3.200 $$\int \frac{\coth ^4(c+d x)}{(a+b \tanh ^2(c+d x))^3} \, dx$$

Optimal. Leaf size=228 $-\frac{\left (8 a^2+55 a b+35 b^2\right ) \coth ^3(c+d x)}{24 a^3 d (a+b)^2}-\frac{\left (-8 a^2 b+8 a^3-55 a b^2-35 b^3\right ) \coth (c+d x)}{8 a^4 d (a+b)^2}+\frac{b^{5/2} \left (63 a^2+90 a b+35 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{8 a^{9/2} d (a+b)^3}+\frac{b (11 a+7 b) \coth ^3(c+d x)}{8 a^2 d (a+b)^2 \left (a+b \tanh ^2(c+d x)\right )}+\frac{b \coth ^3(c+d x)}{4 a d (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{x}{(a+b)^3}$

[Out]

x/(a + b)^3 + (b^(5/2)*(63*a^2 + 90*a*b + 35*b^2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(8*a^(9/2)*(a + b)^
3*d) - ((8*a^3 - 8*a^2*b - 55*a*b^2 - 35*b^3)*Coth[c + d*x])/(8*a^4*(a + b)^2*d) - ((8*a^2 + 55*a*b + 35*b^2)*
Coth[c + d*x]^3)/(24*a^3*(a + b)^2*d) + (b*Coth[c + d*x]^3)/(4*a*(a + b)*d*(a + b*Tanh[c + d*x]^2)^2) + (b*(11
*a + 7*b)*Coth[c + d*x]^3)/(8*a^2*(a + b)^2*d*(a + b*Tanh[c + d*x]^2))

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Rubi [A]  time = 0.368472, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.304, Rules used = {3670, 472, 579, 583, 522, 206, 205} $-\frac{\left (8 a^2+55 a b+35 b^2\right ) \coth ^3(c+d x)}{24 a^3 d (a+b)^2}-\frac{\left (-8 a^2 b+8 a^3-55 a b^2-35 b^3\right ) \coth (c+d x)}{8 a^4 d (a+b)^2}+\frac{b^{5/2} \left (63 a^2+90 a b+35 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{8 a^{9/2} d (a+b)^3}+\frac{b (11 a+7 b) \coth ^3(c+d x)}{8 a^2 d (a+b)^2 \left (a+b \tanh ^2(c+d x)\right )}+\frac{b \coth ^3(c+d x)}{4 a d (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{x}{(a+b)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

x/(a + b)^3 + (b^(5/2)*(63*a^2 + 90*a*b + 35*b^2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(8*a^(9/2)*(a + b)^
3*d) - ((8*a^3 - 8*a^2*b - 55*a*b^2 - 35*b^3)*Coth[c + d*x])/(8*a^4*(a + b)^2*d) - ((8*a^2 + 55*a*b + 35*b^2)*
Coth[c + d*x]^3)/(24*a^3*(a + b)^2*d) + (b*Coth[c + d*x]^3)/(4*a*(a + b)*d*(a + b*Tanh[c + d*x]^2)^2) + (b*(11
*a + 7*b)*Coth[c + d*x]^3)/(8*a^2*(a + b)^2*d*(a + b*Tanh[c + d*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 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 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])

