3.84 \(\int \frac{\text{csch}(x)}{(a+b \sinh (x))^2} \, dx\)

Optimal. Leaf size=85 \[ \frac{2 b \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^2 \left (a^2+b^2\right )^{3/2}}+\frac{b^2 \cosh (x)}{a \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac{\tanh ^{-1}(\cosh (x))}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.215688, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.546, Rules used = {2802, 3001, 3770, 2660, 618, 206} \[ \frac{2 b \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^2 \left (a^2+b^2\right )^{3/2}}+\frac{b^2 \cosh (x)}{a \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac{\tanh ^{-1}(\cosh (x))}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]/(a + b*Sinh[x])^2,x]

[Out]

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

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

Mathematica [A]  time = 0.18121, size = 91, normalized size = 1.07 \[ \frac{\frac{2 b \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}+\frac{a b^2 \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\log \left (\tanh \left (\frac{x}{2}\right )\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]/(a + b*Sinh[x])^2,x]

[Out]

((2*b*(2*a^2 + b^2)*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/(-a^2 - b^2)^(3/2) + Log[Tanh[x/2]] + (a*b^2*C
osh[x])/((a^2 + b^2)*(a + b*Sinh[x])))/a^2

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Maple [B]  time = 0.039, size = 166, normalized size = 2. \begin{align*}{\frac{1}{{a}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-2\,{\frac{{b}^{3}\tanh \left ( x/2 \right ) }{{a}^{2} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) \left ({a}^{2}+{b}^{2} \right ) }}-2\,{\frac{{b}^{2}}{a \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) \left ({a}^{2}+{b}^{2} \right ) }}-4\,{\frac{b}{ \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-2\,{\frac{{b}^{3}}{{a}^{2} \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\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)/(a+b*sinh(x))^2,x)

[Out]

1/a^2*ln(tanh(1/2*x))-2/a^2*b^3/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)/(a^2+b^2)*tanh(1/2*x)-2/a*b^2/(a*tanh(1/2*
x)^2-2*tanh(1/2*x)*b-a)/(a^2+b^2)-4*b/(a^2+b^2)^(3/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))-2/a^2
*b^3/(a^2+b^2)^(3/2)*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)/(a+b*sinh(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.45987, size = 1632, normalized size = 19.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(a+b*sinh(x))^2,x, algorithm="fricas")

[Out]

-(2*a^3*b^2 + 2*a*b^4 - (2*a^2*b^2 + b^4 - (2*a^2*b^2 + b^4)*cosh(x)^2 - (2*a^2*b^2 + b^4)*sinh(x)^2 - 2*(2*a^
3*b + a*b^3)*cosh(x) - 2*(2*a^3*b + a*b^3 + (2*a^2*b^2 + b^4)*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*b +
a^2*b^3)*cosh(x) + (a^4*b + 2*a^2*b^3 + b^5 - (a^4*b + 2*a^2*b^3 + b^5)*cosh(x)^2 - (a^4*b + 2*a^2*b^3 + b^5)*
sinh(x)^2 - 2*(a^5 + 2*a^3*b^2 + a*b^4)*cosh(x) - 2*(a^5 + 2*a^3*b^2 + a*b^4 + (a^4*b + 2*a^2*b^3 + b^5)*cosh(
x))*sinh(x))*log(cosh(x) + sinh(x) + 1) - (a^4*b + 2*a^2*b^3 + b^5 - (a^4*b + 2*a^2*b^3 + b^5)*cosh(x)^2 - (a^
4*b + 2*a^2*b^3 + b^5)*sinh(x)^2 - 2*(a^5 + 2*a^3*b^2 + a*b^4)*cosh(x) - 2*(a^5 + 2*a^3*b^2 + a*b^4 + (a^4*b +
2*a^2*b^3 + b^5)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) - 1) - 2*(a^4*b + a^2*b^3)*sinh(x))/(a^6*b + 2*a^4*b
^3 + a^2*b^5 - (a^6*b + 2*a^4*b^3 + a^2*b^5)*cosh(x)^2 - (a^6*b + 2*a^4*b^3 + a^2*b^5)*sinh(x)^2 - 2*(a^7 + 2*
a^5*b^2 + a^3*b^4)*cosh(x) - 2*(a^7 + 2*a^5*b^2 + a^3*b^4 + (a^6*b + 2*a^4*b^3 + a^2*b^5)*cosh(x))*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{csch}{\left (x \right )}}{\left (a + b \sinh{\left (x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(a+b*sinh(x))**2,x)

[Out]

Integral(csch(x)/(a + b*sinh(x))**2, x)

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Giac [A]  time = 1.48278, size = 192, normalized size = 2.26 \begin{align*} -\frac{{\left (2 \, a^{2} b + b^{3}\right )} \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 )}{{\left (a^{4} + a^{2} b^{2}\right )} \sqrt{a^{2} + b^{2}}} - \frac{2 \,{\left (a b e^{x} - b^{2}\right )}}{{\left (a^{3} + a b^{2}\right )}{\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b\right )}} - \frac{\log \left (e^{x} + 1\right )}{a^{2}} + \frac{\log \left ({\left | e^{x} - 1 \right |}\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(a+b*sinh(x))^2,x, algorithm="giac")

[Out]

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