Integrand size = 10, antiderivative size = 55 \[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=\frac {\coth (x)}{5 \left (a \text {csch}^2(x)\right )^{5/2}}-\frac {4 \coth (x)}{15 a \left (a \text {csch}^2(x)\right )^{3/2}}+\frac {8 \coth (x)}{15 a^2 \sqrt {a \text {csch}^2(x)}} \] Output:
1/5*coth(x)/(a*csch(x)^2)^(5/2)-4/15*coth(x)/a/(a*csch(x)^2)^(3/2)+8/15*co th(x)/a^2/(a*csch(x)^2)^(1/2)
Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=\frac {(150 \cosh (x)-25 \cosh (3 x)+3 \cosh (5 x)) \sqrt {a \text {csch}^2(x)} \sinh (x)}{240 a^3} \] Input:
Integrate[(a*Csch[x]^2)^(-5/2),x]
Output:
((150*Cosh[x] - 25*Cosh[3*x] + 3*Cosh[5*x])*Sqrt[a*Csch[x]^2]*Sinh[x])/(24 0*a^3)
Time = 0.25 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.47, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4610, 209, 209, 208}
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.
\(\displaystyle \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (-a \sec \left (\frac {\pi }{2}+i x\right )^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4610 |
\(\displaystyle -a \int \frac {1}{\left (a \coth ^2(x)-a\right )^{7/2}}d\coth (x)\) |
\(\Big \downarrow \) 209 |
\(\displaystyle -a \left (-\frac {4 \int \frac {1}{\left (a \coth ^2(x)-a\right )^{5/2}}d\coth (x)}{5 a}-\frac {\coth (x)}{5 a \left (a \coth ^2(x)-a\right )^{5/2}}\right )\) |
\(\Big \downarrow \) 209 |
\(\displaystyle -a \left (-\frac {4 \left (-\frac {2 \int \frac {1}{\left (a \coth ^2(x)-a\right )^{3/2}}d\coth (x)}{3 a}-\frac {\coth (x)}{3 a \left (a \coth ^2(x)-a\right )^{3/2}}\right )}{5 a}-\frac {\coth (x)}{5 a \left (a \coth ^2(x)-a\right )^{5/2}}\right )\) |
\(\Big \downarrow \) 208 |
\(\displaystyle -a \left (-\frac {4 \left (\frac {2 \coth (x)}{3 a^2 \sqrt {a \coth ^2(x)-a}}-\frac {\coth (x)}{3 a \left (a \coth ^2(x)-a\right )^{3/2}}\right )}{5 a}-\frac {\coth (x)}{5 a \left (a \coth ^2(x)-a\right )^{5/2}}\right )\) |
Input:
Int[(a*Csch[x]^2)^(-5/2),x]
Output:
-(a*(-1/5*Coth[x]/(a*(-a + a*Coth[x]^2)^(5/2)) - (4*(-1/3*Coth[x]/(a*(-a + a*Coth[x]^2)^(3/2)) + (2*Coth[x])/(3*a^2*Sqrt[-a + a*Coth[x]^2])))/(5*a)) )
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFac tors[Tan[e + f*x], x]}, Simp[b*(ff/f) Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] && !IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(195\) vs. \(2(43)=86\).
Time = 0.14 (sec) , antiderivative size = 196, normalized size of antiderivative = 3.56
method | result | size |
risch | \(\frac {{\mathrm e}^{6 x}}{160 a^{2} \left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}}-\frac {5 \,{\mathrm e}^{4 x}}{96 a^{2} \left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}}+\frac {5 \,{\mathrm e}^{2 x}}{16 a^{2} \left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}}+\frac {5}{16 \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}\, \left ({\mathrm e}^{2 x}-1\right ) a^{2}}-\frac {5 \,{\mathrm e}^{-2 x}}{96 a^{2} \left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}}+\frac {{\mathrm e}^{-4 x}}{160 a^{2} \left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}}\) | \(196\) |
Input:
int(1/(a*csch(x)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/160/a^2*exp(6*x)/(exp(2*x)-1)/(a*exp(2*x)/(exp(2*x)-1)^2)^(1/2)-5/96/a^2 *exp(4*x)/(exp(2*x)-1)/(a*exp(2*x)/(exp(2*x)-1)^2)^(1/2)+5/16/a^2*exp(2*x) /(exp(2*x)-1)/(a*exp(2*x)/(exp(2*x)-1)^2)^(1/2)+5/16/(a*exp(2*x)/(exp(2*x) -1)^2)^(1/2)/(exp(2*x)-1)/a^2-5/96/a^2*exp(-2*x)/(exp(2*x)-1)/(a*exp(2*x)/ (exp(2*x)-1)^2)^(1/2)+1/160/a^2*exp(-4*x)/(exp(2*x)-1)/(a*exp(2*x)/(exp(2* x)-1)^2)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 590 vs. \(2 (43) = 86\).
