3.122 \(\int \frac{\coth ^5(x)}{a+b \text{csch}(x)} \, dx\)

Optimal. Leaf size=70 \[ -\frac{\left (a^2+2 b^2\right ) \text{csch}(x)}{b^3}+\frac{\left (a^2+b^2\right )^2 \log (a+b \text{csch}(x))}{a b^4}+\frac{a \text{csch}^2(x)}{2 b^2}+\frac{\log (\sinh (x))}{a}-\frac{\text{csch}^3(x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.094092, antiderivative size = 70, 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 = {3885, 894} \[ -\frac{\left (a^2+2 b^2\right ) \text{csch}(x)}{b^3}+\frac{\left (a^2+b^2\right )^2 \log (a+b \text{csch}(x))}{a b^4}+\frac{a \text{csch}^2(x)}{2 b^2}+\frac{\log (\sinh (x))}{a}-\frac{\text{csch}^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^5/(a + b*Csch[x]),x]

[Out]

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

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

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 ^5(x)}{a+b \text{csch}(x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (-b^2-x^2\right )^2}{x (a+x)} \, dx,x,b \text{csch}(x)\right )}{b^4}\\ &=-\frac{\operatorname{Subst}\left (\int \left (a^2 \left (1+\frac{2 b^2}{a^2}\right )+\frac{b^4}{a x}-a x+x^2-\frac{\left (a^2+b^2\right )^2}{a (a+x)}\right ) \, dx,x,b \text{csch}(x)\right )}{b^4}\\ &=-\frac{\left (a^2+2 b^2\right ) \text{csch}(x)}{b^3}+\frac{a \text{csch}^2(x)}{2 b^2}-\frac{\text{csch}^3(x)}{3 b}+\frac{\left (a^2+b^2\right )^2 \log (a+b \text{csch}(x))}{a b^4}+\frac{\log (\sinh (x))}{a}\\ \end{align*}

Mathematica [A]  time = 0.123764, size = 83, normalized size = 1.19 \[ \frac{3 a^2 b^2 \text{csch}^2(x)-6 a b \left (a^2+2 b^2\right ) \text{csch}(x)-6 a^2 \left (a^2+2 b^2\right ) \log (\sinh (x))+6 \left (a^2+b^2\right )^2 \log (a \sinh (x)+b)-2 a b^3 \text{csch}^3(x)}{6 a b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^5/(a + b*Csch[x]),x]

[Out]

(-6*a*b*(a^2 + 2*b^2)*Csch[x] + 3*a^2*b^2*Csch[x]^2 - 2*a*b^3*Csch[x]^3 - 6*a^2*(a^2 + 2*b^2)*Log[Sinh[x]] + 6
*(a^2 + b^2)^2*Log[b + a*Sinh[x]])/(6*a*b^4)

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Maple [B]  time = 0.04, size = 219, normalized size = 3.1 \begin{align*}{\frac{1}{24\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}+{\frac{a}{8\,{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}+{\frac{{a}^{2}}{2\,{b}^{3}}\tanh \left ({\frac{x}{2}} \right ) }+{\frac{7}{8\,b}\tanh \left ({\frac{x}{2}} \right ) }-{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{{a}^{3}}{{b}^{4}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b-2\,a\tanh \left ( x/2 \right ) -b \right ) }+2\,{\frac{a\ln \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}b-2\,a\tanh \left ( x/2 \right ) -b \right ) }{{b}^{2}}}+{\frac{1}{a}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b-2\,a\tanh \left ( x/2 \right ) -b \right ) }-{\frac{1}{24\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}}-{\frac{{a}^{2}}{2\,{b}^{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{7}{8\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{a}{8\,{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}-{\frac{{a}^{3}}{{b}^{4}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-2\,{\frac{a\ln \left ( \tanh \left ( x/2 \right ) \right ) }{{b}^{2}}}-{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^5/(a+b*csch(x)),x)

[Out]

1/24/b*tanh(1/2*x)^3+1/8/b^2*tanh(1/2*x)^2*a+1/2/b^3*a^2*tanh(1/2*x)+7/8/b*tanh(1/2*x)-1/a*ln(tanh(1/2*x)+1)+a
^3/b^4*ln(tanh(1/2*x)^2*b-2*a*tanh(1/2*x)-b)+2*a/b^2*ln(tanh(1/2*x)^2*b-2*a*tanh(1/2*x)-b)+1/a*ln(tanh(1/2*x)^
2*b-2*a*tanh(1/2*x)-b)-1/24/b/tanh(1/2*x)^3-1/2/b^3/tanh(1/2*x)*a^2-7/8/b/tanh(1/2*x)+1/8*a/b^2/tanh(1/2*x)^2-
1/b^4*a^3*ln(tanh(1/2*x))-2*a/b^2*ln(tanh(1/2*x))-1/a*ln(tanh(1/2*x)-1)

