∫ coth2 (x)___ (a+b tanh2(x))5∕2 dx

3.254 \(\int \frac{\coth ^2(x)}{(a+b \tanh ^2(x))^{5/2}} \, dx\)

Optimal. Leaf size=131 \[ -\frac{(3 a+2 b) (a+4 b) \coth (x) \sqrt{a+b \tanh ^2(x)}}{3 a^3 (a+b)^2}+\frac{b (7 a+4 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt{a+b \tanh ^2(x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )}{(a+b)^{5/2}}+\frac{b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}} \]

[Out]

ArcTanh[(Sqrt[a + b]*Tanh[x])/Sqrt[a + b*Tanh[x]^2]]/(a + b)^(5/2) + (b*Coth[x])/(3*a*(a + b)*(a + b*Tanh[x]^2

)^(3/2)) + (b*(7*a + 4*b)*Coth[x])/(3*a^2*(a + b)^2*Sqrt[a + b*Tanh[x]^2]) - ((3*a + 2*b)*(a + 4*b)*Coth[x]*Sq

rt[a + b*Tanh[x]^2])/(3*a^3*(a + b)^2)

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Rubi [A] time = 0.241572, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {3670, 472, 579, 583, 12, 377, 206} \[ -\frac{(3 a+2 b) (a+4 b) \coth (x) \sqrt{a+b \tanh ^2(x)}}{3 a^3 (a+b)^2}+\frac{b (7 a+4 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt{a+b \tanh ^2(x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )}{(a+b)^{5/2}}+\frac{b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}} \]

Antiderivative was successfully veriﬁed.

[In]

Int[Coth[x]^2/(a + b*Tanh[x]^2)^(5/2),x]

[Out]

ArcTanh[(Sqrt[a + b]*Tanh[x])/Sqrt[a + b*Tanh[x]^2]]/(a + b)^(5/2) + (b*Coth[x])/(3*a*(a + b)*(a + b*Tanh[x]^2

)^(3/2)) + (b*(7*a + 4*b)*Coth[x])/(3*a^2*(a + b)^2*Sqrt[a + b*Tanh[x]^2]) - ((3*a + 2*b)*(a + 4*b)*Coth[x]*Sq

rt[a + b*Tanh[x]^2])/(3*a^3*(a + b)^2)

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]

:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^

2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ

[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 472

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x

)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(

p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(

p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p

, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 579

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x

_Symbol] :> -Simp[((b*e - a*f)*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*g*n*(b*c - a*d)*(p +

1)), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)*(

m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,

e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]

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 12

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

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]

, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]

/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\coth ^2(x)}{\left (a+b \tanh ^2(x)\right )^{5/2}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tanh (x)\right )\\ &=\frac{b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{-3 a-4 b+4 b x^2}{x^2 \left (1-x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tanh (x)\right )}{3 a (a+b)}\\ &=\frac{b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac{b (7 a+4 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt{a+b \tanh ^2(x)}}+\frac{\operatorname{Subst}\left (\int \frac{(3 a+2 b) (a+4 b)-2 b (7 a+4 b) x^2}{x^2 \left (1-x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tanh (x)\right )}{3 a^2 (a+b)^2}\\ &=\frac{b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac{b (7 a+4 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt{a+b \tanh ^2(x)}}-\frac{(3 a+2 b) (a+4 b) \coth (x) \sqrt{a+b \tanh ^2(x)}}{3 a^3 (a+b)^2}-\frac{\operatorname{Subst}\left (\int -\frac{3 a^3}{\left (1-x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tanh (x)\right )}{3 a^3 (a+b)^2}\\ &=\frac{b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac{b (7 a+4 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt{a+b \tanh ^2(x)}}-\frac{(3 a+2 b) (a+4 b) \coth (x) \sqrt{a+b \tanh ^2(x)}}{3 a^3 (a+b)^2}+\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tanh (x)\right )}{(a+b)^2}\\ &=\frac{b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac{b (7 a+4 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt{a+b \tanh ^2(x)}}-\frac{(3 a+2 b) (a+4 b) \coth (x) \sqrt{a+b \tanh ^2(x)}}{3 a^3 (a+b)^2}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-(a+b) x^2} \, dx,x,\frac{\tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )}{(a+b)^2}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )}{(a+b)^{5/2}}+\frac{b \coth (x)}{3 a (a+b) \left (a+b \tanh ^2(x)\right )^{3/2}}+\frac{b (7 a+4 b) \coth (x)}{3 a^2 (a+b)^2 \sqrt{a+b \tanh ^2(x)}}-\frac{(3 a+2 b) (a+4 b) \coth (x) \sqrt{a+b \tanh ^2(x)}}{3 a^3 (a+b)^2}\\ \end{align*}