Time = 0.09 (sec) , antiderivative size = 590, normalized size of antiderivative = 10.73 \[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate(1/(a*csch(x)^2)^(5/2),x, algorithm="fricas")
Output:
1/480*(3*(e^(2*x) - 1)*sinh(x)^10 - 3*cosh(x)^10 + 30*(cosh(x)*e^(2*x) - c osh(x))*sinh(x)^9 - 5*(27*cosh(x)^2 - (27*cosh(x)^2 - 5)*e^(2*x) - 5)*sinh (x)^8 + 25*cosh(x)^8 - 40*(9*cosh(x)^3 - (9*cosh(x)^3 - 5*cosh(x))*e^(2*x) - 5*cosh(x))*sinh(x)^7 - 10*(63*cosh(x)^4 - 70*cosh(x)^2 - (63*cosh(x)^4 - 70*cosh(x)^2 + 15)*e^(2*x) + 15)*sinh(x)^6 - 150*cosh(x)^6 - 4*(189*cosh (x)^5 - 350*cosh(x)^3 - (189*cosh(x)^5 - 350*cosh(x)^3 + 225*cosh(x))*e^(2 *x) + 225*cosh(x))*sinh(x)^5 - 10*(63*cosh(x)^6 - 175*cosh(x)^4 + 225*cosh (x)^2 - (63*cosh(x)^6 - 175*cosh(x)^4 + 225*cosh(x)^2 + 15)*e^(2*x) + 15)* sinh(x)^4 - 150*cosh(x)^4 - 40*(9*cosh(x)^7 - 35*cosh(x)^5 + 75*cosh(x)^3 - (9*cosh(x)^7 - 35*cosh(x)^5 + 75*cosh(x)^3 + 15*cosh(x))*e^(2*x) + 15*co sh(x))*sinh(x)^3 - 5*(27*cosh(x)^8 - 140*cosh(x)^6 + 450*cosh(x)^4 + 180*c osh(x)^2 - (27*cosh(x)^8 - 140*cosh(x)^6 + 450*cosh(x)^4 + 180*cosh(x)^2 - 5)*e^(2*x) - 5)*sinh(x)^2 + 25*cosh(x)^2 + (3*cosh(x)^10 - 25*cosh(x)^8 + 150*cosh(x)^6 + 150*cosh(x)^4 - 25*cosh(x)^2 + 3)*e^(2*x) - 10*(3*cosh(x) ^9 - 20*cosh(x)^7 + 90*cosh(x)^5 + 60*cosh(x)^3 - (3*cosh(x)^9 - 20*cosh(x )^7 + 90*cosh(x)^5 + 60*cosh(x)^3 - 5*cosh(x))*e^(2*x) - 5*cosh(x))*sinh(x ) - 3)*sqrt(a/(e^(4*x) - 2*e^(2*x) + 1))*e^x/(a^3*cosh(x)^5*e^x + 5*a^3*co sh(x)^4*e^x*sinh(x) + 10*a^3*cosh(x)^3*e^x*sinh(x)^2 + 10*a^3*cosh(x)^2*e^ x*sinh(x)^3 + 5*a^3*cosh(x)*e^x*sinh(x)^4 + a^3*e^x*sinh(x)^5)
\[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=\int \frac {1}{\left (a \operatorname {csch}^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(a*csch(x)**2)**(5/2),x)
Output:
Integral((a*csch(x)**2)**(-5/2), x)
Time = 0.14 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=-\frac {e^{\left (5 \, x\right )}}{160 \, a^{\frac {5}{2}}} + \frac {5 \, e^{\left (3 \, x\right )}}{96 \, a^{\frac {5}{2}}} - \frac {5 \, e^{\left (-x\right )}}{16 \, a^{\frac {5}{2}}} + \frac {5 \, e^{\left (-3 \, x\right )}}{96 \, a^{\frac {5}{2}}} - \frac {e^{\left (-5 \, x\right )}}{160 \, a^{\frac {5}{2}}} - \frac {5 \, e^{x}}{16 \, a^{\frac {5}{2}}} \] Input:
integrate(1/(a*csch(x)^2)^(5/2),x, algorithm="maxima")
Output:
-1/160*e^(5*x)/a^(5/2) + 5/96*e^(3*x)/a^(5/2) - 5/16*e^(-x)/a^(5/2) + 5/96 *e^(-3*x)/a^(5/2) - 1/160*e^(-5*x)/a^(5/2) - 5/16*e^x/a^(5/2)
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=\frac {{\left (150 \, e^{\left (4 \, x\right )} - 25 \, e^{\left (2 \, x\right )} + 3\right )} e^{\left (-5 \, x\right )} + 3 \, e^{\left (5 \, x\right )} - 25 \, e^{\left (3 \, x\right )} + 150 \, e^{x}}{480 \, a^{\frac {5}{2}} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )} \] Input:
integrate(1/(a*csch(x)^2)^(5/2),x, algorithm="giac")
Output:
1/480*((150*e^(4*x) - 25*e^(2*x) + 3)*e^(-5*x) + 3*e^(5*x) - 25*e^(3*x) + 150*e^x)/(a^(5/2)*sgn(e^(3*x) - e^x))
Timed out. \[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (\frac {a}{{\mathrm {sinh}\left (x\right )}^2}\right )}^{5/2}} \,d x \] Input:
int(1/(a/sinh(x)^2)^(5/2),x)
Output:
int(1/(a/sinh(x)^2)^(5/2), x)
Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=\frac {\sqrt {a}\, \left (3 e^{10 x}-25 e^{8 x}+150 e^{6 x}+150 e^{4 x}-25 e^{2 x}+3\right )}{480 e^{5 x} a^{3}} \] Input:
int(1/(a*csch(x)^2)^(5/2),x)
Output:
(sqrt(a)*(3*e**(10*x) - 25*e**(8*x) + 150*e**(6*x) + 150*e**(4*x) - 25*e** (2*x) + 3))/(480*e**(5*x)*a**3)