3.433 \(\int \frac{\cosh ^2(c+d x) \coth (c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=57 \[ -\frac{\left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{a b^2 d}+\frac{\log (\sinh (c+d x))}{a d}+\frac{\sinh (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.128075, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2837, 12, 894} \[ -\frac{\left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{a b^2 d}+\frac{\log (\sinh (c+d x))}{a d}+\frac{\sinh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [A]  time = 0.0817595, size = 48, normalized size = 0.84 \[ \frac{-\left (\frac{a}{b^2}+\frac{1}{a}\right ) \log (a+b \sinh (c+d x))+\frac{\log (\sinh (c+d x))}{a}+\frac{\sinh (c+d x)}{b}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[Sinh[c + d*x]]/a - (a^(-1) + a/b^2)*Log[a + b*Sinh[c + d*x]] + Sinh[c + d*x]/b)/d

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Maple [B]  time = 0.078, size = 178, normalized size = 3.1 \begin{align*} -{\frac{1}{bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{a}{{b}^{2}d}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{a}{{b}^{2}d}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{a}{{b}^{2}d}\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a-2\,\tanh \left ( 1/2\,dx+c/2 \right ) b-a \right ) }-{\frac{1}{da}\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a-2\,\tanh \left ( 1/2\,dx+c/2 \right ) b-a \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.08799, size = 176, normalized size = 3.09 \begin{align*} -\frac{{\left (d x + c\right )} a}{b^{2} d} + \frac{e^{\left (d x + c\right )}}{2 \, b d} - \frac{e^{\left (-d x - c\right )}}{2 \, b 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} - \frac{{\left (a^{2} + b^{2}\right )} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{a b^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 3.20018, size = 533, normalized size = 9.35 \begin{align*} \frac{2 \, a^{2} d x \cosh \left (d x + c\right ) + a b \cosh \left (d x + c\right )^{2} + a b \sinh \left (d x + c\right )^{2} - a b - 2 \,{\left ({\left (a^{2} + b^{2}\right )} \cosh \left (d x + c\right ) +{\left (a^{2} + b^{2}\right )} \sinh \left (d x + c\right )\right )} \log \left (\frac{2 \,{\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 2 \,{\left (b^{2} \cosh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )\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} d x + a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{2 \,{\left (a b^{2} d \cosh \left (d x + c\right ) + a b^{2} d \sinh \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.5483, size = 146, normalized size = 2.56 \begin{align*} \frac{\frac{2 \, a d x}{b^{2}} + \frac{e^{\left (d x + c\right )}}{b} - \frac{e^{\left (-d x - c\right )}}{b} + \frac{2 \, \log \left (e^{\left (d x + c\right )} + 1\right )}{a} + \frac{2 \, \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a} - \frac{2 \,{\left (a^{2} + b^{2}\right )} \log \left ({\left | b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - b \right |}\right )}{a b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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