3.29 \(\int (a \text {csch}^2(x))^{5/2} \, dx\)

Optimal. Leaf size=65 \[ -\frac {3}{8} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \coth (x)}{\sqrt {a \text {csch}^2(x)}}\right )+\frac {3}{8} a^2 \coth (x) \sqrt {a \text {csch}^2(x)}-\frac {1}{4} a \coth (x) \left (a \text {csch}^2(x)\right )^{3/2} \]

[Out]

-3/8*a^(5/2)*arctanh(coth(x)*a^(1/2)/(a*csch(x)^2)^(1/2))-1/4*a*coth(x)*(a*csch(x)^2)^(3/2)+3/8*a^2*coth(x)*(a
*csch(x)^2)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4122, 195, 217, 206} \[ \frac {3}{8} a^2 \coth (x) \sqrt {a \text {csch}^2(x)}-\frac {3}{8} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \coth (x)}{\sqrt {a \text {csch}^2(x)}}\right )-\frac {1}{4} a \coth (x) \left (a \text {csch}^2(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a*Csch[x]^2)^(5/2),x]

[Out]

(-3*a^(5/2)*ArcTanh[(Sqrt[a]*Coth[x])/Sqrt[a*Csch[x]^2]])/8 + (3*a^2*Coth[x]*Sqrt[a*Csch[x]^2])/8 - (a*Coth[x]
*(a*Csch[x]^2)^(3/2))/4

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 4122

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(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
]

Rubi steps

\begin {align*} \int \left (a \text {csch}^2(x)\right )^{5/2} \, dx &=-\left (a \operatorname {Subst}\left (\int \left (-a+a x^2\right )^{3/2} \, dx,x,\coth (x)\right )\right )\\ &=-\frac {1}{4} a \coth (x) \left (a \text {csch}^2(x)\right )^{3/2}+\frac {1}{4} \left (3 a^2\right ) \operatorname {Subst}\left (\int \sqrt {-a+a x^2} \, dx,x,\coth (x)\right )\\ &=\frac {3}{8} a^2 \coth (x) \sqrt {a \text {csch}^2(x)}-\frac {1}{4} a \coth (x) \left (a \text {csch}^2(x)\right )^{3/2}-\frac {1}{8} \left (3 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-a+a x^2}} \, dx,x,\coth (x)\right )\\ &=\frac {3}{8} a^2 \coth (x) \sqrt {a \text {csch}^2(x)}-\frac {1}{4} a \coth (x) \left (a \text {csch}^2(x)\right )^{3/2}-\frac {1}{8} \left (3 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\coth (x)}{\sqrt {a \text {csch}^2(x)}}\right )\\ &=-\frac {3}{8} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \coth (x)}{\sqrt {a \text {csch}^2(x)}}\right )+\frac {3}{8} a^2 \coth (x) \sqrt {a \text {csch}^2(x)}-\frac {1}{4} a \coth (x) \left (a \text {csch}^2(x)\right )^{3/2}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 41, normalized size = 0.63 \[ \frac {1}{64} \sinh (x) \left (a \text {csch}^2(x)\right )^{5/2} \left (6 \left (\cosh (3 x)+4 \sinh ^4(x) \log \left (\tanh \left (\frac {x}{2}\right )\right )\right )-22 \cosh (x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Csch[x]^2)^(5/2),x]

[Out]

((a*Csch[x]^2)^(5/2)*Sinh[x]*(-22*Cosh[x] + 6*(Cosh[3*x] + 4*Log[Tanh[x/2]]*Sinh[x]^4)))/64

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fricas [B]  time = 0.63, size = 1128, normalized size = 17.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*csch(x)^2)^(5/2),x, algorithm="fricas")

[Out]

