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

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Rubi [A]  time = 0.11, antiderivative size = 57, normalized size of antiderivative = 1.00, 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 12

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

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])

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 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 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 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]

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*}

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Mathematica [A]  time = 0.11, 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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fricas [B]  time = 0.42, size = 238, normalized size = 4.18 \[ \frac {a^{3} + a b^{2} - 2 \, {\left (a^{2} b + b^{3}\right )} x \cosh \relax (x) + {\left (a^{3} + a b^{2}\right )} \cosh \relax (x)^{2} + {\left (a^{3} + a b^{2}\right )} \sinh \relax (x)^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + b^{2} \sinh \relax (x)\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {a^{2} \cosh \relax (x)^{2} + a^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cosh \relax (x) + a \sinh \relax (x) + b\right )}}{a \cosh \relax (x)^{2} + a \sinh \relax (x)^{2} + 2 \, b \cosh \relax (x) + 2 \, {\left (a \cosh \relax (x) + b\right )} \sinh \relax (x) - a}\right ) - 2 \, {\left ({\left (a^{2} b + b^{3}\right )} x - {\left (a^{3} + a b^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x)}{2 \, {\left ({\left (a^{4} + a^{2} b^{2}\right )} \cosh \relax (x) + {\left (a^{4} + a^{2} b^{2}\right )} \sinh \relax (x)\right )}} \]

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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giac [A]  time = 0.13, size = 86, normalized size = 1.51 \[ \frac {b^{2} \log \left (\frac {{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{2}} - \frac {b x}{a^{2}} + \frac {e^{\left (-x\right )}}{2 \, a} + \frac {e^{x}}{2 \, a} \]

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

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maple [A]  time = 0.16, size = 92, normalized size = 1.61 \[ -\frac {1}{a \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a^{2}}+\frac {1}{a \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a^{2}}+\frac {2 b^{2} \arctanh \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{2} \sqrt {a^{2}+b^{2}}} \]

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

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maxima [A]  time = 0.40, size = 84, normalized size = 1.47 \[ \frac {b^{2} \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{2}} - \frac {b x}{a^{2}} + \frac {e^{\left (-x\right )}}{2 \, a} + \frac {e^{x}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.58, size = 129, normalized size = 2.26 \[ \frac {{\mathrm {e}}^{-x}}{2\,a}+\frac {{\mathrm {e}}^x}{2\,a}-\frac {b\,x}{a^2}-\frac {b^2\,\ln \left (-\frac {2\,b^2\,{\mathrm {e}}^x}{a^3}-\frac {2\,b^2\,\left (a-b\,{\mathrm {e}}^x\right )}{a^3\,\sqrt {a^2+b^2}}\right )}{a^2\,\sqrt {a^2+b^2}}+\frac {b^2\,\ln \left (\frac {2\,b^2\,\left (a-b\,{\mathrm {e}}^x\right )}{a^3\,\sqrt {a^2+b^2}}-\frac {2\,b^2\,{\mathrm {e}}^x}{a^3}\right )}{a^2\,\sqrt {a^2+b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)/(a + b/sinh(x)),x)

[Out]

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

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

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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