\(\int \frac {\coth ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx\) [179]

Optimal result

Integrand size = 23, antiderivative size = 82 \[ \int \frac {\coth ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {x}{a+b}+\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{5/2} (a+b) d}-\frac {(a-b) \coth (c+d x)}{a^2 d}-\frac {\coth ^3(c+d x)}{3 a d} \]

[Out]

x/(a+b)+b^(5/2)*arctan(b^(1/2)*tanh(d*x+c)/a^(1/2))/a^(5/2)/(a+b)/d-(a-b)*coth(d*x+c)/a^2/d-1/3*coth(d*x+c)^3/
a/d

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3751, 491, 597, 536, 212, 211} \[ \int \frac {\coth ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{5/2} d (a+b)}-\frac {(a-b) \coth (c+d x)}{a^2 d}+\frac {x}{a+b}-\frac {\coth ^3(c+d x)}{3 a d} \]

[In]

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

[Out]

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

Rule 211

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 491

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

Rule 536

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 597

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 3751

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]))

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x^4 \left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = -\frac {\coth ^3(c+d x)}{3 a d}+\frac {\text {Subst}\left (\int \frac {3 (a-b)+3 b x^2}{x^2 \left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{3 a d} \\ & = -\frac {(a-b) \coth (c+d x)}{a^2 d}-\frac {\coth ^3(c+d x)}{3 a d}-\frac {\text {Subst}\left (\int \frac {-3 \left (a^2-a b+b^2\right )-3 (a-b) b x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{3 a^2 d} \\ & = -\frac {(a-b) \coth (c+d x)}{a^2 d}-\frac {\coth ^3(c+d x)}{3 a d}+\frac {\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{(a+b) d}+\frac {b^3 \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tanh (c+d x)\right )}{a^2 (a+b) d} \\ & = \frac {x}{a+b}+\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{5/2} (a+b) d}-\frac {(a-b) \coth (c+d x)}{a^2 d}-\frac {\coth ^3(c+d x)}{3 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.11 \[ \int \frac {\coth ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {\frac {6 \left (c+d x+\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{5/2}}\right )}{a+b}-\frac {(-2 a+3 b+(4 a-3 b) \cosh (2 (c+d x))) \coth (c+d x) \text {csch}^2(c+d x)}{a^2}}{6 d} \]

[In]

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

[Out]

((6*(c + d*x + (b^(5/2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/a^(5/2)))/(a + b) - ((-2*a + 3*b + (4*a - 3*b
)*Cosh[2*(c + d*x)])*Coth[c + d*x]*Csch[c + d*x]^2)/a^2)/(6*d)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.33

method	result	size
derivativedivides	\(-\frac {-\frac {b^{3} \arctan \left (\frac {b \tanh \left (d x +c \right )}{\sqrt {a b}}\right )}{\left (a +b \right ) a^{2} \sqrt {a b}}-\frac {-a +b}{a^{2} \tanh \left (d x +c \right )}+\frac {1}{3 a \tanh \left (d x +c \right )^{3}}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 a +2 b}+\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 a +2 b}}{d}\)	\(109\)
default	\(-\frac {-\frac {b^{3} \arctan \left (\frac {b \tanh \left (d x +c \right )}{\sqrt {a b}}\right )}{\left (a +b \right ) a^{2} \sqrt {a b}}-\frac {-a +b}{a^{2} \tanh \left (d x +c \right )}+\frac {1}{3 a \tanh \left (d x +c \right )^{3}}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 a +2 b}+\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 a +2 b}}{d}\)	\(109\)
risch	\(\frac {x}{a +b}-\frac {2 \left (6 a \,{\mathrm e}^{4 d x +4 c}-3 b \,{\mathrm e}^{4 d x +4 c}-6 \,{\mathrm e}^{2 d x +2 c} a +6 b \,{\mathrm e}^{2 d x +2 c}+4 a -3 b \right )}{3 d \,a^{2} \left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}+\frac {\sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}+a -b}{a +b}\right )}{2 a^{3} \left (a +b \right ) d}-\frac {\sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}-a +b}{a +b}\right )}{2 a^{3} \left (a +b \right ) d}\)	\(190\)

[In]

int(coth(d*x+c)^4/(a+b*tanh(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

-1/d*(-b^3/(a+b)/a^2/(a*b)^(1/2)*arctan(b*tanh(d*x+c)/(a*b)^(1/2))-1/a^2*(-a+b)/tanh(d*x+c)+1/3/a/tanh(d*x+c)^
3-1/(2*a+2*b)*ln(tanh(d*x+c)+1)+1/(2*a+2*b)*ln(tanh(d*x+c)-1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1023 vs. \(2 (72) = 144\).