-1/8*(6*a^2*cosh(x)^7 - 6*(a^2*e^(2*x) - a^2)*sinh(x)^7 - 22*a^2*cosh(x)^5 - 42*(a^2*cosh(x)*e^(2*x) - a^2*cos
h(x))*sinh(x)^6 + 2*(63*a^2*cosh(x)^2 - 11*a^2 - (63*a^2*cosh(x)^2 - 11*a^2)*e^(2*x))*sinh(x)^5 - 22*a^2*cosh(
x)^3 + 10*(21*a^2*cosh(x)^3 - 11*a^2*cosh(x) - (21*a^2*cosh(x)^3 - 11*a^2*cosh(x))*e^(2*x))*sinh(x)^4 + 2*(105
*a^2*cosh(x)^4 - 110*a^2*cosh(x)^2 - 11*a^2 - (105*a^2*cosh(x)^4 - 110*a^2*cosh(x)^2 - 11*a^2)*e^(2*x))*sinh(x
)^3 + 6*a^2*cosh(x) + 2*(63*a^2*cosh(x)^5 - 110*a^2*cosh(x)^3 - 33*a^2*cosh(x) - (63*a^2*cosh(x)^5 - 110*a^2*c
osh(x)^3 - 33*a^2*cosh(x))*e^(2*x))*sinh(x)^2 - 2*(3*a^2*cosh(x)^7 - 11*a^2*cosh(x)^5 - 11*a^2*cosh(x)^3 + 3*a
^2*cosh(x))*e^(2*x) + 3*(a^2*cosh(x)^8 - (a^2*e^(2*x) - a^2)*sinh(x)^8 - 4*a^2*cosh(x)^6 - 8*(a^2*cosh(x)*e^(2
*x) - a^2*cosh(x))*sinh(x)^7 + 4*(7*a^2*cosh(x)^2 - a^2 - (7*a^2*cosh(x)^2 - a^2)*e^(2*x))*sinh(x)^6 + 6*a^2*c
osh(x)^4 + 8*(7*a^2*cosh(x)^3 - 3*a^2*cosh(x) - (7*a^2*cosh(x)^3 - 3*a^2*cosh(x))*e^(2*x))*sinh(x)^5 + 2*(35*a
^2*cosh(x)^4 - 30*a^2*cosh(x)^2 + 3*a^2 - (35*a^2*cosh(x)^4 - 30*a^2*cosh(x)^2 + 3*a^2)*e^(2*x))*sinh(x)^4 - 4
*a^2*cosh(x)^2 + 8*(7*a^2*cosh(x)^5 - 10*a^2*cosh(x)^3 + 3*a^2*cosh(x) - (7*a^2*cosh(x)^5 - 10*a^2*cosh(x)^3 +
3*a^2*cosh(x))*e^(2*x))*sinh(x)^3 + 4*(7*a^2*cosh(x)^6 - 15*a^2*cosh(x)^4 + 9*a^2*cosh(x)^2 - a^2 - (7*a^2*co
sh(x)^6 - 15*a^2*cosh(x)^4 + 9*a^2*cosh(x)^2 - a^2)*e^(2*x))*sinh(x)^2 + a^2 - (a^2*cosh(x)^8 - 4*a^2*cosh(x)^
6 + 6*a^2*cosh(x)^4 - 4*a^2*cosh(x)^2 + a^2)*e^(2*x) + 8*(a^2*cosh(x)^7 - 3*a^2*cosh(x)^5 + 3*a^2*cosh(x)^3 -
a^2*cosh(x) - (a^2*cosh(x)^7 - 3*a^2*cosh(x)^5 + 3*a^2*cosh(x)^3 - a^2*cosh(x))*e^(2*x))*sinh(x))*log((cosh(x)
+ sinh(x) - 1)/(cosh(x) + sinh(x) + 1)) + 2*(21*a^2*cosh(x)^6 - 55*a^2*cosh(x)^4 - 33*a^2*cosh(x)^2 + 3*a^2 -
(21*a^2*cosh(x)^6 - 55*a^2*cosh(x)^4 - 33*a^2*cosh(x)^2 + 3*a^2)*e^(2*x))*sinh(x))*sqrt(a/(e^(4*x) - 2*e^(2*x
) + 1))*e^x/(8*cosh(x)*e^x*sinh(x)^7 + e^x*sinh(x)^8 + 4*(7*cosh(x)^2 - 1)*e^x*sinh(x)^6 + 8*(7*cosh(x)^3 - 3*
cosh(x))*e^x*sinh(x)^5 + 2*(35*cosh(x)^4 - 30*cosh(x)^2 + 3)*e^x*sinh(x)^4 + 8*(7*cosh(x)^5 - 10*cosh(x)^3 + 3
*cosh(x))*e^x*sinh(x)^3 + 4*(7*cosh(x)^6 - 15*cosh(x)^4 + 9*cosh(x)^2 - 1)*e^x*sinh(x)^2 + 8*(cosh(x)^7 - 3*co
sh(x)^5 + 3*cosh(x)^3 - cosh(x))*e^x*sinh(x) + (cosh(x)^8 - 4*cosh(x)^6 + 6*cosh(x)^4 - 4*cosh(x)^2 + 1)*e^x)

