3.105 \(\int \frac {\text {csch}^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx\)

Optimal. Leaf size=134 \[ \frac {\tanh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{a f}-\frac {\coth (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{a f}-\frac {\text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{a f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

[Out]

-coth(f*x+e)*(a+b*sinh(f*x+e)^2)^(1/2)/a/f-(1/(1+sinh(f*x+e)^2))^(1/2)*(1+sinh(f*x+e)^2)^(1/2)*EllipticE(sinh(
f*x+e)/(1+sinh(f*x+e)^2)^(1/2),(1-b/a)^(1/2))*sech(f*x+e)*(a+b*sinh(f*x+e)^2)^(1/2)/a/f/(sech(f*x+e)^2*(a+b*si
nh(f*x+e)^2)/a)^(1/2)+(a+b*sinh(f*x+e)^2)^(1/2)*tanh(f*x+e)/a/f

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3188, 480, 12, 492, 411} \[ \frac {\tanh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{a f}-\frac {\coth (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{a f}-\frac {\text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right )}{a f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Coth[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(a*f)) - (EllipticE[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x
]*Sqrt[a + b*Sinh[e + f*x]^2])/(a*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) + (Sqrt[a + b*Sinh[e +
f*x]^2]*Tanh[e + f*x])/(a*f)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 480

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*e*(m + 1)), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 3188

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[(ff^(m + 1)*Sqrt[Cos[e + f*x]^2])/(f*Cos[e + f*x]), Subst[Int[(x^m*(a + b*ff^2*
x^2)^p)/Sqrt[1 - ff^2*x^2], x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] &&  !In
tegerQ[p]

Rubi steps

\begin {align*} \int \frac {\text {csch}^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx &=\frac {\left (\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {1+x^2} \sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=-\frac {\coth (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{a f}+\frac {\left (\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {b x^2}{\sqrt {1+x^2} \sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{a f}\\ &=-\frac {\coth (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{a f}+\frac {\left (b \sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2} \sqrt {a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{a f}\\ &=-\frac {\coth (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{a f}+\frac {\sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{a f}-\frac {\left (\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\left (1+x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{a f}\\ &=-\frac {\coth (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{a f}-\frac {E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac {b}{a}\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{a f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {\sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{a f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.53, size = 150, normalized size = 1.12 \[ \frac {\sqrt {2} \coth (e+f x) (-2 a-b \cosh (2 (e+f x))+b)+2 i a \sqrt {\frac {2 a+b \cosh (2 (e+f x))-b}{a}} F\left (i (e+f x)\left |\frac {b}{a}\right .\right )-2 i a \sqrt {\frac {2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac {b}{a}\right .\right )}{2 a f \sqrt {2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*(-2*a + b - b*Cosh[2*(e + f*x)])*Coth[e + f*x] - (2*I)*a*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*Elli
pticE[I*(e + f*x), b/a] + (2*I)*a*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticF[I*(e + f*x), b/a])/(2*a*f*
Sqrt[2*a - b + b*Cosh[2*(e + f*x)]])

________________________________________________________________________________________

fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {csch}\left (f x + e\right )^{2}}{\sqrt {b \sinh \left (f x + e\right )^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(csch(f*x + e)^2/sqrt(b*sinh(f*x + e)^2 + a), x)

________________________________________________________________________________________

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(csch(f*x+e)^2/(a+b*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:Erro
r: Bad Argument Type

________________________________________________________________________________________

maple [A]  time = 0.14, size = 189, normalized size = 1.41 \[ -\frac {\sinh \left (f x +e \right ) \sqrt {\frac {b \left (\cosh ^{2}\left (f x +e \right )\right )}{a}+\frac {a -b}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, b \left (\EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )-\EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )\right )+\sqrt {-\frac {b}{a}}\, b \left (\cosh ^{4}\left (f x +e \right )\right )+\left (\sqrt {-\frac {b}{a}}\, a -\sqrt {-\frac {b}{a}}\, b \right ) \left (\cosh ^{2}\left (f x +e \right )\right )}{a \sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}\, \cosh \left (f x +e \right ) \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sinh(f*x+e)*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*b*(EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),
(a/b)^(1/2))-EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2)))+(-1/a*b)^(1/2)*b*cosh(f*x+e)^4+((-1/a*b)^(1/2)
*a-(-1/a*b)^(1/2)*b)*cosh(f*x+e)^2)/a/sinh(f*x+e)/(-1/a*b)^(1/2)/cosh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2)/f

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {csch}\left (f x + e\right )^{2}}{\sqrt {b \sinh \left (f x + e\right )^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(csch(f*x + e)^2/sqrt(b*sinh(f*x + e)^2 + a), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\mathrm {sinh}\left (e+f\,x\right )}^2\,\sqrt {b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csch(e + f*x)**2/sqrt(a + b*sinh(e + f*x)**2), x)

________________________________________________________________________________________