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Maxima [B]  time = 1.04943, size = 257, normalized size = 3.67 \begin{align*} -\frac{2 \,{\left (3 \, a b e^{\left (-2 \, x\right )} - 3 \, a b e^{\left (-4 \, x\right )} - 3 \,{\left (a^{2} + 2 \, b^{2}\right )} e^{\left (-x\right )} + 2 \,{\left (3 \, a^{2} + 4 \, b^{2}\right )} e^{\left (-3 \, x\right )} - 3 \,{\left (a^{2} + 2 \, b^{2}\right )} e^{\left (-5 \, x\right )}\right )}}{3 \,{\left (3 \, b^{3} e^{\left (-2 \, x\right )} - 3 \, b^{3} e^{\left (-4 \, x\right )} + b^{3} e^{\left (-6 \, x\right )} - b^{3}\right )}} + \frac{x}{a} - \frac{{\left (a^{3} + 2 \, a b^{2}\right )} \log \left (e^{\left (-x\right )} + 1\right )}{b^{4}} - \frac{{\left (a^{3} + 2 \, a b^{2}\right )} \log \left (e^{\left (-x\right )} - 1\right )}{b^{4}} + \frac{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (-2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} - a\right )}{a b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^5/(a+b*csch(x)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.83275, size = 3216, normalized size = 45.94 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^5/(a+b*csch(x)),x, algorithm="fricas")

[Out]

