Integrand size = 23, antiderivative size = 119 \[ \int \frac {\coth ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {x}{(a+b)^2}-\frac {b^{3/2} (5 a+3 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{2 a^{5/2} (a+b)^2 d}-\frac {(2 a+3 b) \coth (c+d x)}{2 a^2 (a+b) d}+\frac {b \coth (c+d x)}{2 a (a+b) d \left (a+b \tanh ^2(c+d x)\right )} \] Output:
x/(a+b)^2-1/2*b^(3/2)*(5*a+3*b)*arctan(b^(1/2)*tanh(d*x+c)/a^(1/2))/a^(5/2 )/(a+b)^2/d-1/2*(2*a+3*b)*coth(d*x+c)/a^2/(a+b)/d+1/2*b*coth(d*x+c)/a/(a+b )/d/(a+b*tanh(d*x+c)^2)
Time = 1.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.93 \[ \int \frac {\coth ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {-\frac {2 (c+d x)}{(a+b)^2}+\frac {b^{3/2} (5 a+3 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{5/2} (a+b)^2}+\frac {2 \coth (c+d x)}{a^2}+\frac {b^2 \sinh (2 (c+d x))}{a^2 (a+b) (a-b+(a+b) \cosh (2 (c+d x)))}}{2 d} \] Input:
Integrate[Coth[c + d*x]^2/(a + b*Tanh[c + d*x]^2)^2,x]
Output:
-1/2*((-2*(c + d*x))/(a + b)^2 + (b^(3/2)*(5*a + 3*b)*ArcTan[(Sqrt[b]*Tanh [c + d*x])/Sqrt[a]])/(a^(5/2)*(a + b)^2) + (2*Coth[c + d*x])/a^2 + (b^2*Si nh[2*(c + d*x)])/(a^2*(a + b)*(a - b + (a + b)*Cosh[2*(c + d*x)])))/d
Time = 0.40 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 25, 4153, 25, 374, 25, 445, 25, 397, 218, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\coth ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {1}{\tan (i c+i d x)^2 \left (a-b \tan (i c+i d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {1}{\tan (i c+i d x)^2 \left (a-b \tan (i c+i d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle -\frac {\int -\frac {\coth ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\coth ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 374 |
| \(\displaystyle -\frac {\frac {\int -\frac {\coth ^2(c+d x) \left (-3 b \tanh ^2(c+d x)+2 a+3 b\right )}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{2 a (a+b)}-\frac {b \coth (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {-\frac {\int \frac {\coth ^2(c+d x) \left (-3 b \tanh ^2(c+d x)+2 a+3 b\right )}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{2 a (a+b)}-\frac {b \coth (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle -\frac {-\frac {-\frac {\int -\frac {2 a^2-2 b a-3 b^2+b (2 a+3 b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{a}-\frac {(2 a+3 b) \coth (c+d x)}{a}}{2 a (a+b)}-\frac {b \coth (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {-\frac {\frac {\int \frac {2 a^2-2 b a-3 b^2+b (2 a+3 b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{a}-\frac {(2 a+3 b) \coth (c+d x)}{a}}{2 a (a+b)}-\frac {b \coth (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle -\frac {-\frac {\frac {\frac {2 a^2 \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}-\frac {b^2 (5 a+3 b) \int \frac {1}{b \tanh ^2(c+d x)+a}d\tanh (c+d x)}{a+b}}{a}-\frac {(2 a+3 b) \coth (c+d x)}{a}}{2 a (a+b)}-\frac {b \coth (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {-\frac {\frac {\frac {2 a^2 \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}-\frac {b^{3/2} (5 a+3 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}}{a}-\frac {(2 a+3 b) \coth (c+d x)}{a}}{2 a (a+b)}-\frac {b \coth (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {-\frac {\frac {\frac {2 a^2 \text {arctanh}(\tanh (c+d x))}{a+b}-\frac {b^{3/2} (5 a+3 b) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}}{a}-\frac {(2 a+3 b) \coth (c+d x)}{a}}{2 a (a+b)}-\frac {b \coth (c+d x)}{2 a (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{d}\) |
Input:
Int[Coth[c + d*x]^2/(a + b*Tanh[c + d*x]^2)^2,x]
Output:
-((-1/2*((-((b^(3/2)*(5*a + 3*b)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/ (Sqrt[a]*(a + b))) + (2*a^2*ArcTanh[Tanh[c + d*x]])/(a + b))/a - ((2*a + 3 *b)*Coth[c + d*x])/a)/(a*(a + b)) - (b*Coth[c + d*x])/(2*a*(a + b)*(a + b* Tanh[c + d*x]^2)))/d)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*e*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[b*c*(m + 1) + 2*(b*c - a*d)*(p + 1) + d*b*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
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]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^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] && Ratio nalQ[n]))
Time = 0.22 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.01
| method | result | size |
| derivativedivides | \(-\frac {\frac {b^{2} \left (\frac {\left (\frac {b}{2}+\frac {a}{2}\right ) \tanh \left (d x +c \right )}{a +b \tanh \left (d x +c \right )^{2}}+\frac {\left (5 a +3 b \right ) \arctan \left (\frac {b \tanh \left (d x +c \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{\left (a +b \right )^{2} a^{2}}+\frac {1}{a^{2} \tanh \left (d x +c \right )}-\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}+\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}}{d}\) | \(120\) |
