\(\int \frac {1}{(a \cosh ^2(x))^{5/2}} \, dx\) [127]

Optimal result

Integrand size = 10, antiderivative size = 61 \[ \int \frac {1}{\left (a \cosh ^2(x)\right )^{5/2}} \, dx=\frac {3 \arctan (\sinh (x)) \cosh (x)}{8 a^2 \sqrt {a \cosh ^2(x)}}+\frac {\tanh (x)}{4 a \left (a \cosh ^2(x)\right )^{3/2}}+\frac {3 \tanh (x)}{8 a^2 \sqrt {a \cosh ^2(x)}} \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3283, 3286, 3855} \[ \int \frac {1}{\left (a \cosh ^2(x)\right )^{5/2}} \, dx=\frac {3 \cosh (x) \arctan (\sinh (x))}{8 a^2 \sqrt {a \cosh ^2(x)}}+\frac {3 \tanh (x)}{8 a^2 \sqrt {a \cosh ^2(x)}}+\frac {\tanh (x)}{4 a \left (a \cosh ^2(x)\right )^{3/2}} \]

[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 3283

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 3286

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 3855

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

Rubi steps \begin{align*} \text {integral}& = \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 \arctan (\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] (verified)

Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.59 \[ \int \frac {1}{\left (a \cosh ^2(x)\right )^{5/2}} \, dx=\frac {3 \arctan (\sinh (x)) \cosh (x)+\left (3+2 \text {sech}^2(x)\right ) \tanh (x)}{8 a^2 \sqrt {a \cosh ^2(x)}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(101\) vs. \(2(49)=98\).

Time = 0.19 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.67

method	result	size
default	\(\frac {\sqrt {a \sinh \left (x \right )^{2}}\, \left (-3 \ln \left (\frac {2 \sqrt {-a}\, \sqrt {a \sinh \left (x \right )^{2}}-2 a}{\cosh \left (x \right )}\right ) a \cosh \left (x \right )^{4}+3 \cosh \left (x \right )^{2} \sqrt {a \sinh \left (x \right )^{2}}\, \sqrt {-a}+2 \sqrt {-a}\, \sqrt {a \sinh \left (x \right )^{2}}\right )}{8 a^{3} \cosh \left (x \right )^{3} \sqrt {-a}\, \sinh \left (x \right ) \sqrt {a \cosh \left (x \right )^{2}}}\)	\(102\)
risch	\(\frac {3 \,{\mathrm e}^{6 x}+11 \,{\mathrm e}^{4 x}-11 \,{\mathrm e}^{2 x}-3}{4 a^{2} \left (1+{\mathrm e}^{2 x}\right )^{3} \sqrt {a \left (1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}}}+\frac {3 i \left (1+{\mathrm e}^{2 x}\right ) {\mathrm e}^{-x} \ln \left ({\mathrm e}^{x}+i\right )}{8 a^{2} \sqrt {a \left (1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}}}-\frac {3 i \left (1+{\mathrm e}^{2 x}\right ) {\mathrm e}^{-x} \ln \left ({\mathrm e}^{x}-i\right )}{8 a^{2} \sqrt {a \left (1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}}}\)	\(127\)

[In]

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

[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*cosh(
x)^2*(a*sinh(x)^2)^(1/2)*(-a)^(1/2)+2*(-a)^(1/2)*(a*sinh(x)^2)^(1/2))/(-a)^(1/2)/sinh(x)/(a*cosh(x)^2)^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 837 vs. \(2 (49) = 98\).

Time = 0.27 (sec) , antiderivative size = 837, normalized size of antiderivative = 13.72 \[ \int \frac {1}{\left (a \cosh ^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \]

[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))

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (a \cosh ^2(x)\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23 \[ \int \frac {1}{\left (a \cosh ^2(x)\right )^{5/2}} \, dx=\frac {3 \, e^{\left (7 \, x\right )} + 11 \, e^{\left (5 \, x\right )} - 11 \, e^{\left (3 \, x\right )} - 3 \, e^{x}}{4 \, {\left (a^{\frac {5}{2}} e^{\left (8 \, x\right )} + 4 \, a^{\frac {5}{2}} e^{\left (6 \, x\right )} + 6 \, a^{\frac {5}{2}} e^{\left (4 \, x\right )} + 4 \, a^{\frac {5}{2}} e^{\left (2 \, x\right )} + a^{\frac {5}{2}}\right )}} + \frac {3 \, \arctan \left (e^{x}\right )}{4 \, a^{\frac {5}{2}}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\left (a \cosh ^2(x)\right )^{5/2}} \, dx=\frac {3 \, {\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )}}{16 \, a^{\frac {5}{2}}} - \frac {3 \, {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 20 \, e^{\left (-x\right )} - 20 \, e^{x}}{4 \, {\left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}^{2} a^{\frac {5}{2}}} \]

[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))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a \cosh ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (a\,{\mathrm {cosh}\left (x\right )}^2\right )}^{5/2}} \,d x \]

[In]

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

[Out]

int(1/(a*cosh(x)^2)^(5/2), x)