Time = 0.31 (sec) , antiderivative size = 2368, normalized size of antiderivative = 28.88 \[ \int \frac {\coth ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[1/6*(6*a^2*d*x*cosh(d*x + c)^6 + 36*a^2*d*x*cosh(d*x + c)*sinh(d*x + c)^5 + 6*a^2*d*x*sinh(d*x + c)^6 - 6*(3*
a^2*d*x + 4*a^2 + 2*a*b - 2*b^2)*cosh(d*x + c)^4 + 6*(15*a^2*d*x*cosh(d*x + c)^2 - 3*a^2*d*x - 4*a^2 - 2*a*b +
2*b^2)*sinh(d*x + c)^4 - 6*a^2*d*x + 24*(5*a^2*d*x*cosh(d*x + c)^3 - (3*a^2*d*x + 4*a^2 + 2*a*b - 2*b^2)*cosh
(d*x + c))*sinh(d*x + c)^3 + 6*(3*a^2*d*x + 4*a^2 - 4*b^2)*cosh(d*x + c)^2 + 6*(15*a^2*d*x*cosh(d*x + c)^4 + 3
*a^2*d*x - 6*(3*a^2*d*x + 4*a^2 + 2*a*b - 2*b^2)*cosh(d*x + c)^2 + 4*a^2 - 4*b^2)*sinh(d*x + c)^2 + 3*(b^2*cos
h(d*x + c)^6 + 6*b^2*cosh(d*x + c)*sinh(d*x + c)^5 + b^2*sinh(d*x + c)^6 - 3*b^2*cosh(d*x + c)^4 + 3*(5*b^2*co
sh(d*x + c)^2 - b^2)*sinh(d*x + c)^4 + 3*b^2*cosh(d*x + c)^2 + 4*(5*b^2*cosh(d*x + c)^3 - 3*b^2*cosh(d*x + c))
*sinh(d*x + c)^3 + 3*(5*b^2*cosh(d*x + c)^4 - 6*b^2*cosh(d*x + c)^2 + b^2)*sinh(d*x + c)^2 - b^2 + 6*(b^2*cosh
(d*x + c)^5 - 2*b^2*cosh(d*x + c)^3 + b^2*cosh(d*x + c))*sinh(d*x + c))*sqrt(-b/a)*log(((a^2 + 2*a*b + b^2)*co
sh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^4 + 2*
(a^2 - b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 - b^2)*sinh(d*x + c)^2 + a^2 - 6*
a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2 - b^2)*cosh(d*x + c))*sinh(d*x + c) + 4*((a^2 + a*b)
*cosh(d*x + c)^2 + 2*(a^2 + a*b)*cosh(d*x + c)*sinh(d*x + c) + (a^2 + a*b)*sinh(d*x + c)^2 + a^2 - a*b)*sqrt(-
b/a))/((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(a - b)
*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^2 + 4*((a + b)*cosh(d*x + c)^3 + (a - b
)*cosh(d*x + c))*sinh(d*x + c) + a + b)) - 16*a^2 - 4*a*b + 12*b^2 + 12*(3*a^2*d*x*cosh(d*x + c)^5 - 2*(3*a^2*
d*x + 4*a^2 + 2*a*b - 2*b^2)*cosh(d*x + c)^3 + (3*a^2*d*x + 4*a^2 - 4*b^2)*cosh(d*x + c))*sinh(d*x + c))/((a^3
+ a^2*b)*d*cosh(d*x + c)^6 + 6*(a^3 + a^2*b)*d*cosh(d*x + c)*sinh(d*x + c)^5 + (a^3 + a^2*b)*d*sinh(d*x + c)^
6 - 3*(a^3 + a^2*b)*d*cosh(d*x + c)^4 + 3*(5*(a^3 + a^2*b)*d*cosh(d*x + c)^2 - (a^3 + a^2*b)*d)*sinh(d*x + c)^
4 + 3*(a^3 + a^2*b)*d*cosh(d*x + c)^2 + 4*(5*(a^3 + a^2*b)*d*cosh(d*x + c)^3 - 3*(a^3 + a^2*b)*d*cosh(d*x + c)
