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

Optimal. Leaf size=94 \[ -\frac {\text {csch}^2(x) (a-b \cosh (x))}{2 \left (a^2-b^2\right )}-\frac {a^3 \log (a+b \cosh (x))}{\left (a^2-b^2\right )^2}+\frac {(2 a+b) \log (1-\cosh (x))}{4 (a+b)^2}+\frac {(2 a-b) \log (\cosh (x)+1)}{4 (a-b)^2} \]

[Out]

-1/2*(a-b*cosh(x))*csch(x)^2/(a^2-b^2)+1/4*(2*a+b)*ln(1-cosh(x))/(a+b)^2+1/4*(2*a-b)*ln(1+cosh(x))/(a-b)^2-a^3
*ln(a+b*cosh(x))/(a^2-b^2)^2

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Rubi [A]  time = 0.20, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2721, 1647, 801} \[ -\frac {a^3 \log (a+b \cosh (x))}{\left (a^2-b^2\right )^2}-\frac {\text {csch}^2(x) (a-b \cosh (x))}{2 \left (a^2-b^2\right )}+\frac {(2 a+b) \log (1-\cosh (x))}{4 (a+b)^2}+\frac {(2 a-b) \log (\cosh (x)+1)}{4 (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^3/(a + b*Cosh[x]),x]

[Out]

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

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 2721

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

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 101, normalized size = 1.07 \[ \frac {-8 a^3 \log (a+b \cosh (x))+8 a^3 \log (\sinh (x))-12 a^2 b \log \left (\tanh \left (\frac {x}{2}\right )\right )-(a-b)^2 (a+b) \text {csch}^2\left (\frac {x}{2}\right )+(a-b) (a+b)^2 \text {sech}^2\left (\frac {x}{2}\right )+4 b^3 \log \left (\tanh \left (\frac {x}{2}\right )\right )}{8 (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a - b)^2*(a + b)*Csch[x/2]^2) - 8*a^3*Log[a + b*Cosh[x]] + 8*a^3*Log[Sinh[x]] - 12*a^2*b*Log[Tanh[x/2]] +
4*b^3*Log[Tanh[x/2]] + (a - b)*(a + b)^2*Sech[x/2]^2)/(8*(a - b)^2*(a + b)^2)

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fricas [B]  time = 0.44, size = 839, normalized size = 8.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(a^2*b - b^3)*cosh(x)^3 + 2*(a^2*b - b^3)*sinh(x)^3 - 4*(a^3 - a*b^2)*cosh(x)^2 - 2*(2*a^3 - 2*a*b^2 -
3*(a^2*b - b^3)*cosh(x))*sinh(x)^2 + 2*(a^2*b - b^3)*cosh(x) - 2*(a^3*cosh(x)^4 + 4*a^3*cosh(x)*sinh(x)^3 + a^
3*sinh(x)^4 - 2*a^3*cosh(x)^2 + a^3 + 2*(3*a^3*cosh(x)^2 - a^3)*sinh(x)^2 + 4*(a^3*cosh(x)^3 - a^3*cosh(x))*si
nh(x))*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) + ((2*a^3 + 3*a^2*b - b^3)*cosh(x)^4 + 4*(2*a^3 + 3*a^2*b -
b^3)*cosh(x)*sinh(x)^3 + (2*a^3 + 3*a^2*b - b^3)*sinh(x)^4 + 2*a^3 + 3*a^2*b - b^3 - 2*(2*a^3 + 3*a^2*b - b^3)
*cosh(x)^2 - 2*(2*a^3 + 3*a^2*b - b^3 - 3*(2*a^3 + 3*a^2*b - b^3)*cosh(x)^2)*sinh(x)^2 + 4*((2*a^3 + 3*a^2*b -
b^3)*cosh(x)^3 - (2*a^3 + 3*a^2*b - b^3)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) + 1) + ((2*a^3 - 3*a^2*b + b
^3)*cosh(x)^4 + 4*(2*a^3 - 3*a^2*b + b^3)*cosh(x)*sinh(x)^3 + (2*a^3 - 3*a^2*b + b^3)*sinh(x)^4 + 2*a^3 - 3*a^
2*b + b^3 - 2*(2*a^3 - 3*a^2*b + b^3)*cosh(x)^2 - 2*(2*a^3 - 3*a^2*b + b^3 - 3*(2*a^3 - 3*a^2*b + b^3)*cosh(x)
^2)*sinh(x)^2 + 4*((2*a^3 - 3*a^2*b + b^3)*cosh(x)^3 - (2*a^3 - 3*a^2*b + b^3)*cosh(x))*sinh(x))*log(cosh(x) +
sinh(x) - 1) + 2*(a^2*b - b^3 + 3*(a^2*b - b^3)*cosh(x)^2 - 4*(a^3 - a*b^2)*cosh(x))*sinh(x))/((a^4 - 2*a^2*b
^2 + b^4)*cosh(x)^4 + 4*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)*sinh(x)^3 + (a^4 - 2*a^2*b^2 + b^4)*sinh(x)^4 + a^4 -
2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2 - 2*(a^4 - 2*a^2*b^2 + b^4 - 3*(a^4 - 2*a^2*b^2 + b^4)*c
osh(x)^2)*sinh(x)^2 + 4*((a^4 - 2*a^2*b^2 + b^4)*cosh(x)^3 - (a^4 - 2*a^2*b^2 + b^4)*cosh(x))*sinh(x))

