3.104 \(\int \frac {\text {csch}^4(x)}{a+b \coth (x)} \, dx\)

Optimal. Leaf size=40 \[ -\frac {\left (a^2-b^2\right ) \log (a+b \coth (x))}{b^3}+\frac {a \coth (x)}{b^2}-\frac {\coth ^2(x)}{2 b} \]

[Out]

a*coth(x)/b^2-1/2*coth(x)^2/b-(a^2-b^2)*ln(a+b*coth(x))/b^3

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Rubi [A]  time = 0.07, antiderivative size = 40, normalized size of antiderivative = 1.00, 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 = {3506, 697} \[ -\frac {\left (a^2-b^2\right ) \log (a+b \coth (x))}{b^3}+\frac {a \coth (x)}{b^2}-\frac {\coth ^2(x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^4/(a + b*Coth[x]),x]

[Out]

(a*Coth[x])/b^2 - Coth[x]^2/(2*b) - ((a^2 - b^2)*Log[a + b*Coth[x]])/b^3

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 3506

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

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 50, normalized size = 1.25 \[ \frac {2 \left (a^2-b^2\right ) (\log (\sinh (x))-\log (a \sinh (x)+b \cosh (x)))+2 a b \coth (x)-b^2 \text {csch}^2(x)}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^4/(a + b*Coth[x]),x]

[Out]

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

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fricas [B]  time = 0.42, size = 434, normalized size = 10.85 \[ \frac {2 \, {\left (a b - b^{2}\right )} \cosh \relax (x)^{2} + 4 \, {\left (a b - b^{2}\right )} \cosh \relax (x) \sinh \relax (x) + 2 \, {\left (a b - b^{2}\right )} \sinh \relax (x)^{2} - 2 \, a b - {\left ({\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{4} + 4 \, {\left (a^{2} - b^{2}\right )} \cosh \relax (x) \sinh \relax (x)^{3} + {\left (a^{2} - b^{2}\right )} \sinh \relax (x)^{4} - 2 \, {\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (3 \, {\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{2} - a^{2} + b^{2}\right )} \sinh \relax (x)^{2} + a^{2} - b^{2} + 4 \, {\left ({\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{3} - {\left (a^{2} - b^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x)\right )} \log \left (\frac {2 \, {\left (b \cosh \relax (x) + a \sinh \relax (x)\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + {\left ({\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{4} + 4 \, {\left (a^{2} - b^{2}\right )} \cosh \relax (x) \sinh \relax (x)^{3} + {\left (a^{2} - b^{2}\right )} \sinh \relax (x)^{4} - 2 \, {\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (3 \, {\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{2} - a^{2} + b^{2}\right )} \sinh \relax (x)^{2} + a^{2} - b^{2} + 4 \, {\left ({\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{3} - {\left (a^{2} - b^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x)\right )} \log \left (\frac {2 \, \sinh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right )}{b^{3} \cosh \relax (x)^{4} + 4 \, b^{3} \cosh \relax (x) \sinh \relax (x)^{3} + b^{3} \sinh \relax (x)^{4} - 2 \, b^{3} \cosh \relax (x)^{2} + b^{3} + 2 \, {\left (3 \, b^{3} \cosh \relax (x)^{2} - b^{3}\right )} \sinh \relax (x)^{2} + 4 \, {\left (b^{3} \cosh \relax (x)^{3} - b^{3} \cosh \relax (x)\right )} \sinh \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(a+b*coth(x)),x, algorithm="fricas")

[Out]

