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

Optimal. Leaf size=79 \[ \frac{\left (a^2-b^2\right ) \tanh (x)}{a^3}-\frac{b \left (a^2-b^2\right ) \log (\tanh (x))}{a^4}-\frac{b \left (a^2-b^2\right ) \log (a+b \coth (x))}{a^4}+\frac{b \tanh ^2(x)}{2 a^2}-\frac{\tanh ^3(x)}{3 a} \]

[Out]

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

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Rubi [A]  time = 0.099762, antiderivative size = 79, 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 = {3516, 894} \[ \frac{\left (a^2-b^2\right ) \tanh (x)}{a^3}-\frac{b \left (a^2-b^2\right ) \log (\tanh (x))}{a^4}-\frac{b \left (a^2-b^2\right ) \log (a+b \coth (x))}{a^4}+\frac{b \tanh ^2(x)}{2 a^2}-\frac{\tanh ^3(x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^4/(a + b*Coth[x]),x]

[Out]

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

Rule 3516

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int
[(x^m*(a + x)^n)/(b^2 + x^2)^(m/2 + 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/
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{\text{sech}^4(x)}{a+b \coth (x)} \, dx &=-\left (b \operatorname{Subst}\left (\int \frac{-b^2+x^2}{x^4 (a+x)} \, dx,x,b \coth (x)\right )\right )\\ &=-\left (b \operatorname{Subst}\left (\int \left (-\frac{b^2}{a x^4}+\frac{b^2}{a^2 x^3}+\frac{a^2-b^2}{a^3 x^2}+\frac{-a^2+b^2}{a^4 x}+\frac{a^2-b^2}{a^4 (a+x)}\right ) \, dx,x,b \coth (x)\right )\right )\\ &=-\frac{b \left (a^2-b^2\right ) \log (a+b \coth (x))}{a^4}-\frac{b \left (a^2-b^2\right ) \log (\tanh (x))}{a^4}+\frac{\left (a^2-b^2\right ) \tanh (x)}{a^3}+\frac{b \tanh ^2(x)}{2 a^2}-\frac{\tanh ^3(x)}{3 a}\\ \end{align*}

Mathematica [A]  time = 0.258318, size = 68, normalized size = 0.86 \[ \frac{\left (4 a^3-6 a b^2\right ) \tanh (x)-6 b \left (b^2-a^2\right ) (\log (\cosh (x))-\log (a \sinh (x)+b \cosh (x)))+a^2 \text{sech}^2(x) (2 a \tanh (x)-3 b)}{6 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.053, size = 257, normalized size = 3.3 \begin{align*} 2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{5}}{a \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{5}{b}^{2}}{{a}^{3} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+2\,{\frac{b \left ( \tanh \left ( x/2 \right ) \right ) ^{4}}{{a}^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{4}{3\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-3}}-4\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{3}{b}^{2}}{{a}^{3} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{2}b}{{a}^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+2\,{\frac{\tanh \left ( x/2 \right ) }{a \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-2\,{\frac{\tanh \left ( x/2 \right ){b}^{2}}{{a}^{3} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{b}{{a}^{2}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }-{\frac{{b}^{3}}{{a}^{4}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }-{\frac{b}{{a}^{2}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b+2\,a\tanh \left ( x/2 \right ) +b \right ) }+{\frac{{b}^{3}}{{a}^{4}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b+2\,a\tanh \left ( x/2 \right ) +b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/a/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^5-2/a^3/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^5*b^2+2/a^2/(tanh(1/2*x)^2+1)^3*b*
tanh(1/2*x)^4+4/3/a/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^3-4/a^3/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^3*b^2+2/a^2/(tanh(
1/2*x)^2+1)^3*b*tanh(1/2*x)^2+2/a/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)-2/a^3/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)*b^2+1/
a^2*b*ln(tanh(1/2*x)^2+1)-1/a^4*ln(tanh(1/2*x)^2+1)*b^3-b/a^2*ln(tanh(1/2*x)^2*b+2*a*tanh(1/2*x)+b)+b^3/a^4*ln
(tanh(1/2*x)^2*b+2*a*tanh(1/2*x)+b)

