3.122 \(\int \frac{\coth ^3(x)}{a+b \text{sech}(x)} \, dx\)

Optimal. Leaf size=113 \[ \frac{b^4 \log (a+b \text{sech}(x))}{a \left (a^2-b^2\right )^2}-\frac{1}{4 (a+b) (1-\text{sech}(x))}-\frac{1}{4 (a-b) (\text{sech}(x)+1)}+\frac{(2 a+3 b) \log (1-\text{sech}(x))}{4 (a+b)^2}+\frac{(2 a-3 b) \log (\text{sech}(x)+1)}{4 (a-b)^2}+\frac{\log (\cosh (x))}{a} \]

[Out]

Log[Cosh[x]]/a + ((2*a + 3*b)*Log[1 - Sech[x]])/(4*(a + b)^2) + ((2*a - 3*b)*Log[1 + Sech[x]])/(4*(a - b)^2) +
(b^4*Log[a + b*Sech[x]])/(a*(a^2 - b^2)^2) - 1/(4*(a + b)*(1 - Sech[x])) - 1/(4*(a - b)*(1 + Sech[x]))

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Rubi [A]  time = 0.190496, antiderivative size = 113, 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} \[ \frac{b^4 \log (a+b \text{sech}(x))}{a \left (a^2-b^2\right )^2}-\frac{1}{4 (a+b) (1-\text{sech}(x))}-\frac{1}{4 (a-b) (\text{sech}(x)+1)}+\frac{(2 a+3 b) \log (1-\text{sech}(x))}{4 (a+b)^2}+\frac{(2 a-3 b) \log (\text{sech}(x)+1)}{4 (a-b)^2}+\frac{\log (\cosh (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^3/(a + b*Sech[x]),x]

[Out]

Log[Cosh[x]]/a + ((2*a + 3*b)*Log[1 - Sech[x]])/(4*(a + b)^2) + ((2*a - 3*b)*Log[1 + Sech[x]])/(4*(a - b)^2) +
(b^4*Log[a + b*Sech[x]])/(a*(a^2 - b^2)^2) - 1/(4*(a + b)*(1 - Sech[x])) - 1/(4*(a - b)*(1 + Sech[x]))

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

Mathematica [A]  time = 0.300283, size = 112, normalized size = 0.99 \[ \frac{4 a \left (2 a \left (a^2-2 b^2\right ) \log (\sinh (x))+b \left (3 b^2-a^2\right ) \log \left (\tanh \left (\frac{x}{2}\right )\right )\right )+8 b^4 \log (a \cosh (x)+b)-a (a-b)^2 (a+b) \text{csch}^2\left (\frac{x}{2}\right )+a (a-b) (a+b)^2 \text{sech}^2\left (\frac{x}{2}\right )}{8 a (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*(a - b)^2*(a + b)*Csch[x/2]^2) + 8*b^4*Log[b + a*Cosh[x]] + 4*a*(2*a*(a^2 - 2*b^2)*Log[Sinh[x]] + b*(-a^2
+ 3*b^2)*Log[Tanh[x/2]]) + a*(a - b)*(a + b)^2*Sech[x/2]^2)/(8*a*(a - b)^2*(a + b)^2)

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Maple [A]  time = 0.043, size = 119, normalized size = 1.1 \begin{align*} -{\frac{1}{8\,a-8\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}-{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{8\,a+8\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+{\frac{a}{ \left ( a+b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }+{\frac{3\,b}{2\, \left ( a+b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+{\frac{{b}^{4}}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}a}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b+a+b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.1829, size = 221, normalized size = 1.96 \begin{align*} \frac{b^{4} \log \left (2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} + a\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac{{\left (2 \, a - 3 \, b\right )} \log \left (e^{\left (-x\right )} + 1\right )}{2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac{{\left (2 \, a + 3 \, b\right )} \log \left (e^{\left (-x\right )} - 1\right )}{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac{b e^{\left (-x\right )} - 2 \, a e^{\left (-2 \, x\right )} + b e^{\left (-3 \, x\right )}}{a^{2} - b^{2} - 2 \,{\left (a^{2} - b^{2}\right )} e^{\left (-2 \, x\right )} +{\left (a^{2} - b^{2}\right )} e^{\left (-4 \, x\right )}} + \frac{x}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.7178, size = 2885, normalized size = 25.53 \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*sech(x)),x, algorithm="fricas")

[Out]

