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

Optimal. Leaf size=81 \[ \frac{2 b^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^3 \sqrt{a^2+b^2}}+\frac{\left (a^2-2 b^2\right ) \tanh ^{-1}(\cosh (x))}{2 a^3}+\frac{b \coth (x)}{a^2}-\frac{\coth (x) \text{csch}(x)}{2 a} \]

[Out]

((a^2 - 2*b^2)*ArcTanh[Cosh[x]])/(2*a^3) + (2*b^3*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^3*Sqrt[a^2 +
b^2]) + (b*Coth[x])/a^2 - (Coth[x]*Csch[x])/(2*a)

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Rubi [A]  time = 0.31795, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {2802, 3055, 3001, 3770, 2660, 618, 206} \[ \frac{2 b^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^3 \sqrt{a^2+b^2}}+\frac{\left (a^2-2 b^2\right ) \tanh ^{-1}(\cosh (x))}{2 a^3}+\frac{b \coth (x)}{a^2}-\frac{\coth (x) \text{csch}(x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^3/(a + b*Sinh[x]),x]

[Out]

((a^2 - 2*b^2)*ArcTanh[Cosh[x]])/(2*a^3) + (2*b^3*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^3*Sqrt[a^2 +
b^2]) + (b*Coth[x])/a^2 - (Coth[x]*Csch[x])/(2*a)

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[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*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[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] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.537556, size = 118, normalized size = 1.46 \[ -\frac{4 \left (a^2-2 b^2\right ) \log \left (\tanh \left (\frac{x}{2}\right )\right )+\frac{16 b^3 \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}+a^2 \text{csch}^2\left (\frac{x}{2}\right )+a^2 \text{sech}^2\left (\frac{x}{2}\right )-4 a b \tanh \left (\frac{x}{2}\right )-4 a b \coth \left (\frac{x}{2}\right )}{8 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((16*b^3*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] - 4*a*b*Coth[x/2] + a^2*Csch[x/2]^2 + 4
*(a^2 - 2*b^2)*Log[Tanh[x/2]] + a^2*Sech[x/2]^2 - 4*a*b*Tanh[x/2])/(8*a^3)

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Maple [A]  time = 0.03, size = 108, normalized size = 1.3 \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}{8\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-2}}-{\frac{1}{2\,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}}-2\,{\frac{{b}^{3}}{{a}^{3}\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.9604, size = 2288, normalized size = 28.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*a^3*b + 4*a*b^3 + 2*(a^4 + a^2*b^2)*cosh(x)^3 + 2*(a^4 + a^2*b^2)*sinh(x)^3 - 4*(a^3*b + a*b^3)*cosh(x
)^2 - 2*(2*a^3*b + 2*a*b^3 - 3*(a^4 + a^2*b^2)*cosh(x))*sinh(x)^2 - 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
))*sinh(x))*sqrt(a^2 + b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 + b^2 + 2*(b^2*cosh(x)
+ a*b)*sinh(x) + 2*sqrt(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(
b*cosh(x) + a)*sinh(x) - b)) + 2*(a^4 + a^2*b^2)*cosh(x) - ((a^4 - a^2*b^2 - 2*b^4)*cosh(x)^4 + 4*(a^4 - a^2*b
^2 - 2*b^4)*cosh(x)*sinh(x)^3 + (a^4 - a^2*b^2 - 2*b^4)*sinh(x)^4 + a^4 - a^2*b^2 - 2*b^4 - 2*(a^4 - a^2*b^2 -
2*b^4)*cosh(x)^2 - 2*(a^4 - a^2*b^2 - 2*b^4 - 3*(a^4 - a^2*b^2 - 2*b^4)*cosh(x)^2)*sinh(x)^2 + 4*((a^4 - a^2*
b^2 - 2*b^4)*cosh(x)^3 - (a^4 - a^2*b^2 - 2*b^4)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) + 1) + ((a^4 - a^2*b^
2 - 2*b^4)*cosh(x)^4 + 4*(a^4 - a^2*b^2 - 2*b^4)*cosh(x)*sinh(x)^3 + (a^4 - a^2*b^2 - 2*b^4)*sinh(x)^4 + a^4 -
a^2*b^2 - 2*b^4 - 2*(a^4 - a^2*b^2 - 2*b^4)*cosh(x)^2 - 2*(a^4 - a^2*b^2 - 2*b^4 - 3*(a^4 - a^2*b^2 - 2*b^4)*
cosh(x)^2)*sinh(x)^2 + 4*((a^4 - a^2*b^2 - 2*b^4)*cosh(x)^3 - (a^4 - a^2*b^2 - 2*b^4)*cosh(x))*sinh(x))*log(co
sh(x) + sinh(x) - 1) + 2*(a^4 + a^2*b^2 + 3*(a^4 + a^2*b^2)*cosh(x)^2 - 4*(a^3*b + a*b^3)*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)*co
sh(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{\operatorname{csch}^{3}{\left (x \right )}}{a + b \sinh{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.41891, size = 185, normalized size = 2.28 \begin{align*} -\frac{b^{3} \log \left (\frac{{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} a^{3}} + \frac{{\left (a^{2} - 2 \, b^{2}\right )} \log \left (e^{x} + 1\right )}{2 \, a^{3}} - \frac{{\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | e^{x} - 1 \right |}\right )}{2 \, a^{3}} - \frac{a e^{\left (3 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} + a e^{x} + 2 \, b}{a^{2}{\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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