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Maxima [A]  time = 1.7053, size = 180, normalized size = 2.28 \begin{align*} \frac{2 \,{\left (2 \, a^{2} - 3 \, b^{2} + 3 \,{\left (2 \, a^{2} - a b - 2 \, b^{2}\right )} e^{\left (-2 \, x\right )} - 3 \,{\left (a b + b^{2}\right )} e^{\left (-4 \, x\right )}\right )}}{3 \,{\left (3 \, a^{3} e^{\left (-2 \, x\right )} + 3 \, a^{3} e^{\left (-4 \, x\right )} + a^{3} e^{\left (-6 \, x\right )} + a^{3}\right )}} - \frac{{\left (a^{2} b - b^{3}\right )} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} + a + b\right )}{a^{4}} + \frac{{\left (a^{2} b - b^{3}\right )} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.76842, size = 2187, normalized size = 27.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*(a^2*b - a*b^2)*cosh(x)^4 + 24*(a^2*b - a*b^2)*cosh(x)*sinh(x)^3 + 6*(a^2*b - a*b^2)*sinh(x)^4 + 4*a^3
- 6*a*b^2 + 6*(2*a^3 + a^2*b - 2*a*b^2)*cosh(x)^2 + 6*(2*a^3 + a^2*b - 2*a*b^2 + 6*(a^2*b - a*b^2)*cosh(x)^2)
*sinh(x)^2 + 3*((a^2*b - b^3)*cosh(x)^6 + 6*(a^2*b - b^3)*cosh(x)*sinh(x)^5 + (a^2*b - b^3)*sinh(x)^6 + 3*(a^2
*b - b^3)*cosh(x)^4 + 3*(a^2*b - b^3 + 5*(a^2*b - b^3)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^2*b - b^3)*cosh(x)^3 + 3
*(a^2*b - b^3)*cosh(x))*sinh(x)^3 + a^2*b - b^3 + 3*(a^2*b - b^3)*cosh(x)^2 + 3*(5*(a^2*b - b^3)*cosh(x)^4 + a
^2*b - b^3 + 6*(a^2*b - b^3)*cosh(x)^2)*sinh(x)^2 + 6*((a^2*b - b^3)*cosh(x)^5 + 2*(a^2*b - b^3)*cosh(x)^3 + (
a^2*b - b^3)*cosh(x))*sinh(x))*log(2*(b*cosh(x) + a*sinh(x))/(cosh(x) - sinh(x))) - 3*((a^2*b - b^3)*cosh(x)^6
+ 6*(a^2*b - b^3)*cosh(x)*sinh(x)^5 + (a^2*b - b^3)*sinh(x)^6 + 3*(a^2*b - b^3)*cosh(x)^4 + 3*(a^2*b - b^3 +
5*(a^2*b - b^3)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^2*b - b^3)*cosh(x)^3 + 3*(a^2*b - b^3)*cosh(x))*sinh(x)^3 + a^2
*b - b^3 + 3*(a^2*b - b^3)*cosh(x)^2 + 3*(5*(a^2*b - b^3)*cosh(x)^4 + a^2*b - b^3 + 6*(a^2*b - b^3)*cosh(x)^2)
*sinh(x)^2 + 6*((a^2*b - b^3)*cosh(x)^5 + 2*(a^2*b - b^3)*cosh(x)^3 + (a^2*b - b^3)*cosh(x))*sinh(x))*log(2*co
sh(x)/(cosh(x) - sinh(x))) + 12*(2*(a^2*b - a*b^2)*cosh(x)^3 + (2*a^3 + a^2*b - 2*a*b^2)*cosh(x))*sinh(x))/(a^
4*cosh(x)^6 + 6*a^4*cosh(x)*sinh(x)^5 + a^4*sinh(x)^6 + 3*a^4*cosh(x)^4 + 3*a^4*cosh(x)^2 + 3*(5*a^4*cosh(x)^2
+ a^4)*sinh(x)^4 + a^4 + 4*(5*a^4*cosh(x)^3 + 3*a^4*cosh(x))*sinh(x)^3 + 3*(5*a^4*cosh(x)^4 + 6*a^4*cosh(x)^2
+ a^4)*sinh(x)^2 + 6*(a^4*cosh(x)^5 + 2*a^4*cosh(x)^3 + a^4*cosh(x))*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{4}{\left (x \right )}}{a + b \coth{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.14789, size = 271, normalized size = 3.43 \begin{align*} -\frac{{\left (a^{3} b + a^{2} b^{2} - a b^{3} - b^{4}\right )} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} - a + b \right |}\right )}{a^{5} + a^{4} b} + \frac{{\left (a^{2} b - b^{3}\right )} \log \left (e^{\left (2 \, x\right )} + 1\right )}{a^{4}} - \frac{11 \, a^{2} b e^{\left (6 \, x\right )} - 11 \, b^{3} e^{\left (6 \, x\right )} + 45 \, a^{2} b e^{\left (4 \, x\right )} - 12 \, a b^{2} e^{\left (4 \, x\right )} - 33 \, b^{3} e^{\left (4 \, x\right )} + 24 \, a^{3} e^{\left (2 \, x\right )} + 45 \, a^{2} b e^{\left (2 \, x\right )} - 24 \, a b^{2} e^{\left (2 \, x\right )} - 33 \, b^{3} e^{\left (2 \, x\right )} + 8 \, a^{3} + 11 \, a^{2} b - 12 \, a b^{2} - 11 \, b^{3}}{6 \, a^{4}{\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^3*b + a^2*b^2 - a*b^3 - b^4)*log(abs(a*e^(2*x) + b*e^(2*x) - a + b))/(a^5 + a^4*b) + (a^2*b - b^3)*log(e^(
2*x) + 1)/a^4 - 1/6*(11*a^2*b*e^(6*x) - 11*b^3*e^(6*x) + 45*a^2*b*e^(4*x) - 12*a*b^2*e^(4*x) - 33*b^3*e^(4*x)
+ 24*a^3*e^(2*x) + 45*a^2*b*e^(2*x) - 24*a*b^2*e^(2*x) - 33*b^3*e^(2*x) + 8*a^3 + 11*a^2*b - 12*a*b^2 - 11*b^3
)/(a^4*(e^(2*x) + 1)^3)