3.242 \(\int \frac{\coth ^3(x)}{(a+b \sinh (x))^2} \, dx\)

Optimal. Leaf size=76 \[ \frac{a^2+b^2}{a^3 (a+b \sinh (x))}+\frac{\left (a^2+3 b^2\right ) \log (\sinh (x))}{a^4}-\frac{\left (a^2+3 b^2\right ) \log (a+b \sinh (x))}{a^4}+\frac{2 b \text{csch}(x)}{a^3}-\frac{\text{csch}^2(x)}{2 a^2} \]

[Out]

(2*b*Csch[x])/a^3 - Csch[x]^2/(2*a^2) + ((a^2 + 3*b^2)*Log[Sinh[x]])/a^4 - ((a^2 + 3*b^2)*Log[a + b*Sinh[x]])/
a^4 + (a^2 + b^2)/(a^3*(a + b*Sinh[x]))

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Rubi [A]  time = 0.110462, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2721, 894} \[ \frac{a^2+b^2}{a^3 (a+b \sinh (x))}+\frac{\left (a^2+3 b^2\right ) \log (\sinh (x))}{a^4}-\frac{\left (a^2+3 b^2\right ) \log (a+b \sinh (x))}{a^4}+\frac{2 b \text{csch}(x)}{a^3}-\frac{\text{csch}^2(x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^3/(a + b*Sinh[x])^2,x]

[Out]

(2*b*Csch[x])/a^3 - Csch[x]^2/(2*a^2) + ((a^2 + 3*b^2)*Log[Sinh[x]])/a^4 - ((a^2 + 3*b^2)*Log[a + b*Sinh[x]])/
a^4 + (a^2 + b^2)/(a^3*(a + b*Sinh[x]))

Rule 2721

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

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{\coth ^3(x)}{(a+b \sinh (x))^2} \, dx &=-\operatorname{Subst}\left (\int \frac{-b^2-x^2}{x^3 (a+x)^2} \, dx,x,b \sinh (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (-\frac{b^2}{a^2 x^3}+\frac{2 b^2}{a^3 x^2}+\frac{-a^2-3 b^2}{a^4 x}+\frac{a^2+b^2}{a^3 (a+x)^2}+\frac{a^2+3 b^2}{a^4 (a+x)}\right ) \, dx,x,b \sinh (x)\right )\\ &=\frac{2 b \text{csch}(x)}{a^3}-\frac{\text{csch}^2(x)}{2 a^2}+\frac{\left (a^2+3 b^2\right ) \log (\sinh (x))}{a^4}-\frac{\left (a^2+3 b^2\right ) \log (a+b \sinh (x))}{a^4}+\frac{a^2+b^2}{a^3 (a+b \sinh (x))}\\ \end{align*}

Mathematica [A]  time = 0.197432, size = 73, normalized size = 0.96 \[ \frac{\frac{2 a \left (a^2+b^2\right )}{a+b \sinh (x)}+2 \left (a^2+3 b^2\right ) \log (\sinh (x))-2 \left (a^2+3 b^2\right ) \log (a+b \sinh (x))-a^2 \text{csch}^2(x)+4 a b \text{csch}(x)}{2 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^3/(a + b*Sinh[x])^2,x]

[Out]

(4*a*b*Csch[x] - a^2*Csch[x]^2 + 2*(a^2 + 3*b^2)*Log[Sinh[x]] - 2*(a^2 + 3*b^2)*Log[a + b*Sinh[x]] + (2*a*(a^2
+ b^2))/(a + b*Sinh[x]))/(2*a^4)

