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

Optimal. Leaf size=28 \[ \frac{(a+b) \log (\sinh (c+d x))}{d}-\frac{b \log (\cosh (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.0514288, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {4138, 446, 72} \[ \frac{(a+b) \log (\sinh (c+d x))}{d}-\frac{b \log (\cosh (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0379646, size = 44, normalized size = 1.57 \[ \frac{a (\log (\tanh (c+d x))+\log (\cosh (c+d x)))}{d}-\frac{b (\log (\cosh (c+d x))-\log (\sinh (c+d x)))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 26, normalized size = 0.9 \begin{align*}{\frac{b\ln \left ( \tanh \left ( dx+c \right ) \right ) }{d}}+{\frac{a\ln \left ( \sinh \left ( dx+c \right ) \right ) }{d}} \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*b*ln(tanh(d*x+c))+a*ln(sinh(d*x+c))/d

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Maxima [B]  time = 1.73326, size = 88, normalized size = 3.14 \begin{align*} b{\left (\frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} - \frac{\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d}\right )} + \frac{a \log \left (\sinh \left (d x + c\right )\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]

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

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Fricas [B]  time = 2.09733, size = 178, normalized size = 6.36 \begin{align*} -\frac{a d x + b \log \left (\frac{2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) -{\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 )}{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]

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

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

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Giac [A]  time = 1.17696, size = 76, normalized size = 2.71 \begin{align*} -\frac{a d x -{\left (a e^{\left (2 \, c\right )} + b e^{\left (2 \, c\right )}\right )} e^{\left (-2 \, c\right )} \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) + b \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\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="giac")

[Out]

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