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{\coth ^4(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^4 \left (1-x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b \coth ^3(c+d x)}{4 a (a+b) d \left (a+b \tanh ^2(c+d x)\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{-4 a-7 b+7 b x^2}{x^4 \left (1-x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 a (a+b) d}\\ &=\frac{b \coth ^3(c+d x)}{4 a (a+b) d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{b (11 a+7 b) \coth ^3(c+d x)}{8 a^2 (a+b)^2 d \left (a+b \tanh ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{8 a^2+55 a b+35 b^2-5 b (11 a+7 b) x^2}{x^4 \left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{8 a^2 (a+b)^2 d}\\ &=-\frac{\left (8 a^2+55 a b+35 b^2\right ) \coth ^3(c+d x)}{24 a^3 (a+b)^2 d}+\frac{b \coth ^3(c+d x)}{4 a (a+b) d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{b (11 a+7 b) \coth ^3(c+d x)}{8 a^2 (a+b)^2 d \left (a+b \tanh ^2(c+d x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-3 \left (8 a^3-8 a^2 b-55 a b^2-35 b^3\right )-3 b \left (8 a^2+55 a b+35 b^2\right ) x^2}{x^2 \left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{24 a^3 (a+b)^2 d}\\ &=-\frac{\left (8 a^3-8 a^2 b-55 a b^2-35 b^3\right ) \coth (c+d x)}{8 a^4 (a+b)^2 d}-\frac{\left (8 a^2+55 a b+35 b^2\right ) \coth ^3(c+d x)}{24 a^3 (a+b)^2 d}+\frac{b \coth ^3(c+d x)}{4 a (a+b) d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{b (11 a+7 b) \coth ^3(c+d x)}{8 a^2 (a+b)^2 d \left (a+b \tanh ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{3 \left (8 a^4-8 a^3 b+8 a^2 b^2+55 a b^3+35 b^4\right )+3 b \left (8 a^3-8 a^2 b-55 a b^2-35 b^3\right ) x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{24 a^4 (a+b)^2 d}\\ &=-\frac{\left (8 a^3-8 a^2 b-55 a b^2-35 b^3\right ) \coth (c+d x)}{8 a^4 (a+b)^2 d}-\frac{\left (8 a^2+55 a b+35 b^2\right ) \coth ^3(c+d x)}{24 a^3 (a+b)^2 d}+\frac{b \coth ^3(c+d x)}{4 a (a+b) d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{b (11 a+7 b) \coth ^3(c+d x)}{8 a^2 (a+b)^2 d \left (a+b \tanh ^2(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{(a+b)^3 d}+\frac{\left (b^3 \left (63 a^2+90 a b+35 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tanh (c+d x)\right )}{8 a^4 (a+b)^3 d}\\ &=\frac{x}{(a+b)^3}+\frac{b^{5/2} \left (63 a^2+90 a b+35 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{8 a^{9/2} (a+b)^3 d}-\frac{\left (8 a^3-8 a^2 b-55 a b^2-35 b^3\right ) \coth (c+d x)}{8 a^4 (a+b)^2 d}-\frac{\left (8 a^2+55 a b+35 b^2\right ) \coth ^3(c+d x)}{24 a^3 (a+b)^2 d}+\frac{b \coth ^3(c+d x)}{4 a (a+b) d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac{b (11 a+7 b) \coth ^3(c+d x)}{8 a^2 (a+b)^2 d \left (a+b \tanh ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 3.41783, size = 194, normalized size = 0.85 $\frac{\frac{3 b^{5/2} \left (63 a^2+90 a b+35 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{a^{9/2} (a+b)^3}+\frac{12 b^4 \sinh (2 (c+d x))}{a^3 (a+b)^2 ((a+b) \cosh (2 (c+d x))+a-b)^2}+\frac{3 b^3 (17 a+11 b) \sinh (2 (c+d x))}{a^4 (a+b)^2 ((a+b) \cosh (2 (c+d x))+a-b)}+\frac{8 (9 b-4 a) \coth (c+d x)}{a^4}-\frac{8 \coth (c+d x) \text{csch}^2(c+d x)}{a^3}+\frac{24 (c+d x)}{(a+b)^3}}{24 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((24*(c + d*x))/(a + b)^3 + (3*b^(5/2)*(63*a^2 + 90*a*b + 35*b^2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(a^
(9/2)*(a + b)^3) + (8*(-4*a + 9*b)*Coth[c + d*x])/a^4 - (8*Coth[c + d*x]*Csch[c + d*x]^2)/a^3 + (12*b^4*Sinh[2
*(c + d*x)])/(a^3*(a + b)^2*(a - b + (a + b)*Cosh[2*(c + d*x)])^2) + (3*b^3*(17*a + 11*b)*Sinh[2*(c + d*x)])/(
a^4*(a + b)^2*(a - b + (a + b)*Cosh[2*(c + d*x)])))/(24*d)

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Maple [B]  time = 0.14, size = 2139, 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(coth(d*x+c)^4/(a+b*tanh(d*x+c)^2)^3,x)

[Out]