)*sinh(d*x + c)^3 + 3*(5*(a^3 + a^2*b)*d*cosh(d*x + c)^4 - 6*(a^3 + a^2*b)*d*cosh(d*x + c)^2 + (a^3 + a^2*b)*d
)*sinh(d*x + c)^2 - (a^3 + a^2*b)*d + 6*((a^3 + a^2*b)*d*cosh(d*x + c)^5 - 2*(a^3 + a^2*b)*d*cosh(d*x + c)^3 +
(a^3 + a^2*b)*d*cosh(d*x + c))*sinh(d*x + c)), 1/3*(3*a^2*d*x*cosh(d*x + c)^6 + 18*a^2*d*x*cosh(d*x + c)*sinh
(d*x + c)^5 + 3*a^2*d*x*sinh(d*x + c)^6 - 3*(3*a^2*d*x + 4*a^2 + 2*a*b - 2*b^2)*cosh(d*x + c)^4 + 3*(15*a^2*d*
x*cosh(d*x + c)^2 - 3*a^2*d*x - 4*a^2 - 2*a*b + 2*b^2)*sinh(d*x + c)^4 - 3*a^2*d*x + 12*(5*a^2*d*x*cosh(d*x +
c)^3 - (3*a^2*d*x + 4*a^2 + 2*a*b - 2*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(3*a^2*d*x + 4*a^2 - 4*b^2)*cosh
(d*x + c)^2 + 3*(15*a^2*d*x*cosh(d*x + c)^4 + 3*a^2*d*x - 6*(3*a^2*d*x + 4*a^2 + 2*a*b - 2*b^2)*cosh(d*x + c)^
2 + 4*a^2 - 4*b^2)*sinh(d*x + c)^2 + 3*(b^2*cosh(d*x + c)^6 + 6*b^2*cosh(d*x + c)*sinh(d*x + c)^5 + b^2*sinh(d
*x + c)^6 - 3*b^2*cosh(d*x + c)^4 + 3*(5*b^2*cosh(d*x + c)^2 - b^2)*sinh(d*x + c)^4 + 3*b^2*cosh(d*x + c)^2 +
4*(5*b^2*cosh(d*x + c)^3 - 3*b^2*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*b^2*cosh(d*x + c)^4 - 6*b^2*cosh(d*x +
c)^2 + b^2)*sinh(d*x + c)^2 - b^2 + 6*(b^2*cosh(d*x + c)^5 - 2*b^2*cosh(d*x + c)^3 + b^2*cosh(d*x + c))*sinh(d
*x + c))*sqrt(b/a)*arctan(1/2*((a + b)*cosh(d*x + c)^2 + 2*(a + b)*cosh(d*x + c)*sinh(d*x + c) + (a + b)*sinh(
d*x + c)^2 + a - b)*sqrt(b/a)/b) - 8*a^2 - 2*a*b + 6*b^2 + 6*(3*a^2*d*x*cosh(d*x + c)^5 - 2*(3*a^2*d*x + 4*a^2
+ 2*a*b - 2*b^2)*cosh(d*x + c)^3 + (3*a^2*d*x + 4*a^2 - 4*b^2)*cosh(d*x + c))*sinh(d*x + c))/((a^3 + a^2*b)*d
*cosh(d*x + c)^6 + 6*(a^3 + a^2*b)*d*cosh(d*x + c)*sinh(d*x + c)^5 + (a^3 + a^2*b)*d*sinh(d*x + c)^6 - 3*(a^3
+ a^2*b)*d*cosh(d*x + c)^4 + 3*(5*(a^3 + a^2*b)*d*cosh(d*x + c)^2 - (a^3 + a^2*b)*d)*sinh(d*x + c)^4 + 3*(a^3
+ a^2*b)*d*cosh(d*x + c)^2 + 4*(5*(a^3 + a^2*b)*d*cosh(d*x + c)^3 - 3*(a^3 + a^2*b)*d*cosh(d*x + c))*sinh(d*x
+ c)^3 + 3*(5*(a^3 + a^2*b)*d*cosh(d*x + c)^4 - 6*(a^3 + a^2*b)*d*cosh(d*x + c)^2 + (a^3 + a^2*b)*d)*sinh(d*x
+ c)^2 - (a^3 + a^2*b)*d + 6*((a^3 + a^2*b)*d*cosh(d*x + c)^5 - 2*(a^3 + a^2*b)*d*cosh(d*x + c)^3 + (a^3 + a^2
*b)*d*cosh(d*x + c))*sinh(d*x + c))]

