### 3.48 $$\int \frac{\text{csch}^4(c+d x)}{(a+b \tanh ^2(c+d x))^3} \, dx$$

Optimal. Leaf size=151 $\frac{b (7 a+11 b) \tanh (c+d x)}{8 a^4 d \left (a+b \tanh ^2(c+d x)\right )}+\frac{b (a+b) \tanh (c+d x)}{4 a^3 d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{(a+3 b) \coth (c+d x)}{a^4 d}+\frac{5 \sqrt{b} (3 a+7 b) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{8 a^{9/2} d}-\frac{\coth ^3(c+d x)}{3 a^3 d}$

[Out]

(5*Sqrt[b]*(3*a + 7*b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(8*a^(9/2)*d) + ((a + 3*b)*Coth[c + d*x])/(a^4
*d) - Coth[c + d*x]^3/(3*a^3*d) + (b*(a + b)*Tanh[c + d*x])/(4*a^3*d*(a + b*Tanh[c + d*x]^2)^2) + (b*(7*a + 11
*b)*Tanh[c + d*x])/(8*a^4*d*(a + b*Tanh[c + d*x]^2))

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Rubi [A]  time = 0.200554, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.217, Rules used = {3663, 456, 1259, 1261, 205} $\frac{b (7 a+11 b) \tanh (c+d x)}{8 a^4 d \left (a+b \tanh ^2(c+d x)\right )}+\frac{b (a+b) \tanh (c+d x)}{4 a^3 d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{(a+3 b) \coth (c+d x)}{a^4 d}+\frac{5 \sqrt{b} (3 a+7 b) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{8 a^{9/2} d}-\frac{\coth ^3(c+d x)}{3 a^3 d}$

Antiderivative was successfully veriﬁed.

[In]

Int[Csch[c + d*x]^4/(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

(5*Sqrt[b]*(3*a + 7*b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(8*a^(9/2)*d) + ((a + 3*b)*Coth[c + d*x])/(a^4
*d) - Coth[c + d*x]^3/(3*a^3*d) + (b*(a + b)*Tanh[c + d*x])/(4*a^3*d*(a + b*Tanh[c + d*x]^2)^2) + (b*(7*a + 11
*b)*Tanh[c + d*x])/(8*a^4*d*(a + b*Tanh[c + d*x]^2))

Rule 3663

Int[sin[(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^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rule 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\text{csch}^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{x^4 \left (a+b x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b (a+b) \tanh (c+d x)}{4 a^3 d \left (a+b \tanh ^2(c+d x)\right )^2}-\frac{b \operatorname{Subst}\left (\int \frac{-\frac{4}{a b}+\frac{4 (a+b) x^2}{a^2 b}-\frac{3 (a+b) x^4}{a^3}}{x^4 \left (a+b x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=\frac{b (a+b) \tanh (c+d x)}{4 a^3 d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{b (7 a+11 b) \tanh (c+d x)}{8 a^4 d \left (a+b \tanh ^2(c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-8 a b+8 b (a+2 b) x^2-\frac{b^2 (7 a+11 b) x^4}{a}}{x^4 \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{8 a^3 b d}\\ &=\frac{b (a+b) \tanh (c+d x)}{4 a^3 d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{b (7 a+11 b) \tanh (c+d x)}{8 a^4 d \left (a+b \tanh ^2(c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \left (-\frac{8 b}{x^4}+\frac{8 b (a+3 b)}{a x^2}-\frac{5 b^2 (3 a+7 b)}{a \left (a+b x^2\right )}\right ) \, dx,x,\tanh (c+d x)\right )}{8 a^3 b d}\\ &=\frac{(a+3 b) \coth (c+d x)}{a^4 d}-\frac{\coth ^3(c+d x)}{3 a^3 d}+\frac{b (a+b) \tanh (c+d x)}{4 a^3 d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{b (7 a+11 b) \tanh (c+d x)}{8 a^4 d \left (a+b \tanh ^2(c+d x)\right )}+\frac{(5 b (3 a+7 b)) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tanh (c+d x)\right )}{8 a^4 d}\\ &=\frac{5 \sqrt{b} (3 a+7 b) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{8 a^{9/2} d}+\frac{(a+3 b) \coth (c+d x)}{a^4 d}-\frac{\coth ^3(c+d x)}{3 a^3 d}+\frac{b (a+b) \tanh (c+d x)}{4 a^3 d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{b (7 a+11 b) \tanh (c+d x)}{8 a^4 d \left (a+b \tanh ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.24437, size = 149, normalized size = 0.99 $\frac{\frac{3 \sqrt{a} b \sinh (2 (c+d x)) \left (\left (9 a^2+20 a b+11 b^2\right ) \cosh (2 (c+d x))+9 a^2+6 a b-11 b^2\right )}{((a+b) \cosh (2 (c+d x))+a-b)^2}+15 \sqrt{b} (3 a+7 b) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )-8 \sqrt{a} \coth (c+d x) \left (a \text{csch}^2(c+d x)-2 a-9 b\right )}{24 a^{9/2} d}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Csch[c + d*x]^4/(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

(15*Sqrt[b]*(3*a + 7*b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]] - 8*Sqrt[a]*Coth[c + d*x]*(-2*a - 9*b + a*Csch
[c + d*x]^2) + (3*Sqrt[a]*b*(9*a^2 + 6*a*b - 11*b^2 + (9*a^2 + 20*a*b + 11*b^2)*Cosh[2*(c + d*x)])*Sinh[2*(c +
d*x)])/(a - b + (a + b)*Cosh[2*(c + d*x)])^2)/(24*a^(9/2)*d)

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Maple [B]  time = 0.135, size = 1416, normalized size = 9.4 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^4/(a+b*tanh(d*x+c)^2)^3,x)

