3.144 \(\int \frac{\coth (c+d x)}{a+b \text{sech}^2(c+d x)} \, dx\)

Optimal. Leaf size=46 \[ \frac{\log (\sinh (c+d x))}{d (a+b)}+\frac{b \log \left (a \cosh ^2(c+d x)+b\right )}{2 a d (a+b)} \]

[Out]

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

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Rubi [A]  time = 0.087361, antiderivative size = 46, 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 = {4138, 446, 72} \[ \frac{\log (\sinh (c+d x))}{d (a+b)}+\frac{b \log \left (a \cosh ^2(c+d x)+b\right )}{2 a d (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4138

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Module[{ff =
FreeFactors[Cos[e + f*x], x]}, -Dist[(f*ff^(m + n*p - 1))^(-1), Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*
(ff*x)^n)^p)/x^(m + n*p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n] && IntegerQ[p]

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 \text{sech}^2(c+d x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^3}{\left (1-x^2\right ) \left (b+a x^2\right )} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x}{(1-x) (b+a x)} \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{1}{(-a-b) (-1+x)}-\frac{b}{(a+b) (b+a x)}\right ) \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=\frac{b \log \left (b+a \cosh ^2(c+d x)\right )}{2 a (a+b) d}+\frac{\log (\sinh (c+d x))}{(a+b) d}\\ \end{align*}

Mathematica [A]  time = 0.0890806, size = 42, normalized size = 0.91 \[ \frac{b \log \left (a \sinh ^2(c+d x)+a+b\right )+2 a \log (\sinh (c+d x))}{2 a^2 d+2 a b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Log[Sinh[c + d*x]] + b*Log[a + b + a*Sinh[c + d*x]^2])/(2*a^2*d + 2*a*b*d)

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Maple [B]  time = 0.059, size = 133, normalized size = 2.9 \begin{align*} -{\frac{1}{da}\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+b \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }+{\frac{1}{d \left ( a+b \right ) }\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{1}{da}\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*sech(d*x+c)^2),x)

[Out]

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

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Maxima [B]  time = 1.04251, size = 135, normalized size = 2.93 \begin{align*} \frac{b \log \left (2 \,{\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{2 \,{\left (a^{2} + a b\right )} d} + \frac{d x + c}{a d} + \frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{{\left (a + b\right )} d} + \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{{\left (a + b\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.79042, size = 305, normalized size = 6.63 \begin{align*} -\frac{2 \,{\left (a + b\right )} d x - b \log \left (\frac{2 \,{\left (a \cosh \left (d x + c\right )^{2} + a \sinh \left (d x + c\right )^{2} + a + 2 \, 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 \, a \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*sech(d*x+c)^2),x, algorithm="fricas")

[Out]

-1/2*(2*(a + b)*d*x - b*log(2*(a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + a + 2*b)/(cosh(d*x + c)^2 - 2*cosh(d*x
+ c)*sinh(d*x + c) + sinh(d*x + c)^2)) - 2*a*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 \operatorname{sech}^{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*sech(d*x+c)**2),x)

[Out]

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

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Giac [B]  time = 1.46974, size = 131, normalized size = 2.85 \begin{align*} -\frac{\frac{2 \, d x}{a} - \frac{b \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}{a^{2} + a b} - \frac{2 \, e^{\left (2 \, c\right )} \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right )}{a e^{\left (2 \, c\right )} + b e^{\left (2 \, c\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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