3.127 \(\int \frac {1}{(a \cosh ^2(x))^{5/2}} \, dx\)

Optimal. Leaf size=61 \[ \frac {3 \tanh (x)}{8 a^2 \sqrt {a \cosh ^2(x)}}+\frac {3 \cosh (x) \tan ^{-1}(\sinh (x))}{8 a^2 \sqrt {a \cosh ^2(x)}}+\frac {\tanh (x)}{4 a \left (a \cosh ^2(x)\right )^{3/2}} \]

[Out]

3/8*arctan(sinh(x))*cosh(x)/a^2/(a*cosh(x)^2)^(1/2)+1/4*tanh(x)/a/(a*cosh(x)^2)^(3/2)+3/8*tanh(x)/a^2/(a*cosh(
x)^2)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3204, 3207, 3770} \[ \frac {3 \tanh (x)}{8 a^2 \sqrt {a \cosh ^2(x)}}+\frac {3 \cosh (x) \tan ^{-1}(\sinh (x))}{8 a^2 \sqrt {a \cosh ^2(x)}}+\frac {\tanh (x)}{4 a \left (a \cosh ^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*ArcTan[Sinh[x]]*Cosh[x])/(8*a^2*Sqrt[a*Cosh[x]^2]) + Tanh[x]/(4*a*(a*Cosh[x]^2)^(3/2)) + (3*Tanh[x])/(8*a^2
*Sqrt[a*Cosh[x]^2])

Rule 3204

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(Cot[e + f*x]*(b*Sin[e + f*x]^2)^(p + 1))/(b*f*(
2*p + 1)), x] + Dist[(2*(p + 1))/(b*(2*p + 1)), Int[(b*Sin[e + f*x]^2)^(p + 1), x], x] /; FreeQ[{b, e, f}, x]
&&  !IntegerQ[p] && LtQ[p, -1]

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 40, normalized size = 0.66 \[ \frac {\tanh (x) \left (2 \text {sech}^2(x)+3\right )+6 \cosh (x) \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )}{8 a^2 \sqrt {a \cosh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*ArcTan[Tanh[x/2]]*Cosh[x] + (3 + 2*Sech[x]^2)*Tanh[x])/(8*a^2*Sqrt[a*Cosh[x]^2])

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fricas [B]  time = 1.00, size = 837, normalized size = 13.72 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)^2)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 67, normalized size = 1.10 \[ \frac {3 \, {\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )}}{16 \, a^{\frac {5}{2}}} - \frac {3 \, {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 20 \, e^{\left (-x\right )} - 20 \, e^{x}}{4 \, {\left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}^{2} a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)^2)^(5/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.30, size = 102, normalized size = 1.67 \[ \frac {\sqrt {a \left (\sinh ^{2}\relax (x )\right )}\, \left (-3 \ln \left (\frac {2 \sqrt {-a}\, \sqrt {a \left (\sinh ^{2}\relax (x )\right )}-2 a}{\cosh \relax (x )}\right ) a \left (\cosh ^{4}\relax (x )\right )+3 \sqrt {a \left (\sinh ^{2}\relax (x )\right )}\, \left (\cosh ^{2}\relax (x )\right ) \sqrt {-a}+2 \sqrt {-a}\, \sqrt {a \left (\sinh ^{2}\relax (x )\right )}\right )}{8 a^{3} \cosh \relax (x )^{3} \sqrt {-a}\, \sinh \relax (x ) \sqrt {a \left (\cosh ^{2}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cosh(x)^2)^(5/2),x)

[Out]

1/8/a^3/cosh(x)^3*(a*sinh(x)^2)^(1/2)*(-3*ln(2*((-a)^(1/2)*(a*sinh(x)^2)^(1/2)-a)/cosh(x))*a*cosh(x)^4+3*(a*si
nh(x)^2)^(1/2)*cosh(x)^2*(-a)^(1/2)+2*(-a)^(1/2)*(a*sinh(x)^2)^(1/2))/(-a)^(1/2)/sinh(x)/(a*cosh(x)^2)^(1/2)

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maxima [A]  time = 0.47, size = 75, normalized size = 1.23 \[ \frac {3 \, e^{\left (7 \, x\right )} + 11 \, e^{\left (5 \, x\right )} - 11 \, e^{\left (3 \, x\right )} - 3 \, e^{x}}{4 \, {\left (a^{\frac {5}{2}} e^{\left (8 \, x\right )} + 4 \, a^{\frac {5}{2}} e^{\left (6 \, x\right )} + 6 \, a^{\frac {5}{2}} e^{\left (4 \, x\right )} + 4 \, a^{\frac {5}{2}} e^{\left (2 \, x\right )} + a^{\frac {5}{2}}\right )}} + \frac {3 \, \arctan \left (e^{x}\right )}{4 \, a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)^2)^(5/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cosh(x)^2)^(5/2),x)

[Out]

int(1/(a*cosh(x)^2)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)**2)**(5/2),x)

[Out]

Timed out

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