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Maple [B]  time = 0.056, size = 184, normalized size = 2.4 \begin{align*} -{\frac{1}{8\,{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}-{\frac{b}{{a}^{3}}\tanh \left ({\frac{x}{2}} \right ) }-{\frac{1}{8\,{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+{\frac{1}{{a}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }+3\,{\frac{\ln \left ( \tanh \left ( x/2 \right ) \right ){b}^{2}}{{a}^{4}}}+{\frac{b}{{a}^{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+2\,{\frac{\tanh \left ( x/2 \right ) b}{{a}^{2} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }}+2\,{\frac{\tanh \left ( x/2 \right ){b}^{3}}{{a}^{4} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }}-{\frac{1}{{a}^{2}}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) }-3\,{\frac{\ln \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ){b}^{2}}{{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^3/(a+b*sinh(x))^2,x)

[Out]

-1/8/a^2*tanh(1/2*x)^2-1/a^3*tanh(1/2*x)*b-1/8/a^2/tanh(1/2*x)^2+1/a^2*ln(tanh(1/2*x))+3/a^4*ln(tanh(1/2*x))*b
^2+b/a^3/tanh(1/2*x)+2/a^2*tanh(1/2*x)/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)*b+2/a^4*tanh(1/2*x)/(a*tanh(1/2*x)^
2-2*tanh(1/2*x)*b-a)*b^3-1/a^2*ln(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)-3/a^4*ln(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-
a)*b^2

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Maxima [B]  time = 1.05053, size = 273, normalized size = 3.59 \begin{align*} \frac{2 \,{\left (3 \, a b e^{\left (-2 \, x\right )} - 3 \, a b e^{\left (-4 \, x\right )} +{\left (a^{2} + 3 \, b^{2}\right )} e^{\left (-x\right )} - 2 \,{\left (2 \, a^{2} + 3 \, b^{2}\right )} e^{\left (-3 \, x\right )} +{\left (a^{2} + 3 \, b^{2}\right )} e^{\left (-5 \, x\right )}\right )}}{2 \, a^{4} e^{\left (-x\right )} - 3 \, a^{3} b e^{\left (-2 \, x\right )} - 4 \, a^{4} e^{\left (-3 \, x\right )} + 3 \, a^{3} b e^{\left (-4 \, x\right )} + 2 \, a^{4} e^{\left (-5 \, x\right )} - a^{3} b e^{\left (-6 \, x\right )} + a^{3} b} - \frac{{\left (a^{2} + 3 \, b^{2}\right )} \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a^{4}} + \frac{{\left (a^{2} + 3 \, b^{2}\right )} \log \left (e^{\left (-x\right )} + 1\right )}{a^{4}} + \frac{{\left (a^{2} + 3 \, b^{2}\right )} \log \left (e^{\left (-x\right )} - 1\right )}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^3/(a+b*sinh(x))^2,x, algorithm="maxima")

[Out]

2*(3*a*b*e^(-2*x) - 3*a*b*e^(-4*x) + (a^2 + 3*b^2)*e^(-x) - 2*(2*a^2 + 3*b^2)*e^(-3*x) + (a^2 + 3*b^2)*e^(-5*x
))/(2*a^4*e^(-x) - 3*a^3*b*e^(-2*x) - 4*a^4*e^(-3*x) + 3*a^3*b*e^(-4*x) + 2*a^4*e^(-5*x) - a^3*b*e^(-6*x) + a^
3*b) - (a^2 + 3*b^2)*log(-2*a*e^(-x) + b*e^(-2*x) - b)/a^4 + (a^2 + 3*b^2)*log(e^(-x) + 1)/a^4 + (a^2 + 3*b^2)
*log(e^(-x) - 1)/a^4

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Fricas [B]  time = 2.29207, size = 3726, normalized size = 49.03 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^3/(a+b*sinh(x))^2,x, algorithm="fricas")

[Out]

