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

Optimal. Leaf size=59 \[ \frac{\left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{a^2 b d}-\frac{b \log (\sinh (c+d x))}{a^2 d}-\frac{\text{csch}(c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.122573, antiderivative size = 59, 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} \[ \frac{\left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{a^2 b d}-\frac{b \log (\sinh (c+d x))}{a^2 d}-\frac{\text{csch}(c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csch[c + d*x]/(a*d)) - (b*Log[Sinh[c + d*x]])/(a^2*d) + ((a^2 + b^2)*Log[a + b*Sinh[c + d*x]])/(a^2*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 (c+d x) \coth ^2(c+d x)}{a+b \sinh (c+d x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{b^2 \left (-b^2-x^2\right )}{x^2 (a+x)} \, dx,x,b \sinh (c+d x)\right )}{b^3 d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{-b^2-x^2}{x^2 (a+x)} \, dx,x,b \sinh (c+d x)\right )}{b d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-\frac{b^2}{a x^2}+\frac{b^2}{a^2 x}+\frac{-a^2-b^2}{a^2 (a+x)}\right ) \, dx,x,b \sinh (c+d x)\right )}{b d}\\ &=-\frac{\text{csch}(c+d x)}{a d}-\frac{b \log (\sinh (c+d x))}{a^2 d}+\frac{\left (a^2+b^2\right ) \log (a+b \sinh (c+d x))}{a^2 b d}\\ \end{align*}

Mathematica [A]  time = 0.0859398, size = 52, normalized size = 0.88 \[ \frac{\left (a^2+b^2\right ) \log (a+b \sinh (c+d x))-a b \text{csch}(c+d x)+b^2 (-\log (\sinh (c+d x)))}{a^2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.001, size = 172, normalized size = 2.9 \begin{align*}{\frac{1}{2\,da}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{1}{2\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{b}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{1}{bd}\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{b}{d{a}^{2}}\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)*coth(d*x+c)^2/(a+b*sinh(d*x+c)),x)

[Out]

1/2/d/a*tanh(1/2*d*x+1/2*c)-1/d/b*ln(tanh(1/2*d*x+1/2*c)+1)-1/d/b*ln(tanh(1/2*d*x+1/2*c)-1)-1/2/d/a/tanh(1/2*d
*x+1/2*c)-1/d/a^2*b*ln(tanh(1/2*d*x+1/2*c))+1/d/b*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^
2*b*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 = 1.22645, size = 177, normalized size = 3. \begin{align*} \frac{d x + c}{b d} + \frac{2 \, e^{\left (-d x - c\right )}}{{\left (a e^{\left (-2 \, d x - 2 \, c\right )} - a\right )} d} - \frac{b \log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{2} d} - \frac{b \log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{2} 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^{2} b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.2924, size = 751, normalized size = 12.73 \begin{align*} -\frac{a^{2} d x \cosh \left (d x + c\right )^{2} + a^{2} d x \sinh \left (d x + c\right )^{2} - a^{2} d x + 2 \, a b \cosh \left (d x + c\right ) -{\left ({\left (a^{2} + b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (a^{2} + b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) +{\left (a^{2} + b^{2}\right )} \sinh \left (d x + c\right )^{2} - a^{2} - b^{2}\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 ) +{\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )^{2} - b^{2}\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 \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right )}{a^{2} b d \cosh \left (d x + c\right )^{2} + 2 \, a^{2} b d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{2} b d \sinh \left (d x + c\right )^{2} - a^{2} b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.38761, size = 153, normalized size = 2.59 \begin{align*} -\frac{\frac{d x}{b} + \frac{b \log \left (e^{\left (d x + c\right )} + 1\right )}{a^{2}} + \frac{b \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a^{2}} - \frac{{\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^{2} b} + \frac{2 \, e^{\left (d x + c\right )}}{a{\left (e^{\left (d x + c\right )} + 1\right )}{\left (e^{\left (d x + c\right )} - 1\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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