Mathematica [C] time = 7.45596, size = 246, normalized size = 1.88 \[ \frac{\sqrt{\text{sech}^2(x) ((a+b) \cosh (2 x)+a-b)} \left (\frac{3 \sqrt{2} a^3 \coth (x) \left ((a+b) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\text{csch}^2(x) ((a+b) \cosh (2 x)+a-b)}{b}}}{\sqrt{2}}\right ),1\right )-a \Pi \left (\frac{b}{a+b};\left .\sin ^{-1}\left (\frac{\sqrt{\frac{(a-b+(a+b) \cosh (2 x)) \text{csch}^2(x)}{b}}}{\sqrt{2}}\right )\right |1\right )\right )}{b \sqrt{\frac{\text{csch}^2(x) ((a+b) \cosh (2 x)+a-b)}{b}}}-\frac{(a+b) \left (2 a b^3 \sinh (2 x)+b^2 (9 a+5 b) \sinh (2 x) ((a+b) \cosh (2 x)+a-b)+3 (a+b)^2 \coth (x) ((a+b) \cosh (2 x)+a-b)^2\right )}{((a+b) \cosh (2 x)+a-b)^2}\right )}{3 \sqrt{2} a^3 (a+b)^3} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[Coth[x]^2/(a + b*Tanh[x]^2)^(5/2),x]

[Out]

(Sqrt[(a - b + (a + b)*Cosh[2*x])*Sech[x]^2]*((3*Sqrt[2]*a^3*Coth[x]*((a + b)*EllipticF[ArcSin[Sqrt[((a - b +

(a + b)*Cosh[2*x])*Csch[x]^2)/b]/Sqrt[2]], 1] - a*EllipticPi[b/(a + b), ArcSin[Sqrt[((a - b + (a + b)*Cosh[2*x

])*Csch[x]^2)/b]/Sqrt[2]], 1]))/(b*Sqrt[((a - b + (a + b)*Cosh[2*x])*Csch[x]^2)/b]) - ((a + b)*(3*(a + b)^2*(a

- b + (a + b)*Cosh[2*x])^2*Coth[x] + 2*a*b^3*Sinh[2*x] + b^2*(9*a + 5*b)*(a - b + (a + b)*Cosh[2*x])*Sinh[2*x

]))/(a - b + (a + b)*Cosh[2*x])^2))/(3*Sqrt[2]*a^3*(a + b)^3)

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Maple [F] time = 0.119, size = 0, normalized size = 0. \begin{align*} \int{ \left ({\rm coth} \left (x\right ) \right ) ^{2} \left ( a+b \left ( \tanh \left ( x \right ) \right ) ^{2} \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^2/(a+b*tanh(x)^2)^(5/2),x)

[Out]

int(coth(x)^2/(a+b*tanh(x)^2)^(5/2),x)

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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth \left (x\right )^{2}}{{\left (b \tanh \left (x\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^2/(a+b*tanh(x)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(coth(x)^2/(b*tanh(x)^2 + a)^(5/2), x)

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Fricas [B] time = 18.786, size = 25307, normalized size = 193.18 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^2/(a+b*tanh(x)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*((a^5 + 2*a^4*b + a^3*b^2)*cosh(x)^10 + 10*(a^5 + 2*a^4*b + a^3*b^2)*cosh(x)*sinh(x)^9 + (a^5 + 2*a^4

*b + a^3*b^2)*sinh(x)^10 + (3*a^5 - 2*a^4*b - 5*a^3*b^2)*cosh(x)^8 + (3*a^5 - 2*a^4*b - 5*a^3*b^2 + 45*(a^5 +

2*a^4*b + a^3*b^2)*cosh(x)^2)*sinh(x)^8 + 8*(15*(a^5 + 2*a^4*b + a^3*b^2)*cosh(x)^3 + (3*a^5 - 2*a^4*b - 5*a^3

*b^2)*cosh(x))*sinh(x)^7 + 2*(a^5 - 2*a^4*b + 5*a^3*b^2)*cosh(x)^6 + 2*(a^5 - 2*a^4*b + 5*a^3*b^2 + 105*(a^5 +

2*a^4*b + a^3*b^2)*cosh(x)^4 + 14*(3*a^5 - 2*a^4*b - 5*a^3*b^2)*cosh(x)^2)*sinh(x)^6 + 4*(63*(a^5 + 2*a^4*b +

a^3*b^2)*cosh(x)^5 + 14*(3*a^5 - 2*a^4*b - 5*a^3*b^2)*cosh(x)^3 + 3*(a^5 - 2*a^4*b + 5*a^3*b^2)*cosh(x))*sinh

(x)^5 - a^5 - 2*a^4*b - a^3*b^2 - 2*(a^5 - 2*a^4*b + 5*a^3*b^2)*cosh(x)^4 + 2*(105*(a^5 + 2*a^4*b + a^3*b^2)*c

osh(x)^6 - a^5 + 2*a^4*b - 5*a^3*b^2 + 35*(3*a^5 - 2*a^4*b - 5*a^3*b^2)*cosh(x)^4 + 15*(a^5 - 2*a^4*b + 5*a^3*

b^2)*cosh(x)^2)*sinh(x)^4 + 8*(15*(a^5 + 2*a^4*b + a^3*b^2)*cosh(x)^7 + 7*(3*a^5 - 2*a^4*b - 5*a^3*b^2)*cosh(x

