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

Optimal. Leaf size=78 \[ \frac{\left (a^2-b^2\right ) \coth (x)}{a^3}+\frac{b \left (a^2-b^2\right ) \log (\tanh (x))}{a^4}-\frac{b \left (a^2-b^2\right ) \log (a+b \tanh (x))}{a^4}+\frac{b \coth ^2(x)}{2 a^2}-\frac{\coth ^3(x)}{3 a} \]

[Out]

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

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Rubi [A]  time = 0.0997105, antiderivative size = 78, 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} \[ \frac{\left (a^2-b^2\right ) \coth (x)}{a^3}+\frac{b \left (a^2-b^2\right ) \log (\tanh (x))}{a^4}-\frac{b \left (a^2-b^2\right ) \log (a+b \tanh (x))}{a^4}+\frac{b \coth ^2(x)}{2 a^2}-\frac{\coth ^3(x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^4/(a + b*Tanh[x]),x]

[Out]

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

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

Mathematica [A]  time = 0.239268, size = 70, normalized size = 0.9 \[ \frac{-2 \coth (x) \left (a^3 \text{csch}^2(x)-2 a^3+3 a b^2\right )+6 b \left (a^2-b^2\right ) (\log (\sinh (x))-\log (a \cosh (x)+b \sinh (x)))+3 a^2 b \text{csch}^2(x)}{6 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.045, size = 166, normalized size = 2.1 \begin{align*} -{\frac{1}{24\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}+{\frac{b}{8\,{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}+{\frac{3}{8\,a}\tanh \left ({\frac{x}{2}} \right ) }-{\frac{{b}^{2}}{2\,{a}^{3}}\tanh \left ({\frac{x}{2}} \right ) }-{\frac{b}{{a}^{2}}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( x/2 \right ) b+a \right ) }+{\frac{{b}^{3}}{{a}^{4}}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( x/2 \right ) b+a \right ) }-{\frac{1}{24\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}}+{\frac{3}{8\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{{b}^{2}}{2\,{a}^{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{b}{8\,{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+{\frac{b}{{a}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-{\frac{{b}^{3}}{{a}^{4}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.12194, size = 217, normalized size = 2.78 \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 (-x\right )} + 1\right )}{a^{4}} + \frac{{\left (a^{2} b - b^{3}\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(csch(x)^4/(a+b*tanh(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^(-x) + 1)/a
^4 + (a^2*b - b^3)*log(e^(-x) - 1)/a^4

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Fricas [B]  time = 2.46359, size = 2186, normalized size = 28.03 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(a+b*tanh(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*(a*cosh(x) + b*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*sin
h(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{csch}^{4}{\left (x \right )}}{a + b \tanh{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.2955, size = 273, normalized size = 3.5 \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 ({\left | e^{\left (2 \, x\right )} - 1 \right |}\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(csch(x)^4/(a+b*tanh(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(abs
(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 + 1
1*b^3)/(a^4*(e^(2*x) - 1)^3)