\(\int \frac {1}{\sqrt {a \cosh ^2(x)}} \, dx\) [125]

Optimal result

Integrand size = 10, antiderivative size = 16 \[ \int \frac {1}{\sqrt {a \cosh ^2(x)}} \, dx=\frac {\arctan (\sinh (x)) \cosh (x)}{\sqrt {a \cosh ^2(x)}} \]

[Out]

arctan(sinh(x))*cosh(x)/(a*cosh(x)^2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3286, 3855} \[ \int \frac {1}{\sqrt {a \cosh ^2(x)}} \, dx=\frac {\cosh (x) \arctan (\sinh (x))}{\sqrt {a \cosh ^2(x)}} \]

[In]

Int[1/Sqrt[a*Cosh[x]^2],x]

[Out]

(ArcTan[Sinh[x]]*Cosh[x])/Sqrt[a*Cosh[x]^2]

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 {\cosh (x) \int \text {sech}(x) \, dx}{\sqrt {a \cosh ^2(x)}} \\ & = \frac {\arctan (\sinh (x)) \cosh (x)}{\sqrt {a \cosh ^2(x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {a \cosh ^2(x)}} \, dx=\frac {\arctan (\sinh (x)) \cosh (x)}{\sqrt {a \cosh ^2(x)}} \]

[In]

Integrate[1/Sqrt[a*Cosh[x]^2],x]

[Out]

(ArcTan[Sinh[x]]*Cosh[x])/Sqrt[a*Cosh[x]^2]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(54\) vs. \(2(14)=28\).

Time = 0.15 (sec) , antiderivative size = 55, normalized size of antiderivative = 3.44

method	result	size
default	\(-\frac {\cosh \left (x \right ) \sqrt {a \sinh \left (x \right )^{2}}\, \ln \left (\frac {2 \sqrt {-a}\, \sqrt {a \sinh \left (x \right )^{2}}-2 a}{\cosh \left (x \right )}\right )}{\sqrt {-a}\, \sinh \left (x \right ) \sqrt {a \cosh \left (x \right )^{2}}}\)	\(55\)
risch	\(\frac {i {\mathrm e}^{-x} \left (1+{\mathrm e}^{2 x}\right ) \ln \left ({\mathrm e}^{x}+i\right )}{\sqrt {a \left (1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}}}-\frac {i {\mathrm e}^{-x} \left (1+{\mathrm e}^{2 x}\right ) \ln \left ({\mathrm e}^{x}-i\right )}{\sqrt {a \left (1+{\mathrm e}^{2 x}\right )^{2} {\mathrm e}^{-2 x}}}\)	\(72\)

[In]

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

[Out]

-cosh(x)*(a*sinh(x)^2)^(1/2)/(-a)^(1/2)*ln(2*((-a)^(1/2)*(a*sinh(x)^2)^(1/2)-a)/cosh(x))/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. 35 vs. \(2 (14) = 28\).

Time = 0.26 (sec) , antiderivative size = 186, normalized size of antiderivative = 11.62 \[ \int \frac {1}{\sqrt {a \cosh ^2(x)}} \, dx=\left [-\frac {\sqrt {-a} \log \left (\frac {a \cosh \left (x\right )^{2} - 2 \, \sqrt {a e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} + a} {\left (\cosh \left (x\right ) e^{x} + e^{x} \sinh \left (x\right )\right )} \sqrt {-a} e^{\left (-x\right )} + {\left (a e^{\left (2 \, x\right )} + a\right )} \sinh \left (x\right )^{2} + {\left (a \cosh \left (x\right )^{2} - a\right )} e^{\left (2 \, x\right )} + 2 \, {\left (a \cosh \left (x\right ) e^{\left (2 \, x\right )} + a \cosh \left (x\right )\right )} \sinh \left (x\right ) - a}{{\left (e^{\left (2 \, x\right )} + 1\right )} \sinh \left (x\right )^{2} + \cosh \left (x\right )^{2} + {\left (\cosh \left (x\right )^{2} + 1\right )} e^{\left (2 \, x\right )} + 2 \, {\left (\cosh \left (x\right ) e^{\left (2 \, x\right )} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1}\right )}{a}, \frac {2 \, \sqrt {a e^{\left (4 \, x\right )} + 2 \, a e^{\left (2 \, x\right )} + a} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}{a e^{\left (2 \, x\right )} + a}\right ] \]

[In]

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

[Out]

[-sqrt(-a)*log((a*cosh(x)^2 - 2*sqrt(a*e^(4*x) + 2*a*e^(2*x) + a)*(cosh(x)*e^x + e^x*sinh(x))*sqrt(-a)*e^(-x)
+ (a*e^(2*x) + a)*sinh(x)^2 + (a*cosh(x)^2 - a)*e^(2*x) + 2*(a*cosh(x)*e^(2*x) + a*cosh(x))*sinh(x) - a)/((e^(
2*x) + 1)*sinh(x)^2 + cosh(x)^2 + (cosh(x)^2 + 1)*e^(2*x) + 2*(cosh(x)*e^(2*x) + cosh(x))*sinh(x) + 1))/a, 2*s
qrt(a*e^(4*x) + 2*a*e^(2*x) + a)*arctan(cosh(x) + sinh(x))/(a*e^(2*x) + a)]

Sympy [F]

\[ \int \frac {1}{\sqrt {a \cosh ^2(x)}} \, dx=\int \frac {1}{\sqrt {a \cosh ^{2}{\left (x \right )}}}\, dx \]

[In]

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

[Out]

Integral(1/sqrt(a*cosh(x)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\sqrt {a \cosh ^2(x)}} \, dx=\frac {2 \, \arctan \left (e^{x}\right )}{\sqrt {a}} \]

[In]

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

[Out]

2*arctan(e^x)/sqrt(a)

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {a \cosh ^2(x)}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

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

[In]

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

[Out]

int(1/(a*cosh(x)^2)^(1/2), x)