(6*a^2*b*cosh(x)^4 + 2*(a^3 + 3*a*b^2)*cosh(x)^5 + 2*(a^3 + 3*a*b^2)*sinh(x)^5 - 6*a^2*b*cosh(x)^2 + 2*(3*a^2*
b + 5*(a^3 + 3*a*b^2)*cosh(x))*sinh(x)^4 - 4*(2*a^3 + 3*a*b^2)*cosh(x)^3 + 4*(6*a^2*b*cosh(x) - 2*a^3 - 3*a*b^
2 + 5*(a^3 + 3*a*b^2)*cosh(x)^2)*sinh(x)^3 + 2*(18*a^2*b*cosh(x)^2 + 10*(a^3 + 3*a*b^2)*cosh(x)^3 - 3*a^2*b -
6*(2*a^3 + 3*a*b^2)*cosh(x))*sinh(x)^2 + 2*(a^3 + 3*a*b^2)*cosh(x) - ((a^2*b + 3*b^3)*cosh(x)^6 + (a^2*b + 3*b
^3)*sinh(x)^6 + 2*(a^3 + 3*a*b^2)*cosh(x)^5 + 2*(a^3 + 3*a*b^2 + 3*(a^2*b + 3*b^3)*cosh(x))*sinh(x)^5 - 3*(a^2
*b + 3*b^3)*cosh(x)^4 - (3*a^2*b + 9*b^3 - 15*(a^2*b + 3*b^3)*cosh(x)^2 - 10*(a^3 + 3*a*b^2)*cosh(x))*sinh(x)^
4 - 4*(a^3 + 3*a*b^2)*cosh(x)^3 + 4*(5*(a^2*b + 3*b^3)*cosh(x)^3 - a^3 - 3*a*b^2 + 5*(a^3 + 3*a*b^2)*cosh(x)^2
- 3*(a^2*b + 3*b^3)*cosh(x))*sinh(x)^3 - a^2*b - 3*b^3 + 3*(a^2*b + 3*b^3)*cosh(x)^2 + (15*(a^2*b + 3*b^3)*co
sh(x)^4 + 20*(a^3 + 3*a*b^2)*cosh(x)^3 + 3*a^2*b + 9*b^3 - 18*(a^2*b + 3*b^3)*cosh(x)^2 - 12*(a^3 + 3*a*b^2)*c
osh(x))*sinh(x)^2 + 2*(a^3 + 3*a*b^2)*cosh(x) + 2*(3*(a^2*b + 3*b^3)*cosh(x)^5 + 5*(a^3 + 3*a*b^2)*cosh(x)^4 -
6*(a^2*b + 3*b^3)*cosh(x)^3 + a^3 + 3*a*b^2 - 6*(a^3 + 3*a*b^2)*cosh(x)^2 + 3*(a^2*b + 3*b^3)*cosh(x))*sinh(x
))*log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x))) + ((a^2*b + 3*b^3)*cosh(x)^6 + (a^2*b + 3*b^3)*sinh(x)^6 + 2*(a^
3 + 3*a*b^2)*cosh(x)^5 + 2*(a^3 + 3*a*b^2 + 3*(a^2*b + 3*b^3)*cosh(x))*sinh(x)^5 - 3*(a^2*b + 3*b^3)*cosh(x)^4
- (3*a^2*b + 9*b^3 - 15*(a^2*b + 3*b^3)*cosh(x)^2 - 10*(a^3 + 3*a*b^2)*cosh(x))*sinh(x)^4 - 4*(a^3 + 3*a*b^2)
*cosh(x)^3 + 4*(5*(a^2*b + 3*b^3)*cosh(x)^3 - a^3 - 3*a*b^2 + 5*(a^3 + 3*a*b^2)*cosh(x)^2 - 3*(a^2*b + 3*b^3)*
cosh(x))*sinh(x)^3 - a^2*b - 3*b^3 + 3*(a^2*b + 3*b^3)*cosh(x)^2 + (15*(a^2*b + 3*b^3)*cosh(x)^4 + 20*(a^3 + 3
*a*b^2)*cosh(x)^3 + 3*a^2*b + 9*b^3 - 18*(a^2*b + 3*b^3)*cosh(x)^2 - 12*(a^3 + 3*a*b^2)*cosh(x))*sinh(x)^2 + 2
*(a^3 + 3*a*b^2)*cosh(x) + 2*(3*(a^2*b + 3*b^3)*cosh(x)^5 + 5*(a^3 + 3*a*b^2)*cosh(x)^4 - 6*(a^2*b + 3*b^3)*co
sh(x)^3 + a^3 + 3*a*b^2 - 6*(a^3 + 3*a*b^2)*cosh(x)^2 + 3*(a^2*b + 3*b^3)*cosh(x))*sinh(x))*log(2*sinh(x)/(cos
h(x) - sinh(x))) + 2*(12*a^2*b*cosh(x)^3 + 5*(a^3 + 3*a*b^2)*cosh(x)^4 - 6*a^2*b*cosh(x) + a^3 + 3*a*b^2 - 6*(
2*a^3 + 3*a*b^2)*cosh(x)^2)*sinh(x))/(a^4*b*cosh(x)^6 + a^4*b*sinh(x)^6 + 2*a^5*cosh(x)^5 - 3*a^4*b*cosh(x)^4
- 4*a^5*cosh(x)^3 + 3*a^4*b*cosh(x)^2 + 2*a^5*cosh(x) + 2*(3*a^4*b*cosh(x) + a^5)*sinh(x)^5 - a^4*b + (15*a^4*
b*cosh(x)^2 + 10*a^5*cosh(x) - 3*a^4*b)*sinh(x)^4 + 4*(5*a^4*b*cosh(x)^3 + 5*a^5*cosh(x)^2 - 3*a^4*b*cosh(x) -
a^5)*sinh(x)^3 + (15*a^4*b*cosh(x)^4 + 20*a^5*cosh(x)^3 - 18*a^4*b*cosh(x)^2 - 12*a^5*cosh(x) + 3*a^4*b)*sinh
(x)^2 + 2*(3*a^4*b*cosh(x)^5 + 5*a^5*cosh(x)^4 - 6*a^4*b*cosh(x)^3 - 6*a^5*cosh(x)^2 + 3*a^4*b*cosh(x) + a^5)*
sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{3}{\left (x \right )}}{\left (a + b \sinh{\left (x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**3/(a+b*sinh(x))**2,x)

