### 3.822 $$\int \frac{1}{(\coth ^2(x)+\text{csch}^2(x))^3} \, dx$$

Optimal. Leaf size=54 $x-\frac{\tanh ^3(x)}{2 \left (2-\tanh ^2(x)\right )^2}-\frac{\tanh (x)}{4 \left (2-\tanh ^2(x)\right )}-\frac{7 \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )}{4 \sqrt{2}}$

[Out]

x - (7*ArcTanh[Tanh[x]/Sqrt[2]])/(4*Sqrt[2]) - Tanh[x]^3/(2*(2 - Tanh[x]^2)^2) - Tanh[x]/(4*(2 - Tanh[x]^2))

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Rubi [A]  time = 0.0884717, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 11, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.364, Rules used = {470, 578, 522, 206} $x-\frac{\tanh ^3(x)}{2 \left (2-\tanh ^2(x)\right )^2}-\frac{\tanh (x)}{4 \left (2-\tanh ^2(x)\right )}-\frac{7 \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )}{4 \sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Coth[x]^2 + Csch[x]^2)^(-3),x]

[Out]

x - (7*ArcTanh[Tanh[x]/Sqrt[2]])/(4*Sqrt[2]) - Tanh[x]^3/(2*(2 - Tanh[x]^2)^2) - Tanh[x]/(4*(2 - Tanh[x]^2))

Rule 470

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

Rule 578

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

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

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{1}{\left (\coth ^2(x)+\text{csch}^2(x)\right )^3} \, dx &=\operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right ) \left (2-x^2\right )^3} \, dx,x,\tanh (x)\right )\\ &=-\frac{\tanh ^3(x)}{2 \left (2-\tanh ^2(x)\right )^2}+\frac{1}{4} \operatorname{Subst}\left (\int \frac{x^2 \left (6-2 x^2\right )}{\left (1-x^2\right ) \left (2-x^2\right )^2} \, dx,x,\tanh (x)\right )\\ &=-\frac{\tanh ^3(x)}{2 \left (2-\tanh ^2(x)\right )^2}-\frac{\tanh (x)}{4 \left (2-\tanh ^2(x)\right )}-\frac{1}{8} \operatorname{Subst}\left (\int \frac{-2-6 x^2}{\left (1-x^2\right ) \left (2-x^2\right )} \, dx,x,\tanh (x)\right )\\ &=-\frac{\tanh ^3(x)}{2 \left (2-\tanh ^2(x)\right )^2}-\frac{\tanh (x)}{4 \left (2-\tanh ^2(x)\right )}-\frac{7}{4} \operatorname{Subst}\left (\int \frac{1}{2-x^2} \, dx,x,\tanh (x)\right )+\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (x)\right )\\ &=x-\frac{7 \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )}{4 \sqrt{2}}-\frac{\tanh ^3(x)}{2 \left (2-\tanh ^2(x)\right )^2}-\frac{\tanh (x)}{4 \left (2-\tanh ^2(x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.187356, size = 66, normalized size = 1.22 $\frac{76 x-2 \sinh (2 x)-3 \sinh (4 x)+48 x \cosh (2 x)+4 x \cosh (4 x)-7 \sqrt{2} (\cosh (2 x)+3)^2 \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )}{8 (\cosh (2 x)+3)^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Coth[x]^2 + Csch[x]^2)^(-3),x]

[Out]

(76*x + 48*x*Cosh[2*x] - 7*Sqrt[2]*ArcTanh[Tanh[x]/Sqrt[2]]*(3 + Cosh[2*x])^2 + 4*x*Cosh[4*x] - 2*Sinh[2*x] -
3*Sinh[4*x])/(8*(3 + Cosh[2*x])^2)

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Maple [B]  time = 0.049, size = 145, normalized size = 2.7 \begin{align*} \ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) -\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) +2\,{\frac{-1/8\, \left ( \tanh \left ( x/2 \right ) \right ) ^{7}-5/8\, \left ( \tanh \left ( x/2 \right ) \right ) ^{5}-5/8\, \left ( \tanh \left ( x/2 \right ) \right ) ^{3}-1/8\,\tanh \left ( x/2 \right ) }{ \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{4}+1 \right ) ^{2}}}-{\frac{7\,\sqrt{2}}{32}\ln \left ({ \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}} \right ) }+{\frac{7\,\sqrt{2}}{32}\ln \left ({ \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(coth(x)^2+csch(x)^2)^3,x)

[Out]

ln(tanh(1/2*x)+1)-ln(tanh(1/2*x)-1)+2*(-1/8*tanh(1/2*x)^7-5/8*tanh(1/2*x)^5-5/8*tanh(1/2*x)^3-1/8*tanh(1/2*x))
/(tanh(1/2*x)^4+1)^2-7/32*2^(1/2)*ln((tanh(1/2*x)^2+2^(1/2)*tanh(1/2*x)+1)/(tanh(1/2*x)^2-2^(1/2)*tanh(1/2*x)+
1))+7/32*2^(1/2)*ln((tanh(1/2*x)^2-2^(1/2)*tanh(1/2*x)+1)/(tanh(1/2*x)^2+2^(1/2)*tanh(1/2*x)+1))

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Maxima [B]  time = 1.77191, size = 113, normalized size = 2.09 \begin{align*} \frac{7}{16} \, \sqrt{2} \log \left (-\frac{2 \, \sqrt{2} - e^{\left (-2 \, x\right )} - 3}{2 \, \sqrt{2} + e^{\left (-2 \, x\right )} + 3}\right ) + x - \frac{19 \, e^{\left (-2 \, x\right )} + 57 \, e^{\left (-4 \, x\right )} + 17 \, e^{\left (-6 \, x\right )} + 3}{2 \,{\left (12 \, e^{\left (-2 \, x\right )} + 38 \, e^{\left (-4 \, x\right )} + 12 \, e^{\left (-6 \, x\right )} + e^{\left (-8 \, x\right )} + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(coth(x)^2+csch(x)^2)^3,x, algorithm="maxima")