-1/2*(2*(a^4 - 2*a^2*b^2 + b^4)*x*cosh(x)^4 + 2*(a^4 - 2*a^2*b^2 + b^4)*x*sinh(x)^4 - 2*(a^3*b - a*b^3)*cosh(x
)^3 - 2*(a^3*b - a*b^3 - 4*(a^4 - 2*a^2*b^2 + b^4)*x*cosh(x))*sinh(x)^3 + 4*(a^4 - a^2*b^2 - (a^4 - 2*a^2*b^2
+ b^4)*x)*cosh(x)^2 + 2*(2*a^4 - 2*a^2*b^2 + 6*(a^4 - 2*a^2*b^2 + b^4)*x*cosh(x)^2 - 2*(a^4 - 2*a^2*b^2 + b^4)
*x - 3*(a^3*b - a*b^3)*cosh(x))*sinh(x)^2 + 2*(a^4 - 2*a^2*b^2 + b^4)*x - 2*(a^3*b - a*b^3)*cosh(x) - 2*(b^4*c
osh(x)^4 + 4*b^4*cosh(x)*sinh(x)^3 + b^4*sinh(x)^4 - 2*b^4*cosh(x)^2 + b^4 + 2*(3*b^4*cosh(x)^2 - b^4)*sinh(x)
^2 + 4*(b^4*cosh(x)^3 - b^4*cosh(x))*sinh(x))*log(2*(a*cosh(x) + b)/(cosh(x) - sinh(x))) - ((2*a^4 + a^3*b - 4
*a^2*b^2 - 3*a*b^3)*cosh(x)^4 + 4*(2*a^4 + a^3*b - 4*a^2*b^2 - 3*a*b^3)*cosh(x)*sinh(x)^3 + (2*a^4 + a^3*b - 4
*a^2*b^2 - 3*a*b^3)*sinh(x)^4 + 2*a^4 + a^3*b - 4*a^2*b^2 - 3*a*b^3 - 2*(2*a^4 + a^3*b - 4*a^2*b^2 - 3*a*b^3)*
cosh(x)^2 - 2*(2*a^4 + a^3*b - 4*a^2*b^2 - 3*a*b^3 - 3*(2*a^4 + a^3*b - 4*a^2*b^2 - 3*a*b^3)*cosh(x)^2)*sinh(x
)^2 + 4*((2*a^4 + a^3*b - 4*a^2*b^2 - 3*a*b^3)*cosh(x)^3 - (2*a^4 + a^3*b - 4*a^2*b^2 - 3*a*b^3)*cosh(x))*sinh
(x))*log(cosh(x) + sinh(x) + 1) - ((2*a^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3)*cosh(x)^4 + 4*(2*a^4 - a^3*b - 4*a^2*
b^2 + 3*a*b^3)*cosh(x)*sinh(x)^3 + (2*a^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3)*sinh(x)^4 + 2*a^4 - a^3*b - 4*a^2*b^2
+ 3*a*b^3 - 2*(2*a^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3)*cosh(x)^2 - 2*(2*a^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3 - 3*(2
*a^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3)*cosh(x)^2)*sinh(x)^2 + 4*((2*a^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3)*cosh(x)^3
- (2*a^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) - 1) + 2*(4*(a^4 - 2*a^2*b^2 +
b^4)*x*cosh(x)^3 - a^3*b + a*b^3 - 3*(a^3*b - a*b^3)*cosh(x)^2 + 4*(a^4 - a^2*b^2 - (a^4 - 2*a^2*b^2 + b^4)*x
)*cosh(x))*sinh(x))/(a^5 - 2*a^3*b^2 + a*b^4 + (a^5 - 2*a^3*b^2 + a*b^4)*cosh(x)^4 + 4*(a^5 - 2*a^3*b^2 + a*b^
4)*cosh(x)*sinh(x)^3 + (a^5 - 2*a^3*b^2 + a*b^4)*sinh(x)^4 - 2*(a^5 - 2*a^3*b^2 + a*b^4)*cosh(x)^2 - 2*(a^5 -
2*a^3*b^2 + a*b^4 - 3*(a^5 - 2*a^3*b^2 + a*b^4)*cosh(x)^2)*sinh(x)^2 + 4*((a^5 - 2*a^3*b^2 + a*b^4)*cosh(x)^3
- (a^5 - 2*a^3*b^2 + 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 ^{3}{\left (x \right )}}{a + b \operatorname{sech}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14085, size = 261, normalized size = 2.31 \begin{align*} \frac{b^{4} \log \left ({\left | a{\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, b \right |}\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac{{\left (2 \, a - 3 \, b\right )} \log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{4 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac{{\left (2 \, a + 3 \, b\right )} \log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac{a^{3}{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 2 \, a b^{2}{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 2 \, a^{2} b{\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, b^{3}{\left (e^{\left (-x\right )} + e^{x}\right )} + 4 \, a b^{2}}{2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left ({\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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