3.174 \(\int \frac {\text {csch}^3(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=91 \[ \frac {\text {csch}^2(x) (b-a \cosh (x))}{2 \left (a^2-b^2\right )}+\frac {b^3 \log (a+b \cosh (x))}{\left (a^2-b^2\right )^2}-\frac {(a+2 b) \log (1-\cosh (x))}{4 (a+b)^2}+\frac {(a-2 b) \log (\cosh (x)+1)}{4 (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 91, 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 = {2668, 741, 801} \[ \frac {b^3 \log (a+b \cosh (x))}{\left (a^2-b^2\right )^2}+\frac {\text {csch}^2(x) (b-a \cosh (x))}{2 \left (a^2-b^2\right )}-\frac {(a+2 b) \log (1-\cosh (x))}{4 (a+b)^2}+\frac {(a-2 b) \log (\cosh (x)+1)}{4 (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^3/(a + b*Cosh[x]),x]

[Out]

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

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

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 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(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] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 100, normalized size = 1.10 \[ -\frac {4 a^3 \log \left (\tanh \left (\frac {x}{2}\right )\right )-8 b^3 \log (a+b \cosh (x))-12 a b^2 \log \left (\tanh \left (\frac {x}{2}\right )\right )+(a-b)^2 (a+b) \text {csch}^2\left (\frac {x}{2}\right )+(a-b) (a+b)^2 \text {sech}^2\left (\frac {x}{2}\right )+8 b^3 \log (\sinh (x))}{8 (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((a - b)^2*(a + b)*Csch[x/2]^2 - 8*b^3*Log[a + b*Cosh[x]] + 8*b^3*Log[Sinh[x]] + 4*a^3*Log[Tanh[x/2]] - 1
2*a*b^2*Log[Tanh[x/2]] + (a - b)*(a + b)^2*Sech[x/2]^2)/((a - b)^2*(a + b)^2)

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fricas [B]  time = 0.55, size = 818, normalized size = 8.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(a^3 - a*b^2)*cosh(x)^3 + 2*(a^3 - a*b^2)*sinh(x)^3 - 4*(a^2*b - b^3)*cosh(x)^2 - 2*(2*a^2*b - 2*b^3 -
3*(a^3 - a*b^2)*cosh(x))*sinh(x)^2 + 2*(a^3 - a*b^2)*cosh(x) - 2*(b^3*cosh(x)^4 + 4*b^3*cosh(x)*sinh(x)^3 + b
^3*sinh(x)^4 - 2*b^3*cosh(x)^2 + b^3 + 2*(3*b^3*cosh(x)^2 - b^3)*sinh(x)^2 + 4*(b^3*cosh(x)^3 - b^3*cosh(x))*s
inh(x))*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) - ((a^3 - 3*a*b^2 - 2*b^3)*cosh(x)^4 + 4*(a^3 - 3*a*b^2 - 2
*b^3)*cosh(x)*sinh(x)^3 + (a^3 - 3*a*b^2 - 2*b^3)*sinh(x)^4 + a^3 - 3*a*b^2 - 2*b^3 - 2*(a^3 - 3*a*b^2 - 2*b^3
)*cosh(x)^2 - 2*(a^3 - 3*a*b^2 - 2*b^3 - 3*(a^3 - 3*a*b^2 - 2*b^3)*cosh(x)^2)*sinh(x)^2 + 4*((a^3 - 3*a*b^2 -
2*b^3)*cosh(x)^3 - (a^3 - 3*a*b^2 - 2*b^3)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) + 1) + ((a^3 - 3*a*b^2 + 2*
b^3)*cosh(x)^4 + 4*(a^3 - 3*a*b^2 + 2*b^3)*cosh(x)*sinh(x)^3 + (a^3 - 3*a*b^2 + 2*b^3)*sinh(x)^4 + a^3 - 3*a*b
^2 + 2*b^3 - 2*(a^3 - 3*a*b^2 + 2*b^3)*cosh(x)^2 - 2*(a^3 - 3*a*b^2 + 2*b^3 - 3*(a^3 - 3*a*b^2 + 2*b^3)*cosh(x
)^2)*sinh(x)^2 + 4*((a^3 - 3*a*b^2 + 2*b^3)*cosh(x)^3 - (a^3 - 3*a*b^2 + 2*b^3)*cosh(x))*sinh(x))*log(cosh(x)
+ sinh(x) - 1) + 2*(a^3 - a*b^2 + 3*(a^3 - a*b^2)*cosh(x)^2 - 4*(a^2*b - b^3)*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)*
cosh(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 [B]  time = 0.15, size = 179, normalized size = 1.97 \[ \frac {b^{4} \log \left ({\left | b {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} + \frac {{\left (a - 2 \, b\right )} \log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {{\left (a + 2 \, b\right )} \log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {b^{3} {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 2 \, a^{3} {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a b^{2} {\left (e^{\left (-x\right )} + e^{x}\right )} + 4 \, a^{2} b - 8 \, b^{3}}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left ({\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^4*log(abs(b*(e^(-x) + e^x) + 2*a))/(a^4*b - 2*a^2*b^3 + b^5) + 1/4*(a - 2*b)*log(e^(-x) + e^x + 2)/(a^2 - 2*
a*b + b^2) - 1/4*(a + 2*b)*log(e^(-x) + e^x - 2)/(a^2 + 2*a*b + b^2) + 1/2*(b^3*(e^(-x) + e^x)^2 - 2*a^3*(e^(-
x) + e^x) + 2*a*b^2*(e^(-x) + e^x) + 4*a^2*b - 8*b^3)/((a^4 - 2*a^2*b^2 + b^4)*((e^(-x) + e^x)^2 - 4))