[Out]

7/16*sqrt(2)*log(-(2*sqrt(2) - e^(-2*x) - 3)/(2*sqrt(2) + e^(-2*x) + 3)) + x - 1/2*(19*e^(-2*x) + 57*e^(-4*x)
+ 17*e^(-6*x) + 3)/(12*e^(-2*x) + 38*e^(-4*x) + 12*e^(-6*x) + e^(-8*x) + 1)

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Fricas [B]  time = 2.41457, size = 2363, normalized size = 43.76 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(coth(x)^2+csch(x)^2)^3,x, algorithm="fricas")

[Out]

1/16*(16*x*cosh(x)^8 + 128*x*cosh(x)*sinh(x)^7 + 16*x*sinh(x)^8 + 8*(24*x + 17)*cosh(x)^6 + 8*(56*x*cosh(x)^2
+ 24*x + 17)*sinh(x)^6 + 16*(56*x*cosh(x)^3 + 3*(24*x + 17)*cosh(x))*sinh(x)^5 + 152*(4*x + 3)*cosh(x)^4 + 8*(
140*x*cosh(x)^4 + 15*(24*x + 17)*cosh(x)^2 + 76*x + 57)*sinh(x)^4 + 32*(28*x*cosh(x)^5 + 5*(24*x + 17)*cosh(x)
^3 + 19*(4*x + 3)*cosh(x))*sinh(x)^3 + 8*(24*x + 19)*cosh(x)^2 + 8*(56*x*cosh(x)^6 + 15*(24*x + 17)*cosh(x)^4
+ 114*(4*x + 3)*cosh(x)^2 + 24*x + 19)*sinh(x)^2 + 7*(sqrt(2)*cosh(x)^8 + 8*sqrt(2)*cosh(x)*sinh(x)^7 + sqrt(2
)*sinh(x)^8 + 4*(7*sqrt(2)*cosh(x)^2 + 3*sqrt(2))*sinh(x)^6 + 12*sqrt(2)*cosh(x)^6 + 8*(7*sqrt(2)*cosh(x)^3 +
9*sqrt(2)*cosh(x))*sinh(x)^5 + 2*(35*sqrt(2)*cosh(x)^4 + 90*sqrt(2)*cosh(x)^2 + 19*sqrt(2))*sinh(x)^4 + 38*sqr
t(2)*cosh(x)^4 + 8*(7*sqrt(2)*cosh(x)^5 + 30*sqrt(2)*cosh(x)^3 + 19*sqrt(2)*cosh(x))*sinh(x)^3 + 4*(7*sqrt(2)*
cosh(x)^6 + 45*sqrt(2)*cosh(x)^4 + 57*sqrt(2)*cosh(x)^2 + 3*sqrt(2))*sinh(x)^2 + 12*sqrt(2)*cosh(x)^2 + 8*(sqr
t(2)*cosh(x)^7 + 9*sqrt(2)*cosh(x)^5 + 19*sqrt(2)*cosh(x)^3 + 3*sqrt(2)*cosh(x))*sinh(x) + sqrt(2))*log((3*(2*
sqrt(2) + 3)*cosh(x)^2 - 4*(3*sqrt(2) + 4)*cosh(x)*sinh(x) + 3*(2*sqrt(2) + 3)*sinh(x)^2 + 2*sqrt(2) + 3)/(cos
h(x)^2 + sinh(x)^2 + 3)) + 16*(8*x*cosh(x)^7 + 3*(24*x + 17)*cosh(x)^5 + 38*(4*x + 3)*cosh(x)^3 + (24*x + 19)*
cosh(x))*sinh(x) + 16*x + 24)/(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 4*(7*cosh(x)^2 + 3)*sinh(x)^6 + 1
2*cosh(x)^6 + 8*(7*cosh(x)^3 + 9*cosh(x))*sinh(x)^5 + 2*(35*cosh(x)^4 + 90*cosh(x)^2 + 19)*sinh(x)^4 + 38*cosh
(x)^4 + 8*(7*cosh(x)^5 + 30*cosh(x)^3 + 19*cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 + 45*cosh(x)^4 + 57*cosh(x)^2 +
3)*sinh(x)^2 + 12*cosh(x)^2 + 8*(cosh(x)^7 + 9*cosh(x)^5 + 19*cosh(x)^3 + 3*cosh(x))*sinh(x) + 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(coth(x)**2+csch(x)**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.13614, size = 97, normalized size = 1.8 \begin{align*} -\frac{7}{16} \, \sqrt{2} \log \left (-\frac{2 \, \sqrt{2} - e^{\left (2 \, x\right )} - 3}{2 \, \sqrt{2} + e^{\left (2 \, x\right )} + 3}\right ) + x + \frac{17 \, e^{\left (6 \, x\right )} + 57 \, e^{\left (4 \, x\right )} + 19 \, e^{\left (2 \, x\right )} + 3}{2 \,{\left (e^{\left (4 \, x\right )} + 6 \, e^{\left (2 \, x\right )} + 1\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(coth(x)^2+csch(x)^2)^3,x, algorithm="giac")

[Out]

-7/16*sqrt(2)*log(-(2*sqrt(2) - e^(2*x) - 3)/(2*sqrt(2) + e^(2*x) + 3)) + x + 1/2*(17*e^(6*x) + 57*e^(4*x) + 1
9*e^(2*x) + 3)/(e^(4*x) + 6*e^(2*x) + 1)^2