### 3.186 $$\int \frac{\coth (c+d x)}{(a+b \tanh ^2(c+d x))^2} \, dx$$

Optimal. Leaf size=95 $-\frac{b (2 a+b) \log \left (a+b \tanh ^2(c+d x)\right )}{2 a^2 d (a+b)^2}+\frac{\log (\tanh (c+d x))}{a^2 d}+\frac{b}{2 a d (a+b) \left (a+b \tanh ^2(c+d x)\right )}+\frac{\log (\cosh (c+d x))}{d (a+b)^2}$

[Out]

Log[Cosh[c + d*x]]/((a + b)^2*d) + Log[Tanh[c + d*x]]/(a^2*d) - (b*(2*a + b)*Log[a + b*Tanh[c + d*x]^2])/(2*a^
2*(a + b)^2*d) + b/(2*a*(a + b)*d*(a + b*Tanh[c + d*x]^2))

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Rubi [A]  time = 0.149079, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {3670, 446, 72} $-\frac{b (2 a+b) \log \left (a+b \tanh ^2(c+d x)\right )}{2 a^2 d (a+b)^2}+\frac{\log (\tanh (c+d x))}{a^2 d}+\frac{b}{2 a d (a+b) \left (a+b \tanh ^2(c+d x)\right )}+\frac{\log (\cosh (c+d x))}{d (a+b)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Log[Cosh[c + d*x]]/((a + b)^2*d) + Log[Tanh[c + d*x]]/(a^2*d) - (b*(2*a + b)*Log[a + b*Tanh[c + d*x]^2])/(2*a^
2*(a + b)^2*d) + b/(2*a*(a + b)*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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin{align*} \int \frac{\coth (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x \left (1-x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{(1-x) x (a+b x)^2} \, dx,x,\tanh ^2(c+d x)\right )}{2 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{(a+b)^2 (-1+x)}+\frac{1}{a^2 x}-\frac{b^2}{a (a+b) (a+b x)^2}-\frac{b^2 (2 a+b)}{a^2 (a+b)^2 (a+b x)}\right ) \, dx,x,\tanh ^2(c+d x)\right )}{2 d}\\ &=\frac{\log (\cosh (c+d x))}{(a+b)^2 d}+\frac{\log (\tanh (c+d x))}{a^2 d}-\frac{b (2 a+b) \log \left (a+b \tanh ^2(c+d x)\right )}{2 a^2 (a+b)^2 d}+\frac{b}{2 a (a+b) d \left (a+b \tanh ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.8582, size = 83, normalized size = 0.87 $\frac{\frac{\frac{b \left (\frac{a (a+b)}{a+b \tanh ^2(c+d x)}-(2 a+b) \log \left (a+b \tanh ^2(c+d x)\right )\right )}{(a+b)^2}+2 \log (\tanh (c+d x))}{a^2}+\frac{2 \log (\cosh (c+d x))}{(a+b)^2}}{2 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((2*Log[Cosh[c + d*x]])/(a + b)^2 + (2*Log[Tanh[c + d*x]] + (b*(-((2*a + b)*Log[a + b*Tanh[c + d*x]^2]) + (a*(
a + b))/(a + b*Tanh[c + d*x]^2)))/(a + b)^2)/a^2)/(2*d)