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giac [A]  time = 0.12, size = 75, normalized size = 1.15 \[ \frac {1}{16} \, a^{\frac {5}{2}} {\left (\frac {4 \, {\left (3 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - 20 \, e^{\left (-x\right )} - 20 \, e^{x}\right )}}{{\left ({\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4\right )}^{2}} - 3 \, \log \left (e^{\left (-x\right )} + e^{x} + 2\right ) + 3 \, \log \left (e^{\left (-x\right )} + e^{x} - 2\right )\right )} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*csch(x)^2)^(5/2),x, algorithm="giac")

[Out]

1/16*a^(5/2)*(4*(3*(e^(-x) + e^x)^3 - 20*e^(-x) - 20*e^x)/((e^(-x) + e^x)^2 - 4)^2 - 3*log(e^(-x) + e^x + 2) +
3*log(e^(-x) + e^x - 2))*sgn(e^(3*x) - e^x)

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maple [B]  time = 0.25, size = 123, normalized size = 1.89 \[ \frac {a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}\, \left (3 \,{\mathrm e}^{6 x}-11 \,{\mathrm e}^{4 x}-11 \,{\mathrm e}^{2 x}+3\right )}{4 \left ({\mathrm e}^{2 x}-1\right )^{3}}+\frac {3 a^{2} {\mathrm e}^{-x} \left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}\, \ln \left ({\mathrm e}^{x}-1\right )}{8}-\frac {3 a^{2} {\mathrm e}^{-x} \left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}\, \ln \left ({\mathrm e}^{x}+1\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*csch(x)^2)^(5/2),x)

[Out]

1/4*a^2/(exp(2*x)-1)^3*(a*exp(2*x)/(exp(2*x)-1)^2)^(1/2)*(3*exp(6*x)-11*exp(4*x)-11*exp(2*x)+3)+3/8*a^2*exp(-x
)*(exp(2*x)-1)*(a*exp(2*x)/(exp(2*x)-1)^2)^(1/2)*ln(exp(x)-1)-3/8*a^2*exp(-x)*(exp(2*x)-1)*(a*exp(2*x)/(exp(2*
x)-1)^2)^(1/2)*ln(exp(x)+1)

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maxima [A]  time = 0.54, size = 92, normalized size = 1.42 \[ \frac {3}{8} \, a^{\frac {5}{2}} \log \left (e^{\left (-x\right )} + 1\right ) - \frac {3}{8} \, a^{\frac {5}{2}} \log \left (e^{\left (-x\right )} - 1\right ) + \frac {3 \, a^{\frac {5}{2}} e^{\left (-x\right )} - 11 \, a^{\frac {5}{2}} e^{\left (-3 \, x\right )} - 11 \, a^{\frac {5}{2}} e^{\left (-5 \, x\right )} + 3 \, a^{\frac {5}{2}} e^{\left (-7 \, x\right )}}{4 \, {\left (4 \, e^{\left (-2 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*csch(x)^2)^(5/2),x, algorithm="maxima")

[Out]

3/8*a^(5/2)*log(e^(-x) + 1) - 3/8*a^(5/2)*log(e^(-x) - 1) + 1/4*(3*a^(5/2)*e^(-x) - 11*a^(5/2)*e^(-3*x) - 11*a
^(5/2)*e^(-5*x) + 3*a^(5/2)*e^(-7*x))/(4*e^(-2*x) - 6*e^(-4*x) + 4*e^(-6*x) - e^(-8*x) - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (\frac {a}{{\mathrm {sinh}\relax (x)}^2}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a/sinh(x)^2)^(5/2),x)

[Out]

int((a/sinh(x)^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \operatorname {csch}^{2}{\relax (x )}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*csch(x)**2)**(5/2),x)

[Out]

Integral((a*csch(x)**2)**(5/2), x)

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