Sympy [F]

\[ \int \frac {\coth ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\int \frac {\coth ^{4}{\left (c + d x \right )}}{a + b \tanh ^{2}{\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1038 vs. \(2 (72) = 144\).

Time = 0.42 (sec) , antiderivative size = 1038, normalized size of antiderivative = 12.66 \[ \int \frac {\coth ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/8*(a*b - b^2)*log((a + b)*e^(4*d*x + 4*c) + 2*(a - b)*e^(2*d*x + 2*c) + a + b)/((a^3 + a^2*b)*d) + 1/8*(a*b
- b^2)*log(2*(a - b)*e^(-2*d*x - 2*c) + (a + b)*e^(-4*d*x - 4*c) + a + b)/((a^3 + a^2*b)*d) + 1/16*(a^2*b - 6
*a*b^2 + b^3)*arctan(1/2*((a + b)*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^3 + a^2*b)*sqrt(a*b)*d) - 1/16*(a^2*
b - 6*a*b^2 + b^3)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b))/((a^3 + a^2*b)*sqrt(a*b)*d) - 1/24
*(3*(12*a - b)*e^(4*d*x + 4*c) - 6*(9*a - b)*e^(2*d*x + 2*c) + 22*a - 3*b)/((a^2*e^(6*d*x + 6*c) - 3*a^2*e^(4*
d*x + 4*c) + 3*a^2*e^(2*d*x + 2*c) - a^2)*d) - 1/6*(3*(4*a - b)*e^(4*d*x + 4*c) - 6*(2*a - b)*e^(2*d*x + 2*c)
+ 4*a - 3*b)/((a^2*e^(6*d*x + 6*c) - 3*a^2*e^(4*d*x + 4*c) + 3*a^2*e^(2*d*x + 2*c) - a^2)*d) - 1/24*(6*(9*a -
b)*e^(-2*d*x - 2*c) - 3*(12*a - b)*e^(-4*d*x - 4*c) - 22*a + 3*b)/((3*a^2*e^(-2*d*x - 2*c) - 3*a^2*e^(-4*d*x -
4*c) + a^2*e^(-6*d*x - 6*c) - a^2)*d) - 1/6*(6*(2*a - b)*e^(-2*d*x - 2*c) - 3*(4*a - b)*e^(-4*d*x - 4*c) - 4*
a + 3*b)/((3*a^2*e^(-2*d*x - 2*c) - 3*a^2*e^(-4*d*x - 4*c) + a^2*e^(-6*d*x - 6*c) - a^2)*d) + 1/4*(6*(a + b)*e
^(-2*d*x - 2*c) - 3*b*e^(-4*d*x - 4*c) - 2*a - 3*b)/((3*a^2*e^(-2*d*x - 2*c) - 3*a^2*e^(-4*d*x - 4*c) + a^2*e^
(-6*d*x - 6*c) - a^2)*d) + 1/4*b*log((a + b)*e^(4*d*x + 4*c) + 2*(a - b)*e^(2*d*x + 2*c) + a + b)/(a^2*d) - 1/
4*b*log(2*(a - b)*e^(-2*d*x - 2*c) + (a + b)*e^(-4*d*x - 4*c) + a + b)/(a^2*d) + 1/4*(2*a - b)*log(e^(2*d*x +
2*c) - 1)/(a^2*d) - 1/2*b*log(e^(2*d*x + 2*c) - 1)/(a^2*d) - 1/4*(2*a - b)*log(e^(-2*d*x - 2*c) - 1)/(a^2*d) +
1/2*b*log(e^(-2*d*x - 2*c) - 1)/(a^2*d) - 1/4*(a*b - b^2)*arctan(1/2*((a + b)*e^(2*d*x + 2*c) + a - b)/sqrt(a
*b))/(sqrt(a*b)*a^2*d) - 3/8*(a*b + b^2)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b))/(sqrt(a*b)*a
^2*d) + 1/4*(a*b - b^2)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a*b))/(sqrt(a*b)*a^2*d)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (72) = 144\).

