3.444 \(\int \frac {\coth ^2(e+f x)}{\sqrt {a+a \sinh ^2(e+f x)}} \, dx\)

Optimal. Leaf size=25 \[ -\frac {\coth (e+f x)}{f \sqrt {a \cosh ^2(e+f x)}} \]

[Out]

-coth(f*x+e)/f/(a*cosh(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3176, 3207, 2606, 8} \[ -\frac {\coth (e+f x)}{f \sqrt {a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Coth[e + f*x]^2/Sqrt[a + a*Sinh[e + f*x]^2],x]

[Out]

-(Coth[e + f*x]/(f*Sqrt[a*Cosh[e + f*x]^2]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3176

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

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
]])

Rubi steps

\begin {align*} \int \frac {\coth ^2(e+f x)}{\sqrt {a+a \sinh ^2(e+f x)}} \, dx &=\int \frac {\coth ^2(e+f x)}{\sqrt {a \cosh ^2(e+f x)}} \, dx\\ &=\frac {\cosh (e+f x) \int \coth (e+f x) \text {csch}(e+f x) \, dx}{\sqrt {a \cosh ^2(e+f x)}}\\ &=-\frac {(i \cosh (e+f x)) \operatorname {Subst}(\int 1 \, dx,x,-i \text {csch}(e+f x))}{f \sqrt {a \cosh ^2(e+f x)}}\\ &=-\frac {\coth (e+f x)}{f \sqrt {a \cosh ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 25, normalized size = 1.00 \[ -\frac {\coth (e+f x)}{f \sqrt {a \cosh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[e + f*x]^2/Sqrt[a + a*Sinh[e + f*x]^2],x]

[Out]

-(Coth[e + f*x]/(f*Sqrt[a*Cosh[e + f*x]^2]))

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fricas [B]  time = 0.51, size = 170, normalized size = 6.80 \[ -\frac {2 \, \sqrt {a e^{\left (4 \, f x + 4 \, e\right )} + 2 \, a e^{\left (2 \, f x + 2 \, e\right )} + a} {\left (\cosh \left (f x + e\right ) e^{\left (f x + e\right )} + e^{\left (f x + e\right )} \sinh \left (f x + e\right )\right )} e^{\left (-f x - e\right )}}{a f \cosh \left (f x + e\right )^{2} + {\left (a f e^{\left (2 \, f x + 2 \, e\right )} + a f\right )} \sinh \left (f x + e\right )^{2} - a f + {\left (a f \cosh \left (f x + e\right )^{2} - a f\right )} e^{\left (2 \, f x + 2 \, e\right )} + 2 \, {\left (a f \cosh \left (f x + e\right ) e^{\left (2 \, f x + 2 \, e\right )} + a f \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^2/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

-2*sqrt(a*e^(4*f*x + 4*e) + 2*a*e^(2*f*x + 2*e) + a)*(cosh(f*x + e)*e^(f*x + e) + e^(f*x + e)*sinh(f*x + e))*e
^(-f*x - e)/(a*f*cosh(f*x + e)^2 + (a*f*e^(2*f*x + 2*e) + a*f)*sinh(f*x + e)^2 - a*f + (a*f*cosh(f*x + e)^2 -
a*f)*e^(2*f*x + 2*e) + 2*(a*f*cosh(f*x + e)*e^(2*f*x + 2*e) + a*f*cosh(f*x + e))*sinh(f*x + e))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^2/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep^3*exp(exp(1))^3+t_nostep*exp(exp(1)))]index.cc index_m operator + Error: Bad Argument Value

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maple [A]  time = 0.16, size = 32, normalized size = 1.28 \[ -\frac {\cosh \left (f x +e \right )}{\sinh \left (f x +e \right ) \sqrt {a \left (\cosh ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(f*x+e)^2/(a+a*sinh(f*x+e)^2)^(1/2),x)

[Out]

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

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maxima [B]  time = 0.53, size = 101, normalized size = 4.04 \[ \frac {\frac {\arctan \left (e^{\left (-f x - e\right )}\right )}{\sqrt {a}} + \frac {\sqrt {a} e^{\left (-f x - e\right )}}{a e^{\left (-2 \, f x - 2 \, e\right )} - a}}{f} - \frac {\arctan \left (e^{\left (-f x - e\right )}\right )}{\sqrt {a} f} + \frac {\sqrt {a} e^{\left (-f x - e\right )}}{{\left (a e^{\left (-2 \, f x - 2 \, e\right )} - a\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^2/(a+a*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

(arctan(e^(-f*x - e))/sqrt(a) + sqrt(a)*e^(-f*x - e)/(a*e^(-2*f*x - 2*e) - a))/f - arctan(e^(-f*x - e))/(sqrt(
a)*f) + sqrt(a)*e^(-f*x - e)/((a*e^(-2*f*x - 2*e) - a)*f)

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mupad [B]  time = 0.11, size = 76, normalized size = 3.04 \[ -\frac {4\,{\mathrm {e}}^{3\,e+3\,f\,x}\,\sqrt {a+a\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}}{a\,f\,\left ({\mathrm {e}}^{2\,e+2\,f\,x}-1\right )\,\left ({\mathrm {e}}^{e+f\,x}+{\mathrm {e}}^{3\,e+3\,f\,x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(e + f*x)^2/(a + a*sinh(e + f*x)^2)^(1/2),x)

[Out]

-(4*exp(3*e + 3*f*x)*(a + a*(exp(e + f*x)/2 - exp(- e - f*x)/2)^2)^(1/2))/(a*f*(exp(2*e + 2*f*x) - 1)*(exp(e +
f*x) + exp(3*e + 3*f*x)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth ^{2}{\left (e + f x \right )}}{\sqrt {a \left (\sinh ^{2}{\left (e + f x \right )} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)**2/(a+a*sinh(f*x+e)**2)**(1/2),x)

[Out]

Integral(coth(e + f*x)**2/sqrt(a*(sinh(e + f*x)**2 + 1)), x)

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