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giac [A]  time = 0.13, size = 178, normalized size = 1.89 \[ -\frac {a^{3} b \log \left ({\left | b {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} + \frac {{\left (2 \, a - b\right )} \log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {{\left (2 \, a + b\right )} \log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {a^{3} {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 2 \, a^{2} b {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, b^{3} {\left (e^{\left (-x\right )} + e^{x}\right )} - 4 \, a b^{2}}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left ({\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 97, normalized size = 1.03 \[ -\frac {\tanh ^{2}\left (\frac {x}{2}\right )}{8 \left (a -b \right )}-\frac {a^{3} \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{8 \left (a +b \right ) \tanh \left (\frac {x}{2}\right )^{2}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right ) a}{\left (a +b \right )^{2}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right ) b}{2 \left (a +b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 156, normalized size = 1.66 \[ -\frac {a^{3} \log \left (2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} + b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (2 \, a - b\right )} \log \left (e^{\left (-x\right )} + 1\right )}{2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {{\left (2 \, a + b\right )} \log \left (e^{\left (-x\right )} - 1\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {b e^{\left (-x\right )} - 2 \, a e^{\left (-2 \, x\right )} + b e^{\left (-3 \, x\right )}}{a^{2} - b^{2} - 2 \, {\left (a^{2} - b^{2}\right )} e^{\left (-2 \, x\right )} + {\left (a^{2} - b^{2}\right )} e^{\left (-4 \, x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.52, size = 291, normalized size = 3.10 \[ \frac {\frac {2\,\left (a\,b^2-a^3\right )}{{\left (a^2-b^2\right )}^2}+\frac {{\mathrm {e}}^x\,\left (a^2\,b-b^3\right )}{{\left (a^2-b^2\right )}^2}}{{\mathrm {e}}^{2\,x}-1}-\frac {\frac {2\,a}{a^2-b^2}-\frac {2\,b\,{\mathrm {e}}^x}{a^2-b^2}}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}+\frac {\ln \left ({\mathrm {e}}^x+1\right )\,\left (2\,a-b\right )}{2\,a^2-4\,a\,b+2\,b^2}-\frac {a^3\,\ln \left (b^7\,{\mathrm {e}}^{2\,x}-16\,a^6\,b+b^7-6\,a^2\,b^5+9\,a^4\,b^3-32\,a^7\,{\mathrm {e}}^x-6\,a^2\,b^5\,{\mathrm {e}}^{2\,x}+9\,a^4\,b^3\,{\mathrm {e}}^{2\,x}+2\,a\,b^6\,{\mathrm {e}}^x-16\,a^6\,b\,{\mathrm {e}}^{2\,x}-12\,a^3\,b^4\,{\mathrm {e}}^x+18\,a^5\,b^2\,{\mathrm {e}}^x\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {\ln \left ({\mathrm {e}}^x-1\right )\,\left (2\,a+b\right )}{2\,a^2+4\,a\,b+2\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^3/(a + b*cosh(x)),x)

[Out]

((2*(a*b^2 - a^3))/(a^2 - b^2)^2 + (exp(x)*(a^2*b - b^3))/(a^2 - b^2)^2)/(exp(2*x) - 1) - ((2*a)/(a^2 - b^2) -
(2*b*exp(x))/(a^2 - b^2))/(exp(4*x) - 2*exp(2*x) + 1) + (log(exp(x) + 1)*(2*a - b))/(2*a^2 - 4*a*b + 2*b^2) -
(a^3*log(b^7*exp(2*x) - 16*a^6*b + b^7 - 6*a^2*b^5 + 9*a^4*b^3 - 32*a^7*exp(x) - 6*a^2*b^5*exp(2*x) + 9*a^4*b
^3*exp(2*x) + 2*a*b^6*exp(x) - 16*a^6*b*exp(2*x) - 12*a^3*b^4*exp(x) + 18*a^5*b^2*exp(x)))/(a^4 + b^4 - 2*a^2*
b^2) + (log(exp(x) - 1)*(2*a + b))/(4*a*b + 2*a^2 + 2*b^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth ^{3}{\relax (x )}}{a + b \cosh {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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