[Out]

-15/8/d/a^2*b/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^
(1/2)-a-2*b)*a)^(1/2))+3/2/d/a^4*tanh(1/2*d*x+1/2*c)*b+3/2/d/a^4/tanh(1/2*d*x+1/2*c)*b-35/8/d*b^3/a^4/(b*(a+b)
)^(1/2)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))
-35/8/d*b^3/a^4/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))
^(1/2)+a+2*b)*a)^(1/2))+9/4/d/a^2*b/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2
*b+a)^2*tanh(1/2*d*x+1/2*c)^7-25/4/d/a^3*b^2/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh
(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))+3/8/d/a^3*tanh(1/2*d*x+1/2*c)+3/8/d/a^3/tanh(1/2*d*x+1/2*
c)-25/4/d/a^3*b^2/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+
b))^(1/2)-a-2*b)*a)^(1/2))-15/8/d/a^2*b/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*
d*x+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))+27/4/d/a^2*b/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*
a+4*tanh(1/2*d*x+1/2*c)^2*b+a)^2*tanh(1/2*d*x+1/2*c)^3+67/4/d/a^3*b^2/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+
1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)^2*tanh(1/2*d*x+1/2*c)^3+9/4/d/a^2*b/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/
2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)^2*tanh(1/2*d*x+1/2*c)+15/8/d/a^3*b/((2*(b*(a+b))^(1/2)-a-2*b)*a)
^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))-15/8/d/a^3*b/((2*(b*(a+b))^(1/2)+a+2
*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))-35/8/d*b^2/a^4/((2*(b*(a+b))^(1
/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))+11/d*b^3/a^4/(tanh(1/2*d
*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)^2*tanh(1/2*d*x+1/2*c)^5+11/d*b^3/a^4/(tan
h(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)^2*tanh(1/2*d*x+1/2*c)^3+13/4/d*b^2
/a^3/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)^2*tanh(1/2*d*x+1/2*c)+13/
4/d*b^2/a^3/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)^2*tanh(1/2*d*x+1/2
*c)^7+27/4/d/a^2*b/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)^2*tanh(1/2*
d*x+1/2*c)^5+67/4/d/a^3*b^2/(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)^2*
tanh(1/2*d*x+1/2*c)^5-1/24/d/a^3*tanh(1/2*d*x+1/2*c)^3-1/24/d/a^3/tanh(1/2*d*x+1/2*c)^3+35/8/d*b^2/a^4/((2*(b*
(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4/(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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(csch(d*x+c)^4/(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{csch}^{4}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)**4/(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Integral(csch(c + d*x)**4/(a + b*tanh(c + d*x)**2)**3, x)

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Giac [B]  time = 1.91782, size = 549, normalized size = 3.64 \begin{align*} \frac{\frac{15 \,{\left (3 \, a b e^{\left (2 \, c\right )} + 7 \, b^{2} e^{\left (2 \, c\right )}\right )} \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt{a b}}\right ) e^{\left (-2 \, c\right )}}{\sqrt{a b} a^{4}} - \frac{6 \,{\left (9 \, a^{3} b e^{\left (6 \, d x + 6 \, c\right )} + 7 \, a^{2} b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 13 \, a b^{3} e^{\left (6 \, d x + 6 \, c\right )} - 11 \, b^{4} e^{\left (6 \, d x + 6 \, c\right )} + 27 \, a^{3} b e^{\left (4 \, d x + 4 \, c\right )} + 15 \, a^{2} b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 5 \, a b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 33 \, b^{4} e^{\left (4 \, d x + 4 \, c\right )} + 27 \, a^{3} b e^{\left (2 \, d x + 2 \, c\right )} + 37 \, a^{2} b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 23 \, a b^{3} e^{\left (2 \, d x + 2 \, c\right )} - 33 \, b^{4} e^{\left (2 \, d x + 2 \, c\right )} + 9 \, a^{3} b + 29 \, a^{2} b^{2} + 31 \, a b^{3} + 11 \, b^{4}\right )}}{{\left (a^{5} + a^{4} b\right )}{\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}^{2}} + \frac{16 \,{\left (9 \, b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a e^{\left (2 \, d x + 2 \, c\right )} - 18 \, b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a + 9 \, b\right )}}{a^{4}{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{24 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4/(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

1/24*(15*(3*a*b*e^(2*c) + 7*b^2*e^(2*c))*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))
*e^(-2*c)/(sqrt(a*b)*a^4) - 6*(9*a^3*b*e^(6*d*x + 6*c) + 7*a^2*b^2*e^(6*d*x + 6*c) - 13*a*b^3*e^(6*d*x + 6*c)
- 11*b^4*e^(6*d*x + 6*c) + 27*a^3*b*e^(4*d*x + 4*c) + 15*a^2*b^2*e^(4*d*x + 4*c) + 5*a*b^3*e^(4*d*x + 4*c) + 3
3*b^4*e^(4*d*x + 4*c) + 27*a^3*b*e^(2*d*x + 2*c) + 37*a^2*b^2*e^(2*d*x + 2*c) - 23*a*b^3*e^(2*d*x + 2*c) - 33*
b^4*e^(2*d*x + 2*c) + 9*a^3*b + 29*a^2*b^2 + 31*a*b^3 + 11*b^4)/((a^5 + a^4*b)*(a*e^(4*d*x + 4*c) + b*e^(4*d*x
+ 4*c) + 2*a*e^(2*d*x + 2*c) - 2*b*e^(2*d*x + 2*c) + a + b)^2) + 16*(9*b*e^(4*d*x + 4*c) - 6*a*e^(2*d*x + 2*c
) - 18*b*e^(2*d*x + 2*c) + 2*a + 9*b)/(a^4*(e^(2*d*x + 2*c) - 1)^3))/d