63/8/d*b^3/(a+b)^3/a^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))-63/8/d*b^3/(a+b)^3/a^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))-125/8/d*b^5/(a+b)^3/a^3/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arc
tanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))-63/8/d*b^3/(a+b)^3/a/(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))-153/8/d*b^4/(a+b
)^3/a^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))-63/8/d*b^3/(a+b)^3/a/(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))-153/8/d*b^4/(a+b)^3/a^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))+1/d/(a+b)^3*ln(tanh(1/2*d*x+1/2*
c)+1)-1/d/(a+b)^3*ln(tanh(1/2*d*x+1/2*c)-1)+143/4/d*b^5/(a+b)^3/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)^3+45/4/d*b^4/(a+b)^3/a^3/((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))-45/4/d*b^4/(a+b)^3/a^3/((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))+15/2/d*b^4/(a+b)^
3/a^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)^7-
35/8/d*b^5/(a+b)^3/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))-125/8/d*b^5/(a+b)^3/a^3/(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))+3/2/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
+11/d*b^6/(a+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^6/(a+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)^3-5/8/d/a^3*tanh(1/2*d*x+1/2*c)-5/8/d/a^3/tanh(1/2*d*x+1/2*c)+17/4/d*b^3/(a+b)^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/a*tanh(1/2*d*x+1/2*c)^7+51/4
/d*b^3/(a+b)^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/a*tanh(1/2*d*
x+1/2*c)^5+75/2/d*b^4/(a+b)^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/a^2*tanh(1/2*d*x+1/2*c)^5+51/4/d*b^3/(a+b)^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/a*tanh(1/2*d*x+1/2*c)^3+75/2/d*b^4/(a+b)^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/a^2*tanh(1/2*d*x+1/2*c)^3+17/4/d*b^3/(a+b)^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/a*tanh(1/2*d*x+1/2*c)-35/8/d*b^6/(a+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))+13/4/d
*b^5/(a+b)^3/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+35/8/d*b^5/(a+b)^3/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))-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+13/4/d*b^5/(a+b)^
3/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)+15
/2/d*b^4/(a+b)^3/a^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)+143/4/d*b^5/(a+b)^3/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)^5-35/8/d*b^6/(a+b)^3/a^4/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arct
anh(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(coth(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(coth(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{\coth ^{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(coth(d*x+c)**4/(a+b*tanh(d*x+c)**2)**3,x)

[Out]

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

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Giac [B]  time = 1.45776, size = 683, normalized size = 3. \begin{align*} \frac{{\left (63 \, a^{2} b^{3} + 90 \, a b^{4} + 35 \, b^{5}\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 )}{8 \,{\left (a^{7} d + 3 \, a^{6} b d + 3 \, a^{5} b^{2} d + a^{4} b^{3} d\right )} \sqrt{a b}} + \frac{d x + c}{a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d} - \frac{17 \, a^{3} b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 7 \, a^{2} b^{4} e^{\left (6 \, d x + 6 \, c\right )} - 21 \, a b^{5} e^{\left (6 \, d x + 6 \, c\right )} - 11 \, b^{6} e^{\left (6 \, d x + 6 \, c\right )} + 51 \, a^{3} b^{3} e^{\left (4 \, d x + 4 \, c\right )} - a^{2} b^{4} e^{\left (4 \, d x + 4 \, c\right )} + 29 \, a b^{5} e^{\left (4 \, d x + 4 \, c\right )} + 33 \, b^{6} e^{\left (4 \, d x + 4 \, c\right )} + 51 \, a^{3} b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 37 \, a^{2} b^{4} e^{\left (2 \, d x + 2 \, c\right )} - 47 \, a b^{5} e^{\left (2 \, d x + 2 \, c\right )} - 33 \, b^{6} e^{\left (2 \, d x + 2 \, c\right )} + 17 \, a^{3} b^{3} + 45 \, a^{2} b^{4} + 39 \, a b^{5} + 11 \, b^{6}}{4 \,{\left (a^{7} d + 3 \, a^{6} b d + 3 \, a^{5} b^{2} d + a^{4} b^{3} d\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{2 \,{\left (6 \, a e^{\left (4 \, d x + 4 \, c\right )} - 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 )} + 4 \, a - 9 \, b\right )}}{3 \, a^{4} d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(63*a^2*b^3 + 90*a*b^4 + 35*b^5)*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a
^7*d + 3*a^6*b*d + 3*a^5*b^2*d + a^4*b^3*d)*sqrt(a*b)) + (d*x + c)/(a^3*d + 3*a^2*b*d + 3*a*b^2*d + b^3*d) - 1
/4*(17*a^3*b^3*e^(6*d*x + 6*c) + 7*a^2*b^4*e^(6*d*x + 6*c) - 21*a*b^5*e^(6*d*x + 6*c) - 11*b^6*e^(6*d*x + 6*c)
+ 51*a^3*b^3*e^(4*d*x + 4*c) - a^2*b^4*e^(4*d*x + 4*c) + 29*a*b^5*e^(4*d*x + 4*c) + 33*b^6*e^(4*d*x + 4*c) +
51*a^3*b^3*e^(2*d*x + 2*c) + 37*a^2*b^4*e^(2*d*x + 2*c) - 47*a*b^5*e^(2*d*x + 2*c) - 33*b^6*e^(2*d*x + 2*c) +
17*a^3*b^3 + 45*a^2*b^4 + 39*a*b^5 + 11*b^6)/((a^7*d + 3*a^6*b*d + 3*a^5*b^2*d + a^4*b^3*d)*(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) - 2/3*(6*a*e^(4*d*x + 4*c) - 9*b*
e^(4*d*x + 4*c) - 6*a*e^(2*d*x + 2*c) + 18*b*e^(2*d*x + 2*c) + 4*a - 9*b)/(a^4*d*(e^(2*d*x + 2*c) - 1)^3)