3.176 $$\int \frac{\coth (c+d x)}{a+b \tanh ^2(c+d x)} \, dx$$

Optimal. Leaf size=60 $-\frac{b \log \left (a+b \tanh ^2(c+d x)\right )}{2 a d (a+b)}+\frac{\log (\cosh (c+d x))}{d (a+b)}+\frac{\log (\tanh (c+d x))}{a d}$

[Out]

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

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Rubi [A]  time = 0.101497, antiderivative size = 60, 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} $-\frac{b \log \left (a+b \tanh ^2(c+d x)\right )}{2 a d (a+b)}+\frac{\log (\cosh (c+d x))}{d (a+b)}+\frac{\log (\tanh (c+d x))}{a d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0539794, size = 54, normalized size = 0.9 $\frac{-b \log \left (a+b \tanh ^2(c+d x)\right )+2 (a+b) \log (\tanh (c+d x))+2 a \log (\cosh (c+d x))}{2 a d (a+b)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.076, size = 121, normalized size = 2. \begin{align*} -{\frac{1}{d \left ( a+b \right ) }\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{b}{2\,da \left ( a+b \right ) }\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}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{1}{d \left ( a+b \right ) }\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),x)

[Out]

-1/d/(a+b)*ln(tanh(1/2*d*x+1/2*c)+1)-1/2/d*b/a/(a+b)*ln(tanh(1/2*d*x+1/2*c)^4*a+2*tanh(1/2*d*x+1/2*c)^2*a+4*ta
nh(1/2*d*x+1/2*c)^2*b+a)+1/d/a*ln(tanh(1/2*d*x+1/2*c))-1/d/(a+b)*ln(tanh(1/2*d*x+1/2*c)-1)

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Maxima [A]  time = 1.09107, size = 136, normalized size = 2.27 \begin{align*} -\frac{b \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^{2} + a b\right )} d} + \frac{d x + c}{{\left (a + b\right )} d} + \frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} + \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{a 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),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.0569, size = 319, normalized size = 5.32 \begin{align*} -\frac{2 \, a d x + b \log \left (\frac{2 \,{\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{2} +{\left (a + b\right )} \sinh \left (d x + c\right )^{2} + a - b\right )}}{\cosh \left (d x + c\right )^{2} - 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2}}\right ) - 2 \,{\left (a + b\right )} \log \left (\frac{2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )}{2 \,{\left (a^{2} + a b\right )} 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),x, algorithm="fricas")

[Out]

-1/2*(2*a*d*x + b*log(2*((a + b)*cosh(d*x + c)^2 + (a + b)*sinh(d*x + c)^2 + a - b)/(cosh(d*x + c)^2 - 2*cosh(
d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)) - 2*(a + b)*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))))/(
(a^2 + a*b)*d)

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

[Out]

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

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Giac [A]  time = 1.21516, size = 138, normalized size = 2.3 \begin{align*} -\frac{b \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^{2} d + a b d\right )}} - \frac{d x + c}{a d + b d} + \frac{\log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right )}{a 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),x, algorithm="giac")

[Out]

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