3.499 \(\int \frac{\coth ^2(e+f x)}{(a+b \sinh ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=237 \[ \frac{\text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{a^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{2 \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{a^2 f}-\frac{2 \coth (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{a^2 f}-\frac{2 \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{a^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\coth (e+f x)}{a f \sqrt{a+b \sinh ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.270323, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3196, 469, 583, 531, 418, 492, 411} \[ \frac{2 \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{a^2 f}-\frac{2 \coth (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{a^2 f}+\frac{\text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{a^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{2 \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{a^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\coth (e+f x)}{a f \sqrt{a+b \sinh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Coth[e + f*x]^2/(a + b*Sinh[e + f*x]^2)^(3/2),x]

[Out]

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

Rule 3196

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_), 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)/(1 - ff^2*x^2)^((m + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2]
&&  !IntegerQ[p]

Rule 469

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)/(a*e*n*(p + 1)), x] + Dist[1/(a*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)
^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m + n*(p + 1) + 1) + d*(m + n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ[{
a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[0, q, 1] && IntBinomialQ[a, b, c
, d, e, m, n, p, q, x]

Rule 583

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

Rule 531

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Dist[
e, Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x] + Dist[f, Int[x^n*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, n, p, q}, x]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*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[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

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 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]

Rubi steps

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

Mathematica [C]  time = 0.835849, size = 153, normalized size = 0.65 \[ \frac{i \sqrt{2} a \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )-2 \coth (e+f x) (a+b \cosh (2 (e+f x))-b)-2 i \sqrt{2} a \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{a^2 f \sqrt{4 a+2 b \cosh (2 (e+f x))-2 b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[e + f*x]^2/(a + b*Sinh[e + f*x]^2)^(3/2),x]

[Out]

(-2*(a - b + b*Cosh[2*(e + f*x)])*Coth[e + f*x] - (2*I)*Sqrt[2]*a*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*Elli
pticE[I*(e + f*x), b/a] + I*Sqrt[2]*a*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticF[I*(e + f*x), b/a])/(a^
2*f*Sqrt[4*a - 2*b + 2*b*Cosh[2*(e + f*x)]])

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Maple [A]  time = 0.282, size = 219, normalized size = 0.9 \begin{align*} -{\frac{1}{\sinh \left ( fx+e \right ){a}^{2}\cosh \left ( fx+e \right ) f} \left ( -\sinh \left ( fx+e \right ) \sqrt{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{b \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a-b}{a}}} \left ( a{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) -2\,b{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) +2\,b{\it EllipticE} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) \right ) +2\,\sqrt{-{\frac{b}{a}}}b \left ( \cosh \left ( fx+e \right ) \right ) ^{4}+ \left ( \sqrt{-{\frac{b}{a}}}a-2\,\sqrt{-{\frac{b}{a}}}b \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{-{\frac{b}{a}}}}}{\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(f*x+e)^2/(a+b*sinh(f*x+e)^2)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth \left (f x + e\right )^{2}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^2/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(coth(f*x + e)^2/(b*sinh(f*x + e)^2 + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sinh \left (f x + e\right )^{2} + a} \coth \left (f x + e\right )^{2}}{b^{2} \sinh \left (f x + e\right )^{4} + 2 \, a b \sinh \left (f x + e\right )^{2} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^2/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sinh(f*x + e)^2 + a)*coth(f*x + e)^2/(b^2*sinh(f*x + e)^4 + 2*a*b*sinh(f*x + e)^2 + a^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{2}{\left (e + f x \right )}}{\left (a + b \sinh ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)**2/(a+b*sinh(f*x+e)**2)**(3/2),x)

[Out]

Integral(coth(e + f*x)**2/(a + b*sinh(e + f*x)**2)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth \left (f x + e\right )^{2}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(f*x+e)^2/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate(coth(f*x + e)^2/(b*sinh(f*x + e)^2 + a)^(3/2), x)