3.141 \(\int \frac{\coth ^3(x)}{a+b \tanh (x)} \, dx\)

Optimal. Leaf size=76 \[ -\frac{b x}{a^2-b^2}+\frac{\left (a^2+b^2\right ) \log (\sinh (x))}{a^3}+\frac{b^4 \log (a \cosh (x)+b \sinh (x))}{a^3 \left (a^2-b^2\right )}+\frac{b \coth (x)}{a^2}-\frac{\coth ^2(x)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.309112, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {3569, 3649, 3652, 3530, 3475} \[ -\frac{b x}{a^2-b^2}+\frac{\left (a^2+b^2\right ) \log (\sinh (x))}{a^3}+\frac{b^4 \log (a \cosh (x)+b \sinh (x))}{a^3 \left (a^2-b^2\right )}+\frac{b \coth (x)}{a^2}-\frac{\coth ^2(x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^3/(a + b*Tanh[x]),x]

[Out]

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

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3652

Int[((A_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Simp[((a*(A*c - c*C) - b*(A*d - C*d))*x)/((a^2 + b^2)*(c^2 + d^2)), x] + (Dist[
(A*b^2 + a^2*C)/((b*c - a*d)*(a^2 + b^2)), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] - Dist[(c^2*C
+ A*d^2)/((b*c - a*d)*(c^2 + d^2)), Int[(d - c*Tan[e + f*x])/(c + d*Tan[e + f*x]), x], x]) /; FreeQ[{a, b, c,
d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.184197, size = 91, normalized size = 1.2 \[ \frac{2 a b \left (a^2-b^2\right ) \coth (x)+\left (a^2 b^2-a^4\right ) \text{csch}^2(x)-2 a^3 b x+2 a^4 \log (\sinh (x))+2 b^4 \log (a \cosh (x)+b \sinh (x))-2 b^4 \log (\sinh (x))}{2 a^3 (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 134, normalized size = 1.8 \begin{align*} -{\frac{1}{8\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}}+{\frac{b}{2\,{a}^{2}}\tanh \left ({\frac{x}{2}} \right ) }-{\frac{1}{a-b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{{b}^{4}}{{a}^{3} \left ( a+b \right ) \left ( a-b \right ) }\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( x/2 \right ) b+a \right ) }-{\frac{1}{8\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }+{\frac{{b}^{2}}{{a}^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }+{\frac{b}{2\,{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{1}{a+b}\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)^3/(a+b*tanh(x)),x)

[Out]

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

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Maxima [A]  time = 1.27848, size = 163, normalized size = 2.14 \begin{align*} \frac{b^{4} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{5} - a^{3} b^{2}} + \frac{2 \,{\left ({\left (a + b\right )} e^{\left (-2 \, x\right )} - b\right )}}{2 \, a^{2} e^{\left (-2 \, x\right )} - a^{2} e^{\left (-4 \, x\right )} - a^{2}} + \frac{x}{a + b} + \frac{{\left (a^{2} + b^{2}\right )} \log \left (e^{\left (-x\right )} + 1\right )}{a^{3}} + \frac{{\left (a^{2} + b^{2}\right )} \log \left (e^{\left (-x\right )} - 1\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.83733, size = 1539, normalized size = 20.25 \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*tanh(x)),x, algorithm="fricas")

[Out]

-((a^4 + a^3*b)*x*cosh(x)^4 + 4*(a^4 + a^3*b)*x*cosh(x)*sinh(x)^3 + (a^4 + a^3*b)*x*sinh(x)^4 + 2*a^3*b - 2*a*
b^3 + 2*(a^4 - a^3*b - a^2*b^2 + a*b^3 - (a^4 + a^3*b)*x)*cosh(x)^2 + 2*(a^4 - a^3*b - a^2*b^2 + a*b^3 + 3*(a^
4 + a^3*b)*x*cosh(x)^2 - (a^4 + a^3*b)*x)*sinh(x)^2 + (a^4 + a^3*b)*x - (b^4*cosh(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*sinh(x))/(cosh(x) - sinh(x))) - ((a^4 - b^4)*cosh(x)^4 + 4*(a^4 - b^4)*cosh
(x)*sinh(x)^3 + (a^4 - b^4)*sinh(x)^4 + a^4 - b^4 - 2*(a^4 - b^4)*cosh(x)^2 - 2*(a^4 - b^4 - 3*(a^4 - b^4)*cos
h(x)^2)*sinh(x)^2 + 4*((a^4 - b^4)*cosh(x)^3 - (a^4 - b^4)*cosh(x))*sinh(x))*log(2*sinh(x)/(cosh(x) - sinh(x))
) + 4*((a^4 + a^3*b)*x*cosh(x)^3 + (a^4 - a^3*b - a^2*b^2 + a*b^3 - (a^4 + a^3*b)*x)*cosh(x))*sinh(x))/(a^5 -
a^3*b^2 + (a^5 - a^3*b^2)*cosh(x)^4 + 4*(a^5 - a^3*b^2)*cosh(x)*sinh(x)^3 + (a^5 - a^3*b^2)*sinh(x)^4 - 2*(a^5
- a^3*b^2)*cosh(x)^2 - 2*(a^5 - a^3*b^2 - 3*(a^5 - a^3*b^2)*cosh(x)^2)*sinh(x)^2 + 4*((a^5 - a^3*b^2)*cosh(x)
^3 - (a^5 - a^3*b^2)*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 \tanh{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17969, size = 131, normalized size = 1.72 \begin{align*} \frac{b^{4} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{5} - a^{3} b^{2}} - \frac{x}{a - b} + \frac{{\left (a^{2} + b^{2}\right )} \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right )}{a^{3}} - \frac{2 \,{\left (a b +{\left (a^{2} - a b\right )} e^{\left (2 \, x\right )}\right )}}{a^{3}{\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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