)^5 + 5*(a^5 - 2*a^4*b + 5*a^3*b^2)*cosh(x)^3 - (a^5 - 2*a^4*b + 5*a^3*b^2)*cosh(x))*sinh(x)^3 - (3*a^5 - 2*a^

4*b - 5*a^3*b^2)*cosh(x)^2 + (45*(a^5 + 2*a^4*b + a^3*b^2)*cosh(x)^8 + 28*(3*a^5 - 2*a^4*b - 5*a^3*b^2)*cosh(x

)^6 - 3*a^5 + 2*a^4*b + 5*a^3*b^2 + 30*(a^5 - 2*a^4*b + 5*a^3*b^2)*cosh(x)^4 - 12*(a^5 - 2*a^4*b + 5*a^3*b^2)*

cosh(x)^2)*sinh(x)^2 + 2*(5*(a^5 + 2*a^4*b + a^3*b^2)*cosh(x)^9 + 4*(3*a^5 - 2*a^4*b - 5*a^3*b^2)*cosh(x)^7 +

6*(a^5 - 2*a^4*b + 5*a^3*b^2)*cosh(x)^5 - 4*(a^5 - 2*a^4*b + 5*a^3*b^2)*cosh(x)^3 - (3*a^5 - 2*a^4*b - 5*a^3*b

^2)*cosh(x))*sinh(x))*sqrt(a + b)*log(-((a*b^2 + b^3)*cosh(x)^8 + 8*(a*b^2 + b^3)*cosh(x)*sinh(x)^7 + (a*b^2 +

b^3)*sinh(x)^8 - 2*(a*b^2 + 2*b^3)*cosh(x)^6 - 2*(a*b^2 + 2*b^3 - 14*(a*b^2 + b^3)*cosh(x)^2)*sinh(x)^6 + 4*(

14*(a*b^2 + b^3)*cosh(x)^3 - 3*(a*b^2 + 2*b^3)*cosh(x))*sinh(x)^5 + (a^3 - a^2*b + 4*a*b^2 + 6*b^3)*cosh(x)^4

+ (70*(a*b^2 + b^3)*cosh(x)^4 + a^3 - a^2*b + 4*a*b^2 + 6*b^3 - 30*(a*b^2 + 2*b^3)*cosh(x)^2)*sinh(x)^4 + 4*(1

4*(a*b^2 + b^3)*cosh(x)^5 - 10*(a*b^2 + 2*b^3)*cosh(x)^3 + (a^3 - a^2*b + 4*a*b^2 + 6*b^3)*cosh(x))*sinh(x)^3

+ a^3 + 3*a^2*b + 3*a*b^2 + b^3 + 2*(a^3 - 3*a*b^2 - 2*b^3)*cosh(x)^2 + 2*(14*(a*b^2 + b^3)*cosh(x)^6 - 15*(a*

b^2 + 2*b^3)*cosh(x)^4 + a^3 - 3*a*b^2 - 2*b^3 + 3*(a^3 - a^2*b + 4*a*b^2 + 6*b^3)*cosh(x)^2)*sinh(x)^2 + sqrt

(2)*(b^2*cosh(x)^6 + 6*b^2*cosh(x)*sinh(x)^5 + b^2*sinh(x)^6 - 3*b^2*cosh(x)^4 + 3*(5*b^2*cosh(x)^2 - b^2)*sin

h(x)^4 + 4*(5*b^2*cosh(x)^3 - 3*b^2*cosh(x))*sinh(x)^3 - (a^2 - 2*a*b - 3*b^2)*cosh(x)^2 + (15*b^2*cosh(x)^4 -

18*b^2*cosh(x)^2 - a^2 + 2*a*b + 3*b^2)*sinh(x)^2 - a^2 - 2*a*b - b^2 + 2*(3*b^2*cosh(x)^5 - 6*b^2*cosh(x)^3

- (a^2 - 2*a*b - 3*b^2)*cosh(x))*sinh(x))*sqrt(a + b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 + a - b)/(co

sh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*(2*(a*b^2 + b^3)*cosh(x)^7 - 3*(a*b^2 + 2*b^3)*cosh(x)^5 + (a^3

- a^2*b + 4*a*b^2 + 6*b^3)*cosh(x)^3 + (a^3 - 3*a*b^2 - 2*b^3)*cosh(x))*sinh(x))/(cosh(x)^6 + 6*cosh(x)^5*sinh

(x) + 15*cosh(x)^4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)

^6)) + 3*((a^5 + 2*a^4*b + a^3*b^2)*cosh(x)^10 + 10*(a^5 + 2*a^4*b + a^3*b^2)*cosh(x)*sinh(x)^9 + (a^5 + 2*a^4