Time = 0.35 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.79 \[ \int \frac {\coth ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {\frac {3 \, b^{3} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{{\left (a^{3} + a^{2} b\right )} \sqrt {a b}} + \frac {3 \, {\left (d x + c\right )}}{a + b} - \frac {2 \, {\left (6 \, a e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a e^{\left (2 \, d x + 2 \, c\right )} + 6 \, b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a - 3 \, b\right )}}{a^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{3 \, d} \]

[In]

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

[Out]

1/3*(3*b^3*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^3 + a^2*b)*sqrt(a*b)) + 3
*(d*x + c)/(a + b) - 2*(6*a*e^(4*d*x + 4*c) - 3*b*e^(4*d*x + 4*c) - 6*a*e^(2*d*x + 2*c) + 6*b*e^(2*d*x + 2*c)
+ 4*a - 3*b)/(a^2*(e^(2*d*x + 2*c) - 1)^3))/d

Mupad [B] (verification not implemented)

Time = 2.21 (sec) , antiderivative size = 519, normalized size of antiderivative = 6.33 \[ \int \frac {\coth ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx=\frac {x}{a+b}+\frac {\mathrm {atan}\left (\left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (\frac {4\,b^3}{a^2\,d\,{\left (a+b\right )}^3\,\left (a^3+b\,a^2\right )\,\sqrt {b^5}}+\frac {\left (a^4\,d\,\sqrt {b^5}-a^2\,b^2\,d\,\sqrt {b^5}\right )\,\left (a-b\right )}{b^3\,{\left (a+b\right )}^2\,\left (a^3+b\,a^2\right )\,\sqrt {a^7\,d^2+2\,a^6\,b\,d^2+a^5\,b^2\,d^2}\,\sqrt {a^5\,d^2\,{\left (a+b\right )}^2}}\right )+\frac {\left (a-b\right )\,\left (a^4\,d\,\sqrt {b^5}+2\,a^3\,b\,d\,\sqrt {b^5}+a^2\,b^2\,d\,\sqrt {b^5}\right )}{b^3\,{\left (a+b\right )}^2\,\left (a^3+b\,a^2\right )\,\sqrt {a^7\,d^2+2\,a^6\,b\,d^2+a^5\,b^2\,d^2}\,\sqrt {a^5\,d^2\,{\left (a+b\right )}^2}}\right )\,\left (\frac {a^4\,\sqrt {a^7\,d^2+2\,a^6\,b\,d^2+a^5\,b^2\,d^2}}{2}+a^3\,b\,\sqrt {a^7\,d^2+2\,a^6\,b\,d^2+a^5\,b^2\,d^2}+\frac {a^2\,b^2\,\sqrt {a^7\,d^2+2\,a^6\,b\,d^2+a^5\,b^2\,d^2}}{2}\right )\right )\,\sqrt {b^5}}{\sqrt {a^7\,d^2+2\,a^6\,b\,d^2+a^5\,b^2\,d^2}}-\frac {8}{3\,a\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )}-\frac {4\,\left (a^2+b\,a\right )}{a^2\,d\,\left (a+b\right )\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {2\,\left (2\,a^2+a\,b-b^2\right )}{a^2\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )\,\left (a+b\right )} \]

[In]

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

[Out]

x/(a + b) + (atan((exp(2*c)*exp(2*d*x)*((4*b^3)/(a^2*d*(a + b)^3*(a^2*b + a^3)*(b^5)^(1/2)) + ((a^4*d*(b^5)^(1
/2) - a^2*b^2*d*(b^5)^(1/2))*(a - b))/(b^3*(a + b)^2*(a^2*b + a^3)*(a^7*d^2 + 2*a^6*b*d^2 + a^5*b^2*d^2)^(1/2)
*(a^5*d^2*(a + b)^2)^(1/2))) + ((a - b)*(a^4*d*(b^5)^(1/2) + 2*a^3*b*d*(b^5)^(1/2) + a^2*b^2*d*(b^5)^(1/2)))/(
b^3*(a + b)^2*(a^2*b + a^3)*(a^7*d^2 + 2*a^6*b*d^2 + a^5*b^2*d^2)^(1/2)*(a^5*d^2*(a + b)^2)^(1/2)))*((a^4*(a^7
*d^2 + 2*a^6*b*d^2 + a^5*b^2*d^2)^(1/2))/2 + a^3*b*(a^7*d^2 + 2*a^6*b*d^2 + a^5*b^2*d^2)^(1/2) + (a^2*b^2*(a^7
*d^2 + 2*a^6*b*d^2 + a^5*b^2*d^2)^(1/2))/2))*(b^5)^(1/2))/(a^7*d^2 + 2*a^6*b*d^2 + a^5*b^2*d^2)^(1/2) - 8/(3*a
*d*(3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1)) - (4*(a*b + a^2))/(a^2*d*(a + b)*(exp(4*c
+ 4*d*x) - 2*exp(2*c + 2*d*x) + 1)) - (2*(a*b + 2*a^2 - b^2))/(a^2*d*(exp(2*c + 2*d*x) - 1)*(a + b))