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*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])

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Rubi [A]  time = 0.0385913, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, 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*}

Mathematica [A]  time = 0.0350681, 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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Maple [B]  time = 0.062, size = 102, normalized size = 1.7 \begin{align*}{\frac{1}{8\,{a}^{3} \left ( \cosh \left ( x \right ) \right ) ^{3}\sinh \left ( x \right ) }\sqrt{a \left ( \sinh \left ( x \right ) \right ) ^{2}} \left ( -3\,\ln \left ( 2\,{\frac{\sqrt{-a}\sqrt{a \left ( \sinh \left ( x \right ) \right ) ^{2}}-a}{\cosh \left ( x \right ) }} \right ) a \left ( \cosh \left ( x \right ) \right ) ^{4}+3\,\sqrt{a \left ( \sinh \left ( x \right ) \right ) ^{2}} \left ( \cosh \left ( x \right ) \right ) ^{2}\sqrt{-a}+2\,\sqrt{-a}\sqrt{a \left ( \sinh \left ( x \right ) \right ) ^{2}} \right ){\frac{1}{\sqrt{-a}}}{\frac{1}{\sqrt{a \left ( \cosh \left ( x \right ) \right ) ^{2}}}}} \end{align*}

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 = 1.80192, size = 101, normalized size = 1.66 \begin{align*} \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}}} \end{align*}

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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Fricas [B]  time = 2.33545, size = 2461, normalized size = 40.34 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.1452, size = 100, normalized size = 1.64 \begin{align*} \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 \, \sqrt{a}{\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 20 \, \sqrt{a}{\left (e^{\left (-x\right )} - e^{x}\right )}}{4 \,{\left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}^{2} a^{3}} \end{align*}

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*sqrt(a)*(e^(-x) - e^x)^3 + 20*sqrt(a)*(e^(-x)
- e^x))/(((e^(-x) - e^x)^2 + 4)^2*a^3)