[Out]

Integral(coth(x)**3/(a + b*sinh(x))**2, x)

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Giac [B]  time = 1.15317, size = 257, normalized size = 3.38 \begin{align*} \frac{{\left (a^{2} + 3 \, b^{2}\right )} \log \left ({\left | -e^{\left (-x\right )} + e^{x} \right |}\right )}{a^{4}} - \frac{{\left (a^{2} b + 3 \, b^{3}\right )} \log \left ({\left | -b{\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{a^{4} b} + \frac{a^{2} b{\left (e^{\left (-x\right )} - e^{x}\right )} + 3 \, b^{3}{\left (e^{\left (-x\right )} - e^{x}\right )} - 4 \, a^{3} - 8 \, a b^{2}}{{\left (b{\left (e^{\left (-x\right )} - e^{x}\right )} - 2 \, a\right )} a^{4}} - \frac{3 \, a^{2}{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 9 \, b^{2}{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 8 \, a b{\left (e^{\left (-x\right )} - e^{x}\right )} + 4 \, a^{2}}{2 \, a^{4}{\left (e^{\left (-x\right )} - e^{x}\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^3/(a+b*sinh(x))^2,x, algorithm="giac")

[Out]

(a^2 + 3*b^2)*log(abs(-e^(-x) + e^x))/a^4 - (a^2*b + 3*b^3)*log(abs(-b*(e^(-x) - e^x) + 2*a))/(a^4*b) + (a^2*b
*(e^(-x) - e^x) + 3*b^3*(e^(-x) - e^x) - 4*a^3 - 8*a*b^2)/((b*(e^(-x) - e^x) - 2*a)*a^4) - 1/2*(3*a^2*(e^(-x)
- e^x)^2 + 9*b^2*(e^(-x) - e^x)^2 + 8*a*b*(e^(-x) - e^x) + 4*a^2)/(a^4*(e^(-x) - e^x)^2)