-1/3*(3*b^4*x*cosh(x)^6 + 3*b^4*x*sinh(x)^6 + 6*(a^3*b + 2*a*b^3)*cosh(x)^5 + 6*(3*b^4*x*cosh(x) + a^3*b + 2*a
*b^3)*sinh(x)^5 - 3*b^4*x - 3*(3*b^4*x + 2*a^2*b^2)*cosh(x)^4 + 3*(15*b^4*x*cosh(x)^2 - 3*b^4*x - 2*a^2*b^2 +
10*(a^3*b + 2*a*b^3)*cosh(x))*sinh(x)^4 - 4*(3*a^3*b + 4*a*b^3)*cosh(x)^3 + 4*(15*b^4*x*cosh(x)^3 - 3*a^3*b -
4*a*b^3 + 15*(a^3*b + 2*a*b^3)*cosh(x)^2 - 3*(3*b^4*x + 2*a^2*b^2)*cosh(x))*sinh(x)^3 + 3*(3*b^4*x + 2*a^2*b^2
)*cosh(x)^2 + 3*(15*b^4*x*cosh(x)^4 + 3*b^4*x + 2*a^2*b^2 + 20*(a^3*b + 2*a*b^3)*cosh(x)^3 - 6*(3*b^4*x + 2*a^
2*b^2)*cosh(x)^2 - 4*(3*a^3*b + 4*a*b^3)*cosh(x))*sinh(x)^2 + 6*(a^3*b + 2*a*b^3)*cosh(x) - 3*((a^4 + 2*a^2*b^
2 + b^4)*cosh(x)^6 + 6*(a^4 + 2*a^2*b^2 + b^4)*cosh(x)*sinh(x)^5 + (a^4 + 2*a^2*b^2 + b^4)*sinh(x)^6 - 3*(a^4
+ 2*a^2*b^2 + b^4)*cosh(x)^4 - 3*(a^4 + 2*a^2*b^2 + b^4 - 5*(a^4 + 2*a^2*b^2 + b^4)*cosh(x)^2)*sinh(x)^4 - a^4
- 2*a^2*b^2 - b^4 + 4*(5*(a^4 + 2*a^2*b^2 + b^4)*cosh(x)^3 - 3*(a^4 + 2*a^2*b^2 + b^4)*cosh(x))*sinh(x)^3 + 3
*(a^4 + 2*a^2*b^2 + b^4)*cosh(x)^2 + 3*(5*(a^4 + 2*a^2*b^2 + b^4)*cosh(x)^4 + a^4 + 2*a^2*b^2 + b^4 - 6*(a^4 +
2*a^2*b^2 + b^4)*cosh(x)^2)*sinh(x)^2 + 6*((a^4 + 2*a^2*b^2 + b^4)*cosh(x)^5 - 2*(a^4 + 2*a^2*b^2 + b^4)*cosh
(x)^3 + (a^4 + 2*a^2*b^2 + b^4)*cosh(x))*sinh(x))*log(2*(a*sinh(x) + b)/(cosh(x) - sinh(x))) + 3*((a^4 + 2*a^2
*b^2)*cosh(x)^6 + 6*(a^4 + 2*a^2*b^2)*cosh(x)*sinh(x)^5 + (a^4 + 2*a^2*b^2)*sinh(x)^6 - 3*(a^4 + 2*a^2*b^2)*co
sh(x)^4 - 3*(a^4 + 2*a^2*b^2 - 5*(a^4 + 2*a^2*b^2)*cosh(x)^2)*sinh(x)^4 - a^4 - 2*a^2*b^2 + 4*(5*(a^4 + 2*a^2*
b^2)*cosh(x)^3 - 3*(a^4 + 2*a^2*b^2)*cosh(x))*sinh(x)^3 + 3*(a^4 + 2*a^2*b^2)*cosh(x)^2 + 3*(5*(a^4 + 2*a^2*b^
2)*cosh(x)^4 + a^4 + 2*a^2*b^2 - 6*(a^4 + 2*a^2*b^2)*cosh(x)^2)*sinh(x)^2 + 6*((a^4 + 2*a^2*b^2)*cosh(x)^5 - 2
*(a^4 + 2*a^2*b^2)*cosh(x)^3 + (a^4 + 2*a^2*b^2)*cosh(x))*sinh(x))*log(2*sinh(x)/(cosh(x) - sinh(x))) + 6*(3*b
^4*x*cosh(x)^5 + 5*(a^3*b + 2*a*b^3)*cosh(x)^4 + a^3*b + 2*a*b^3 - 2*(3*b^4*x + 2*a^2*b^2)*cosh(x)^3 - 2*(3*a^
3*b + 4*a*b^3)*cosh(x)^2 + (3*b^4*x + 2*a^2*b^2)*cosh(x))*sinh(x))/(a*b^4*cosh(x)^6 + 6*a*b^4*cosh(x)*sinh(x)^
5 + a*b^4*sinh(x)^6 - 3*a*b^4*cosh(x)^4 + 3*a*b^4*cosh(x)^2 - a*b^4 + 3*(5*a*b^4*cosh(x)^2 - a*b^4)*sinh(x)^4
+ 4*(5*a*b^4*cosh(x)^3 - 3*a*b^4*cosh(x))*sinh(x)^3 + 3*(5*a*b^4*cosh(x)^4 - 6*a*b^4*cosh(x)^2 + a*b^4)*sinh(x
)^2 + 6*(a*b^4*cosh(x)^5 - 2*a*b^4*cosh(x)^3 + a*b^4*cosh(x))*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{5}{\left (x \right )}}{a + b \operatorname{csch}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**5/(a+b*csch(x)),x)

[Out]

Integral(coth(x)**5/(a + b*csch(x)), x)

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Giac [B]  time = 1.15052, size = 230, normalized size = 3.29 \begin{align*} -\frac{{\left (a^{3} + 2 \, a b^{2}\right )} \log \left ({\left | -e^{\left (-x\right )} + e^{x} \right |}\right )}{b^{4}} + \frac{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | -a{\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, b \right |}\right )}{a b^{4}} + \frac{11 \, a^{3}{\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 22 \, a b^{2}{\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 12 \, a^{2} b{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 24 \, b^{3}{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 12 \, a b^{2}{\left (e^{\left (-x\right )} - e^{x}\right )} + 16 \, b^{3}}{6 \, b^{4}{\left (e^{\left (-x\right )} - e^{x}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^5/(a+b*csch(x)),x, algorithm="giac")

[Out]

-(a^3 + 2*a*b^2)*log(abs(-e^(-x) + e^x))/b^4 + (a^4 + 2*a^2*b^2 + b^4)*log(abs(-a*(e^(-x) - e^x) + 2*b))/(a*b^
4) + 1/6*(11*a^3*(e^(-x) - e^x)^3 + 22*a*b^2*(e^(-x) - e^x)^3 + 12*a^2*b*(e^(-x) - e^x)^2 + 24*b^3*(e^(-x) - e
^x)^2 + 12*a*b^2*(e^(-x) - e^x) + 16*b^3)/(b^4*(e^(-x) - e^x)^3)