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Maple [B]  time = 0.102, size = 325, normalized size = 3.4 \begin{align*} -{\frac{1}{d \left ( a+b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-2\,{\frac{{b}^{2} \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d \left ( a+b \right ) ^{2}a \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+4\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a \right ) }}-2\,{\frac{{b}^{3} \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d \left ( a+b \right ) ^{2}{a}^{2} \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+4\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a \right ) }}-{\frac{b}{d \left ( a+b \right ) ^{2}a}\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}a+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+4\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a \right ) }-{\frac{{b}^{2}}{2\,d \left ( a+b \right ) ^{2}{a}^{2}}\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}a+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+4\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a \right ) }+{\frac{1}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{1}{d \left ( a+b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.12821, size = 317, normalized size = 3.34 \begin{align*} \frac{2 \, b^{2} e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3} + 2 \,{\left (a^{4} + a^{3} b - a^{2} b^{2} - a b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} +{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} - \frac{{\left (2 \, a b + b^{2}\right )} \log \left (2 \,{\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} +{\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \,{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} d} + \frac{d x + c}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{2} d} + \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{2} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 3.14217, size = 2678, normalized size = 28.19 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(a^3 + a^2*b)*d*x*cosh(d*x + c)^4 + 8*(a^3 + a^2*b)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(a^3 + a^2*b
)*d*x*sinh(d*x + c)^4 + 2*(a^3 + a^2*b)*d*x - 4*(a*b^2 - (a^3 - a^2*b)*d*x)*cosh(d*x + c)^2 + 4*(3*(a^3 + a^2*
b)*d*x*cosh(d*x + c)^2 - a*b^2 + (a^3 - a^2*b)*d*x)*sinh(d*x + c)^2 + ((2*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)
^4 + 4*(2*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)*sinh(d*x + c)^3 + (2*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^4 + 2
*a^2*b + 3*a*b^2 + b^3 + 2*(2*a^2*b - a*b^2 - b^3)*cosh(d*x + c)^2 + 2*(2*a^2*b - a*b^2 - b^3 + 3*(2*a^2*b + 3
*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((2*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^3 + (2*a^2*b - a*b
^2 - b^3)*cosh(d*x + c))*sinh(d*x + c))*log(2*((a + b)*cosh(d*x + c)^2 + (a + b)*sinh(d*x + c)^2 + a - b)/(cos
h(d*x + c)^2 - 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)) - 2*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x
+ c)^4 + 4*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*si
nh(d*x + c)^4 + a^3 + 3*a^2*b + 3*a*b^2 + b^3 + 2*(a^3 + a^2*b - a*b^2 - b^3)*cosh(d*x + c)^2 + 2*(a^3 + a^2*b
- a*b^2 - b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^3 + 3*a^2*b + 3*a*
b^2 + b^3)*cosh(d*x + c)^3 + (a^3 + a^2*b - a*b^2 - b^3)*cosh(d*x + c))*sinh(d*x + c))*log(2*sinh(d*x + c)/(co
sh(d*x + c) - sinh(d*x + c))) + 8*((a^3 + a^2*b)*d*x*cosh(d*x + c)^3 - (a*b^2 - (a^3 - a^2*b)*d*x)*cosh(d*x +
c))*sinh(d*x + c))/((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)^4 + 4*(a^5 + 3*a^4*b + 3*a^3*b^2 + a
^2*b^3)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*d*sinh(d*x + c)^4 + 2*(a^5 + a
^4*b - a^3*b^2 - a^2*b^3)*d*cosh(d*x + c)^2 + 2*(3*(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)^2 + (
a^5 + a^4*b - a^3*b^2 - a^2*b^3)*d)*sinh(d*x + c)^2 + (a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*d + 4*((a^5 + 3*a^
4*b + 3*a^3*b^2 + a^2*b^3)*d*cosh(d*x + c)^3 + (a^5 + a^4*b - a^3*b^2 - a^2*b^3)*d*cosh(d*x + c))*sinh(d*x + c
))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.21831, size = 279, normalized size = 2.94 \begin{align*} -\frac{{\left (2 \, a b + b^{2}\right )} \log \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 \,{\left (a^{4} d + 2 \, a^{3} b d + a^{2} b^{2} d\right )}} - \frac{d x + c}{a^{2} d + 2 \, a b d + b^{2} d} + \frac{2 \, b^{2} e^{\left (2 \, d x + 2 \, c\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 )}{\left (a + b\right )}^{2} a d} + \frac{\log \left ({\left | -e^{\left (2 \, d x + 2 \, c\right )} + 1 \right |}\right )}{a^{2} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a*b + b^2)*log(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)/(a^4*d + 2*a^3*b*d + a^2*b^2*d) - (d*x + c)/(a^2*d + 2*a*b*d + b^2*d) + 2*b^2*e^(2*d*x + 2*c)/((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)*(a + b)^2*a*d) + log(abs(-e^(
2*d*x + 2*c) + 1))/(a^2*d)