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maple [A]  time = 0.09, size = 97, normalized size = 1.07 \[ \frac {\tanh ^{2}\left (\frac {x}{2}\right )}{8 a -8 b}+\frac {b^{3} \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{8 \left (a +b \right ) \tanh \left (\frac {x}{2}\right )^{2}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right ) a}{2 \left (a +b \right )^{2}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right ) b}{\left (a +b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 154, normalized size = 1.69 \[ \frac {b^{3} \log \left (2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} + b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a - 2 \, b\right )} \log \left (e^{\left (-x\right )} + 1\right )}{2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {{\left (a + 2 \, b\right )} \log \left (e^{\left (-x\right )} - 1\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {a e^{\left (-x\right )} - 2 \, b e^{\left (-2 \, x\right )} + a e^{\left (-3 \, x\right )}}{a^{2} - b^{2} - 2 \, {\left (a^{2} - b^{2}\right )} e^{\left (-2 \, x\right )} + {\left (a^{2} - b^{2}\right )} e^{\left (-4 \, x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^3*log(2*a*e^(-x) + b*e^(-2*x) + b)/(a^4 - 2*a^2*b^2 + b^4) + 1/2*(a - 2*b)*log(e^(-x) + 1)/(a^2 - 2*a*b + b^
2) - 1/2*(a + 2*b)*log(e^(-x) - 1)/(a^2 + 2*a*b + b^2) - (a*e^(-x) - 2*b*e^(-2*x) + a*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 = 1.48, size = 291, normalized size = 3.20 \[ \frac {\frac {2\,\left (a^2\,b-b^3\right )}{{\left (a^2-b^2\right )}^2}+\frac {{\mathrm {e}}^x\,\left (a\,b^2-a^3\right )}{{\left (a^2-b^2\right )}^2}}{{\mathrm {e}}^{2\,x}-1}+\frac {\frac {2\,b}{a^2-b^2}-\frac {2\,a\,{\mathrm {e}}^x}{a^2-b^2}}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}+\frac {b^3\,\ln \left (16\,b^7\,{\mathrm {e}}^{2\,x}-a^6\,b+16\,b^7-9\,a^2\,b^5+6\,a^4\,b^3-2\,a^7\,{\mathrm {e}}^x-9\,a^2\,b^5\,{\mathrm {e}}^{2\,x}+6\,a^4\,b^3\,{\mathrm {e}}^{2\,x}+32\,a\,b^6\,{\mathrm {e}}^x-a^6\,b\,{\mathrm {e}}^{2\,x}-18\,a^3\,b^4\,{\mathrm {e}}^x+12\,a^5\,b^2\,{\mathrm {e}}^x\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {\ln \left ({\mathrm {e}}^x-1\right )\,\left (a+2\,b\right )}{2\,a^2+4\,a\,b+2\,b^2}+\frac {\ln \left ({\mathrm {e}}^x+1\right )\,\left (a-2\,b\right )}{2\,a^2-4\,a\,b+2\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^3*(a + b*cosh(x))),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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