| default | \(-\frac {\frac {b^{2} \left (\frac {\left (\frac {b}{2}+\frac {a}{2}\right ) \tanh \left (d x +c \right )}{a +b \tanh \left (d x +c \right )^{2}}+\frac {\left (5 a +3 b \right ) \arctan \left (\frac {b \tanh \left (d x +c \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{\left (a +b \right )^{2} a^{2}}+\frac {1}{a^{2} \tanh \left (d x +c \right )}-\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}+\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}}{d}\) | \(120\) |
| risch | \(\frac {x}{a^{2}+2 a b +b^{2}}-\frac {2 a^{3} {\mathrm e}^{4 d x +4 c}+6 a^{2} b \,{\mathrm e}^{4 d x +4 c}+5 a \,b^{2} {\mathrm e}^{4 d x +4 c}+3 b^{3} {\mathrm e}^{4 d x +4 c}+4 a^{3} {\mathrm e}^{2 d x +2 c}+4 a^{2} b \,{\mathrm e}^{2 d x +2 c}-4 a \,b^{2} {\mathrm e}^{2 d x +2 c}-6 b^{3} {\mathrm e}^{2 d x +2 c}+2 a^{3}+6 a^{2} b +7 a \,b^{2}+3 b^{3}}{a^{2} d \left ({\mathrm e}^{2 d x +2 c}-1\right ) \left (a +b \right )^{2} \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 a \,{\mathrm e}^{2 d x +2 c}-2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )}+\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-a +2 \sqrt {-a b}+b}{a +b}\right ) b}{4 a^{2} \left (a +b \right )^{2} d}+\frac {3 \sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-a +2 \sqrt {-a b}+b}{a +b}\right )}{4 a^{3} \left (a +b \right )^{2} d}-\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a +2 \sqrt {-a b}-b}{a +b}\right ) b}{4 a^{2} \left (a +b \right )^{2} d}-\frac {3 \sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a +2 \sqrt {-a b}-b}{a +b}\right )}{4 a^{3} \left (a +b \right )^{2} d}\) | \(439\) |
Input:
int(coth(d*x+c)^2/(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
Output:
-1/d*(b^2/(a+b)^2/a^2*((1/2*b+1/2*a)*tanh(d*x+c)/(a+b*tanh(d*x+c)^2)+1/2*( 5*a+3*b)/(a*b)^(1/2)*arctan(b*tanh(d*x+c)/(a*b)^(1/2)))+1/a^2/tanh(d*x+c)- 1/2/(a+b)^2*ln(1+tanh(d*x+c))+1/2/(a+b)^2*ln(-1+tanh(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 1702 vs. \(2 (105) = 210\).
Time = 0.18 (sec) , antiderivative size = 3725, normalized size of antiderivative = 31.30 \[ \int \frac {\coth ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:
integrate(coth(d*x+c)^2/(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\coth ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\int \frac {\coth ^{2}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(coth(d*x+c)**2/(a+b*tanh(d*x+c)**2)**2,x)
Output:
Integral(coth(c + d*x)**2/(a + b*tanh(c + d*x)**2)**2, x)
Leaf count of result is larger than twice the leaf count of optimal. 976 vs. \(2 (105) = 210\).
Time = 0.29 (sec) , antiderivative size = 976, normalized size of antiderivative = 8.20 \[ \int \frac {\coth ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:
integrate(coth(d*x+c)^2/(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")
Output:
-1/4*(2*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^4 + 2*a^3*b + a^2*b^2)*d) + 1/4*(2*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^4 + 2*a^3*b + a^2 *b^2)*d) + 1/8*(3*a^2*b - 4*a*b^2 - 3*b^3)*arctan(1/2*((a + b)*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^4 + 2*a^3*b + a^2*b^2)*sqrt(a*b)*d) - 1/8*(3* a^2*b - 4*a*b^2 - 3*b^3)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqr t(a*b))/((a^4 + 2*a^3*b + a^2*b^2)*sqrt(a*b)*d) + 1/4*(2*a^3 + 5*a^2*b + 6 *a*b^2 + 3*b^3 + (2*a^3 + 7*a^2*b + 3*b^3)*e^(4*d*x + 4*c) + 2*(2*a^3 + 2* a^2*b + a*b^2 - 3*b^3)*e^(2*d*x + 2*c))/((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2* b^3 - (a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*e^(6*d*x + 6*c) - (a^5 - a^4*b - 5*a^3*b^2 - 3*a^2*b^3)*e^(4*d*x + 4*c) + (a^5 - a^4*b - 5*a^3*b^2 - 3*a ^2*b^3)*e^(2*d*x + 2*c))*d) - 1/4*(2*a^3 + 5*a^2*b + 6*a*b^2 + 3*b^3 + 2*( 2*a^3 + 2*a^2*b + a*b^2 - 3*b^3)*e^(-2*d*x - 2*c) + (2*a^3 + 7*a^2*b + 3*b ^3)*e^(-4*d*x - 4*c))/((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3 + (a^5 - a^4*b - 5*a^3*b^2 - 3*a^2*b^3)*e^(-2*d*x - 2*c) - (a^5 - a^4*b - 5*a^3*b^2 - 3* a^2*b^3)*e^(-4*d*x - 4*c) - (a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*e^(-6*d* x - 6*c))*d) - 1/2*(2*a^2 + 5*a*b + 3*b^2 + 2*(2*a^2 - 3*b^2)*e^(-2*d*x - 2*c) + (2*a^2 + 3*a*b + 3*b^2)*e^(-4*d*x - 4*c))/((a^4 + 2*a^3*b + a^2*b^2 + (a^4 - 2*a^3*b - 3*a^2*b^2)*e^(-2*d*x - 2*c) - (a^4 - 2*a^3*b - 3*a^2*b ^2)*e^(-4*d*x - 4*c) - (a^4 + 2*a^3*b + a^2*b^2)*e^(-6*d*x - 6*c))*d) +...
Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (105) = 210\).
Time = 0.20 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.82 \[ \int \frac {\coth ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {\frac {{\left (5 \, a b^{2} + 3 \, b^{3}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {a b}} - \frac {2 \, {\left (d x + c\right )}}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (2 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 5 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 4 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 6 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{3} + 6 \, a^{2} b + 7 \, a b^{2} + 3 \, b^{3}\right )}}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} {\left (a e^{\left (6 \, d x + 6 \, c\right )} + b e^{\left (6 \, d x + 6 \, c\right )} + a e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b e^{\left (4 \, d x + 4 \, c\right )} - a e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b e^{\left (2 \, d x + 2 \, c\right )} - a - b\right )}}}{2 \, d} \] Input:
integrate(coth(d*x+c)^2/(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")
Output:
-1/2*((5*a*b^2 + 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^4 + 2*a^3*b + a^2*b^2)*sqrt(a*b)) - 2*(d*x + c)/(a ^2 + 2*a*b + b^2) + 2*(2*a^3*e^(4*d*x + 4*c) + 6*a^2*b*e^(4*d*x + 4*c) + 5 *a*b^2*e^(4*d*x + 4*c) + 3*b^3*e^(4*d*x + 4*c) + 4*a^3*e^(2*d*x + 2*c) + 4 *a^2*b*e^(2*d*x + 2*c) - 4*a*b^2*e^(2*d*x + 2*c) - 6*b^3*e^(2*d*x + 2*c) + 2*a^3 + 6*a^2*b + 7*a*b^2 + 3*b^3)/((a^4 + 2*a^3*b + a^2*b^2)*(a*e^(6*d*x + 6*c) + b*e^(6*d*x + 6*c) + a*e^(4*d*x + 4*c) - 3*b*e^(4*d*x + 4*c) - a* e^(2*d*x + 2*c) + 3*b*e^(2*d*x + 2*c) - a - b)))/d
Timed out. \[ \int \frac {\coth ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\int \frac {{\mathrm {coth}\left (c+d\,x\right )}^2}{{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^2} \,d x \] Input:
int(coth(c + d*x)^2/(a + b*tanh(c + d*x)^2)^2,x)
Output:
int(coth(c + d*x)^2/(a + b*tanh(c + d*x)^2)^2, x)
Time = 0.32 (sec) , antiderivative size = 1911, normalized size of antiderivative = 16.06 \[ \int \frac {\coth ^2(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:
int(coth(d*x+c)^2/(a+b*tanh(d*x+c)^2)^2,x)
Output:
( - 5*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sq rt(b))/sqrt(a))*a**3*b + 7*e**(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**2*b**2 + 21*e**(6*c + 6*d*x)*sqrt( b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a*b**3 + 9*e **(6*c + 6*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/ sqrt(a))*b**4 - 5*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt (a + b) - sqrt(b))/sqrt(a))*a**3*b + 27*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a)*a tan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**2*b**2 - 27*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a)) *a*b**3 - 27*e**(4*c + 4*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**4 + 5*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a)*atan((e** (c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**3*b - 27*e**(2*c + 2*d*x)*sqr t(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**2*b**2 + 27*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqr t(b))/sqrt(a))*a*b**3 + 27*e**(2*c + 2*d*x)*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**4 + 5*sqrt(b)*sqrt(a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**3*b - 7*sqrt(b)*sqrt(a)*atan((e* *(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a**2*b**2 - 21*sqrt(b)*sqrt(a)* atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*a*b**3 - 9*sqrt(b)*sqrt (a)*atan((e**(c + d*x)*sqrt(a + b) - sqrt(b))/sqrt(a))*b**4 + 5*e**(6*c...