(2*(a*b - b^2)*cosh(x)^2 + 4*(a*b - b^2)*cosh(x)*sinh(x) + 2*(a*b - b^2)*sinh(x)^2 - 2*a*b - ((a^2 - b^2)*cosh
(x)^4 + 4*(a^2 - b^2)*cosh(x)*sinh(x)^3 + (a^2 - b^2)*sinh(x)^4 - 2*(a^2 - b^2)*cosh(x)^2 + 2*(3*(a^2 - b^2)*c
osh(x)^2 - a^2 + b^2)*sinh(x)^2 + a^2 - b^2 + 4*((a^2 - b^2)*cosh(x)^3 - (a^2 - b^2)*cosh(x))*sinh(x))*log(2*(
b*cosh(x) + a*sinh(x))/(cosh(x) - sinh(x))) + ((a^2 - b^2)*cosh(x)^4 + 4*(a^2 - b^2)*cosh(x)*sinh(x)^3 + (a^2
- b^2)*sinh(x)^4 - 2*(a^2 - b^2)*cosh(x)^2 + 2*(3*(a^2 - b^2)*cosh(x)^2 - a^2 + b^2)*sinh(x)^2 + a^2 - b^2 + 4
*((a^2 - b^2)*cosh(x)^3 - (a^2 - b^2)*cosh(x))*sinh(x))*log(2*sinh(x)/(cosh(x) - sinh(x))))/(b^3*cosh(x)^4 + 4
*b^3*cosh(x)*sinh(x)^3 + b^3*sinh(x)^4 - 2*b^3*cosh(x)^2 + b^3 + 2*(3*b^3*cosh(x)^2 - b^3)*sinh(x)^2 + 4*(b^3*
cosh(x)^3 - b^3*cosh(x))*sinh(x))

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giac [B]  time = 0.12, size = 106, normalized size = 2.65 \[ -\frac {{\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} - a + b \right |}\right )}{a b^{3} + b^{4}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right )}{b^{3}} - \frac {2 \, {\left (a b - {\left (a b - b^{2}\right )} e^{\left (2 \, x\right )}\right )}}{b^{3} {\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(a+b*coth(x)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.12, size = 116, normalized size = 2.90 \[ -\frac {\tanh ^{2}\left (\frac {x}{2}\right )}{8 b}+\frac {a \tanh \left (\frac {x}{2}\right )}{2 b^{2}}-\frac {\ln \left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b +2 a \tanh \left (\frac {x}{2}\right )+b \right ) a^{2}}{b^{3}}+\frac {\ln \left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b +2 a \tanh \left (\frac {x}{2}\right )+b \right )}{b}-\frac {1}{8 b \tanh \left (\frac {x}{2}\right )^{2}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right ) a^{2}}{b^{3}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{b}+\frac {a}{2 b^{2} \tanh \left (\frac {x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)^4/(a+b*coth(x)),x)

[Out]

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

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maxima [B]  time = 0.33, size = 110, normalized size = 2.75 \[ \frac {2 \, {\left ({\left (a + b\right )} e^{\left (-2 \, x\right )} - a\right )}}{2 \, b^{2} e^{\left (-2 \, x\right )} - b^{2} e^{\left (-4 \, x\right )} - b^{2}} - \frac {{\left (a^{2} - b^{2}\right )} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} + a + b\right )}{b^{3}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (e^{\left (-x\right )} + 1\right )}{b^{3}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (e^{\left (-x\right )} - 1\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(a+b*coth(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.44, size = 88, normalized size = 2.20 \[ \frac {2\,\left (a-b\right )}{b^2\,\left ({\mathrm {e}}^{2\,x}-1\right )}-\frac {2}{b\,\left ({\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1\right )}-\frac {\ln \left (b-a+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )\,\left (a+b\right )\,\left (a-b\right )}{b^3}+\frac {\ln \left ({\mathrm {e}}^{2\,x}-1\right )\,\left (a+b\right )\,\left (a-b\right )}{b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^4*(a + b*coth(x))),x)

[Out]

(2*(a - b))/(b^2*(exp(2*x) - 1)) - 2/(b*(exp(4*x) - 2*exp(2*x) + 1)) - (log(b - a + a*exp(2*x) + b*exp(2*x))*(
a + b)*(a - b))/b^3 + (log(exp(2*x) - 1)*(a + b)*(a - b))/b^3

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {csch}^{4}{\relax (x )}}{a + b \coth {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)**4/(a+b*coth(x)),x)

[Out]

Integral(csch(x)**4/(a + b*coth(x)), x)

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