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

Optimal. Leaf size=57 \[ -\frac{2 b^2 \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^2 \sqrt{a^2+b^2}}-\frac{b x}{a^2}+\frac{\cosh (x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.106813, antiderivative size = 57, 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 = {3853, 12, 3783, 2660, 618, 206} \[ -\frac{2 b^2 \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{a^2 \sqrt{a^2+b^2}}-\frac{b x}{a^2}+\frac{\cosh (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]/(a + b*Csch[x]),x]

[Out]

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

Rule 3853

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(a*f*n), x] - Dist[1/(a*d*n), Int[((d*Csc[e + f*x])^(n + 1)*Simp[b*n - a*(n + 1)*Csc[e
+ f*x] - b*(n + 1)*Csc[e + f*x]^2, x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 -
b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3783

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

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

Mathematica [A]  time = 0.104826, size = 61, normalized size = 1.07 \[ \frac{b \left (\frac{2 b \tan ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}-x\right )+a \cosh (x)}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]/(a + b*Csch[x]),x]

[Out]

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

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Maple [A]  time = 0.033, size = 92, normalized size = 1.6 \begin{align*}{\frac{1}{a} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{b}{{a}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+2\,{\frac{{b}^{2}}{{a}^{2}\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,\tanh \left ( x/2 \right ) b-2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-{\frac{1}{a} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{b}{{a}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.93253, size = 640, normalized size = 11.23 \begin{align*} \frac{a^{3} + a b^{2} - 2 \,{\left (a^{2} b + b^{3}\right )} x \cosh \left (x\right ) +{\left (a^{3} + a b^{2}\right )} \cosh \left (x\right )^{2} +{\left (a^{3} + a b^{2}\right )} \sinh \left (x\right )^{2} + 2 \,{\left (b^{2} \cosh \left (x\right ) + b^{2} \sinh \left (x\right )\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{a^{2} \cosh \left (x\right )^{2} + a^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + a^{2} + 2 \, b^{2} + 2 \,{\left (a^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt{a^{2} + b^{2}}{\left (a \cosh \left (x\right ) + a \sinh \left (x\right ) + b\right )}}{a \cosh \left (x\right )^{2} + a \sinh \left (x\right )^{2} + 2 \, b \cosh \left (x\right ) + 2 \,{\left (a \cosh \left (x\right ) + b\right )} \sinh \left (x\right ) - a}\right ) - 2 \,{\left ({\left (a^{2} b + b^{3}\right )} x -{\left (a^{3} + a b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )}{2 \,{\left ({\left (a^{4} + a^{2} b^{2}\right )} \cosh \left (x\right ) +{\left (a^{4} + a^{2} b^{2}\right )} \sinh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^3 + a*b^2 - 2*(a^2*b + b^3)*x*cosh(x) + (a^3 + a*b^2)*cosh(x)^2 + (a^3 + a*b^2)*sinh(x)^2 + 2*(b^2*cosh
(x) + b^2*sinh(x))*sqrt(a^2 + b^2)*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) + a^2 + 2*b^2 + 2*(a^2*c
osh(x) + a*b)*sinh(x) - 2*sqrt(a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x)^2 + a*sinh(x)^2 + 2*b*cosh(x
) + 2*(a*cosh(x) + b)*sinh(x) - a)) - 2*((a^2*b + b^3)*x - (a^3 + a*b^2)*cosh(x))*sinh(x))/((a^4 + a^2*b^2)*co
sh(x) + (a^4 + a^2*b^2)*sinh(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(x)/(a + b*csch(x)), x)

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Giac [A]  time = 1.1766, size = 116, normalized size = 2.04 \begin{align*} \frac{b^{2} \log \left (\frac{{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} a^{2}} - \frac{b x}{a^{2}} + \frac{e^{\left (-x\right )}}{2 \, a} + \frac{e^{x}}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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