### 3.215 $$\int \coth ^2(x) \sqrt{a+b \tanh ^2(x)} \, dx$$

Optimal. Leaf size=48 $\sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a+b} \tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )-\coth (x) \sqrt{a+b \tanh ^2(x)}$

[Out]

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

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Rubi [A]  time = 0.0926206, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.294, Rules used = {3670, 475, 12, 377, 206} $\sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a+b} \tanh (x)}{\sqrt{a+b \tanh ^2(x)}}\right )-\coth (x) \sqrt{a+b \tanh ^2(x)}$

Antiderivative was successfully veriﬁed.

[In]

Int[Coth[x]^2*Sqrt[a + b*Tanh[x]^2],x]

[Out]

Sqrt[a + b]*ArcTanh[(Sqrt[a + b]*Tanh[x])/Sqrt[a + b*Tanh[x]^2]] - Coth[x]*Sqrt[a + b*Tanh[x]^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 475

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

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

Mathematica [C]  time = 0.105273, size = 42, normalized size = 0.88 $-\coth (x) \sqrt{a+b \tanh ^2(x)} \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{(a+b) \tanh ^2(x)}{b \tanh ^2(x)+a}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Coth[x]^2*Sqrt[a + b*Tanh[x]^2],x]

[Out]

-(Coth[x]*Hypergeometric2F1[-1/2, 1, 1/2, ((a + b)*Tanh[x]^2)/(a + b*Tanh[x]^2)]*Sqrt[a + b*Tanh[x]^2])

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Maple [F]  time = 0.153, size = 0, normalized size = 0. \begin{align*} \int \left ({\rm coth} \left (x\right ) \right ) ^{2}\sqrt{a+b \left ( \tanh \left ( x \right ) \right ) ^{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^2*(a+b*tanh(x)^2)^(1/2),x)

[Out]

int(coth(x)^2*(a+b*tanh(x)^2)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \tanh \left (x\right )^{2} + a} \coth \left (x\right )^{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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.88513, size = 4456, normalized size = 92.83 \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)^(1/2),x, algorithm="fricas")

[Out]

[1/4*((cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*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*(14*(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)*sinh(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)/(cosh(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)) + (cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt(a + b)*log(((a + b
)*cosh(x)^4 + 4*(a + b)*cosh(x)*sinh(x)^3 + (a + b)*sinh(x)^4 + 2*a*cosh(x)^2 + 2*(3*(a + b)*cosh(x)^2 + a)*si
nh(x)^2 + sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*sqrt(a + b)*sqrt(((a + b)*cosh(x)^2 + (a + b
)*sinh(x)^2 + a - b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 4*((a + b)*cosh(x)^3 + a*cosh(x))*sinh(x)
+ a + b)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)) - 4*sqrt(2)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2
+ a - b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1), -1/2*(
(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt(-a - b)*arctan(sqrt(2)*(b*cosh(x)^2 + 2*b*cosh(x)*sinh(x)
+ b*sinh(x)^2 - a - b)*sqrt(-a - b)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x)^2 + a - b)/(cosh(x)^2 - 2*cosh(
x)*sinh(x) + sinh(x)^2))/((a*b + b^2)*cosh(x)^4 + 4*(a*b + b^2)*cosh(x)*sinh(x)^3 + (a*b + b^2)*sinh(x)^4 + (a
^2 - a*b - 2*b^2)*cosh(x)^2 + (6*(a*b + b^2)*cosh(x)^2 + a^2 - a*b - 2*b^2)*sinh(x)^2 + a^2 + 2*a*b + b^2 + 2*
(2*(a*b + b^2)*cosh(x)^3 + (a^2 - a*b - 2*b^2)*cosh(x))*sinh(x))) + (cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2
- 1)*sqrt(-a - b)*arctan(sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*sqrt(-a - b)*sqrt(((a + b)*c
osh(x)^2 + (a + b)*sinh(x)^2 + a - b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/((a + b)*cosh(x)^4 + 4*(a +
b)*cosh(x)*sinh(x)^3 + (a + b)*sinh(x)^4 + 2*(a - b)*cosh(x)^2 + 2*(3*(a + b)*cosh(x)^2 + a - b)*sinh(x)^2 +
4*((a + b)*cosh(x)^3 + (a - b)*cosh(x))*sinh(x) + a + b)) + 2*sqrt(2)*sqrt(((a + b)*cosh(x)^2 + (a + b)*sinh(x
)^2 + a - b)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)))/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**2*(a+b*tanh(x)**2)**(1/2),x)

[Out]

Integral(sqrt(a + b*tanh(x)**2)*coth(x)**2, x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^2*(a+b*tanh(x)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError