)))/(b^4*d*cosh(d*x + c)^6 + 6*b^4*d*cosh(d*x + c)*sinh( d*x + c)^5 + b^4*d*sinh(d*x + c)^6 - 3*b^4*d*cosh(d*x + c)^4 + 3*b^4*d*cos h(d*x + c)^2 - b^4*d + 3*(5*b^4*d*cosh(d*x + c)^2 - b^4*d)*sinh(d*x + c)^4 + 4*(5*b^4*d*cosh(d*x + c)^3 - 3*b^4*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3 *(5*b^4*d*cosh(d*x + c)^4 - 6*b^4*d*cosh(d*x + c)^2 + b^4*d)*sinh(d*x + c) ^2 + 6*(b^4*d*cosh(d*x + c)^5 - 2*b^4*d*cosh(d*x + c)^3 + b^4*d*cosh(d*x + c))*sinh(d*x + c))
Timed out. \[ \int \frac {1}{(b \sinh (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(b*sinh(d*x+c))**(7/2),x)
Output:
Timed out
\[ \int \frac {1}{(b \sinh (c+d x))^{7/2}} \, dx=\int { \frac {1}{\left (b \sinh \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate(1/(b*sinh(d*x+c))^(7/2),x, algorithm="maxima")
Output:
integrate((b*sinh(d*x + c))^(-7/2), x)
\[ \int \frac {1}{(b \sinh (c+d x))^{7/2}} \, dx=\int { \frac {1}{\left (b \sinh \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate(1/(b*sinh(d*x+c))^(7/2),x, algorithm="giac")
Output:
integrate((b*sinh(d*x + c))^(-7/2), x)
Timed out. \[ \int \frac {1}{(b \sinh (c+d x))^{7/2}} \, dx=\int \frac {1}{{\left (b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{7/2}} \,d x \] Input:
int(1/(b*sinh(c + d*x))^(7/2),x)
Output:
int(1/(b*sinh(c + d*x))^(7/2), x)
\[ \int \frac {1}{(b \sinh (c+d x))^{7/2}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\sinh \left (d x +c \right )}}{\sinh \left (d x +c \right )^{4}}d x \right )}{b^{4}} \] Input:
int(1/(b*sinh(d*x+c))^(7/2),x)
Output:
(sqrt(b)*int(sqrt(sinh(c + d*x))/sinh(c + d*x)**4,x))/b**4